Electricidad y magnetismo Raymond A. Serway 3ed Sol

http://www.ĞůƐŽůƵĐŝŽŶĂƌŝŽ͘ďůŽŐƐƉŽƚ.com /,%52681,9(5,67$5,26 <62/8&,21$5,26'( 08&+26'((6726/,%526 /2662/8&,21$5,26 &217,(1(172'26/26 (-(5&,&,26'(//,%52 5(68(/726<(;3/,&$'26 '()250$&/$5$ 9,6,7$1263$5$ '(6$5*$/26*5$7,6 Chapter 23 Solutions 23.1 (a)   10.0 grams electrons  24 23 atoms   47.0 = 2.62 × 10 N=  6.02 × 10 mol   atom   107.87 grams mol   (b) # electrons added = or 23.2 Q 1.00 × 10 −3 C = = 6.25 × 1015 e 1.60 × 10 -19 C electron 2.38 electrons for every 10 9 already present ( )( ) 2 (a) 8.99 × 10 9 N ⋅ m 2/ C 2 1.60 × 10 −19 C k qq Fe = e 21 2 = (3.80 × 10 − 10 m)2 r (b) 6.67 × 10 −11 N ⋅ m 2 kg 2 (1.67 × 10 − 27 kg)2 G m1m2 Fg = = = 1.29 × 10 − 45 N (3.80 × 10 −10 m)2 r2 ( = 1.59 × 10 − 9 N (repulsion) ) The electric force is larger by 1.24 × 10 36 times (c) If ke q = m 23.3 q1q2 mm = G 12 2 2 r r G = ke with q1 = q2 = q and m1 = m2 = m, then 6.67 × 10 −11 N ⋅ m 2 / kg 2 = 8.61 × 10 −11 C / kg 8.99 × 10 9 N ⋅ m 2 / C 2 If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains  70, 000 grams   protons  molecules   10 N≈  ≈ 2.3 × 10 28 protons 6.02 × 10 23     mol molecule   18 grams mol  With an excess of 1% electrons over protons, each person has a charge q = (0.01)(1.6 × 10 −19 C)(2.3 × 10 28 ) = 3.7 × 107 C So F = ke 7 2 q1q2 9 (3.7 × 10 ) = (9 × 10 ) N = 4 × 10 25 N ~ 1026 N 0.6 2 r2 This force is almost enough to lift a "weight" equal to that of the Earth: Mg = (6 × 10 24 kg)(9.8 m s 2 ) = 6 × 10 25 N ~ 1026 N © 2000 by Harcourt, Inc. All rights reserved. 2 Chapter 23 Solutions We find the equal-magnitude charges on both spheres: 23.4 F = ke q2 q1q2 = ke 2 2 r r q=r so F 1.00 × 10 4 N = (1.00 m ) = 1.05 × 10 −3 C ke 8.99 × 10 9 N ⋅ m 2/ C 2 The number of electron transferred is then ( N xfer = 1.05 × 10 −3 C ) (1.60 × 10 −19 ) C / e − = 6.59 × 1015 electrons The whole number of electrons in each sphere is   10.0 g 23 − 24 − Ntot =   6.02 × 10 atoms / mol 47 e / atom = 2.62 × 10 e  107.87 g / mol  ( )( ) The fraction transferred is then f= Ntot  6.59 × 1015 =  = 2.51 × 10–9 2.62 × 1024 ( )( = 2.51 charges in every billion 8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C qq F = ke 1 2 2 = 2 r 2(6.37 × 106 m) 23.5 *23.6 N xfer (a) ] 2 23 2 = 514 kN The force is one of attraction. The distance r in Coulomb's law is the distance between centers. The magnitude of the force is F= (b) [ ) (6.02 × 10 ) ( )( ) 2 12.0 × 10 −9 C 18.0 × 10 −9 C ke q1q2  9 N⋅m  = 8.99 × 10 = 2.16 × 10 − 5 N   (0.300 m)2 C2  r2  The net charge of − 6.00 × 10 −9 C will be equally split between the two spheres, or − 3.00 × 10 −9 C on each. The force is one of repulsion, and its magnitude is ( )( ) 2 3.00 × 10 −9 C 3.00 × 10 −9 C ke q1q2  9 N⋅m  F= =  8.99 × 10 =  (0.300 m)2 r2 C2   8.99 × 10 −7 N Chapter 23 Solutions 23.7 F1 = ke q1q2 (8.99 × 10 9 N ⋅ m 2/ C 2 )(7.00 × 10 −6 C)(2.00 × 10 −6 C) = = 0.503 N r2 (0.500 m)2 F2 = k e q1q2 (8.99 × 10 9 N ⋅ m 2 / C 2 )(7.00 × 10 −6 C)(4.00 × 10 −6 C) = = 1.01 N (0.500 m)2 r2 3 Fx = (0.503 + 1.01) cos 60.0° = 0.755 N Fy = (0.503 − 1.01) sin 60.0° = − 0.436 N F = (0.755 N)i − (0.436 N)j = 0.872 N at an angle of 330° Goal Solution Three point charges are located at the corners of an equilateral triangle as shown in Figure P23.7. Calculate the net electric force on the 7.00− µ C charge. G: Gather Information: The 7.00− µ C charge experiences a repulsive force F1 due to the 2.00− µ C charge, and an attractive force F 2 due to the −4.00− µ C charge, where F2 = 2F1. If we sketch these force vectors, we find that the resultant appears to be about the same magnitude as F2 and is directed to the right about 30.0° below the horizontal. O: Organize : We can find the net electric force by adding the two separate forces acting on the 7.00− µ C charge. These individual forces can be found by applying Coulomb’s law to each pair of charges. A: Analyze: F1 The force on the 7.00− µ C charge by the 2.00− µ C charge is (8.99 × 10 = 9 )( )( N ⋅ m 2/ C 2 7.00 × 10 −6 C 2.00 × 10 −6 C (0.500 m) 2 ) (cos60°i + sin 60°j) = F 1 = (0.252i + 0.436j) N Similarly, the force on the 7.00− µ C by the −4.00− µ C charge is ( )( ) −6 −6  N ⋅ m 2  7.00 × 10 C − 4.00 × 10 C F 2 = − 8.99 × 10 9 (cos60°i − sin 60° j) = (0.503i − 0.872j) N  C2   (0.500 m)2 Thus, the total force on the 7.00− µ C , expressed as a set of components, is F = F1 + F 2 = (0.755 i − 0.436 j) N = 0.872 N at 30.0° below the +x axis L: Learn: Our calculated answer agrees with our initial estimate. An equivalent approach to this problem would be to find the net electric field due to the two lower charges and apply F=qE to find the force on the upper charge in this electric field. © 2000 by Harcourt, Inc. All rights reserved. 4 Chapter 23 Solutions Let the third bead have charge Q and be located distance x from the left end of the rod. This bead will experience a net force given by *23.8 F= ke ( 3q)Q x2 i+ ke ( q)Q ( d − x )2 ( −i) The net force will be zero if 3 1 x = , or d − x = 3 x 2 ( d − x )2 This gives an equilibrium position of the third bead of stable if the third bead has positive charge . The equilibrium is *23.9 k ee 2 (1.60 × 10–19 C)2 = (8.99 × 109 N ⋅ m2/C 2) = 8.22 × 10–8 N 2 r (0.529 × 10–10 m)2 (a) F= (b) We have F = mv 2 r from which v = Fr = m (8.22 × 10 The top charge exerts a force on the negative charge 23.10 to the left, at an angle of tan −1 ( d / 2x ) to the x-axis. force  2 k qQ  2 e  d 4 + x2  ( (a)  (− x)i    d2 4 + x2  ) (   = ma 1/2   ) −8 )( N 0.529 × 10 −10 m 9.11 × 10 −31 ke qQ ( d 2)2 + x 2 or for x << d 2, kg )= 2.19 × 106 m/s which is directed upward and The two positive charges together exert a≈ − 2 ke qQ x md 3 / 8 The acceleration is equal to a negative constant times the excursion from equilibrium, as i n 16 ke qQ a = −ω 2 x, so we have Simple Harmonic Motion with ω 2 = . md 3 T= (b) x = 0.634d 2π = ω π 2 vmax = ω A = md 3 ke qQ 4a , where m is the mass of the object with charge −Q. ke qQ md 3 Chapter 23 Solutions For equilibrium, F e = −F g , or qE = −mg( −j) . Thus, E = 23.11 23.12 *23.13 mg j. q (a) E= (9.11 × 10 −31 kg)(9.80 m s 2 ) mg j= j = − ( 5.58 × 10 −11 N C) j q −1.60 × 10 −19 C (b) E= 1.67 × 10 −27 kg 9.80 m s 2 mg j= j= q 1.60 × 10 −19 C ( ( 5 ) ( )( ∑ Fy = 0: QE j + mg(− j) = 0 ∴ m= ) ) (1.02 × 10 −7 ) NC j (24.0 × 10 -6 C)(610 N / C) QE = = 1.49 grams g 9.80 m / s 2 The point is designated in the sketch. The magnitudes of the electric fields, E1, (due to the –2.50 × 10–6 C charge) and E2 (due to the 6.00 × 10–6 C charge) are E1 = (8.99 × 109 N · m2/C 2)(2.50 × 10–6 C) k eq = d2 r2 (1) E2 = (8.99 × 109 N · m2/C 2)(6.00 × 10–6 C) k eq = (d + 1.00 m)2 r2 (2) Equate the right sides of (1) and (2) to get (d + 1.00 m)2 = 2.40d 2 or d + 1.00 m = ±1.55d which yields d = 1.82 m or d = – 0.392 m The negative value for d is unsatisfactory because that locates a point between the charges where both fields are in the same direction. Thus, d = 1.82 m to the left of the -2.50 µC charge. 23.14 If we treat the concentrations as point charges, E + = ke q  N ⋅ m 2  ( 40.0 C) =  8.99 × 10 9 −j = 3.60 × 10 5 N / C ( −j) (downward) 2 2  2 ( ) r C   (1000 m ) E − = ke 2 q  9 N ⋅ m  ( 40.0 C) = 8.99 × 10 (−j) = 3.60 × 105 N / C (−j) (downward)   r2  C 2  (1000 m )2 E = E + + E − = 7.20 × 10 5 N / C downward © 2000 by Harcourt, Inc. All rights reserved. 6 Chapter 23 Solutions *23.15 (a) ( )( ) ( )( ) 8.99 × 10 9 7.00 × 10 −6 ke q E1 = 2 = = 2.52 × 10 5 N C 2 r (0.500) E2 = 8.99 × 10 9 4.00 × 10 −6 ke q = = 1.44 × 10 5 N C r2 (0.500)2 Ex = E2 − E1 cos 60° = 1.44 × 10 5 − 2.52 × 10 5 cos 60.0° = 18.0 × 10 3 N C Ey = −E1 sin 60.0° = −2.52 × 10 5 sin 60.0° = −218 × 10 3 N C E = [18.0i − 218 j] × 10 3 N C = [18.0i − 218 j] kN C (b) *23.16 (a) ( ) F = qE = 2.00 × 10 −6 C (18.0i − 218 j) × 10 3 N C = ( 36.0i − 436 j) × 10 −3 N = E1 = ke q1 E2 = ke q2 r12 r22 (8.99 × 10 )(3.00 × 10 ) (− j) = − (2.70 × 10 (− j) = −9 9 (0.100)2 (− i) = 3 (8.99 × 10 )(6.00 × 10 ) (− i) = − (5.99 × 10 −9 9 (0.300)2 ( ) NC j ) ( 2 ) NC i ) E = E 2 + E1 = − 5.99 × 10 2 N C i − 2.70 × 10 3 N C j (b) ( ) F = qE = 5.00 × 10 −9 C ( −599i − 2700 j) N C ( ) F = − 3.00 × 10 −6 i − 13.5 × 10 −6 j N = 23.17 (− 3.00 i − 13.5 j ) µN (a) The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce fields that cancel out. (b) You may need to review vector addition in Chapter Three. The magnitude of the field at point P due to each of the charges along the base of the triangle is E = ke q a 2 . The direction of the field in each case is along the line joining the charge in question to point P as shown in the diagram at the right. The x components add to zero, leaving E= ke q kq sin 60.0°) j + e2 (sin 60.0°) j = 2 ( a a 3 ke q j a2 (36.0i − 436 j ) mN Chapter 23 Solutions 7 Goal Solution Three equal positive charges q are at the corners of an equilateral triangle of side a, as shown in Figure P23.17. (a) Assume that the three charges together create an electric field. Find the location of a point (other than ∞) where the electric field is zero. (Hint: Sketch the field lines in the plane of the charges.) (b) What are the magnitude and direction of the electric field at P due to the two charges at the base? The electric field has the general appearance shown by the black arrows in the figure to the right. This drawing indicates that E = 0 at the center of the triangle, since a small positive charge placed at the center of this triangle will be pushed away from each corner equally strongly. This fact could be verified by vector addition as in part (b) below. G: The electric field at point P should be directed upwards and about twice the magnitude of the electric field due to just one of the lower charges as shown in Figure P23.17. For part (b), we must ignore the effect of the charge at point P , because a charge cannot exert a force on itself. O: The electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges: E = E1 + E2 A: (b) The electric field from a point charge is As shown in the solution figure above, E1 = ke q to the right and upward at 60° a2 q to the left and upward at 60° a2 q q q E = E1 + E2 = ke 2 (cos60°i + sin 60°j) + ( − cos60°i + sin 60° j) = ke 2 2(sin 60° j) = 1.73ke 2 j a a a E 2 = ke [ L: ] [ ] The net electric field at point P is indeed nearly twice the magnitude due to a single charge and is entirely vertical as expected from the symmetry of the configuration. In addition to the center of the triangle, the electric field lines in the figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically. 23.18 (a) E= ke q (8.99 × 10 9 )(2.00 × 10 −6 ) = = 14, 400 N C (1.12)2 r2 Ex = 0 so (b) and Ey = 2(14, 400) sin 26.6° = 1.29 × 10 4 N C E = 1.29 × 10 4 j N C F = Eq = (1.29 × 10 4 j)(−3.00 × 10 −6 ) = −3.86 × 10 −2 j N © 2000 by Harcourt, Inc. All rights reserved. 8 23.19 Chapter 23 Solutions (a) E= k ( 2q) k ( 3q) k ( 4q) ke q1 k q k q ~ + e 22 ~2 + e 23 ~3 = e 2 i + e 2 (i cos 45.0° + j sin 45.0°) + e 2 j 2 1 r1 r2 r3 a 2a a E = 3.06 (b) ke q kq kq i + 5.06 e2 j = 5.91 e2 at 58.8° a2 a a F = qE = 5.91 ke q 2 at 58.8° a2 The magnitude of the field at (x, y) due to charge q at (x0 , y0 ) is given by E = ke q r 2 where r is the distance from (x0 , y0 ) to (x, y). Observe the geometry in the diagram at the right. From triangle ABC, r 2 = (x − x0 )2 + (y − y0 )2 , or 23.20 r = (x − x0 )2 + (y − y0 )2 , sin θ = (y − y0 ) , r Ex = Ecos θ = ke q (x − x0 ) ke q(x − x0 ) = 2 r r [(x − x0 )2 + (y − y0 )2 ]3/2 and Ey = Esin θ = ke q (y − y0 ) ke q(y − y0 ) = r r2 [(x − x0 )2 + (y − y0 )2 ]3/2 E= k eq (x – a)2 (a) One of the charges creates at P a field E= – (x − x0 ) r k eq (x – (–a))2 E≈ When x is much, much greater than a, we find 23.22 cos θ = Thus, The electric field at any point x is 23.21 and = k eq(4ax) (x 2 – a 2)2 (4a)(k eq) x3 ke Q/n R 2 + x2 at an angle θ to the x-axis as shown. When all the charges produce field, for n > 1, the components perpendicular to the x-axis add to zero. The total field is (b) nke (Q/n)i keQxi cos θ = 2 2 2 R +x (R + x 2)3/2 A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n. Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field. . Chapter 23 Solutions 23.23 E=∑ 23.24 E= − ke qi  π 2 ke q ke q kq kq kq 1 1  ~ = e2 (− i)+ e 2 (− i) + e 2 (− i) + . . . = 1+ 2 + 3 + ... = − i 2 2   r a (2a) (3a) 6 a2 2 3 a (8.99 × 109)(22.0 × 10–6) k (Q / l)l ke λ l keQ = e = = (0.290)(0.140 + 0.290) d(l+ d) d(l+ d) d(l+ d) E = 1.59 × 106 N/C , E=∫ 23.25 ke dq x2 directed toward the rod . where dq = λ0 dx ∞ dx  1 2 = k e  – x x0 x ⌠ E = ke λ0 ⌡ E = ∫ dE = ∫ 23.26 ∞ x0 E= 23.27 ∞ = x0 k eλ 0 x0 The direction is –i or left for λ0 > 0  1 ∞ −3  ke λ 0 x0 dx( −i )  x i λ = − k λ x i x dx = − k  − 2 0 0 0 0 e e ∫x0   x3    2x k exQ (8.99 × 109)(75.0 × 10–6)x 6.74 × 105 x = 2 2 3/2 = 2 2 3/2 (x + a ) (x + 0.100 ) (x + 0.0100)3/2 2 (a) At x = 0.0100 m, E = 6.64 × 106 i N/C = 6.64 i MN/C (b) At x = 0.0500 m, E = 2.41 × 107 i N/C = 24.1 i MN/C (c) At x = 0.300 m, E = 6.40 × 106 i N/C = 6.40 i MN/C (d) At x = 1.00 m, E = 6.64 × 105 i N/C = 0.664 i MN/C © 2000 by Harcourt, Inc. All rights reserved. ∞ x0   =  ke λ 0 (− i) 2x0 9 10 Chapter 23 Solutions E= 23.28 ke Q x (x + a2 )3/2 2 For a maximum, x 2 + a2 − 3x 2 = 0   1 3x 2 dE = Qke  2 − =0 2 2 5/2 2 3/2 dx (x + a )   (x + a ) a 2 x= or Substituting into the expression for E gives E= 2ke Q ke Qa k Q = e = 3 3 a2 2( 23 a2 )3/2 3 3 a2 2 Q 6 3 π e0 a2   x E = 2 π ke σ  1 −  2 2  x +R  23.29  E = 2 π 8.99 × 10 9 7.90 × 10 −3  1 −   ( )( )    = 4.46 × 108  1 − 2  x 2 + (0.350)   x (a) At x = 0.0500 m, E = 3.83 × 108 N C = 383 MN C (b) At x = 0.100 m, E = 3.24 × 108 N C = 324 MN C (c) At x = 0.500 m, E = 8.07 × 107 N C = 80.7 MN C (d) At x = 2.00 m, 23.30 = (a) From Example 23.9: σ=   x 2 + 0.123  x E = 6.68 × 108 N C = 6.68 MN C  E = 2 π ke σ  1 −    x2 + R2  x Q = 1.84 × 10 −3 C m 2 πR 2 E = (1.04 × 108 N C)(0.900) = 9.36 × 107 N C = 93.6 MN/C appx: E = 2 π ke σ = 104 MN/C (about 11% high) (b)  E = (1.04 × 108 N / C)  1 −  appx: E = ke  8  = (1.04 × 10 N C)(0.00496) = 0.516 MN/C 2 2 30.0 + 3.00 cm  30.0 cm −6 Q 9 5.20 × 10 = (8.99 × 10 ) = 0.519 MN/C (about 0.6% high) r2 (0.30)2 Chapter 23 Solutions 23.31 The electric field at a distance x is  Ex = 2 π ke σ 1 −  This is equivalent to  Ex = 2 π ke σ 1 −   For large x, R 2 x 2 << 1 and 2 1 1+ R 2   x 2  R2 R2 ≈ 1+ x2 2x 2  1 Ex = 2 π ke σ  1 − 2  (2x 2 ) 1+ R  Substitute σ = Q/π R2, Ex = [ 1 1 ≈ 2 , so 2 x +R 2 x 2 [1+ R Ex ≈ ( keQ 1 x 2 2 ) ] ( )  1+ R 2 (2x 2 ) − 1  = 2 π ke σ  1+ R 2 (2x 2 )  [ ]  2 R2 = k Q e x + 2    (2x 2 ) ] keQ for a disk at large distances x2 The sheet must have negative charge to repel the negative charge on the Styrofoam. The magnitude of the upward electric force must equal the magnitude of the downward gravitational force for the Styrofoam to "float" (i.e., Fe = F g ). Thus, 23.33   x + R  x 2 so But for x > > R, 23.32 1+ 11  σ  2e0 mg −qE = mg, or −q   = mg which gives σ = − q 2e  0 Due to symmetry Ey = ∫ dEy = 0, and E x = ∫ dE sin θ = ke ∫ dq sin θ r2 π where dq = λ ds = λr dθ , so that, Ex = ke λ r q where λ = L Ex = 2ke qπ 2(8.99 × 109 N · m2/C 2)(7.50 × 10–6 C)π = L2 (0.140 m)2 and r = L π . Thus, π ∫0 sin θ dθ = 2k λ ke λ (− cos θ ) = e r r 0 Solving, E = Ex = 2.16 × 107 N/C Since the rod has a negative charge, E = (–2.16 × 107 i) N/C = –21.6 i MN/C © 2000 by Harcourt, Inc. All rights reserved. 12 23.34 Chapter 23 Solutions (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has charge Qdx/h and produces, at the chosen point, a field dE = Q dx ke x i 2 3/2 h (x + R ) 2 The total field is ∫ dE E= =∫ all charge d+h d keQ x dx k Qi d + h 2 i = e ∫ (x + R 2 )− 3/2 2x dx 2 2 3/2 2h x = d h(x + R ) k Q i (x 2 + R 2 )− 1/2 E= e 2h (− 1/ 2) (b) So, 2 π keQ dx π R2h E= ∫ ) charge Q dx/h, and charge-   x  1 − (x 2 + R 2 )1/2  i   all charge dE = ∫ d+h x=d 2 keQ dx R2 h   x  1 − (x 2 + R 2 )1/2  i    d+ h 1 (x 2 + R 2 )1/2 2 keQ i  d + h 1 d + h(x 2 + R 2 )− 1/2 2x dx  = 2 keQ i x dx − −  2 ∫x = d  2 1/ 2 R 2 h  ∫d R 2 h  d E= 2 keQ i R 2h 2 keQ i R 2h ( d + h − d − (d + h)2 + R 2  ) 1/2 d + h d   + (d 2 + R 2 )1/2   (  h + (d 2 + R 2 )1/2 − (d + h)2 + R 2  ) 1/2   ke dq and is directed (x + y 2 ) along the line joining the element of length to point P . By symmetry, The electric field at point P due to each element of length dx, is dE = Ex = ∫ dEx = 0 and since dq = λ dx, E = Ey = ∫ dEy = ∫ dE cos θ where cos θ = Therefore, (b)   1/2   E= E= (a) ( x=d Think of the cylinder as a stack of disks, each with thickness dx, per-area σ = Q dx / πR 2 h. One disk produces a field dE = 23.35  keQ i  1 1 = − 2 2 1/2  h (d + R ) (d + h)2 + R 2  d+h dx = (x + y 2 )3 2 2 y (x + y 2 )1 2 2 2ke λ sinθ 0 y For a bar of infinite length, θ → 90° and Ey = 2ke λ y 2 Chapter 23 Solutions *23.36 (a) The whole surface area of the cylinder is A = 2 π r 2 + 2 π rL = 2 π r(r + L) . ( ) Q = σA = 15.0 × 10 −9 C m 2 2 π (0.0250 m )[0.0250 m + 0.0600 m ] = 2.00 × 10 −10 C (b) For the curved lateral surface only, A = 2 πrL. ( ) Q = σA = 15.0 × 10 −9 C m 2 2 π (0.0250 m )(0.0600 m ) = 1.41 × 10 −10 C (c) *23.37 (a) ( ) Q = ρ V = ρ π r 2 L = 500 × 10 −9 C m 3 π (0.0250 m ) (0.0600 m ) = 5.89 × 10 −11 C 2 Every object has the same volume, V = 8(0.0300 m ) = 2.16 × 10 −4 m 3 . 3 ( )( ) For each, Q = ρ V = 400 × 10 −9 C m 3 2.16 × 10 −4 m 3 = 8.64 × 10 −11 C (b) We must count the 9.00 cm 2 squares painted with charge: (i) 6 × 4 = 24 squares ( ) ( ) ( ) ( ) ( ) ( ) 4.59 × 10 −10 C ( ) ( ) 4.32 × 10 −10 C Q = σA = 15.0 × 10 −9 C m 2 24.0 9.00 × 10 −4 m 2 = 3.24 × 10 −10 C (ii) 34 squares exposed Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = 4.59 × 10 −10 C (iii) 34 squares Q = σA = 15.0 × 10 −9 C m 2 34.0 9.00 × 10 −4 m 2 = (iv) 32 squares Q = σA = 15.0 × 10 −9 C m 2 32.0 9.00 × 10 −4 m 2 = (c) (i) total edge length: = 24 × (0.0300 m ) ( ) Q = λ = 80.0 × 10 −12 C m 24 × (0.0300 m ) = ( ) ( ) 5.76 × 10 −11 C (ii) Q = λ = 80.0 × 10 −12 C m 44 × (0.0300 m ) = 1.06 × 10 −10 C (iii) Q = λ = 80.0 × 10 −12 C m 64 × (0.0300 m ) = 1.54 × 10 −10 C © 2000 by Harcourt, Inc. All rights reserved. 13 14 Chapter 23 Solutions (iv) ( ) Q = λ = 80.0 × 10 −12 C m 40 × (0.0300 m ) = 0.960 × 10 −10 C Chapter 23 Solutions 22.38 22.39 23.40 (a) q1 − 6 = = q2 18 (b) q1 is negative, q2 is positive − 1 3 F = qE = ma a = 23.41 qE m qEt m v = v i + at v= electron: ve = (1.602 × 10 −19 )(520)(48.0 × 10 −9 ) = 4.39 × 106 m/s 9.11 × 10 −31 in a direction opposite to the field proton: vp = (1.602 × 10 −19 )(520)(48.0 × 10 −9 ) = 2.39 × 103 m/s 1.67 × 10 −27 in the same direction as the field 23.42 (a) (b) a = qE (1.602 × 10 −19 )(6.00 × 10 5 ) = = 5.76 × 1013 m s so −27 m (1.67 × 10 ) v = vi + 2a(x − xi ) 0 = vi 2 + 2(−5.76 × 1013 )(0.0700) (c) a= v i = 2.84 × 106 i m s v = vi + at 0 = 2.84 × 106 + (−5.76 × 1013 )t t = 4.93 × 10 −8 s © 2000 by Harcourt, Inc. All rights reserved. −5.76 × 1013 i m s 2 15 16 23.43 Chapter 23 Solutions ( ) (a) 1.602 × 10 −19 (640) qE a= = = m 1.67 × 10 −27 (b) v = v i + at ( ) 6.14 × 1010 m/s2 1.20 × 106 = (6.14 × 1010)t t = 1.95 × 10-5 s (c) x − xi = 21 ( vi + v )t ( )( ) x = 21 1.20 × 106 1.95 × 10 −5 = 11.7 m (d) 23.44 K = 21 mv 2 = 12 (1.67 × 10 − 27 kg)(1.20 × 106 m / s)2 = 1.20 × 10-15 J The required electric field will be in the direction of motion . We know that Work = ∆K So, 1 2 –Fd = – 2 m v i (since the final velocity = 0) 1 This becomes Eed = mvi 2 2 E = 23.45 1.60 × 10–17 J (1.60 × 10–19 C)(0.100 m) E= or 1 2 2 mv i ed = 1.00 × 103 N/C (in direction of electron's motion) The required electric field will be in the direction of motion . 1 2 Work done = ∆K so, –Fd = – 2 m v i which becomes eEd = K and E= K ed (since the final velocity = 0) Chapter 23 Solutions 17 Goal Solution The electrons in a particle beam each have a kinetic energy K . What are the magnitude and direction of the electric field that stops these electrons in a distance of d? G: We should expect that a larger electric field would be required to stop electrons with greater kinetic energy. Likewise, E must be greater for a shorter stopping distance, d. The electric field should be i n the same direction as the motion of the negatively charged electrons in order to exert an opposing force that will slow them down. O: The electrons will experience an electrostatic force F = qE. Therefore, the work done by the electric field can be equated with the initial kinetic energy since energy should be conserved. A: The work done on the charge is and Assuming v is in the + x direction, E is therefore in the direction of the electron beam: L: W = F ⋅ d = qE ⋅ d Ki + W = K f = 0 K + ( −e )E ⋅ di = 0 eE ⋅ ( di ) = K K E= i ed As expected, the electric field is proportional to K , and inversely proportional to d. The direction of the electric field is important; if it were otherwise the electron would speed up instead of slowing down! If the particles were protons instead of electrons, the electric field would need to be directed opposite to v in order for the particles to slow down. 23.46 (a) 2 The acceleration is given by v 2 = v i + 2a(x – xi)or Solving, a=– Now ∑ F = ma: –mgj + qE = – Therefore   mv2 qE =  – 2h + m g j   v 2 = 0 + 2a(–h) v2 2h mv 2 j 2h Gravity alone would give the bead downward impact velocity ( ) 2 9.80 m / s 2 ( 5.00 m ) = 9.90 m / s To change this to 21.0 m/s down, a downward electric field must exert a downward electric force. (b) q= m E  v 2  2h – g   =  1.00 × 10–3 kg  N · s2   (21.0 m/s)2  kg · m  2(5.00 m) – 9.80 m/s2 = 3.43 µC 4 1.00 × 10 N/C    © 2000 by Harcourt, Inc. All rights reserved. 18 23.47 Chapter 23 Solutions x 0.0500 = = 1.11 × 10-7 s = 111 ns v 4.50 × 10 5 (a) t= (b) ay = qE (1.602 × 10 −19 )(9.60 × 10 3 ) = = 9.21 × 1011 m / s 2 m (1.67 × 10 − 27 ) y − yi = vy i t + 21 ay t 2 y = 21 (9.21 × 1011 )(1.11 × 10 −7 )2 = 5.67 × 10-3 m = 5.67 mm (c) v x = 4.50 × 105 m/s vy = vy i + ay = (9.21 × 1011)(1.11 × 10-7) = 1.02 × 105 m/s ay = 23.48 t= 2vi sin θ ay t= 2(8.20 × 10 5 )sin 30.0° = 1.20 × 10-8 s) = 12.0 ns 6.86 × 1013 (b) h= vi 2 sin 2 θ (8.20 × 10 5 )2 sin 2 30.0° = = 1.23 mm 2ay 2(6.86 × 1013 ) (c) R= vi 2 sin 2 θ (8.20 × 10 5 )2 sin 60.0° = = 4.24 mm 2ay 2(6.86 × 1013 ) (a) 23.49 qE (1.602 × 10 −19 )(390) = = 6.86 × 1013 m / s 2 − 31 m (9.11 × 10 ) from projectile motion equations vi = 9.55 × 103 m/s (a) ay = eE (1.60 × 10 −19 )(720) = = 6.90 × 1010 m s 2 m (1.67 × 10 −27 ) R= vi 2 sin 2θ = 1.27 × 10-3 m ay sin 2θ = 0.961 (b) t= R R = vix vi cos θ θ = 36.9° (9.55 × 10 3 )2 sin 2θ = 1.27 × 10 −3 6.90 × 1010 so that 90.0° – θ = 53.1° If θ = 36.9°, t = 167 ns If θ = 53.1°, t = 221 ns Chapter 23 Solutions *23.50 (a) 19 The field, E1, due to the 4.00 × 10–9 C charge is in the –x direction. E1 = ke q r2 = (8.99 × 109 N · m2/C 2)(− 4.00 × 10–9 C) i = −5.75i N/C (2.50 m)2 Likewise, E2 and E3, due to the 5.00 × 10–9 C charge and the 3.00 × 10–9 C charge are E2 = ke q r2 = (8.99 × 109 N · m2/C 2)(5.00 × 10–9 C) i (2.00 m)2 = 11.2 N/C E3 = (8.99 × 109 N · m2/C 2)(3.00 × 10–9 C) i = 18.7 N/C (1.20 m)2 ER = E1 + E2 + E3 = 24.2 N/C (b) in +x direction. E1 = ke q r2 = ( −8.46 N / C)(0.243i + 0.970j) E2 = ke q r2 = (11.2 N / C)( +j) E3 = ke q r2 = ( 5.81 N / C)( −0.371i + 0.928j) Ex = E1x + E3x = – 4.21i N/C ER = 9.42 N/C Ey = E1y + E2y + E3y = 8.43j N/C θ = 63.4° above –x axis ( ) 1.60 × 10 −19 C (640 N / C) qE The proton moves with acceleration ap = = = 6.13 × 1010 m s 2 −27 m 1.67 × 10 kg 23.51 ae = while the e− has acceleration (a) (1.60 × 10 −19 ) C (640 N/C) 9.11 × 10 −31 kg = 1.12 × 1014 m s 2 = 1836 ap We want to find the distance traveled by the proton (i.e., d = 21 apt 2 ), knowing: ( 4.00 cm = 21 apt 2 + 21 ae t 2 = 1837 21 apt 2 4.00 cm Thus, d = 21 apt 2 = = 1837 ) 21.8 µ m © 2000 by Harcourt, Inc. All rights reserved. 20 Chapter 23 Solutions (b) The distance from the positive plate to where the meeting occurs equals the distance the sodium ion travels (i.e., dNa = 21 aNat 2 ). This is found from: 1  eE  2 1  eE  2 t + t 4.00 cm = 21 aNat 2 + 21 aClt 2 : 4.00 cm = 2  22.99 u  2  35.45 u  23.52 4.00 cm = 21 aNat 2 + 21 (0.649aNa )t 2 = 1.65 21 aNat 2 so dNa = 21 aNat 2 = ∑Fy = 0 and T cos 15.0° = 1.96 × 10–2 N So T = 2.03 × 10–2 N or (a) q= ) 4.00 cm = 2.43 cm 1.65 From the free-body diagram shown, From ∑Fx = 0, we have 23.53 ( This may be written as qE = T sin 15.0° (2.03 × 10–2 N) sin 15.0° T sin 15.0° = = 5.25 × 10–6 C = 5.25 µC E 1.00 × 103 N/C Let us sum force components to find ∑Fx = qEx – T sin θ = 0, and ∑Fy = qEy + T cos θ – mg = 0 Combining these two equations, we get q= (b) mg (Ex cot θ + Ey) = (1.00 × 10-3)(9.80) = 1.09 × 10–8 C = 10.9 nC (3.00 cot 37.0° + 5.00) × 105 From the two equations for ∑Fx and ∑Fy we also find T= qEx –3 sin 37.0° = 5.44 × 10 N = 5.44 mN Free Body Diagram for Goal Solution Chapter 23 Solutions 21 Goal Solution A charged cork ball of mass 1.00 g is suspended on a light string in the presence of a uniform electric field, as shown in Fig. P23.53. When E = ( 3.00i + 5.00j) × 10 5 N / C , the ball is in equilibrium at θ = 37.0°. Find (a) the charge on the ball and (b) the tension in the string. G: (a) Since the electric force must be in the same direction as E, the ball must be positively charged. If we examine the free body diagram that shows the three forces acting on the ball, the sum of which must be zero, we can see that the tension is about half the magnitude of the weight. O: The tension can be found from applying Newton's second law to this statics problem (electrostatics, in this case!). Since the force vectors are in two dimensions, we must apply ΣF = ma to both the x and y directions. A: Applying Newton's Second Law in the x and y directions, and noting that ΣF = T + qE + F g = 0, ΣFx = qEx − T sin 37.0° = 0 (1) ΣFy = qEy + T cos 37.0° − mg = 0 (2) We are given Ex = 3.00 × 10 5 N / C and Ey = 5.00 × 10 5 N / C; substituting T from (1) into (2): q= mg   Ex  Ey + tan 37.0°    = (1.00 × 10 −3 kg)(9.80 m / s 2 ) = 1.09 × 10 −8 C  3.00  5  5.00 + tan 37.0°  × 10 N / C   (b) Using this result for q in Equation (1), we find that the tension is L: 23.54 T= qEx = 5.44 × 10 −3 N sin 37.0° The tension is slightly more than half the weight of the ball ( F g = 9.80 × 10 −3 N) so our result seems reasonable based on our initial prediction. (a) Applying the first condition of equilibrium to the ball gives: and qEx qA = sin θ sin θ ΣFx = qEx − T sin θ = 0 or T= ΣFy = qEy + T cos θ − mg = 0 or qB + T cos θ = mg Substituting from the first equation into the second gives: q( A cot θ + B) = mg , or (b) q= mg ( A cot θ + B) Substituting the charge into the equation obtained from ΣFx yields T= mgA mg  A  =   A cos θ + Bsin θ ( A cot θ + B) sin θ © 2000 by Harcourt, Inc. All rights reserved. Chapter 23 Solutions 22 Goal Solution A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field, as shown in Figure P23.53. When E = ( Ai + Bj) N / C , where A and B are positive numbers, the ball is i n equilibrium at the angle θ . Find (a) the charge on the ball and (b) the tension in the string. G: This is the general version of the preceding problem. The known quantities are A , B, m, g , and θ . The unknowns are q and T . O: The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 53. A: Again, Newton's second law: −T sin θ + qA = 0 (1) and + T cos θ + qB − mg = 0 (2) (a) (b) Substituting T = qA , into Eq. (2), sin θ qA cos θ + qB = mg sin θ Isolating q on the left, q= mg A cot θ + B) ( Substituting this value into Eq. (1), T= ( A cos θ + Bsin θ ) mgA L : If we had solved this general problem first, we would only need to substitute the appropriate values in the equations for q and T to find the numerical results needed for problem 53. If you find this problem more difficult than problem 53, the little list at the Gather step is useful. It shows what symbols to think of as known data, and what to consider unknown. The list is a guide for deciding what to solve for in the Analysis step, and for recognizing when we have an answer. 23.55 F= ke q1q2 r2 F1 = (8.99 × 109)(10.0 × 10–6)2 = 40.0 N (0.150)2 F3 = (8.99 × 109)(10.0 × 10–6)2 = 2.50 N (0.600)2 F2 = tan θ = 15.0 60.0 θ = 14.0° (8.99 × 109)(10.0 × 10–6)2 = 2.35 N (0.619)2 Fx = –F3 – F2 cos 14.0° = –2.50 – 2.35 cos 14.0° = – 4.78 N Fy = –F1 – F2 sin 14.0° = – 40.0 – 2.35 sin 14.0° = – 40.6 N Fnet = 2 2 Fx + Fy = Fy – 40.6 tan φ = F = – 4.78 x (– 4.78)2 + (– 40.6)2 = 40.9 N φ = 263° Chapter 23 Solutions 23.56 From Fig. A: d cos 30.0° = 15.0 cm, From Fig. B: θ = sin −1 Fq mg or d= 15.0 cm cos 30.0°   15.0 cm d   = sin −1   = 20.3°  50.0 cm   50.0 cm(cos 30.0°)  = tan θ Figure A or Fq = mg tan 20.3° (1) From Fig. C:   ke q 2 Fq = 2F cos 30.0° = 2  cos 30.0° 2  (0.300 m )  (2) Equating equations (1) and (2),   ke q 2 2 cos 30.0° = mg tan 20.3° 2  (0.300 m )  q2 = mg(0.300 m ) tan 20.3° 2ke cos 30.0° q2 = (2.00 × 10 kg)(9.80 m s )(0.300 m) tan 20.3° 2(8.99 × 10 N ⋅ m C ) cos 30.0° Figure B 2 −3 2 2 9 2 2 Figure C q = 4.2 0 × 10 −14 C 2 = 2.05 × 10 −7 C= 0.205 µ C 23.57 Charge Q/2 resides on each block, which repel as point charges: F= ke(Q/2)(Q/2) = k(L – L i) L2 Q = 2L 23.58 23 k ( L − Li ) = 2(0.400 m ) ke (100 N / m)(0.100 m) (8.99 × 10 9 N ⋅ m 2 / C2 ) = 26.7 µC Charge Q /2 resides on each block, which repel as point charges: F = Solving for Q , Q = 2L k ( L − Li ) ke © 2000 by Harcourt, Inc. All rights reserved. ke (Q 2)(Q 2) L2 = k(L − L i ) 24 Chapter 23 Solutions According to the result of Example 23.7, the lefthand rod creates this field at a distance d from its righthand end: *23.59 k eQ E = d(2a + d) dx keQQ dF = 2a d(d + 2a) F= k eQ 2 2a F = b ∫ x = b – 2a x(xdx + 2a) = k eQ 2  1 2a + x  b – ln 2a  2a x  b – 2a +k eQ 2  2a + b b  k eQ 2 b2 k e Q 2  b2  – ln + ln = ln = ln    2 2 2 b b – 2a  (b – 2a)(b + 2a) 4a 2  4a 2  4a   b – 4a  The charge moves with acceleration of magnitude a given by *23.60 (a) qE 1.60 × 10–19 C (1.00 N/C) = 1.76 × 1011 m/s2 a= m = 9.11 × 10–31 kg Then v = v i + at = 0 + at gives 23.61 ∑F = ma = q E t= v 3.00 × 107 m/s = = 171 µs a 1.76 × 1011 m/s2 (b) vm (3.00 × 107 m/s)(1.67 × 10–27 kg) v = 0.313 s t = a = qE = (1.60 × 10–19 C)(1.00 N/C) (c) From t = vm , as E increases, t gets shorter qE 90.0° in inverse proportion. Q = ∫ λ dl = ∫ –90.0° λ 0 cos θ Rd θ = λ 0 R sin θ Q = 12.0 µC = (2λ 0 )(0.600) m = 12.0 µC 90.0° –90.0° = λ 0 R [1 – (–1)] = 2λ 0 R λ 0 = 10.0 µC/m so (  ( 3.00 µ C) λ 0 cos 2 θ Rdθ 1  ( 3.00 µ C)(λ dl)  1 dFy =    cos θ =  4 π e0  4 π e0  R2 R2 )    90.0° N · m2 (3.00 × 10–6 C)(10.0 × 10–6 C/m)  Fy = ∫ –90.0°  8.99 × 109 cos2 θ d θ  (0.600 m) C2   Fy = 8.99( 30.0) 10 −3 N 0.600 ( ( Fy = (0.450 N ) 1 π 2 1 4 π /2 )∫( − π /2 1 2 1 2 ) + cos 2θ dθ π /2 ) + sin 2θ − π /2 = 0.707 N Downward. Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0. Chapter 23 Solutions r = 2(0.100 m )sin10.0° = 3.47 × 10 −2 m At equilibrium, the distance between the charges is 23.62 25 Now consider the forces on the sphere with charge +q , and use ΣFy = 0: ΣFy = 0: T cos 10.0° = mg, or T = ΣFx = 0: Fnet = F2 − F1 = T sin10.0° mg cos 10.0° (1) (2) Fnet is the net electrical force on the charged sphere. from (2) by use of (1). Fnet = ( )( Eliminate T ) mg sin 10.0° = mg tan 10.0° = 2.00 × 10 −3 kg 9.80 m / s 2 tan 10.0° = 3.46 × 10 −3 N cos 10.0° Fnet is the resultant of two forces, F1 and F 2 . F 1 is the attractive force on +q exerted by –q, and F 2 is the force exerted on +q by the external electric field. Fnet = F2 – F1 or F2 = Fnet + F1 ( (5.00 × 10 C)(5.00 × 10 /C ) (3.47 × 10 m) −8 F1 = 8.99 × 10 N ⋅ m 9 2 2 −3 2 −8 C ) = 1.87 × 10 −2 N Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10 −3 N + 1.87 × 10 −2 N = 2.21 × 10 − 2 N and F2 = qE, or 23.63 (a) E= F2 2.21 × 10 − 2 N = = 4.43 × 105 N/C = 443 kN/C q 5.00 × 10 − 8 C From the 2Q charge we have Combining these we find (b) Fe − T 2 sin θ 2 = 0 and mg − T 2 cos θ 2 = 0 T sin θ 2 Fe = 2 = tan θ 2 mg T 2 cos θ 2 From the Q charge we have Fe − T1 sin θ1 = 0 and mg − T1 cos θ1 = 0 Combining these we find T sin θ1 Fe = 1 = tan θ1 mg T1 cos θ1 Fe = or θ2 = θ1 ke 2QQ 2keQ 2 = r2 r2 If we assume θ is small then . Substitute expressions for F e and tan θ into either equation found in part (a) and solve for r. Fe = tan θ mg then and solving for r we find © 2000 by Harcourt, Inc. All rights reserved. 26 Chapter 23 Solutions At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be located by determining the angle θ corresponding to equilibrium. In terms of lengths s, 1 a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an attractive force 2 23.64 keQ q (s + 21 a 3)2 . The other two charges exert equal repulsive forces of magnitude keQq r 2 . The horizontal components of the two repulsive forces add, balancing the attractive force,   2 cos θ 1 =0 Fnet = keQq − 2 2 1 (s + 2 a 3)   r From Figure P23.64, r= 1a 2 sin θ s = 21 a cot θ The equilibrium condition, in terms of θ , is Thus the equilibrium value of θ is Fnet = 1    4 k Qq 2 cos θ sin 2θ −  =0  a2  e  ( 3 + cot θ )2  2 cos θ sin 2 θ ( 3 + cot θ )2 = 1. One method for solving for θ is to tabulate the left side. To three significant figures the value of θ corresponding to equilibrium is 81.7°. The distance from the origin to the equilibrium position is x = 21 a( 3 + cot 81.7°) = 0.939a 23.65 (a) θ 2 cos θ sin 2 θ ( 3 + cot θ )2 60° 70° 80° 90° 81° 81.5° 81.7° 4 2.654 1.226 0 1.091 1.024 0.997 The distance from each corner to the center of the square is ( L 2 )2 + ( L 2)2 = L 2 The distance from each positive charge to −Q is then z 2 + L2 2 . Each positive charge exerts a force directed along the line joining q and −Q, of magnitude keQq z + L2 2 2 The line of force makes an angle with the z-axis whose cosine is The four charges together exert forces whose x and y components add to zero, while the z-components add to z z + L2 2 2 F= − (z 4keQ q z 2 + L2 2 ) 32 k Chapter 23 Solutions (b) For z << L, the magnitude of this force is Fz ≈ − 4keQqz ( L 2) 2 32 Therefore, the object’s vertical acceleration is of the form with ω 2 = 27  4( 2)3 2 keQq  = −  z = maz L3   az = − ω 2 z 4( 2) keQq keQq 128 = mL3 mL3 32 Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by T= 23.66 (a) 2π 2π = ω (128)1 4 mL3 = keQq π (8)1 4 mL3 keQq qE   , F = qE + mg = m g +  m The total non-contact force on the cork ball is: which is constant and directed downward. Therefore, it behaves like a simple pendulum i n the presence of a modified uniform gravitational field with a period given by: T = 2π (b) (a) qE g + m 0.500 m = 2π 9.80 m / s 2 (2.00 × 10 + (2.00 × 10 −6 )( −6 )( C 1.00 × 10 5 N / C 1.00 × 10 − 3 kg Yes . Without gravity in part (a), we get T = 2π 23.67 L 0.500 m ) T = 2π C 1.00 × 10 5 N / C 1.00 × 10 − 3 kg ) = 0.307 s L qE m = 0.314 s (a 2.28% difference). Due to symmetry the field contribution from each negative charge is equal and opposite to each other. Therefore, their contribution to the net field is zero. The field contribution of the +q charge is E= ke q ke q 4k q = = e r 2 ( 3 a 2 4) 3a 2 in the negative y direction, i.e., E = y x − 4ke q j 3a2 © 2000 by Harcourt, Inc. All rights reserved. 28 Chapter 23 Solutions (b) If Fe = 0, then E at P must equal zero. In order for the field to cancel at P , the − 4q must be above + q on the y-axis. Then, E=0=− ke q (1.00 m) 2 + ke (4q) , which reduces to y 2 = 4.00 m 2 . y2 Thus, y = ± 2.00 m . Only the positive answer is acceptable since the − 4q must be located above + q. Therefore, the − 4q must be placed 2.00 meters above point P along the + y − axis . The bowl exerts a normal force on each bead, directed along the radius line or at 60.0° above the horizontal. Consider the free-body diagram of the bead on the left: 23.68 ΣFy = nsin 60.0° −mg = 0 , n= Also, ΣFx = −Fe + ncos 60.0° = 0, or ke q 2 mg mg = ncos 60.0° = = 2 tan 60.0° 3 R Thus, 23.69 (a) mg sin 60.0° or q=  mg  R   ke 3  n Fe mg 12 There are 7 terms which contribute: 3 are s away (along sides) 3 are 1 is 2 s away (face diagonals) and sin θ = 3 s away (body diagonal) and sin φ = 1 = cos θ 2 1 3 The component in each direction is the same by symmetry. F= (b) ke q 2 s2 2 1   1+ 2 2 + 3 3  (i + j + k) = F = F 2x + F 2y + F 2z = 3.29 60.0˚ ke q 2 (1.90)(i + j + k) s2 ke q 2 away from the origin s2 Chapter 23 Solutions 23.70 (a) Zero contribution from the same face due to symmetry, opposite face contributes 4  ke q  sin φ  r2  E=4 (b) *23.71 r= where 2 2  s  s + + s2 = 1.5 s = 1.22 s  2  2 ke q s kq 4 ke q = = 2.18 e2 s (1.22)3 s2 r3 sin φ = s/r The direction is the k direction.   k λ −x i + 0.150 m j dx −x i + 0.150 m j ) e (  = 2 3 2 2 2 x + (0.150 m )  x 2 + (0.150 m )2  x 2 + (0.150 m ) ke dq dE = E=∫ all charge [ dE = ke λ ∫ 0.400 m x=0  +i  E = ke λ  2 2  x + (0.150 m )  ] dE x (−x i + 0.150 m j)dx [x 2 0.400 m 0 + (0.150 m ) E = ( −1.36i + 1.96 j ) × 10 3 N C = By symmetry  ∑ Ex = 0. q ∑ Ey = ke  (a2 + y 2 ) sin θ x ] 2 3 2 (0.150 m) j x + (0.150 m)2 x 2 + (0.150 m)2 0.400 m  0      But sin θ = (−1.36i + 1.96 j ) kN C Using the distances as labeled, + q 2q  sin θ − 2  2 (a + y ) y  2  y 1 , so E = ∑ Ey = 2ke q  2 − 2 2 3 2 y  (a 2 +y 2 )  (a + y ) y Expand (a2 +y 2 )− 3 2 as (a2 + y 2 )− 3 2 = y −3 − (3 2)a2 y −5 + . . . Therefore, for a << y, we can ignore terms in powers higher than 2, and we have  1  3  a2 1 E = 2ke q  2 − − 2  or 4   2 y y  y y 0.150 m  N ⋅ m2   −9 C  E =  8.99 × 10 9  35.0 × 10 m  [ i ( 2.34 − 6.67 ) m + j (6.24 − 0) m ] C2    23.72 29  k 3qa 2  E = − e 4  j y   © 2000 by Harcourt, Inc. All rights reserved. dq 30 23.73 Chapter 23 Solutions E = Ex = The field on the axis of the ring is calculated in Example 23.8, ke xQ (x 2 + a2 )3/2 The force experienced by a charge –q placed along the axis of the ring is   x F = −keQq  2 2 3/2   (x + a )  and when x << a, this becomes F=  keQq  x  a3  This expression for the force is in the form of Hooke's law, 23.74 with an effective spring constant of k = keQq a3 Since ω = 2 π f = k m , we have f= 1 2π keQq ma3 The electrostatic forces exerted on the two charges result in a net torque τ = −2Fa sin θ = −2Eqa sin θ . For small θ, sin θ ≈ θ and using p = 2qa, we have τ = –Epθ. The torque produces an angular acceleration given by τ = Iα = I Combining these two expressions for torque, we have d 2θ  Ep  + θ =0 dt 2  I  This equation can be written in the form Ep d 2θ = − ω 2θ where ω 2 = 2 I dt This is the same form as Equation 13.17 and the frequency of oscillation is found by comparison with Equation 13.19, or f= 1 2π d 2θ dt 2 pE = I 1 2π 2qaE I Chapter 24 Solutions 24.1 (a) ΦE = EA cos θ = (3.50 × 103)(0.350 × 0.700) cos 0° = 858 N · m2/C (b) θ = 90.0° (c) ΦE = (3.50 × 103)(0.350 × 0.700) cos 40.0° = 657 N · m2/C ΦE = 0 24.2 ΦE = EA cos θ = (2.00 × 104 N/C)(18.0 m2)cos 10.0° = 355 kN · m2/C 24.3 ΦE = EA cos θ A = π r 2 = π (0.200)2 = 0.126 m2 5.20 × 105 = E (0.126) cos 0° E = 4.14 × 106 N/C = 4.14 MN/C The uniform field enters the shell on one side and exits on the other so the total flux is zero . 24.4 24.5 (a) A′ = (10.0 cm)( 30.0 cm) 3 0.0 cm A′ = 300 cm 2 = 0.0300 m 2 Φ E, A ′ = EA′ cos θ ( 0.0 cm ) 0.0˚ Φ E, A ′ = 7.80 × 10 4 (0.0300) cos 180° Φ E, A ′ = (b) − 2.34 kN ⋅ m 2 C ( ) Φ E, A = EA cos θ = 7.80 × 10 4 ( A) cos 60.0°  10.0 cm  2 2 A = ( 30.0 cm )( w ) = ( 30.0 cm )  = 600 cm = 0.0600 m  cos 60.0°  ( ) Φ E, A = 7.80 × 10 4 (0.0600) cos 60° = (c) + 2.34 kN ⋅ m 2 C The bottom and the two triangular sides all lie parallel to E, so Φ E = 0 for each of these. Thus, © 2000 by Harcourt, Inc. All rights reserved. Chapter 24 Solutions Φ E, total = − 2.34 kN ⋅ m 2 C + 2.34 kN ⋅ m 2 C + 0 + 0 + 0 = 0 © 2000 by Harcourt, Inc. All rights reserved. 33 34 24.6 Chapter 24 Solutions (a) Φ E = E ⋅ A = (ai + b j) ⋅ A i = aA (b) Φ E = (ai + bj) ⋅ Aj = bA (c) Φ E = (ai + bj) ⋅ Ak = 0 Only the charge inside radius R contributes to the total flux. 24.7 Φ E = q / e0 Φ E = EA cos θ through the base 24.8 Φ E = ( 52.0)( 36.0) cos 180° = –1.87 kN · m2/C Note the same number of electric field lines go through the base as go through the pyramid's surface (not counting the base). For the slanting surfaces, Φ E = +1.87 kN ⋅ m 2 / C 24.9 The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side of the cone is Φ E = ∫ E ⋅ dA = E R h . This is the same as the flux that exits the right side of the cone. Note that for a uniform field only the cross sectional area matters, not shape. *24.10 (a) E= k eQ r2 8.90 × 102 = (8.99 × 109)Q , (0.750)2 But Q is negative since E points inward. Q = – 5.56 × 10–8 C = – 55.6 nC 24.11 (b) The negative charge has a spherically symmetric charge distribution. (a) ΦE = (b) Since the net electric flux is negative, more lines enter than leave the surface. qin ( +5.00 µ C − 9.00 µ C + 27.0 µ C − 84.0 µ C) = = – 6.89 × 106 N · m2/C = – 6.89 MN · m2/C e0 8.85 × 10 −12 C 2 / N ⋅ m 2 Chapter 24 Solutions ΦE = 24.12 qin e0 Through S1 Through S2 Φ E = 24.13 (a) Q e0 −2Q + Q − Q 2Q = − e0 e0 ΦE = Through S4 ΦE = 0 One-half of the total flux created by the charge q goes through the plane. Thus, q 1 1 q  Φ E, total =   = 2e0 2 2  e0  The square looks like an infinite plane to a charge very close to the surface. Hence, q 2e0 The plane and the square look the same to the charge. The flux through the curved surface is equal to the flux through the flat circle, E0 π r 2 . 24.14 24.15 − + Q−Q = 0 e0 Φ E, square ≈ Φ E, plane = (c) −2Q + Q = e0 Through S3 Φ E, plane = (b) ΦE = (a) +Q 2 e0 (b) –Q 2 e0 Simply consider half of a closed sphere. (from ΦΕ, total = ΦΕ, dome + ΦΕ, flat = 0) © 2000 by Harcourt, Inc. All rights reserved. 35 Chapter 24 Solutions 36 Goal Solution A point charge Q is located just above the center of the flat face of a hemisphere of radius R, as shown i n Figure P24.15. What is the electric flux (a) through the curved surface and (b) through the flat face? G: From Gauss’s law, the flux through a sphere with a point charge in it should be Q e0 , so we should expect the electric flux through a hemisphere to be half this value: Φ curved = Q 2e0 . Since the flat section appears like an infinite plane to a point just above its surface so that half of all the field lines from the point charge are intercepted by the flat surface, the flux through this section should also equal Q 2e0 . O: We can apply the definition of electric flux directly for part (a) and then use Gauss’s law to find the flux for part (b). A : (a) With δ very small, all points on the hemisphere are nearly at distance R from the charge, so the field everywhere on the curved surface is keQ / R 2 radially outward (normal to the surface). Therefore, the flux is this field strength times the area of half a sphere:  Q Φ curved = ∫ E ⋅ dA = Elocal Ahemisphere = ke 2  R  ( 21 )(4πR2 ) = 4π1e Q( 2 π ) = 0 Q 2e0 (b) The closed surface encloses zero charge so Gauss's law gives Φ curved + Φ flat = 0 or Φ flat = −Φ curved = −Q 2e0 L : The direct calculations of the electric flux agree with our predictions, except for the negative sign i n part (b), which comes from the fact that the area unit vector is defined as pointing outward from an enclosed surface, and in this case, the electric field has a component in the opposite direction (down). 24.16 24.17 qin 12.0 × 10 −6 = = 1.36 × 106 N ⋅ m 2 / C = 1.36 MN · m2/C e0 8.85 × 10 −12 (a) Φ E, shell = (b) Φ E, half shell = 21 (1.36 × 106 N ⋅ m 2 / C) = 6.78 × 10 5 N ⋅ m 2 / C = 678 kN · m2/C (c) the same number of field lines will pass through each surface, no matter how the No, radius changes. From Gauss's Law, Φ E = ∫ E ⋅ dA = Thus, ΦE = qin . e0 Q 0.0462 × 10 −6 C = = e0 8.85 × 10 -12 C 2 N ⋅ m 2 5.22 kN ⋅ m 2 C Chapter 24 Solutions 24.18 37 If R ≤ d, the sphere encloses no charge and Φ E = qin / e0 = 0 If R > d, the length of line falling within the sphere is 2 R 2 − d 2 so 24.19 ΦΕ = 2λ R 2 − d 2 e0 (ΦE )one face = (ΦE )one face = 24.20 ( ) Q−6 q 6e0 Q − 6 q (5.00 − 6.00) × 10 − 6 C ⋅ N ⋅ m 2 = = − 18.8 kN ⋅ m 2/C 6e0 6 × 8.85 × 10 −12 C 2 Q−6 q 6e0 When R < d, the cylinder contains no charge and Φ Ε = 0 . When R > d, 24.22 ) The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q / e0 , of which one-sixth goes through each face: (ΦE )one face = 24.21 ( The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q / e0 , of which one-sixth goes through each face: ΦE = Φ E, hole = E ⋅ A hole = qin λL = e0 e0 ( ) ( )( ) (  8.99 × 10 9 N ⋅ m 2 C 2 10.0 × 10 −6 C   keQ  2 π r =  π 1.00 × 10 −3 m  2  R2    0.100 m ( )   Φ E, hole = 28.2 N ⋅ m 2 C © 2000 by Harcourt, Inc. All rights reserved. ) 2 38 Chapter 24 Solutions Chapter 24 Solutions ΦE = 24.23 (a) qin 170 × 10 −6 C = = 1.92 × 107 N ⋅ m 2 C e0 8.85 × 10 -12 C 2 N ⋅ m 2 (ΦE )one face = 61 ΦE = (ΦE )one face 24.24 39 1.92 × 107 N ⋅ m 2 C 6 = 3.20 MN ⋅ m 2 C (b) Φ E = 19.2 MN ⋅ m 2 C (c) The answer to (a) would change because the flux through each face of the cube would not be equal with an unsymmetrical charge distribution. The sides of the cube nearer the charge would have more flux and the ones farther away would have less. The answer to (b) would remain the same, since the overall flux would remain the same. (a) ΦE = qin e0 8.60 × 10 4 = qin 8.85 × 10 −12 qin = 7.61 × 10 − 7 C = 761 nC (b) (c) 24.25 Since the net flux is positive, the net charge must be positive . It can have any distribution. The net charge would have the same magnitude but be negative. No charge is inside the cube. The net flux through the cube is zero. Positive flux comes out through the three faces meeting at g. These three faces together fill solid angle equal to one1 eighth of a sphere as seen from q, and together pass flux (q e0 ) . Each face containing a 8 intercepts equal flux going into the cube: 0 = Φ E, net = 3Φ E, abcd + q / 8e0 Φ E, abcd = −q / 24e0 © 2000 by Harcourt, Inc. All rights reserved. 40 Chapter 24 Solutions The charge distributed through the nucleus creates a field at the surface equal to that of a point charge at its center: E = ke q r 2 24.26 E= (8.99 × 109 Nm2/C 2)(82 × 1.60 × 10–19 C) [(208)1/3 1.20 × 10–15 m] 2 E = 2.33 × 1021 N/C 24.27 away from the nucleus (a) E= ke Qr = 0 a3 (b) E= ke Qr a3 (c) E= ke Q (8.99 × 109)(26.0 × 10–6) = = 1.46 MN/C 2 r (0.400)2 (d) E = = (8.99 × 109)(26.0 × 10–6)(0.100) = 365 kN/C (0.400)3 (8.99 × 109)(26.0 × 10–6) ke Q = = 649 kN/C 2 (0.600)2 r The direction for each electric field is radially outward. *24.28 (a) E= 2ke λ r 3.60 × 104 = 2(8.99 × 109)(Q/2.40) (0.190) Q = + 9.13 × 10–7 C = +913 nC (b) 24.29 E=0 ∫ ο qin E · dA = e 0 = ρ 2 E2π rl = e0 l π r ρ ∫ ρ dV = e lπr2 e0 0 Chapter 24 Solutions 41 ρ E = 2 e r away from the axis 0 Goal Solution Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ . electric field at distance r from the axis where r < R. Find the G: According to Gauss’s law, only the charge enclosed within the gaussian surface of radius r needs to be considered. The amount of charge within the gaussian surface will certainly increase as ρ and r increase, but the area of this gaussian surface will also increase, so it is difficult to predict which of these two competing factors will more strongly affect the electric field strength. O: We can find the general equation for E from Gauss’s law. A : If ρ is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r , contained inside the charged rod. Its volume is π r 2 L and it encloses charge ρπ r 2 L. The circular end caps have no electric flux through them; there E ⋅ dA = EdA cos 90.0° = 0. The curved surface has E ⋅ dA = EdA cos 0° , and E must be the same strength everywhere over the curved surface. ∫ Gauss’s law, E ⋅ dA = q , becomes e0 E ∫ dA = Curved Surface ρπ r 2 L e0 Now the lateral surface area of the cylinder is 2 π rL : E= Thus, E( 2 π r )L = ρr radially away from the cylinder axis 2e0 ρπ r 2 L e0 L : As we expected, the electric field will increase as ρ increases, and we can now see that E is also proportional to r . For the region outside the cylinder ( r > R), we should expect the electric field to decrease as r increases, just like for a line of charge. ( σ = 8.60 × 10 −6 C / cm 2 24.30 E= ) 100mcm  8.60 × 10 −2 σ = = 2e0 2 8.85 × 10 −12 ( ) 2 = 8.60 × 10 −2 C / m 2 4.86 × 10 9 N / C The field is essentially uniform as long as the distance from the center of the wall to the field point is much less than the dimensions of the wall. 24.31 (a) E=0 © 2000 by Harcourt, Inc. All rights reserved. 42 Chapter 24 Solutions (b) E= keQ (8.99 × 10 9 )(32.0 × 10 −6 ) = = 7.19 MN/C r2 (0.200)2 Chapter 24 Solutions The distance between centers is 2 × 5.90 × 10–15 m. Each produces a field as if it were a point charge at its center, and each feels a force as if all its charge were a point at its center. 24.32 k eq 1q 2 r2 F= N · m2 (46)2 (1.60 × 10–19 C)2  =  8.99 × 109 = 3.50 × 103 N = 3.50 kN  C 2  (2 × 5.90 × 10–15 m)2  Consider two balloons of diameter 0.2 m , each with mass 1 g , hanging apart with a 0.05 m separation on the ends of strings making angles of 10˚ with the vertical. *24.33 (a) ΣFy = T cos 10° − mg = 0 ⇒ T = mg cos 10° ΣFx = T sin 10° − Fe = 0 ⇒ Fe = T sin 10° , so  mg  2 Fe =   sin 10° = mg tan 10° = (0.001 kg ) 9.8 m s tan 10°  cos 10°  ( Fe ≈ 2 × 10 −3 N (b) ) ~10 -3 N or 1 mN ke q 2 r2 Fe = 2 × 10 −3 (8.99 × 10 N≈ 9 ( ) N ⋅ m 2 C2 q2 (0.25 m) q ≈ 1.2 × 10 −7 C 24.34 43 2 ~10 −7 C or 100 nC )( ) 8.99 × 10 9 N ⋅ m 2 C 2 1.2 × 10 −7 C ke q ≈ ≈ 1.7 × 10 4 N C 2 r2 0.25 m ( ) (c) E= (d) ΦE = q 1.2 × 10 −7 C ≈ = 1.4 × 10 4 N ⋅ m 2 C e0 8.85 × 10 −12 C 2 N ⋅ m 2 (a) ρ= Q 4 π a3 3 = ~10 kN C ~ 10 kN ⋅ m 2 C 5.70 × 10 −6 = 2.13 × 10 −2 C / m 3 3 4 π (0.0400) 3 ( ) ( )( ) ( ) ( )( ) 3 qin = ρ 43 π r 3 = 2.13 × 10 −2 43 π (0.0200) = 7.13 × 10 −7 C = 713 nC (b) 3 qin = ρ 43 π r 3 = 2.13 × 10 −2 43 π (0.0400) = 5.70 µC © 2000 by Harcourt, Inc. All rights reserved. 44 24.35 Chapter 24 Solutions (a) )[( ( ) 9 2 2 2.00 × 10 −6 C 7.00 m 2ke λ 2 8.99 × 10 N ⋅ m C E= = r 0.100 m ] E = 51.4 kN/C, radially outward (b) Φ E = EA cos θ = E(2 π r )cos 0˚ ( ) Φ E = 5.14 × 10 4 N C 2 π (0.100 m )(0.0200 m )(1.00) = 646 N ⋅ m 2 C Note that the electric field in each case is directed radially inward, toward the filament. 24.36 ( )( ) ( )( ) ( )( ) −6 9 2 2 2ke λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C = = 16.2 MN C r 0.100 m (a) E= (b) −6 9 2 2 2ke λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C E= = = 8.09 MN C r 0.200 m (c) E= −6 9 2 2 2ke λ 2 8.99 × 10 N ⋅ m C 90.0 × 10 C = = 1.62 MN C r 1.00 m 24.37 E= 9.00 × 10 − 6 C / m 2 σ = = 508 kN/C, upward 2e0 2(8.85 × 10 −12 C 2 / N ⋅ m 2 ) 24.38 From Gauss's Law, EA = Q e0 Q σ = A = e0 E = (8.85 × 10-12)(130)= 1.15 × 10-9 C/m2 = 1.15 nC/m2 ∫ ο 24.39 E dA = E(2π rl ) = (a) r = 3.00 cm E = 0 (b) r = 10.0 cm E= (c) r = 100 cm E= qin e0 E= qin/l 2π e0 r = λ 2π e0 r inside the conductor 30.0 × 10–9 2π (8.85 × 10–12)(0.100) = 5400 N/C, outward 30.0 × 10–9 = 540 N/C, outward 2π (8.85 × 10–12)(1.00) Chapter 24 Solutions 45 Just above the aluminum plate (a conductor), the electric field is E = σ ′ e0 where the charge Q is divided equally between the upper and lower surfaces of the plate: *24.40 σ′ = Thus (Q 2) = A Q 2A and E= Q 2e0 A For the glass plate (an insulator), E = σ / 2e0 where σ = Q / A since the entire charge Q is o n the upper surface. Therefore, E = Q 2e0 A The electric field at a point just above the center of the upper surface is the same for each of the plates. E= *24.41 (a) Q , vertically upward in each case (assuming Q > 0) 2e0 A E = σ e0 σ = (8.00 × 104)(8.85 × 10–12) = 7.08 × 10-7 C/m2 σ = 708 nC/m2 , positive on one face and negative on the other. (b) Q σ= A Q = σA = (7.08 × 10–7) (0.500)2 C Q = 1.77 × 10–7 C = 177 nC , positive on one face and negative on the other. 24.42 Use Gauss's Law to evaluate the electric field in each region, recalling that the electric field is zero everywhere within conducting materials. The results are: E = 0 inside the sphere and inside the shell E = ke Q between sphere and shell, directed radially inward r2 E = ke 2Q outside the shell, directed radially inward r2 Charge –Q is on the outer surface of the sphere . Charge +Q is on the inner surface of the shell , © 2000 by Harcourt, Inc. All rights reserved. 46 Chapter 24 Solutions and +2Q is on the outer surface of the shell. Chapter 24 Solutions 47 The charge divides equally between the identical spheres, with charge Q/2 on each. Then they repel like point charges at their centers: 24.43 F= k e Q2 ke(Q/2)(Q/2) 8.99 × 109 N · m2(60.0 × 10-6 C)2 = = = 2.00 N 4 C 2(2.01 m)2 4(L + 2R)2 (L + R + R)2 The electric field on the surface of a conductor varies inversely with the radius of curvature of the surface. Thus, the field is most intense where the radius of curvature is smallest and viseversa. The local charge density and the electric field intensity are related by *24.44 E= (a) σ e0 σ = e0E or Where the radius of curvature is the greatest, ( )( ) σ = e0Emin = 8.85 × 10 −12 C 2 N ⋅ m 2 2.80 × 10 4 N C = 248 nC m 2 (b) Where the radius of curvature is the smallest, ( )( ) σ = e0Emax = 8.85 × 10 −12 C 2 N ⋅ m 2 5.60 × 10 4 N C = 496 nC m 2 24.45 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length = – λ. qin 0 = λ + qin ⇒ = –λ Outside surface: The total charge on the metal cylinder is 2λ l = qin + qout . qout = 2λ l + λ l so the outside charge/length = 3λ (b) E= 2ke (3λ) r = 6ke λ 3λ = r 2 π e0 r © 2000 by Harcourt, Inc. All rights reserved. 48 24.46 Chapter 24 Solutions ( )( ) (a) 8.99 × 10 9 6.40 × 10 −6 keQ E= 2 = = 2.56 MN/C, radially inward r (0.150)2 (b) E=0 Chapter 24 Solutions 24.47 (a) 49 The charge density on each of the surfaces (upper and lower) of the plate is: −8  q  1 (4.00 × 10 C) = σ=1 = 8.00 × 10 − 8 C / m 2 = 80.0 nC / m 2 2  A  2 (0.500 m)2 24. 48 (b)  8.00 × 10 − 8 C / m 2  σ E= k= k= −12 2  e0  C / N ⋅ m 2   8.85 × 10 (c) E= (a) The charge +q at the center induces charge −q on the inner surface of the conductor, where its surface density is: ( − 9.04 kN / C) k σa = (b) 24.50 −q 4 π a2 The outer surface carries charge Q + q with density σb = 24.49 (9.04 kN / C) k (a) E=0 (b) E= (c) E=0 (d) E= Q+q 4π b2 ( )( ) ( )( ) 8.99 × 10 9 8.00 × 10 −6 keQ = = 7.99 × 107 N / C = 79.9 MN/C r2 (0.0300)2 8.99 × 10 9 4.00 × 10 −6 keQ = = 7.34 × 106 N / C = 7.34 MN/C r2 (0.0700)2 An approximate sketch is given at the right. Note that the electric field lines should be perpendicular to the conductor both inside and outside. © 2000 by Harcourt, Inc. All rights reserved. 50 24.51 Chapter 24 Solutions (a) Uniform E, pointing radially outward, so ΦE = EA. ds = Rd θ , and the circumference is 2π r = 2π R sin θ θ θ 0 0 The arc length is θ A = ∫ 2 π rds = ∫ (2 π R sin θ )Rdθ = 2 π R 2 ∫ sin θ dθ = 2 π R 2 (− cos θ ) 0 = 2 π R 2 (1 − cos θ ) ΦE = 1 Q Q ⋅ 2 π R 2 (1 − cos θ ) = (1 − cos θ ) 2 4 π e0 R 2e0 [independent of R!] (b) For θ = 90.0° (hemisphere): ΦE = Q (1 − cos 90°) = 2e0 (c) For θ = 180° (entire sphere): ΦE = Q (1 − cos 180°) = 2e0 *24.52 In general, E = ay i + bz j + cx k In the xy plane, z = 0 and E = ay i + cx k Φ E = ch ∫ w x=0 *24.53 (a) (b) x dx = ch x 2 = x=0 chw 2 Q e0 [Gauss's Law] z y=0 y=h x=0 Φ E = ∫ E ⋅ dA = ∫ ( ay i + cx k ) ⋅ k dA 2 w Q 2e0 x=w 2 x y dA = hdx qin = +3Q − Q = +2Q The charge distribution is spherically symmetric and qin > 0 . Thus, the field is directed radially outward . (c) E= ke qin 2keQ = r2 r2 for r ≥ c (d) Since all points within this region are located inside conducting material, E = 0 (e) Φ E = ∫ E ⋅ dA = 0 ⇒ qin = e0 Φ E = 0 (f) qin = + 3Q (g) E= ke qin 3keQ = r2 r2 (radially outward) for a ≤ r < b for b < r < c . Chapter 24 Solutions (h) 51  +3Q   4 r3  qin = ρ V =  4 3  π r 3 = +3Q 3  a  3 π a  3 ke qin ke  r3  r = 2  +3Q 3  = 3keQ 3 2 a r  a  r (i) E= (radially outward) for 0 ≤ r ≤ a (j) From part (d), E = 0 for b < r < c . Thus, for a spherical gaussian surface with b < r < c , qin = +3Q + qinner = 0 where qinner is the charge on the inner surface of the conducting shell. This yields qinner = − 3Q (k) Since the total charge on the conducting shell is E qnet = qouter + qinner = − Q , we have qouter = − Q − qinner = − Q − ( − 3Q) = +2Q (l) a b c r The sphere with large charge creates a strong field to polarize the other sphere. That means it pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side. This patch of opposite charge is smaller in amount but located in a stronger external field, so it can feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker field on the other side. 24.54 24.55 This is shown in the figure to the right. (a) ∫ E ⋅ dA = E( 4 π r For r < a, 2 )=q qin = ρ in ( 4 π r3 3 For a < r < b and c < r, For b ≤ r ≤ c, (b) e0 ) so E= qin = Q E = 0, since E = 0 pr 3e0 so that E = Q 4 π r 2e0 inside a conductor. Let q 1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero. q1 + Q = 0 Therefore, and σ1 = q1 4π b 2 = –Q 4π b 2 Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphere is uncharged, we require q1 + q2 = 0 and σ2 = q1 = 4π c2 Q 4π c2 © 2000 by Harcourt, Inc. All rights reserved. 52 Chapter 24 Solutions ∫ E ⋅ dA = E( 4 π r 24.56 (a) (−3.60 × 10 2 ) = qe in 0 ) N C 4 π (0.100 m ) = 3 Q = − 4.00 × 10 −9 C = (b) Q 2 8.85 × 10 −12 − 4.00 nC We take Q′ to be the net charge on the hollow sphere. Outside c, (+ 2.00 × 10 2 ) N C 4 π (0.500 m ) = 2 Q + Q′ 8.85 × 10 −12 C 2 N ⋅ m 2 Q + Q′ = + 5.56 × 10 −9 C, so Q′ = + 9.56 × 10 −9 C = (c) ( a < r < b) C2 N ⋅ m 2 ( r > c) + 9.56 nC For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface of the hollow sphere. Thus, Q1 = −Q = + 4.00 nC Then, if Q2 is the total charge Q2 = Q′ − Q1 = 9.56 nC − 4.00 nC = 24.57 on the outer surface of the hollow sphere, + 5.56 nC The field direction is radially outward perpendicular to the axis. The field strength depends on r but not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of radius r and length L. If r < a , ΦE = E= qin e0 λ 2 π r e0 If a < r < b, and or E( 2 π rL) = E= If r > b , λ 2 π r e0 E( 2 π rL) = E= λL e0 (r < a) ( e0 ( λ + ρπ r 2 − a2 2 π r e0 E( 2 π rL) = ) λ L + ρπ r 2 − a2 L ) ( ( a < r < b) ) λ L + ρπ b 2 − a2 L e0 Chapter 24 Solutions E= ( λ + ρπ b 2 − a2 2 π r e0 ) (r > b) © 2000 by Harcourt, Inc. All rights reserved. 53 54 Chapter 24 Solutions Consider the field due to a single sheet and let E+ and E– represent the fields due to the positive and negative sheets. The field at any distance from each sheet has a magnitude given by Equation 24.8: 24.58 E+ = E– = σ 2e0 (a) To the left of the positive sheet, E+ is directed toward the left and E– toward the right and the net field over this region is E = 0 . (b) In the region between the sheets, E+ and E– are both directed toward the right and the net field is E= (c) σ toward the right e0 To the right of the negative sheet, E+ and E– are again oppositely directed and E = 0 . The magnitude of the field due to each sheet given by Equation 24.8 is 24.59 E= (a) In the region to the left of the pair of sheets, both fields are directed toward the left and the net field is E= (b) σ to the left e0 In the region between the sheets, the fields due to the individual sheets are oppositely directed and the net field is E= (c) σ directed perpendicular to the sheet. 2e0 0 In the region to the right of the pair of sheets, both fields are directed toward the right and the net field is E= σ to the right e0 Chapter 24 Solutions 55 Goal Solution Repeat the calculations for Problem 58 when both sheets have positive uniform charge densities of value σ. Note: The new problem statement would be as follows: Two infinite, nonconducting sheets of charge are parallel to each other, as shown in Figure P24.58. Both sheets have positive uniform charge densities σ. Calculate the value of the electric field at points (a) to the left of, (b) in between, and (c) to the right of the two sheets. G: When both sheets have the same charge density, a positive test charge at a point midway between them will experience the same force in opposite directions from each sheet. Therefore, the electric field here will be zero. (We should ask: can we also conclude that the electron will experience equal and oppositely directed forces everywhere in the region between the plates?) Outside the sheets the electric field will point away and should be twice the strength due to one sheet of charge, so E = σ / e0 in these regions. O: The principle of superposition can be applied to add the electric field vectors due to each sheet of charge. A : For each sheet, the electric field at any point is E = σ (2e0 ) directed away from the sheet. (a) At a point to the left of the two parallel sheets E = E1( −i) + E2 ( −i) = 2E( −i) = − (b) At a point between the two sheets E = E1i + E2 ( −i) = 0 (c) At a point to the right of the two parallel sheets E = E1i + E2i = 2E i = σ i e0 σ i e0 L : We essentially solved this problem in the Gather information step, so it is no surprise that these results are what we expected. A better check is to confirm that the results are complementary to the case where the plates are oppositely charged (Problem 58). 24.60 The resultant field within the cavity is the superposition of two fields, one E + due to a uniform sphere of positive charge of radius 2a , and the other E − due to a sphere of negative charge of radius a centered within the cavity. 4 π r 3ρ = 4 π r 2E+ 3 e0 – so E+ = ρr ρr = 3e0 3e0 4 π r13ρ = 4 π r12E− so 3 e0 E− = ρ r1 (− 3e0 E− = −ρ(r − a) 3e0 Since r = a + r 1 , E = E+ + E− = Thus, 1)= −ρ r1 3e0 ρr ρr ρa ρa ρa − + = = 0i + j 3e0 3e0 3e0 3e0 3e0 Ex = 0 and Ey = ρa 3e0 at all points within the cavity. © 2000 by Harcourt, Inc. All rights reserved. 56 Chapter 24 Solutions First, consider the field at distance r < R from the center of a uniform sphere of positive charge (Q = +e ) with radius R. 24.61 (4π r )E = qe 2 = in 0 (a) 3 4 ρ V  +e  3 π r =4  e0  3 π R 3  e0   e E= r directed outward 3  4 π e0 R  so The force exerted on a point charge q = −e located at distance r from the center is then   e2   e F = qE = − e r = −  4 π e R 3  r = −Kr 3  4 π e0 R    0 e2 ke e 2 = 4 π e0 R 3 R3 (b) K= (c)  k e2   k e2  Fr = me ar = −  e 3  r, so ar = −  e 3  r = − ω 2 r  R   me R  f= Thus, the motion is simple harmonic with frequency (d) f = 2.47 × 10 15 1 Hz = 2π (8.99 × 10 9 )( N ⋅ m 2 C 2 1.60 × 10 −19 C (9.11 × 10 −31 ) ) ω 1 = 2π 2π ke e 2 me R 3 2 kg R 3 which yields R 3 = 1.05 × 10 −30 m 3 , or R = 1.02 × 10 −10 m = 102 pm The electric field throughout the region is directed along x; therefore, E will be perpendicular to dA over the four faces of the surface which are perpendicular to the yz plane, and E will be parallel to dA over the two faces which are parallel to the yz plane. Therefore, 24.62 ( ) ( ) ( ) ( ) Φ E = − Ex x=a A + Ex x=a+c A = − 3 + 2a 2 ab + 3 + 2(a + c)2 ab = 2abc(2a + c) Substituting the given values for a, b, and c, we find Φ E = 0.269 N · m2/C Q = ∈0 ΦE = 2.38 × 10-12 C = 2.38 pC ∫ E ⋅ dA = E(4 π r 24.63 2 )= qin e0 R (a) For r > R, qin = ∫ Ar 2 (4 π r 2 )dr = 4 π 0 r (b) For r < R, qin = ∫ Ar 2 (4 π r 2 )dr = 0 AR 5 5 4 π Ar 5 5 and and E= E= AR 5 5e0 r 2 Ar 3 5e0 Chapter 24 Solutions 24.64 57 The total flux through a surface enclosing the charge Q is Q/ e0 . The flux through the disk is Φ disk = ∫ E ⋅ dA where the integration covers the area of the disk. We must evaluate this integral and set it equal to 41 Q/ e0 to find how b and R are related. In the figure, take dA to be the area of an annular ring of radius s and width ds. The flux through dA is E · dA = E dA cos θ = E (2π sds) cos θ The magnitude of the electric field has the same value at all points within the annular ring, E= 1 Q 1 Q = 4 π e0 r 2 4 π e0 s2 + b 2 cos θ = and b b = r (s2 + b 2 )1/2 Integrate from s = 0 to s = R to get the flux through the entire disk. Φ E, disk = Qb 2e0 R ∫0 [ s ds Qb −(s2 + b 2 )1/2 = (s + b 2 )3/2 2e0 2 ] R 0 = Q 2e0 The flux through the disk equals Q/4 e0 provided that This is satisfied if R = 24.65 r qin 1 a = 4 π r 2 dr e0 e0 ∫0 r E4 π r 2 = 4π a r 2 4π a r dr = ∫ e0 0 e0 2 a 2e0 1 b = . 2 1/2 2 (R + b ) 2 3b . ∫ E ⋅ dA = E=   b 1 − 2 2 1/2  (R + b )   r = constant magnitude (The direction is radially outward from center for positive a; radially inward for negative a.) © 2000 by Harcourt, Inc. All rights reserved. 58 Chapter 24 Solutions In this case the charge density is not uniform, and Gauss's law is written as 24.66 We use a gaussian surface which is a cylinder of radius r, length charge distribution. (a) (b) (a) r so inside the cylinder, E= ρ0 r  2r  a−  2e0 3b  E= ρ0 R 2  2R  a−  3b  2e0 r When r > R, Gauss's law becomes E( 2 π rl) = 24.67 , and is coaxial with the ρ0  r a − dV . The element of volume is a cylindrical ∫  b e0 0 shell of radius r, length l, and thickness dr so that dV = 2 π rl dr. When r < R, this becomes E( 2 π rl) =  2 π r 2lρ0   a r  E( 2 π rl) =    2 − 3b  e0   ` R ρ0  r a − ( 2 π rldr ) ∫  b e0 0 or outside the cylinder, Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with one end in the yz plane and the other end containing the point x : Use Gauss's law: y q ∫ E ⋅ dA = ein0 gaussian surface By symmetry, the electric field is zero in the yz plane and is perpendicular to dA over the wall of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap containing the point x : q ∫ E ⋅ dA = ein0 or EA = a= x z x ρ ( Ax ) e0 so that at distance x from the mid-line of the slab, (b) 1 ∫ E ⋅ dA = e0 ∫ ρ dV . E= ρx e0 F ( −e )E = −  ρe  x =   me me  mee0  The acceleration of the electron is of the form a = −ω 2 x Thus, the motion is simple harmonic with frequency f= with ω 1 = 2π 2π ρe mee0 ω= ρe mee0 Chapter 24 Solutions 59 Consider the gaussian surface described in the solution to problem 67. 24.68 (a) For x > d , 2 dq = ρ dV = ρ A dx = C Ax 2 dx 1 ∫ E ⋅ dA = e0 ∫ dq EA = E= (b) 24.69 (a) Cd 3 24e0 For − E= CA e0 d/2 ∫ x 2 dx = 0 1  CA   d 3    3  e0   8  E= or d d <x< 2 2 d Cd 3 i for x > ; 2 24e0 ∫ E ⋅ dA = C x3 i for x > 0; 3e0 E=− d Cd 3 i for x < − 2 24e0 x CA 2 C Ax 3 1 dq = x dx = e0 ∫ 3e0 e0 ∫0 E=− Cx 3 i for x < 0 3e0 Gm r2 A point mass m creates a gravitational acceleration g=− The flux of this field through a sphere is ∫ g ⋅ dA = − at a distance r. ( ) Gm 4 π r 2 = − 4 π Gm r2 Since the r has divided out, we can visualize the field as unbroken field lines. The same flux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates field to make its own contribution to the net flux according to ∫ g ⋅ dA = − 4 π Gmin (b) Take a spherical gaussian surface of radius r. The field is inward so ∫ g ⋅ dA = g 4 π r and 2 cos 180° = − g 4 π r 2 − 4 π Gmin = − 4G 43 π r 3ρ Then, − g 4 π r 2 = − 4 π G 43 π r 3ρ Or, since ρ = ME / 43 π RE3 , and g= g = 43 π rρ G MEGr 3 RE or g= MEGr inward RE3 © 2000 by Harcourt, Inc. All rights reserved. Chapter 25 Solutions ∆V = –14.0 V and 25.1 Q = –N A e = – (6.02 × 1023)(1.60 × 10–19 C) = – 9.63 × 104 C W ∆V = Q , so W = Q(∆V) = (– 9.63 × 104 C)(–14.0 J/C) = 1.35 ΜJ 7.37 × 10-17 = q(115) ∆K = q∆V 25.2 q = 6.41 × 10-19 C W = ∆K = q∆V 25.3 1 2 mv 2 = e(120 V) = 1.92 × 10–17 J Thus, v = 3.84 × 10 −17 J m (a) For a proton, this becomes v= 3.84 × 10 −17 J = 1.52 × 105 m/s = 152 km/s 1.67 × 10 −27 kg (b) If an electron, v= 3.84 × 10 −17 J = 6.49 × 106 m/s = 6.49 Mm/s 9.11 × 10 −31 kg Goal Solution (a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V. (b) Calculate the speed of an electron that is accelerated through the same potential difference. G: Since 120 V is only a modest potential difference, we might expect that the final speed of the particles will be substantially less than the speed of light. We should also expect the speed of the electron to be significantly greater than the proton because, with me << mp , an equal force on both particles will result in a much greater acceleration for the electron. O: Conservation of energy can be applied to this problem to find the final speed from the kinetic energy of the particles. (Review this work-energy theory of motion from Chapter 8 if necessary.) © 2000 by Harcourt, Inc. All rights reserved. Chapter 25 Solutions A: 57 (a) Energy is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V: Ki + U i + ∆Enc = K f + U f 0 + qV + 0 = 21 mvp 2 + 0  1J  1 = (1.67 × 10 −27 kg)vp 2 (1.60 × 10 −19 C)(120 V)  1 V ⋅ C 2 vp = 1.52 × 10 5 m / s (b) The electron will gain speed in moving the other way, from Vi = 0 to V f = 120 V: Ki + U i + ∆Enc = K f + U f 0 + 0 + 0 = 21 mve 2 + qV 0 = 21 (9.11 × 10 −31 kg)ve 2 + (−1.60 × 10 −19 C)(120 J / C) ve = 6.49 × 106 m / s L: 25.4 Both of these speeds are significantly less than the speed of light as expected, which also means that we were justified in not using the relativistic kinetic energy formula. (For precision to three significant digits, the relativistic formula is only needed if v is greater than about 0.1 c.) 1 For speeds larger than one-tenth the speed of light, 2 mv 2 gives noticeably wrong answers for kinetic energy, so we use    2 1 1 K = mc 2  − 1 = 9.11 × 10 −31 kg 3.00 × 108 m / s  − 1 = 7.47 × 10–15 J 2 2 2  1 − 0.400   1− v /c  ( )( ) Energy is conserved during acceleration: Ki + U i + ∆E = Kf + U f 0 + qV i + 0 = 7.47 × 10–15 J + qV f The change in potential is V f – V i : V f – Vi = –7.47 × 10–15 J – 7.47 × 10–15 J = = + 46.7 kV q –1.60 × 10–19 C The positive answer means that the electron speeds up in moving toward higher potential. 25.5 W = ∆K = − q ∆V ( )( 0 − 21 9.11 × 10 −31 kg 4.20 × 10 5 m / s ) 2 ( ) = − − 1.60 × 10 −19 C ∆V From which, ∆V = – 0.502 V © 2000 by Harcourt, Inc. All rights reserved. 58 *25.6 Chapter 25 Solutions (a) We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm). ∆U = – (work done) ∆U = −( work from origin to (20.0 cm, 0) ) − ( work from (20.0 cm, 0) to (20.0 cm, 50.0 cm) ) Note that the last term is equal to 0 because the force is perpendicular to the displacement. ∆U = – (qEx)(∆x) = – (12.0 × 10–6 C)(250 V/m)(0.200 m) = – 6.00 × 10–4 J (b) E= *25.7 *25.8 ∆V = 6.00 × 10–4 J ∆U = – = – 50.0 J/C = – 50.0 V q 12.0 × 10–6 C 3 ∆V 25.0 × 10 J/C = = 1.67 × 106 N/C = 1.67 MN/C 1.50 × 10–2 m d (a) ∆V = Ed = (5.90 × 103 V/m)(0.0100 m) = 59.0 V (b) 1 2 1 2 mv 2f = q(∆V); (9.11 × 10–31) vf2 = (1.60 × 10–19)(59.0) vf = 4.55 × 106 m/s 25.9 1 2 ( ) ∆U = − m v f 2 − vi 2 = − ∆U = q ∆V: ∆V = – 38.9 V 1 9.11 × 10 −31 2 ( kg  1.40 × 10 5 m / s  + 6.23 × 10–18 = (–1.60 × 10–19)∆V The origin is at higher potential. ) − (3.70 × 10 2 6 ) 2 m / s  = 6.23 × 10 −18 J  Chapter 25 Solutions B C B A A C 59 VB − V A = − ∫ E ⋅ ds = − ∫ E ⋅ ds − ∫ E ⋅ ds *25.10 VB − V A = (−Ecos180°) 0.500 ∫ 0.400 dy − (Ecos 90.0°) ∫ dx −0.200 −0.300 VB – VA = (325)(0.800) = + 260 V 25.11 (a) Arbitrarily choose V = 0 at x = 0 . Then at other points, V = − Ex and U e = QV = −QEx . Between the endpoints of the motion, (K + U s + U e )i = (K + U s + U e ) f 2 0 + 0 + 0 = 0 + 21 kxmax − QExmax so the block comes to rest when the spring is stretched by an amount xmax = (b) At equilibrium, x= (c) ( )( ) 5 −6 2QE 2 50.0 × 10 C 5.00 × 10 V m = = 0.500 m 100 N m k ( ΣFx = − Fs + Fe = 0 or kx = QE. Thus, the equilibrium position is at )( ) 50.0 × 10 −6 C 5.00 × 10 5 N C QE = = 0.250 m k 100 N m The equation of motion for the block is ΣFx = −kx + QE = m so the equation of motion becomes: d 2 ( x ′ + QE k ) QE   + QE = m −k x ′ + , or  k  dt 2 QE d2x QE . Let x ′ = x − , or x = x ′ + 2 k k dt d2x′  k x′ =− 2  m dt ( ) This is the equation for simple harmonic motion ax ′ = − ω 2 x ′ , with ω = k m . The period of the motion is then T= (d) 4.00 kg 2π m = 2π = 2π = 1.26 s k ω 100 N m (K + U s + U e )i + ∆E = (K + U s + U e ) f 2 0 + 0 + 0 − µ k mgxmax = 0 + 21 kxmax − QExmax © 2000 by Harcourt, Inc. All rights reserved. 60 Chapter 25 Solutions xmax 25.12 (a) [( )( ) )] ( −6 5 2 2(QE − µ k mg) 2 50.0 × 10 C 5.00 × 10 N C − 0.200( 4.00 kg ) 9.80 m s = = = 0.343 m k 100 N m Arbitrarily choose V = 0 at 0. Then at other points V = − Ex and U e = QV = −QEx . Between the endpoints of the motion, (K + U s + U e )i = (K + U s + U e ) f 2 0 + 0 + 0 = 0 + 21 kxmax − QExmax xmax = so 2QE k ΣFx = − Fs + Fe = 0 or kx = QE. So the equilibrium position is at x = (b) At equilibrium, (c) The block's equation of motion is ΣFx = − kx + QE = m the equation of motion becomes: d 2 ( x ′ + QE k ) QE   + QE = m −k x ′ + , or  k  dt 2 QE k d2x QE QE . Let x ′ = x − , or x = x ′ + , so 2 k k dt d2x′  k x′ =− 2  m dt ( ) This is the equation for simple harmonic motion ax ′ = − ω 2 x ′ , with ω = k m The period of the motion is then (d) T= 2π m = 2π k ω (K + U s + U e )i + ∆E = (K + U s + U e ) f 2 0 + 0 + 0 − µ k mgxmax = 0 + 21 kxmax − QExmax xmax = 25.13 2(QE − µ k mg) k For the entire motion, y − yi = vyi t + 21 ay t 2 0 − 0 = vi t + 21 ay t 2 ∑ Fy = may : −mg − qE = − E= For the upward flight: 2vi t so ay = − and E =− and ymax = 41 vi t 2mvi t m  2vi  −g  q t m  2vi  −g j  q t 2 2 vyf = vyi + 2ay (y − yi )  2v  0 = vi2 + 2 − i (ymax − 0)  t  Chapter 25 Solutions ∆V = ∫ y max 0 ∆V = E ⋅ dy = + m  2vi  −g y  q t y max 0 = ( m  2vi  − g 41 vi t  q t [ ) ] 2.00 kg  2(20.1 m s)  − 9.80 m s 2  41 (20.1 m s)( 4.10 s) = 40.2 kV   5.00 × 10 −6 C  4.10 s © 2000 by Harcourt, Inc. All rights reserved. 61 62 Chapter 25 Solutions Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length, V = − Ed and U e = − λ LEd 25.14 (a) (K + U)i = (K + U) f 0 + 0 = 21 µ Lv 2 − λ LEd v= (b) 2λ Ed = µ 2(40.0 × 10 − 6 C / m)(100 N / C)(2.00 m) = 0.400 m/s (0.100 kg / m) The same. Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position of the charge is – E · s = –EL cos θ. At the final point a, V = –EL. Suppose the table is frictionless: (K + U)i = (K + U) f 25.15 0 − qEL cos θ = 21 mv 2 − qEL v= = *25.16 (a) 2(2.00 × 10 − 6 C)(300 N / C)(1.50 m)(1 − cos 60.0°) = 0.300 m/s 0.0100 kg The potential at 1.00 cm is V 1 = ke (b) 2 qEL(1 − cos θ ) = m q (8.99 × 109 N · m2/C 2)(1.60 × 10–19 C) = = 1.44 × 10–7 V r 1.00 × 10–2 m The potential at 2.00 cm is V 2 = ke q (8.99 × 109 N · m2/C 2)(1.60 × 10–19 C) = = 0.719 × 10–7 V r 2.00 × 10–2 m Thus, the difference in potential between the two points is ∆V = V 2 – V 1 = –7.19 × 10–8 V (c) The approach is the same as above except the charge is – 1.60 × 10–19 C. This changes the sign of all the answers, with the magnitudes remaining the same. That is, the potential at 1.00 cm is –1.44 × 10–7 V The potential at 2.00 cm is – 0.719 × 10–7 V, so ∆V = V 2 – V 1 = 7.19 × 10–8 V . Chapter 25 Solutions 25.17 (a) Since the charges are equal and placed symmetrically, F = 0 (b) Since F = qE = 0, E = 0 (c) q N · m2  2.00 × 10–6 C   V = 2k e r = 2  8.99 × 109  C 2   0.800 m   V = 4.50 × 104 V = 45.0 kV 25.18 (a) Ex = ke q1 ke q2 + 2 x (x – 2.00)2 =0 becomes –2q   +q E x = ke  2 + =0 (x – 2.00)2 x 2qx2 = q(x – 2.00)2 Dividing by ke, x2 + 4.00x – 4.00 = 0 Therefore E = 0 when x = −4.00 ± 16.0 + 16.0 = – 4.83 m 2 (Note that the positive root does not correspond to a physically valid situation.) (b) V= ke q1 ke q2 + =0 x (2.00 – x) +q 2q  =0 V = ke  –  x (2.00 – x) or Again solving for x, For 0 ≤ x ≤ 2.00 V = 0 when x = 0.667 m q and x For x < 0 25.19 2qx = q(2.00 – x) = –2q 2–x x = –2.00 m (a) U= ke q1q2 −(8.99 × 10 9 )(1.60 × 10 −19 )2 = r 0.0529 × 10 −9 (b) U= ke q1q2 r = = – 4.35 × 10–18 J = –27.2 eV – (8.99 × 109)(1.60 × 10–19)2 22(0.0529 × 10–9) = – 6.80 eV © 2000 by Harcourt, Inc. All rights reserved. 63 64 Chapter 25 Solutions (c) U= ke q1q2 r = – ke e 2 ∞ = 0 Goal Solution The Bohr model of the hydrogen atom states that the single electron can exist only in certain allowed orbits around the proton. The radius of each Bohr orbit is r = n2 (0.0529 nm) where n = 1, 2, 3, . . . . Calculate the electric potential energy of a hydrogen atom when the electron is in the (a) first allowed orbit, n = 1; (b) second allowed orbit, n = 2; and (c) when the electron has escaped from the atom ( r = ∞). Express your answers in electron volts. G: We may remember from chemistry that the lowest energy level for hydrogen is E1 = −13.6 eV, and higher energy levels can be found from En = E1 / n2 , so that E2 = −3.40 eV and E∞ = 0 eV . (see section 42.2) Since these are the total energies (potential plus kinetic), the electric potential energy alone should be lower (more negative) because the kinetic energy of the electron must be positive. U = ke O: The electric potential energy is given by A: q1q2 r (a) For the first allowed Bohr orbit,  − 4.35 × 10 −18 J N ⋅ m 2  (−1.60 × 10 −19 C)(1.60 × 10 −19 C) −18 U =  8.99 × 10 9 = − 4.35 × 10 J = = −27.2 eV  (0.0529 × 10 −9 m) C2  1.60 × 10 −19 J / eV  (b) For the second allowed orbit, (−1.60 × 10 −19 C)(1.60 × 10 −19 C) U = (8.99 × 10 9 N ⋅ m 2 / C 2 ) = −1.088 × 10 −18 J = −6.80 eV 2 2 (0.0529 × 10 −9 m) When the electron is at r = ∞, (c) ( U = 8.99 × 10 9 N ⋅ m 2/ C 2 −1.60 × 10 )( −19 )( C 1.60 × 10 −19 C ∞ )=0J L : The potential energies appear to be twice the magnitude of the total energy values, so apparently the kinetic energy of the electron has the same absolute magnitude as the total energy. *25.20 (a) U= (5.00 × 10–9 C)(– 3.00 × 10–9 C)(8.99 × 109 V · m/C) qQ = = – 3.86 × 10–7 J (0.350 m) 4 π e0 r The minus sign means it takes 3.86 × 10–7 J to pull the two charges apart from 35 cm to a much larger separation. (b) V= Q1 Q2 (5.00 × 10 −9 C)(8.99 × 10 9 V ⋅ m / C) (−3.00 × 10 −9 C)(8.99 × 10 9 V ⋅ m / C) + = + 0.175 m 0.175 m 4 π e0 r1 4 π e0 r2 V = 103 V Chapter 25 Solutions 25.21 qi V=∑ k ri i –1 1 1  V = (8.99 × 109)(7.00 × 10–6)  – + 0.0100 0.0387 0.0100 V = –1.10 × 107 C = –11.0 MV © 2000 by Harcourt, Inc. All rights reserved. 65 66 Chapter 25 Solutions U e = q4V 1 + q4V 2 + q4V 3 = q4 *25.22 ( U e = 10.0 × 10 −6 C ) (8.99 × 10 2 q 1 + q2 + q3 4π ∈0 r 1 r 2 r 3  9 1  1 1 + + N ⋅ m 2/ C 2   0.600 m 0.150 m    (0.600 m)2 + (0.150 m)2  ) 1 U e= 8.95 J U = U1 + U2 + U3 + U4 25.23 U = 0 + U 12 + (U 13 + U 23) + (U 14 + U 24 + U 34) U =0+ U= 2 keQ 2 keQ 2  1 1  kQ   + + 1 + e  1 + + 1      s s s 2 2 ke Q keQ 2  2   = 5.41 s 4 + s  2 2 ( An alternate way to get the term 4 + 2 diagonal pairs. *25.24 (a) V= ( ) 2 is to recognize that there are 4 side pairs and 2 face )( )  8.99 × 10 9 N ⋅ m 2 C 2 2.00 × 10 −6 C  ke q1 ke q2  k q + = 2 e = 2  2 2  r    r1 r2 + 0.500 m 1.00 m ( ) ( )   y P 2.00 µC V = 3.22 × 10 4 V = 32.2 kV (-1.00 m, 0) (b) *25.25 ( ) J  = −9.65 × 10 −2 J U = qV = −3.00 × 10 −6 C 3.22 × 10 4  C Each charge creates equal potential at the center. The total potential is:  k ( −q)  5ke q V = 5 e = − R  R  (0, 0.500 m) 2.00 µC (1.00 m, 0) x Chapter 25 Solutions *25.26 (a) 67 Each charge separately creates positive potential everywhere. The total potential produced by the three charges together is then the sum of three positive terms. There is no point located at a finite distance from the charges, where this total potential is zero. 25.27 2ke q ke q ke q + = a a a (b) V= (a) Conservation of momentum: 0 = m 1 v 1 i + m 2 v 2 (–i) or v 2 = By conservation of energy, 0+ k e (–q 1 )q 2 d ke q1q2 r1 + r2 and 1 m 1v 1 m2 1 2 2 = 2 m 1v 1 + 2 m 2v 2 ke q1q2 – d = 1 2 2 m 1v 1 + k e(–q 1 )q 2 (r1 + r2) 2 2 + 1 m 1v 1 2 m2 v1 = 2m2 ke q1q2  1 1 −   m1 ( m1 + m2 )  r1 + r2 d  v1 = 2(0.700 kg ) 8.99 × 10 9 N ⋅ m 2/ C 2 2 × 10 −6 C 3 × 10 −6 C  1  1 − = 10.8 m/s −3  1.00 m 0.100 kg 0.800 kg 8 × 10 m ( )( ) ( )( )( ) m 1v 1 (0.100 kg)(10.8 m/s) = = 1.55 m/s v2 = m 0.700 kg 2 25.28 (b) If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) . (a) Conservation of momentum: 0 = m1v1 i + m2 v2 (− i) By conservation of energy, 0+ (b) v2 = m1v1 / m2 ke (− q1 )q2 1 k (− q1 )q2 = 2 m1v 12 + 21 m2 v 22 + e d (r1 + r2 ) ke q1q2 kqq 1 1 m12 v12 − e 1 2 = m1v 12 + r1 + r2 2 2 m2 d and v1 = or 2m2 ke q1q2  1 1 −   m1(m1 + m2 )  r1 + r2 d  m  v2 =  1  v1 =  m2  2m1ke q1q2  1 1 −   m2 (m1 + m2 )  r1 + r2 d  If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) . © 2000 by Harcourt, Inc. All rights reserved. 68 Chapter 25 Solutions 25.29 k Q V= e r ( so For V = 100 V, 50.0 V, and 25.0 V, The radii are inversely proportional 25.30 (a) V (x) = V (x) = ) r = 0.720 m, 1.44 m, and 2.88 m to the potential. ke ( +Q) ke ( +Q) ke Q1 ke Q2 + = + 2 r1 r2 x 2 + a2 x 2 + ( −a)  ke Q  = a  x 2 + a2  2ke Q V (x) = ( ke Q a) (b) )( 8.99 × 10 9 N ⋅ m 2 C 2 8.00 × 10 −9 C ke Q 72.0 V ⋅ m r= = = V V V   2 (x a) + 1  2 2 ( x a) 2 + 1 V( y) = ke Q1 ke Q2 ke ( +Q) ke ( −Q) + = + r1 r2 y−a y+a V( y) = ke Q  1 1  −  a  y a − 1 y a + 1   1 V( y) 1  =  − (ke Q a)  y a − 1 y a + 1  25.31 Using conservation of energy, we have K f + U f = Ki + U i . But Ui = ke qα qgold ri , and ri ≈ ∞ . Thus, U i = 0. Also K f = 0 ( v f = 0 at turning point), so U f = Ki , rmin = 2ke qα qgold mα vα2 = ( or ke qα qgold rmin 2(8.99 × 10 9 N ⋅ m 2 / C 2 )(2)(79) 1.60 × 10 -19 C (6.64 × 10 − 27 kg)(2.00 × 107 m / s)2 ) = 21 mα vα2 2 = 2.74 × 10 −14 m = 27.4 fm Chapter 25 Solutions 25.32 25.33 Using conservation of energy we have: ke eQ k eQ = e + 21 mv 2 r1 r2 which gives: v= 1 2ke eQ  1 −   r2  m  r1 or v= (2)(8.99 × 10 9 N ⋅ m 2 / C 2 )(−1.60 × 10 −19 C)(10 − 9 C)  1 1  − -31   0.0300 m 0.0200 m 9.11 × 10 kg Thus, v= U=∑ ke qi q j ri j 7.26 × 106 m / s , summed over all pairs of (i, j ) where i ≠ j  q( − 2q) ( − 2q)( 3q) ( 2q)( 3q) q( 2q) U = ke  + + + + a b a  b q( 3q) a2 + b 2 + 2q( − 2q)   a2 + b 2  6 6 2 3 4   −2 U = ke q 2  − + + + −  0.400 0.200 0.400 0.200 0.447 0.447  ( )( ) 2 4 4 1  = – 3.96 J U = 8.99 × 10 9 6.00 × 10 − 6  − −  0.400 0.200 0.447  25.34 69 Each charge moves off on its diagonal line. All charges have equal speeds. ∑ (K + U)i 0+ = ∑ (K + U) f ( ) 4 ke q 2 2 ke q 2 4 k q 2 2 ke q 2 + = 4 21 mv 2 + e + L 2L 2L 2 2L 1  ke q 2  = 2 mv 2  2 +  2 L 1  ke q 2  v = 1 +   8  mL © 2000 by Harcourt, Inc. All rights reserved. 70 Chapter 25 Solutions A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12 25.35 face diagonal pairs separated by U= ke q 2 s ke q 2 12 4   12 + + 22.8 =  s 2 3  V = a + bx = 10.0 V + ( −7.00 V m )x 25.36 (a) (b) At x = 0, V = 10.0 V At x = 3.00 m, V = − 11.0 V At x = 6.00 m, V = − 32.0 V E=− dV = −b = −( −7.00 V m ) = 7.00 N C in + x direction dx V = 5x – 3x2y + 2yz2 25.37 Evaluate E at (1, 0 – 2) Ex = – ∂V = – 5 + 6xy ∂x Ey = – ∂V = +3x 2 – 2z2 = 3(1)2 – 2(–2)2 = – 5 ∂y Ez = – ∂V = – 4yz ∂z (a) For r < R Er = − (b) dV = dr For r ≥ R Er = − V= = – 5 + 6(1)(0) = – 5 = – 4(0)(–2) = 0 E = Ex 2 + Ey 2 + Ez 2 = 25.38 2 s, and 4 interior diagonal pairs separated ( −5)2 + ( −5)2 + 02 keQ R 0 V= keQ r dV  k Q = − − e2 =  r  dr keQ r2 = 7.07 N/C 3 s. Chapter 25 Solutions 25.39 Ey = − Ey = 25.40 71 2 2   ∂V ∂ k Q l+ l + y  = −  e ln    ∂y y ∂y  l     y2 keQ  keQ 1 − = 2 2 2 2 ly  l + y + l l + y  y l2 + y 2   Ex = Ey = Ez = 0 . Inside the sphere, ( ∂V ∂ =− V0 − E0 z + E0 a3 z(x 2 + y 2 + z 2 )− 3/2 ∂x ∂x Outside, Ex = − So Ex = − 0 + 0 + E0 a3 z(− 3 / 2)(x 2 + y 2 + z 2 )− 5/2 (2x) Ey = − [ ] ( ∂V ∂ =− V0 − E0 z + E0 a3 z(x 2 + y 2 + z 2 )− 3/2 ∂y ∂y = 3E0 a3 xz(x 2 + y 2 + z 2 )− 5/2 ) Ey= − E0 a3 z(− 3 / 2)(x 2 + y 2 + z 2 )− 5/2 2y = 3E0 a3 yz(x 2 + y 2 + z 2 ) Ez = − ) − 5/2 ∂V = E0 − E0 a3 z(− 3 / 2)(x 2 + y 2 + z 2 )− 5/2 (2z) − E0 a3 (x 2 + y 2 + z 2 )− 3/2 ∂z Ez = E0 + E0 a3 (2z 2 − x 2 − y 2 )(x 2 + y 2 + z 2 )− 5/2 *25.41 ∆V = V 2R − V0 = *25.42 V = ∫ dV = keQ R + ( 2R) 2 2 − k eQ keQ keQ  1  = − 1 = – 0.553  R   R R 5 dq 1 ∫ 4 π e0 r All bits of charge are at the same distance from O, so ( ) 2 −7.50 × 10 −6 C 1 Q  9 N⋅m  V= = 8.99 × 10 = –1.51 MV  4 π e0 R  C 2  (0.140 m / π ) © 2000 by Harcourt, Inc. All rights reserved. 72 25.43 25.44 Chapter 25 Solutions (a) λ C [α] =  x  = m   (b) L ⌠λ dx x dx L   ⌠d q   V = ke ⌡ = k = k α = ke α L − d ln 1 +  e⌡ r e ∫ r  d (d + x)  0 V=∫ α x dx ke dq = ke ∫ 2 2 r b + (L 2 − x) L − x. 2 Let z = V = ke α ∫ V=− 1 C ⋅   = m  m2 x= Then L − z, 2 (L 2 − z)(−dz) = − keα L b2 + z2 2 ke α L  ln ( L 2 − x ) + 2  ∫ dx = −dz and dz b2 + z2 (L 2 − x)2 + b2   L 25.45 dV = 2 2 ke α L  b + (L 4) − L ln  2  b 2 + (L2 4) + L  ke dq b a 25.46 V = ke ∫ r dr r +x all charge 2 = −R − 3R + L 0 (L 2 − L)2 + b2 − (L 2)2 + b2  2  2  2 π ke σ  x 2 + b 2 − x 2 + a 2    3R λ dx − R λ dx λ ds dq = ke ∫ + ke ∫ + ke ∫ R − 3R semicircle r −x R x V = − ke λ ln(− x) V = ke λ ln 2 ke α L ln(z + z 2 + b 2 ) + ke α z 2 + b 2 2 ( L 2 − x )2 + b 2 where dq = σ dA = σ 2 π r dr r2 + x2 V = 2 πσ ke ∫ + ke α b2 + z2 = − 0 2  2 ke α L  L 2 − L + ( L 2 ) + b  + ke α  V=− ln 2  2  2   L 2 + ( L 2) + b  V= − z dz + ke α ∫ 3R ke λ π R + ke λ ln x R R 3R + ke λπ + ke λ ln 3 = ke λ ( π + 2 ln 3) R  Chapter 25 Solutions Substituting given values into 25.47 ke q , r q = 2.50 × 10 −7 C, Substituting 25.48 V= 7.50 × 103 V = N= (8.99 × 109 N·m2/C2) q (0.300 m) 2.50 × 10-7 C = 1.56 × 1012 electrons 1.60 × 10-19 C/e− q1 + q2 = 20.0 µC so q1 = 20.0 µC – q2 r1 q1 = r2 q2 so 20.0 µC – q2 4.00 cm = q2 6.00 cm 6.00(20.0 µC – q2) = 4.00q2 ; Therefore q2 = 12.0 µC Solving, (a) E1 = 73 and ( )( ) ( )( ) q1 = 20.0 µC – 12.0 µC = 8.00 µC 8.99 × 10 9 8.00 × 10 −6 ke q1 = = 4.50 × 107 V / m = 45.0 MV/m 2 r12 0.0400 ( ) 8.99 × 10 9 12.0 × 10 −6 ke q2 E2 = 2 = = 3.00 × 107 V / m = 30.0 MV/m 2 r2 (0.0600) 25.49 (b) V1 = V 2 = ke q2 = 1.80 MV r2 (a) E= 0 ; V= (b) E= ke q (8.99 × 10 9 )(26.0 × 10 −6 ) = = 5.84 MN/C (0.200)2 r2 V= ke q (8.99 × 10 9 )(26.0 × 10 −6 ) = = 1.17 MV r (0.200) E= ke q (8.99 × 10 9 )(26.0 × 10 −6 ) = = 11.9 MN/C away R2 (0.140)2 V= ke q = 1.67 MV R (c) 25.50 ke q (8.99 × 10 9 )(26.0 × 10 −6 ) = = 1.67 MV 0.140 R away No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric field in the wire to push more charge to the outer sphere. Charge Q is on the outer sphere. Therefore, zero charge is on the inner sphere and 10.0 µC is on the outer sphere . © 2000 by Harcourt, Inc. All rights reserved. Chapter 25 Solutions 74 25.51 (a) Emax = 3.00 × 106 V / m = keQ keQ 1 1 = = Vmax r r r r2 Vmax = Emax r = 3.00 × 106 (0.150) = 450 kV keQmax = Emax r2 (b) Qmax = keQmax   = Vmax  or r   Emax r 2 3.00 × 106 (0.150)2 = = 7.51 µC ke 8.99 × 10 9 Goal Solution Consider a Van de Graaff generator with a 30.0-cm-diameter dome operating in dry air. (a) What is the maximum potential of the dome? (b) What is the maximum charge on the dome? G: Van de Graaff generators produce voltages that can make your hair stand on end, somewhere on the order of about 100 kV (see the Puzzler at beginning of Chapter 25). With these high voltages, the maximum charge on the dome is probably more than typical point charge values of about 1 µ C . The maximum potential and charge will be limited by the electric field strength at which the air surrounding the dome will ionize. This critical value is determined by the dielectric strength of air which, from page 789 or from Table 26.1, is Ecritical = 3 × 106 V / m . An electric field stronger than this will cause the air to act like a conductor instead of an insulator. This process is called dielectric breakdown and may be seen as a spark. O: From the maximum allowed electric field, we can find the charge and potential that would create this situation. Since we are only given the diameter of the dome, we will assume that the conductor is spherical, which allows us to use the electric field and potential equations for a spherical conductor. With these equations, it will be easier to do part (b) first and use the result for part (a). A : (b) (a) For a spherical conductor with total charge Q , ( ) E = Q= 3.00 × 106 V / m (0.150 m ) Er 2 = (1 N ⋅ m / V ⋅ C) = 7.51 µC ke 8.99 × 10 9 N ⋅ m 2 / C 2 V= keQ (8.99 × 10 9 N ⋅ m 2 / C 2 )(7.51 × 10 −6 C) = = 450 kV r 0.150 m keQ r2 2 L : These calculated results seem reasonable based on our predictions. The voltage is about 4000 times larger than the 120 V found from common electrical outlets, but the charge is similar in magnitude to many of the static charge problems we have solved earlier. This implies that most of these charge configurations would have to be in a vacuum because the electric field near these point charges would be strong enough to cause sparking in air. (Example: A charged ball with Q = 1 µ C and r = 1 mm would have an electric field near its surface of ( )( ) 9 × 10 9 N ⋅ m 2/ C 2 1 × 10 −6 C keQ E= 2 = = 9 × 10 9 V / m r (0.001 m)2 which is well beyond the dielectric breakdown of air!) Chapter 25 Solutions V= 25.52 (b) r= (a) q = 25.53 ke q r ke q r2 V Since E = r , V 6.00 × 105 V = = 0.200 m E 3.00 × 106 V/m and Vr k e = 13.3 µC U = qV = k e *25.54 (a) and E = 75 q1q2 r 12 = (8.99 × 109) (38)(54)(1.60 × 10–19)2 (5.50 + 6.20) × 10–15 = 4.04 × 10–11 J = 253 MeV To make a spark 5 mm long in dry air between flat metal plates requires potential difference ( )( ) V = Ed = 3.0 × 106 V m 5.0 × 10 −3 m = 1.5 × 10 4 V ~10 4 V (b) Suppose your surface area is like that of a 70-kg cylinder with the density of water and radius 12 cm. Its length would be given by 70 × 10 3 cm 3 = π (12 cm ) l 2 l = 1.6 m The lateral surface area is A = 2 π r l = 2 π (0.12 m )(1.6 m ) = 1.2 m 2 The electric field close to your skin is described by E= σ Q = , so e0 Ae0  C2  N  1.2 m 2  8.85 × 10 −12 Q = EA ∈0 = 3.0 × 106   C N ⋅ m2   ( 25.55 (a) ) ~10 −5 C 2 1  1 V = ke Q  – + x x – a x + a 2k e Qa 2 x(x – a) – 2(x + a)(x – a) + x(x + a) = 3 V = ke Q  x(x + a)(x – a)   x – xa 2 (b) V= 2k e Qa 2 x3 for a << 1 x © 2000 by Harcourt, Inc. All rights reserved. 76 25.56 Chapter 25 Solutions (a) Ex = − (b) Ex = (2keQa2 )(3x 2 − a2 ) dV d  2k Qa2  = −  3 e 2 = dx dx  x − xa  (x 3 − xa 2 )2 Ey = Ez = 0 and [ 2(8.99 × 10 9 N ⋅ m 2/ C 2 )(3 × 10 −6 C)(2 × 10 −3 m)2 3(6 × 10 −3 m)2 − (2 × 10 −3 m)2 [(6 × 10 −3 m) − (6 × 10 3 −3 m)(2 × 10 −3 m) ] 2 2 ] Ex = 609 × 106 N/C = 609 MN/C 25.57 (a) (b) E= Q 4 π∈0 r 2 V= Q 4 π∈0 r r= V 3000 V = = 6.00 m 500 V / m E V = −3000 V = Q= 25.58 Q 4 π∈0 (6.00 m) −3000 V (6.00 m) = – 2.00 µC (8.99 × 10 9 V ⋅ m / C) From Example 25.5, the potential created by the ring at the electron's starting point is Vi = keQ xi2 +a 2 = ke ( 2 π λ a) xi2 + a2 V f = 2 π ke λ . From conservation of energy, while at the center, it is ( 0 + ( −eVi ) = 21 me v 2f + −eV f v 2f = v 2f ( ) 4 π eke λ 2e V f − Vi = me me ( )( )   a  1 −  xi2 + a2   )( ) 4 π 1.60 × 10 −19 8.99 × 10 9 1.00 × 10 −7  = 1 −  9.11 × 10 −31  v f = 1.45 × 107 m/s   (0.100)2 + (0.200)2  0.200 Chapter 25 Solutions 25.59 (a) 77 Take the origin at the point where we will find the potential. One ring, of width dx, has charge Q dx/h and, according to Example 25.5, creates potential keQ dx dV = h x 2 + R2 The whole stack of rings creates potential V = ∫ dV all charge (b) =∫ d+h d keQ dx h x2 + R2 = ke Q  ln  x + x 2 + R 2  h d+h d = 2 2 ke Q  d + h + (d + h) + R  ln     h d + d2 + R2   A disk of thickness dx has charge Q dx/h and charge-per-area Q dx/π R 2 h. Example 25.6, it creates potential dV = 2 π ke According to Q dx  x 2 + R 2 − x π R2h  Integrating, V=∫ d+ h d V= 2k Q 2keQ  2 x + R 2 dx − x dx = 2e 2  R h R h  d + h + (d + h)2 + R 2   keQ  2 2 2 2 2 2 (d + h) (d + h) + R − d d + R − 2 dh − h ln + R   2 2   R 2 h  d + d + R   The positive plate by itself creates a field 25.60 d+h 1 x2 R2  2 2 ln  x + x 2 + R 2  −  x x +R +  2  2 2 d E= 36.0 × 10–9 C/m2 kN σ = = 2.03 C –12 2 2 2 ∈0 2(8.85 × 10 C /N · m ) away from the + plate. The negative plate by itself creates the same size field and between the plates it is in the same direction. Together the plates create a uniform field 4.07 kN/C in the space between. (a) Take V = 0 at the negative plate. The potential at the positive plate is then V − 0 = −∫ 12.0 cm 0 ( −4.07 kN / C) dx The potential difference between the plates is V = (4.07 × 103 N/C)(0.120 m) = 488 V (b)  1 mv 2 + q V 2  1 =  2 mv 2 + q V   i 1 f 2 qV = (1.60 × 10–19 C)(488 V) = 2 m v f = 7.81 × 10–17 J (c) v f = 306 km/s © 2000 by Harcourt, Inc. All rights reserved. 78 Chapter 25 Solutions 2 (d) v f 2 = v i + 2a(x – xi) (3.06 × 105 m/s)2 = 0 + 2a(0.120 m) a = 3.90 × 1011 m/s2 (e) ∑F = ma = (1.67 × 10–27 kg)(3.90 × 1011 m/s2) = 6.51 × 10–16 N (f) F 6.51 × 10–16 N = 4.07 kN/C E=q = 1.60 × 10–19 C Q ke q W = ∫ V dq where V = R ; 25.61 Therefore, W = 0 25.62 (a) k eQ 2 2R B VB − V A = − ∫ E ⋅ ds and the field at distance r from a uniformly A 2k λ λ E= = e 2 π e0 r r −λ ra charged rod (where r > radius of charged rod) is rb λ In this case, the field between the central wire and the coaxial cylinder is directed perpendicular to the line of charge so that VB − V A = − ∫ rb ra (b) r  2ke λ dr = 2ke λ ln a  , r  rb  or r  ∆V = 2ke λ ln a   rb  From part (a), when the outer cylinder is considered to be at zero potential, the potential at a distance r from the axis is V = 2ke λ ln  ra   r The field at r is given by E = − But, from part (a), 2ke λ =  r   r  2k λ ∂V = − 2ke λ   − 2a = e ∂r r  ra   r  ∆V . ln(r a rb ) Therefore, E= ∆V  1 ln(r a rb )  r  Chapter 25 Solutions 25.63 V2 – V1 = 25.64 r2 r2 r1 r1 V 2 − V1 = − ∫ E ⋅ dr = − ∫ 79 λ dr 2 π ε 0r r  −λ ln 2  2 π∈0  r1  ke ( q) ke ( −2q) + r1 r2 For the given charge distribution, V ( x, y, z) = where r1 = The surface on which V ( x, y, z) = 0 is given by  1 2 ke q −  = 0, or 2r1 = r2  r1 r2  This gives: 4( x + R) + 4y 2 + 4z 2 = x 2 + y 2 + z 2 which may be written in the form: x2 + y 2 + z2 + ( x + R )2 + y 2 + z 2 and r2 = x 2 + y 2 + z 2 2  4 2 8  R x + (0) y + (0) z + R =0 3  3  [1] The general equation for a sphere of radius a centered at ( x0 , y0 , z0 ) is: (x − x0 )2 + ( y − y0 )2 + ( z − z0 )2 − a2 = 0 or ( ) x 2 + y 2 + z 2 + ( −2x0 )x + ( −2y0 ) y + ( −2z0 )z + x02 + y02 + z02 − a2 = 0 [2] Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 is indeed a sphere and that: −2x0 = 8 R; 3 4 Thus, x0 = − R, 3 −2y0 = 0; y0 = z0 = 0, −2z0 = 0; and a2 = x02 + y02 + z02 − a2 = 4 2 R 3  16 4  2 4 2 − R = R .  9 3 9 The equipotential surface is therefore a sphere centered at  4  − R, 0, 0  3  © 2000 by Harcourt, Inc. All rights reserved. , having a radius 2 R 3 80 25.65 Chapter 25 Solutions (a) (b) EA = 0 From Gauss's law, EB = ke −8 qA 9 (1.00 × 10 ) = (8.99 × 10 ) = r2 r2 EC = ke −9 (qA + qB ) 9 (− 5.00 × 10 ) = (8.99 × 10 ) = r2 r2 VC = k e  89.9  V/m  r2   45.0  − 2 V/m  r  (qA + qB ) (− 5.00 × 10 − 9 ) = (8.99 × 10 9 ) = r r ∴ At r2 , V = −  45.0  V −  r  45.0 = − 150 V 0.300 Inside r2 , VB = − 150 V + ∫ r r2 89.9 1  89.9  1  = − 450 + V dr = − 150 + 89.9 − 2    r 0.300 r  r 89.9 = + 150 V 0.150 ∴ At r1 , V = − 450 + 25.66 (no charge within) so V A = + 150 V From Example 25.5, the potential at the center of the ring is Vi = keQ R and the potential at an infinite distance from the ring is V f = 0. Thus, the initial and final potential energies of the point charge are: U i = QVi = keQ 2 R and U f = QV f = 0 From conservation of energy, K f + U f = Ki + U i or 25.67 1 Mv 2 f 2 +0=0+ keQ 2 R The sheet creates a field E1 = giving vf = 2keQ 2 MR σ i for x > 0 . Along the x − axis, the line of charge creates a 2 ∈0 field E2 = λ λ away = (− i) for x < 3.00 m 2 π r ∈0 2 π ∈0 (3.00 m − x) The total field along the x − axis in the region 0 < x < 3.00 m is then  σ  λ E = E1 + E2 =  − i 3.00 − x 2 ∈ 2 π ∈ )  0(  0 Chapter 25 Solutions (a) The potential at point x follows from V − V0 = − V = V0 − x ∫0 E ⋅ idx = − x σ λ  ∫0  2 ∈0 − 2π ∈0 (3.00 − x)  dx σx λ x   − ln 1 − 3.00  2 ∈0 2 π ∈0  V = 1.00 kV − (25.0 × 10 −9 C / m 2 )x 80.0 × 10 −9 C / m x   − ln 1 − −12 2 2 −12 2 2  3.00  C / N ⋅ m ) 2 π (8.85 × 10 C /N ⋅ m ) 2(8.85 × 10  x  kV   x − (1.44 kV) ln  1.00 − V = 1.00 kV − 1.41   m 3.00 m   (b) At x = 0.800 m, ( V = ke a+L ∫ a 25.69 (a) ) U = QV = 2.00 × 10 −9 C ( 316 J C) = 6.33 × 10 −7 J = 633 nJ and 25.68 V = 316 V Er = − λ dx x2 + b2 = ke λ ln  x + (x 2 + b 2 )    a+L a  a + L + ( a + L)2 + b 2   = ke λ ln  2 2   a + a + b   2ke p cos θ ∂V = ∂r r3 In spherical coordinates, the θ component of the gradient is Therefore, Eθ = − 1 ∂  . r  ∂θ  k p sin θ 1  ∂V  = e 3 r  ∂θ  r For r >> a, Er (0°) = 2ke p r3 and Er (90°) = 0 , Eθ (0°) = 0 These results are reasonable for r >> a . However, for r → 0, E(0) → ∞ . (b) V= ke py (x 2 + y 2 )3/2 Ey = − and Ex = − 3ke pxy ∂V = ∂ x (x 2 + y 2 )5/2 k p(2y 2 − x 2 ) ∂V = e 2 ∂y (x + y 2 )5/2 © 2000 by Harcourt, Inc. All rights reserved. and Eθ (90°) = ke p r3 81 82 25.70 25.71 Chapter 25 Solutions (a) EA > EB since E = ∆V ∆s (6 – 2) V ∆V =– = 200 N/C 2 cm ∆s (b) EB = – (c) The figure is shown to the right, with sample field lines sketched in. down For an element of area which is a ring of radius r and width dr, dV = ke dq r2 + x2 dq = σ dA = Cr (2π r dr) and R V = C(2 π ke ) ∫ 0 25.72 r 2 dr r2 + x2    x = C( π ke ) R R 2 + x 2 + x 2 ln   R + R 2 + x 2    dU = V dq where the potential V = ke q . r The element of charge in a shell is dq = ρ (volume element) charge q in a sphere of radius r is r  4π r 3  q = 4 πρ ∫ r 2 dr = ρ    3  0 Substituting this into the expression for dU, we have dU =  4 π r 3   1  16 π 2  2 4  ke q  2 ρ (4 π r dr) = k dq = keρ  e  3  ρ r dr   r     3  r  16 π 2  2 R 4  16 π 2  2 5 U = ∫ dU = ke  ρ ∫ r dr = ke   ρ R  3  0  15  But the total charge, Q = ρ 43 πR 3 . Therefore, U= 3 keQ 2 5 R or dq = ρ (4π r 2 dr) and the Chapter 25 Solutions *25.73 (a) ∆V 1 ln(r a rb ) r From Problem 62, E= We require just outside the central wire 5.50 × 106 or m (110 m )r ln 0.850  =1 r  -1 V 50.0 × 10 3 V  1  =   m  0.850 m   rb  ln  rb   b b We solve by homing in on the required value (110 m ) -1 rb (m) 0.0100 0.00100 0.00150 0.00145 0.00143 0.00142  0.850 m  rb ln  rb   4.89 0.740 1.05 1.017 1.005 0.999 Thus, to three significant figures, (b) At r a , E = rb = 1.42 mm 50.0 kV 1   = 9.20 kV m ln(0.850 m 0.00142 m )  0.850 m  © 2000 by Harcourt, Inc. All rights reserved. 83 Chapter 26 Solutions *26.1 26.2 (a) Q = C (∆V) = (4.00 × 10–6 F)(12.0 V) = 4.80 × 10–5 C = 48.0 µC (b) Q = C (∆V) = (4.00 × 10–6 F)(1.50 V) = 6.00 × 10–6 C = 6.00 µC (a) C= (b) ∆V = (4.90 × 10 4 N / C)(0.210 m)2 = 0.240 µ C 8.99 × 10 9 N ⋅ m 2 / C 2 ke q : r2 (a) σ= q 0.240 × 10 −6 = = 1.33 µC/m2 A 4 π (0.120)2 (b) C = 4 π e0 r = 4 π (8.85 × 10 −12 )(0.120) = 13.3 pF (a) C = 4π e0 R R= 26.5 Q 100 × 10 − 6 C = = 100 V C 1.00 × 10 − 6 F E= 26.3 26.4 10.0 × 10 − 6 C Q = = 1.00 × 10 − 6 F = 1.00 µF ∆V 10.0 V q= C = ke C = (8.99 × 109 N · m2/C 2)(1.00 × 10–12 F) = 8.99 mm 4π e0 4π (8.85 × 10–12 C 2)(2.00 × 10–3 m) = 0.222 pF N · m2 (b) C = 4π e0 R = (c) Q = CV = (2.22 × 10–13 F)(100 V) = 2.22 × 10–11 C (a) Q1 R1 = Q2 R2  R  Q1 + Q2 =  1 + 1  Q2 = 3.50Q2 = 7.00 µ C R2   Q2 = 2.00 µC (b) V1 = V 2 = Q1= 5.00 µC 5.00 µ C Q1 Q2 = = = 8.99 × 10 4 V = 89.9 kV −1 9 C1 C2 8.99 × 10 m F (0.500 m) ( ) © 2000 by Harcourt, Inc. All rights reserved. Chapter 26 Solutions C= *26.6 κ e0 A (1.00)(8.85 × 10–12 C 2)(1.00 × 103 m)2 = = 11.1 nF d N · m2(800 m) The potential between ground and cloud is ∆V = Ed = (3.00 × 106 N/C)(800 m) = 2.40 × 109 V Q = C (∆V) = (11.1 × 10-9 C/V)(2.40 × 109 V) = 26.6 C 26.7 (a) (b) ∆V = Ed E= 20.0 V = 11.1 kV/m 1.80 × 10–3 m E= σ e0 σ = (1.11 × 104 N/C)(8.85 × 10–12 C 2/N · m2) = 98.3 nC/m2 (c) ( (d) ∆V = 26.8 )( ) 8.85 × 10 −12 C 2 / N ⋅ m 2 7.60 cm 2 (1.00 m / 100 cm ) e0 A C= = = 3.74 pF d 1.80 × 10 −3 m Q C 2 Q = (20.0 V)(3.74 × 10–12 F) = 74.7 pC C= κ e0 A = 60.0 × 10 −15 F d d= −12 21.0 × 10 −12 κ e0 A (1) 8.85 × 10 = C 60.0 × 10 −15 ( )( ) d = 3.10 × 10 −9 m = 3.10 nm 26.9 Q= e ( ∆V ) Q =σ = 0 d A e0 A ( ∆V ) d ( ) 8.85 × 10 −12 C 2 N ⋅ m 2 (150 V ) e0 ( ∆V ) d= = = 4.42 µm σ 30.0 × 10 − 9 C cm 2 1.00 × 10 4 cm 2 m 2 ( )( ) © 2000 by Harcourt, Inc. All rights reserved. 85 86 Chapter 26 Solutions With θ = π , the plates are out of mesh and the overlap area is zero. With θ = 0, the overlap area is that of a semi-circle, π R 2 2 . By proportion, the effective area of a single sheet of charge is ( π − θ )R 2 2. 26.10 d When there are two plates in each comb, the number of adjoining sheets of positive and negative charge is 3, as shown in the sketch. When there are N plates on each comb, the number of parallel capacitors is 2N − 1 and the total capacitance is C = ( 2N − 1) 26.11 e0 Aeffective l (a) C= (b) Method 1: b 2k e ln   a λ=q/l = distance = = (2N − 1)e0 ( π − θ )R 2 2 d2 50.0 7.27 2(8.99 × 109) ln  2.58 = - + + + + - + + + + + + + + - (2N − 1) ∈0 ( π − θ )R 2 d = 2.68 nF b ∆V = 2k e λ ln   a  8.10 × 10–6 C = 1.62 × 10–7 C/m 50.0 m 7.27 = 3.02 kV ∆V = 2(8.99 × 10 9)(1.62 × 10–7) ln  2.58 Method 2: 26.12 ∆V = Q 8.10 × 10–6 = = 3.02 kV C 2.68 × 10–9 Let the radii be b and a with b = 2a. Put charge Q on the inner conductor and – Q on the outer. Electric field exists only in the volume between them. The potential of the inner sphere is V a = keQ a; that of the outer is Vb = keQ b . Then V a − Vb = ke Q ke Q Q  b − a − =   4 π e0  ab  a b Here C = 4 π e0 2a2 = 8 π e0 a a The intervening volume is a= C= and 4 π e0 ab Q = b−a V a − Vb C 8 π e0 4 3 4 3 Volume = π b 3 − π a3 = 7 Volume = The outer sphere is 360 km in diameter. ( 4 π a3 3 ) = 7 ( π ) 8 πC e 4 3 3 3 3 0 3 = 7C 3 384 π 2e0 3 7 (20.0 × 10 − 6 C 2 / N⋅m)3 = 2.13 × 1016 m3 384 π 2 (8.85 × 10 −12 C 2 / N⋅m 2 )3 Chapter 26 Solutions ΣFy = 0: T cos θ − mg = 0 26.13 Dividing, tan θ = ΣFx = 0: T sin θ − Eq = 0 so E= mg tan θ q mgd tan θ (350 × 10 − 6 kg)(9.80 m / s 2 )(4.00 × 10 − 2 m) tan 15.0° = = 1.23 kV q 30.0 × 10 −9 C ∆V = Ed = ΣFy = 0: T cos θ − mg = 0 26.14 Dividing, tan θ = 26.15 Eq , mg 87 Eq , mg so ΣFx = 0: T sin θ − Eq = 0 E= mg tan θ q (a) C= ab (0.0700)(0.140) = = 15.6 pF ke (b − a) (8.99 × 10 9 )(0.140 − 0.0700) (b) C= Q ∆V ∆V = mgd tan θ q ∆V = Ed = and Q 4.00 × 10 −6 C = = 256 kV C 15.6 × 10 −12 F Goal Solution An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and 14.0 cm, respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of 4.00 µ C on the capacitor? G: Since the separation between the inner and outer shells is much larger than a typical electronic capacitor with d ~ 0.1 mm and capacitance in the microfarad range, we might expect the capacitance of this spherical configuration to be on the order of picofarads, (based on a factor of about 700 times larger spacing between the conductors). The potential difference should be sufficiently low to prevent sparking through the air that separates the shells. O: The capacitance can be found from the equation for spherical shells, and the voltage can be found from Q = C∆V . A : (a) For a spherical capacitor with inner radius a and outer radius b, C= (b) ab (0.0700 m)(0.140 m) = = 1.56 × 10 −11 F = 15.6 pF k(b − a) 8.99 × 10 9 N ⋅ m 2 C 2 (0.140 − 0.0700) m ∆V = ( ) Q (4.00 × 10 −6 C) = = 2.56 × 10 5 V = 256 kV C 1.56 × 10 -11 F L : The capacitance agrees with our prediction, but the voltage seems rather high. We can check this voltage by approximating the configuration as the electric field between two charged parallel plates separated by d = 7.00 cm, so E~ ∆V 2.56 × 10 5 V = = 3.66 × 106 V / m d 0.0700 m ( ) This electric field barely exceeds the dielectric breakdown strength of air 3 × 106 V / m , so it may not even be possible to place 4.00 µ C of charge on this capacitor! © 2000 by Harcourt, Inc. All rights reserved. 88 Chapter 26 Solutions ( )( ) 26.16 C = 4 π e0 R = 4 π 8.85 × 10 −12 C N ⋅ m 2 6.37 × 106 m = 7.08 × 10 −4 F *26.17 (a) Capacitors in parallel add. Thus, the equivalent capacitor has a value of Ceq = C1 + C2 = 5.00 µF + 12.0 µF = 17.0 µF (b) The potential difference across each branch is the same and equal to the voltage of the battery. ∆V = 9.00 V (c) *26.18 (a) Q 5 = C (∆V) = (5.00 µF)(9.00 V) = 45.0 µC and In series capacitors add as 1 1 1 1 1 C eq = C 1 + C 2 = 5.00 µF + 12.0 µF (c) Q12 = C (∆V) = (12.0 µF)(9.00 V) = 108 µC and C eq = 3.53 µF The charge on the equivalent capacitor is Q eq = C eq (∆V) = (3.53 µF)(9.00 V) = 31.8 µC Each of the series capacitors has this same charge on it. So (b) The voltage across each is 31.8 µC Q1 ∆V 1 = C = 5.00 µF = 6.35 V 1 26.19 Q 1 = Q 2 = 31.8 µC 1 Cp = C1 + C2 C s = and ∆V 2 = Q2 31.8 µC = = 2.65 V C2 12.0 µF 1 1 + C1 C2 1 1 1 Cp – C1 + C1 C s = C 1 + C p – C 1 = C 1(C p – C 1) Substitute C 2 = C p – C 1 2 Simplifying, C 1 – C 1 C p + C p C s = 0 C1 = Cp ± Cp 2 − 4CpCs 2 1 2 = Cp ± 1 C 2 4 p − CpCs We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign, we would get the same two answers with their names interchanged.) 1 2 C1 = Cp + 1 C 2 4 p 1 2 − CpCs = C2 = Cp − C1 = Cp − 1 C 2 4 p 1 2 (9.00 pF ) + 1 1 4 (9.00 pF )2 − (9.00 pF )(2.00 pF ) = − CpCs = 2 (9.00 pF) – 1.50 pF = 3.00 pF 6.00 pF Chapter 26 Solutions Cp = C1 + C2 26.20 Substitute 89 1 1 1 = + Cs C1 C2 and Cp − C1 + C1 1 1 1 = + = Cs C1 Cp − C1 C1  Cp − C1    C2 = Cp − C1 : C12 − C1Cp + CpCs = 0 Simplifying, C1 = and Cp ± Cp2 − 4CpCs 2 = 1 2 Cp + 1 4 Cp2 − CpCs where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values for the capacitances, with the names reversed). Then, from C2 = Cp − C1 C2 = 26.21 (a) 1 2 Cp − 1 4 2 Cp − CpCs 1 1 1 = + Cs = 2.50 µF 15.0 3.00 Cs Cp = 2.50 + 6.00 = 8.50 µF  1 1  Ceq =  +   8.50 µ F 20.0 µ F  (b) −1 = 5.96 µF Q = ( ∆V )C = (15.0 V)(5.96 µF) = 89.5 µC ∆V = on 20.0 µF Q 89.5 µC = = 4.47 V C 20.0 µF 15.0 – 4.47 = 10.53 V Q = ( ∆V )C= (10.53)(6.00 µF) = 63.2 µC on 6.00 µF 89.5 – 63.2 = 26.3 µC on 15.0 µF and 3.00 µF 26.22 The circuit reduces first according to the rule for capacitors in series, as shown in the figure, then according to the rule for capacitors in parallel, shown below. 1 1  11  = C = 1.83C Ceq = C 1 + +  2 3 6 © 2000 by Harcourt, Inc. All rights reserved. ⇒ 90 Chapter 26 Solutions so Q 6.00 × 10–6 = 20.0 Q1 = 120 µC – Q2 and ∆V = 120 – Q 2 Q2 = C2 C1 or 120 – Q2 Q2 = 6.00 3.00 C= 26.23 Q ∆V and Q = 120 µC Q C (3.00)(120 – Q2) = (6.00)Q2 Q2 = 360 = 40.0 µC 9.00 Q1 = 120 µC – 40.0 µC = 80.0 µC *26.24 (a) In series , to reduce the effective capacitance: 1 1 1 = + 32.0 µ F 34.8 µ F Cs Cs = (b) 1 = 398 µ F 2.51 × 10 −3 µ F In parallel , to increase the total capacitance: 29.8 µ F + Cp = 32.0 µ F Cp = 2.20 µ F With switch closed, distance d' = 0.500d and capacitance C ′ = 26.25 (a) Q = C ′(∆V) = 2C(∆V) = 2(2.00 × 10 − 6 F)(100 V) = 400 µC (b) The force stretching out one spring is F= e0 A 2e0 A = = 2C d d′ 2C 2 (∆V) 2 2C(∆V) 2 Q2 4C 2 (∆V) 2 = = = (e0 A / d)d 2e0 A 2e0 A d One spring stretches by distance x = d /4, so k= F 2C(∆V) 2  4  8C(∆V) 2 8(2.00 × 10 − 6 F)(100 V) 2 = = = 2.50 kN/m =  d x d (8.00 × 10 − 3 m) 2 d2 Chapter 26 Solutions 26.26 Positive charge on A will induce equal negative charges on B, D, and F, and equal positive charges on C and E. The nesting spheres form three capacitors in series. From Example 26.3, CAB = CCD = R(2R) 2R ab = = ke (b − a) ke ke R (3R)( 4R) = 12R ke R Ceq = ke ke R ( 5R)( 6R) CEF = 26.27 91 = 30R ke 60 R 1 = ke / 2R + ke / 12R + ke / 30R 37 ke nC = nC = = 100C n 100 n/C n2 = 100 so and n = 10 Goal Solution A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group? G: Since capacitors in parallel add and ones in series add as inverses, 2 capacitors in parallel would have a capacitance 4 times greater than if they were in series, and 3 capacitors would give a ratio Cp / Cs = 9 , so maybe n = Cp / Cs = 100 = 10. O: The ratio reasoning above seems like an efficient way to solve this problem, but we should check the answer with a more careful analysis based on the general relationships for series and parallel combinations of capacitors. A : Call C the capacitance of one capacitor and n the number of capacitors. The equivalent capacitance for n capacitors in parallel is Cp = C1 + C2 + . . . + Cn = nC The relationship for n capacitors in series is Therefore Cp Cs = nC = n2 C/n or n= Cp Cs 1 1 1 n 1 = + +...+ = Cn C Cs C1 C2 = 100 = 10 L : Our prediction appears to be correct. A qualitative reason that Cp / Cs = n2 is because the amount of charge that can be stored on the capacitors increases according to the area of the plates for a parallel combination, but the total charge remains the same for a series combination. © 2000 by Harcourt, Inc. All rights reserved. 92 Chapter 26 Solutions Cs = 26.28 1   1 +  5.00 10.0  −1 = 3.33 µ F Cp1 = 2(3.33) + 2.00 = 8.66 µF Cp2 = 2(10.0) = 20.0 µF Ceq = 1   1 +  8.66 20.0  −1 = 6.04 µF ( ) Qeq = Ceq ( ∆V ) = 6.04 × 10 −6 F (60.0 V ) = 3.62 × 10 −4 C 26.29 Qp1 = Qeq , so ( ∆V p1 = Qeq Cp1 = ) ( 3.62 × 10 −4 C = 41.8 V 8.66 × 10 −6 F ) Q3 = C3 ∆V p1 = 2.00 × 10 −6 F ( 41.8 V ) = 83.6 µ C Cs = 26.30 1   1 +  5.00 7.00  −1 = 2.92 µ F Cp = 2.92 + 4.00 + 6.00 = 12.9 µF 1 1 1 1 *26.31 (a) U = 2 C (∆V)2 = 2 (3.00 µF)(12.0 V) 2 = 216 µ J (b) U = 2 C (∆V)2 = 2 (3.00 µF)(6.00 V) 2 = 54.0 µ J 1 U = 2 C (∆V)2 *26.32 The circuit diagram is shown at the right. (a) Cp = C1 + C2 = 25.0 µF + 5.00 µF = 30.0 µF 1 U = 2 (30.0 × 10–6)(100) 2 = 0.150 J (b) 1 1 Cs =  C + C   1 2 –1 1 1 =  25.0 µF + 5.00 µF    1 U = 2 C (∆V)2 ∆V = 2U = C (0.150)(2) 4.17 × 10 −6 = 268 V –1 = 4.17 µF Chapter 26 Solutions *26.33 Use U = 1 Q2 2 C 26.34 u= e0 A d and C = If d2 = 2d1, C2 = 1 C . 2 1 93 Therefore, the stored energy doubles . U 1 = e0E 2 V 2 1.00 × 10 −7 1 = (8.85 × 10 −12 )(3000)2 2 V V = 2.51 × 10 −3 m 3 ( = 2.51 × 10 −3 m 3 L = ) 1000 m  3 2.51 L Q2 d  Q2  d  Q2x  dU = = = dx dx  2c  dx  2e0 A  2 e0 A 26.35 W = U = ∫ F dx 26.36 Plate a experiences force – kx i from the spring and force QE i due to the electric field created by plate b according to E = σ / 2e0 = Q / 2 Ae0 . Then, F= so Q2 kx = 2A e 0 x= Q2 2Ae0 k where A is the area of one plate. 26.37 The energy transferred is W = 21 Q( ∆V ) = 21 (50.0 C)(1.00 × 108 V) = 2.50 × 10 9 J and 1% of this (or W' = 2.50 × 107 J) is absorbed by the tree. If m is the amount of water boiled away, then W' = m(4186 J/kg °C)(100 °C – 30.0 °C) + m(2.26 × 106 J/kg) = 2.50 × 107 J giving m = 9.79 kg © 2000 by Harcourt, Inc. All rights reserved. 94 Chapter 26 Solutions 26.38 U= 1 R kQ kQ 2 C( ∆V ) where C = 4 π e0 R = and ∆V = e − 0 = e 2 ke R R U= 1  R   keQ  keQ 2 =   2  ke   R  2R 2 keQ 2 = mc 2 2R 26.39 R= ke e 2 (8.99 × 10 9 N ⋅ m 2 / C)(1.60 × 10 −19 C)2 = = 1.40 fm 2 2 mc 2(9.11 × 10 − 31 kg)(3.00 × 108 m / s)2 *26.40 C= κ ∈0 A 4.90(8.85 × 10–12 F/m)(5.00 × 10–4 m2) = = 1.08 × 10–11 F = 10.8 pF d 2.00 × 10–3 m *26.41 (a) C= κ ∈0 A 2.10(8.85 × 10–12 F/m)(1.75 × 10–4 m2) = = 8.13 × 10–11 F = 81.3 pF d 4.00 × 10–5 m (b) ∆Vmax = Emax d = (60.0 × 106 V/m)(4.00 × 10–5 m) = 2.40 kV Qmax = C (∆Vmax), but ∆Vmax = Emax d *26.42 κ ∈0 A d κ ∈0 A Thus, Qmax = d (Emax d) = κ ∈0 AEmax Also, C = (a) With air between the plates, κ = 1.00 and Emax = 3.00 × 106 V/m. Therefore, Qmax = κ ∈0 AEmax = (8.85 × 10–12 F/m)(5.00 × 10–4 m2)(3.00 × 106 V/m) = 13.3 nC (b) With polystyrene between the plates, κ = 2.56 and Emax = 24.0 × 106 V/m. Qmax = κ ∈0 AEmax = 2.56(8.85 × 10–12 F/m)(5.00 × 10–4 m2)(24.0 × 106 V/m) = 272 nC 26.43 C= κ ∈0 A d or Chapter 26 Solutions = 1.04 m © 2000 by Harcourt, Inc. All rights reserved. 95 96 Chapter 26 Solutions Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between them. Suppose the plastic has κ ≅ 3, Emax ~ 107 V/m and thickness 1 mil = 2.54 cm/1000. Then, *26.44 C= κ ∈0 A 3(8.85 × 10–12 C2/N · m2)(0.4 m2) ~ ~ 10–6 F d 2.54 × 10–5 m ( ) V  2.54 × 10 −5 m ~ 10 2 V ∆Vmax = Emax d ~ 107  m *26.45 (a) (b) With air between the plates, we find C0 = Q ∆V 48.0 µC = 12.0 V = 4.00 µF When Teflon is inserted, the charge remains the same (48.0 µC) because the plates are isolated. However, the capacitance, and hence the voltage, changes. The new capacitance is C' = κ C 0 = 2.10(4.00 µF) = 8.40 µF (c) The voltage on the capacitor now is ∆V' = Q 48.0 µC C' = 8.40 µF = 5.71 V and the charge is 48.0 µC Originally, C =∈0 A / d = Q /(∆V)i 26.46 (a) The charge is the same before and after immersion, with value Q = ∈0 A(∆V)i / d . Q= (b) (8.85 × 10 −12 C 2 )(25.0 × 10 − 4 m 2 )(250 V) = 369 pC N ⋅ m 2 (1.50 × 10 − 2 m) Cf = 80.0(8.85 × 10 −12 C 2 )(25.0 × 10 − 4 m 2 ) = 118 pF N ⋅ m 2 (1.50 × 10 − 2 m) (∆V) f = (c) C f = κ ∈0 A / d = Q /(∆V)f Finally, ∈ A(∆V)i d (∆V)i 250 V Qd = 0 = = = 3.12 V 80.0 κ ∈0 A κ ∈0 A d κ ∈0 A(∆V)i2 2d Originally, U = 21 C(∆V)i2 = Finally, U f = 21 C f (∆V) f = So, ∆U = ∆U = U f − U = − 2 κ ∈0 A(∆V) i2 ∈0 A(∆V) i2 = 2d κ 2d κ 2 ∈0 A(∆V)i2 (κ − 1) 2d κ (− 8.85 × 10 −12 C 2)( 25.0 × 10 − 4 m 2 )(250 V)2 (79.0) = – 45.5 nJ N ⋅ m 2 2(1.50 × 10 − 2 m)80 Chapter 26 Solutions k (b − a) ke (c − b) 1 1 1 = + = e + C  κ 1ab   κ 2bc  κ 1ab κ 2bc  k (b − a)   k (c − b)   e   e  26.47 C= 26.48 97 κ 1κ 2 abc 1 = = ke (b − a) ke (c − b) ke κ 2 (bc − ac) + ke κ 1 (ac − ab) + κ 1 ab κ 2bc 4 π κ 1κ 2 abc∈0 κ 2bc − κ 1ab + (κ 1 − κ 2 )ac κ ∈0 A (173)(8.85 × 10 −12 )(1.00 × 10 −4 m 2 ) = = 1.53 nF d 0.100 × 10 −3 m (a) C = κ C0 = (b) The battery delivers the free charge Q = C (∆V) = (1.53 × 10-9 F)(12.0 V) = 18.4 nC (c) The surface density of free charge is σ= Q 18.4 × 10 −9 C = = 1.84 × 10-4 C/m2 A 1.00 × 10 −4 m 2 The surface density of polarization charge is 1 1  σ p = σ 1 −  = σ 1 − = 1.83 × 10-4 C/m2   κ 173  (d) We have E = E0/κ and E0 = ∆V/d ; hence, E= 26.49 ∆V 12.0 V = = 694 V/m κ d (173)(1.00 × 10 −4 m) The given combination of capacitors is equivalent to the circuit diagram shown to the right. Put charge Q on point A . Then, Q = (40.0 µ F)∆V AB = (10.0 µ F)∆VBC = (40.0 µ F)∆VCD So, ∆VBC = 4 ∆V AB = 4 ∆VCD , and the center capacitor will break down first, at ∆VBC = 15.0 V. When this occurs, ∆VAB = ∆VCD = 41 ( ∆VBC ) = 3.75 V and V AD = V AB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V © 2000 by Harcourt, Inc. All rights reserved. 98 Chapter 26 Solutions *26.50 (a) The displacement from negative to positive charge is 2a = ( −1.20 i + 1.10 j) mm − (1.40 i − 1.30 j) mm = ( −2.60i + 2.40 j) × 10 −3 m The electric dipole moment is ( ) p = 2aq = 3.50 × 10 −9 C ( −2.60i + 2.40 j) × 10 −3 m = (b) [ (−9.10i + 8.40 j) × 10−12 C ⋅ m ] [ τ = p × E = ( −9.10i + 8.40 j) × 10 −12 C ⋅ m × (7.80i − 4.90 j) × 10 3 N C ] τ = ( + 44.6 k − 65.5k ) × 10 −9 N ⋅ m = −2.09 × 10 −8 N ⋅ m k (c) [ ][ U = −p ⋅ E = − ( −9.10i + 8.40 j) × 10 −12 C ⋅ m ⋅ (7.80i − 4.90 j) × 10 3 N C ] U = (71.0 + 41.2) × 10 −9 J = 112 nJ (d) p = (9.10)2 + (8.40)2 × 10 −12 E = (7.80)2 + ( 4.90)2 × 103 U max = p E = 114 nJ, C ⋅ m = 12.4 × 10 −12 C ⋅ m N C = 9.21 × 10 3 N C U min = − 114 nJ U max − U min = 228 nJ *26.51 (a) Let x represent the coordinate of the negative charge. Then x + 2acos θ is the coordinate of the positive charge. The force on the negative charge is F − = −qE( x ) i. The force on the positive charge is F- (b) The balloon creates field along the x − axis of Thus, ( F = F− + F+ = q dE dE (2acos θ ) i = p cos θ i dx dx ke q i. x2 dE ( −2)ke q = dx x3 At x = 16.0 cm , ( )( ) 9 −6 MN dE ( −2) 8.99 × 10 2.00 × 10 = = − 8.78 3 C⋅m dx (0.160) ) θ E dE F + = +qE( x + 2acos θ ) i ≅ qE( x ) i + q ( 2acos θ ) i dx The force on the dipole is altogether F+ p N   cos 0˚ i = – 55.3 i mN F = 6.30 × 10 −9 C ⋅ m − 8.78 × 106  C ⋅ m Chapter 26 Solutions qin ∈0 2π r E = 26.52 so r2 r2 r1 r1 ∆V = − ∫ E ⋅ dr = ∫ E= r  λ λ dr = ln 1  2 π r ∈0 2 π ∈0  r2  99 λ 2 π r ∈0 λ max = Emax rinner 2 π∈0 V   25.0  (0.100 × 10 −3 m) ln ∆V = 1.20 × 106    0.200  m *26.53 (a) Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A′ << A parallel to the sheet. The side wall of the cylinder passes no flux of electric field since this surface is everywhere parallel to the field. Gauss’s law becomes EA′ + EA′ = (b) Q A′ , ∈A so E= Q 2∈A Q ∈A Assume that the field is in the positive x − direction. Then, the potential of the positive plate relative to the negative plate is ∆V = − ∫ + plate − plate E ⋅ ds = − ∫ + plate − plate Q i ⋅ ( − i dx ) = ∈A (d) Capacitance is defined by: C = 26.54 directed away from the positive sheet. In the space between the sheets, each creates field Q 2 ∈ A away from the positive and toward the negative sheet. Together, they create a field of E= (c) ∆Vmax = 579 V –1 1 1 +  2.00 + 4.00   –1 1  1 C= + 3.00 6.00 (c) Qac = Cac (∆Vac) = (2.00 µF)(90.0 V) = 180 µC ( Qd ∈A ∈ A κ ∈0 A Q Q = = = d ∆V Qd ∈ A d (a) Therefore, + = 3.33 µF Q3 = Q6 = 180 µC ) Qdf = Cdf ∆V df = (1.33 µ F )(90.0 V ) = 120 µC © 2000 by Harcourt, Inc. All rights reserved. 100 Chapter 26 Solutions (b) ∆V 3 = 180 µC Q3 = = 60.0 V 3.00 µF C3 ∆V 6 = Q6 180 µC C 6 = 6.00 µF = 30.0 V ∆V 2 = 120 µC Q2 C 2 = 2.00 µF = 60.0 V ∆V 4 = Q4 120 µC C 4 = 4.00 µF = 30.0 V 1 1 (d) UT = 2 C eq (∆V)2 = 2 (3.33 × 10–6)(90.0 V) 2 = 13.4 mJ *26.55 The electric field due to the charge on the positive wire is perpendicular to the wire, radial, and of magnitude E+ = λ 2 π ∈0 r The potential difference between wires due to the presence of this charge is ∆V1 = − ∫ +wire − wire E ⋅ dr = − λ 2 π ∈0 d ∫ D−d dr λ  D − d ln = r 2 π ∈0  d  The presence of the linear charge density − λ on the negative wire makes an identical contribution to the potential difference between the wires. Therefore, the total potential difference is ∆V = 2( ∆V1 ) = λ  D − d ln π∈0  d  and the capacitance of this system of two wires, each of length , is The capacitance per unit length is: Chapter 26 Solutions 101 26.56 (a) We use Equation 26.11 to find the potential energy. As we will see, the potential difference ∆V changes as the dielectric is withdrawn. The initial and final energies are 1 Q 2 Ui = 2  C   i and Uf = 1 Q 2   2 Cf  But the initial capacitance (with the dielectric) is Ci = κ C f. Therefore, 1 Q 2 Uf = 2 κ    Ci  Since the work done by the external force in removing the dielectric equals the change i n potential energy, we have Q 2 1 W = Uf – Ui = 2 κ  C   i – 1 Q 2  2  C i  = 1 Q 2  (κ – 1) 2  C i  To express this relation in terms of potential difference ∆V i , we substitute Q = Ci (∆V i), and evaluate: 1 1 W = 2 C i (∆V i)2(κ – 1) = 2 (2.00 × 10–9 F)(100 V) 2(5.00 – 1.00) = 4.00 × 10–5 J The positive result confirms that the final energy of the capacitor is greater than the initial energy. The extra energy comes from the work done o n the system by the external force that pulled out the dielectric. (b) Q The final potential difference across the capacitor is ∆V f = C f Substituting C f = Ci and Q = Ci (∆Vi) gives κ ∆Vf = κ ∆Vi = (5.00)(100 V) = 500 V Even though the capacitor is isolated and its charge remains constant, the potential difference across the plates does increase in this case. κ = 3.00, Emax = 2.00 × 108 V/m = ∆Vmax /d 26.57 For 26.58 (a) C1 = C= κ e0 A = 0.250 × 10–6 F, d A= (0.250 × 10–6)(4000) Cd C( ∆Vmax ) = = = 0.188 m2 (3.00)(8.85 × 10–12)(2.00 × 108) κ e0 κ e0Emax κ1 e0 A/2 ; d 1 + 1 C 2 C 3 –1 = C2 = κ2 e0 A/2 ; d/2 C3 = κ3 e0 A/2 d/2 e0 A  κ 2 κ 3  C 2C 3  = d  C2 + C3 κ 2 + κ 3  1 1 C = C1 +  + C  C 2 3 –1 = e0 A  κ 1 κ 2κ 3  d  2 + κ 2 + κ 3 © 2000 by Harcourt, Inc. All rights reserved. 102 Chapter 26 Solutions (b) Using the given values we find: Ctotal = 1.76 × 10–12 F = 1.76 pF Chapter 26 Solutions 103 26.59 The system may be considered to be two capacitors in series: C1 = e0 A t1 and e0 A t2 C2 = 1 1 1 t1 + t2 C = C1 + C2 = e0 A e0 A e0 A C= t +t = s – d 1 2 Goal Solution A conducting slab of a thickness d and area A is inserted into the space between the plates of a parallelplate capacitor with spacing s and surface area A, as shown in Figure P26.59. The slab is not necessarily halfway between the capacitor plates. What is the capacitance of the system? G: It is difficult to predict an exact relationship for the capacitance of this system, but we can reason that C should increase if the distance between the slab and plates were decreased (until they touched and formed a short circuit). So maybe C ∝ 1/ ( s − d) . Moving the metal slab does not change the amount of charge the system can store, so the capacitance should therefore be independent of the slab position. The slab must have zero net charge, with each face of the plate holding the same magnitude of charge as the outside plates, regardless of where the slab is between the plates. O: If the capacitor is charged with + Q on the top plate and −Q on the bottom plate, then free charges will move across the conducting slab to neutralize the electric field inside it, with the top face of the slab carrying charge −Q and the bottom face carrying charge + Q. Then the capacitor and slab combination is electrically equivalent to two capacitors in series. (We are neglecting the slight fringing effect of the electric field near the edges of the capacitor.) Call x the upper gap, so that s − d − x is the distance between the lower two surfaces. A : For the upper capacitor, C1 = ∈0 A x and the lower has C2 = So the combination has C= ∈0 A s−d−x ∈ A 1 1 = = 0 1 1 x s−d−x s−d + + C1 C2 ∈0 A ∈0 A L : The equivalent capacitance is inversely proportional to ( s − d) as expected, and is also proportional to A . This result is the same as for the special case in Example 26.9 when the slab is just halfway between the plates; the only critical factor is the thickness of the slab relative to the plate spacing. © 2000 by Harcourt, Inc. All rights reserved. 104 Chapter 26 Solutions 26.60 (a) (b) Put charge Q on the sphere of radius a and – Q on the other sphere. Relative to V = 0 at infinity, k eQ k eQ – the potential at the surface of a is Va = d a k eQ –k eQ + d b and the potential of b is Vb = The difference in potential is Va – Vb = and C= k eQ k eQ k eQ k eQ a + b – d – d   Q 4 π∈0 =   Va – Vb  (1 a) + (1 b) − ( 2 d)  As d → ∞, 1/d becomes negligible compared to 1/a. Then, C= 4 π ∈0 1 a+1 b and 1 1 1 = + C 4π ∈0 a 4π ∈0 b as for two spheres in series. Note that the potential difference between the plates is held constant at 26.61 Ci = q0 ∆Vi and C f = But C f = κ Ci , Thus, κ = 26.62 so qf ∆Vi = q0 + q ∆Vi  q  q0 + q = κ 0  ∆Vi  ∆Vi  q0 + q q or κ = 1 + q0 q0 (a) (b) (c) U = 21 C( ∆V )2 = F = − dU = dx to the left ∆Vi by the battery. Chapter 26 Solutions 105 (d) F= (2000)2 (8.85 × 10 −12 )(0.0500)(4.50 − 1) = 1.55 × 10 −3 N 2(2.00 × 10 −3 ) © 2000 by Harcourt, Inc. All rights reserved. 106 Chapter 26 Solutions The portion of the capacitor nearly filled by metal has capacitance κ ∈0 ( x) d → ∞ and stored *26.63 energy Q 2 2C → 0 . The unfilled portion has capacitance ∈0 portion is Q = ( −x)Q0 / . (a) The stored energy is (b) F= − ( −x ) d. The charge on this dU = dx F= to the right (c) (d) 26.64 Gasoline: Battery: Capacitor:   1.00 m 3   1.00 gal Btu   J J  = 5.25 × 107  126 000 gal   1054 Btu    -3 3  kg    3.786 × 10 m   670 kg  (12.0 J C)(100 C s)( 3600 s) = 2.70 × 10 5 J kg 16.0 kg 1 0.100 2( F )(12.0 V ) 0.100 kg 2 = 72.0 J kg Gasoline has 194 times the specific energy content of the battery and 727 000 times that of the capacitor Chapter 26 Solutions 107 26.65 Call the unknown capacitance Cu Q = Cu (∆Vi ) = (Cu + C)(∆V f ) Cu = C(∆V f ) (∆Vi ) − (∆V f ) = (10.0 µ F)( 30.0 V) = 4.29 µF (100 V − 30.0 V) Goal Solution An isolated capacitor of unknown capacitance has been charged to a potential difference of 100 V. W h e n the charged capacitor is then connected in parallel to an uncharged 10.0-µ F capacitor, the voltage across the combination is 30.0 V. Calculate the unknown capacitance. G: The voltage of the combination will be reduced according to the size of the added capacitance. (Example: If the unknown capacitance were C = 10.0 µ F, then ∆V1 = 50.0 V because the charge is now distributed evenly between the two capacitors.) Since the final voltage is less than half the original, we might guess that the unknown capacitor is about 5.00 µ F. O: We can use the relationships for capacitors in parallel to find the unknown capacitance, along with the requirement that the charge on the unknown capacitor must be the same as the total charge o n the two capacitors in parallel. A : We name our ignorance and call the unknown capacitance Cu . The charge originally deposited o n each plate, + on one, − on the other, is Q = Cu ∆V = Cu (100 V ) Now in the new connection this same conserved charge redistributes itself between the two capacitors according to Q = Q1 + Q2 . Q1 = Cu ( 30.0 V ) and Q2 = (10.0 µ F )( 30.0 V ) = 300 µ C We can eliminate Q and Q1 by substitution: Cu (100 V ) = Cu ( 30.0 V ) + 300 µ C so Cu = 300 µ C = 4.29 µ F 70.0 V L : The calculated capacitance is close to what we expected, so our result seems reasonable. In this and other capacitance combination problems, it is important not to confuse the charge and voltage of the system with those of the individual components, especially if they have different values. Careful attention must be given to the subscripts to avoid this confusion. It is also important to not confuse the variable “ C ” for capacitance with the unit of charge, “ C” for coulombs. © 2000 by Harcourt, Inc. All rights reserved. 108 Chapter 26 Solutions Put five 6.00 pF capacitors in series. 26.66 The potential difference across any one of the capacitors will be: ∆V = ∆Vmax 1000 V = = 200 V 5 5 and the equivalent capacitance is:  1 1  = 5  Ceq  6.00 pF  or Ceq = 6.00 pF = 1.20 pF 5 When ∆V db = 0, ∆V bc = ∆V dc , and 26.67 Also, ∆V ba = ∆V da or Q2 Q3 = C2 C3 Q1 Q4 = C1 C4 C Q Q  From these equations we have C2 =  3   2   4  C1  C4   Q1   Q3  However, from the properties of capacitors in series, we have Therefore, C  9.00 C2 =  3  C1 = (4.00 µ F) = 3.00 µF 12.0  C4  Let C = the capacitance of an individual capacitor, and Cs represent the equivalent capacitance of the group in series. While being charged in parallel, each capacitor receives charge 26.68 Q = C ∆Vchg = (5.00 × 10-4 F)(800 V) = 0.400 C While being discharged in series, ∆Vdisch = Q Q 0.400 C = = = 8.00 kV Cs C / 10 5.00 × 10 −5 F or 10 times the original voltage. 26.69 Q 1 = Q 2 and Q 3 = Q 4 (a) C0 = ∈0 A Q0 = d ∆V0 When the dielectric is inserted at constant voltage, Chapter 26 Solutions 109 C = κ C0 = U= Q ; ∆V0 U0 = C0 (∆V0 )2 2 C(∆V0 )2 κ C0 (∆V0 )2 = 2 2 and U =κ U0 The extra energy comes from (part of the) electrical work done by the battery in separating the extra charge. (b) Q0 = C0 ∆V0 and Q = C ∆V0 = κ C0 ∆V0 so Q /Q 0 = κ © 2000 by Harcourt, Inc. All rights reserved. 110 Chapter 26 Solutions 26.70 (a) A slice of width (dx) at coordinate x in 0 ≤ x ≤ L has thickness x d /L filled with dielectric κ 2, and d – x d /L is filled with the material having constant κ 1. This slice has a capacitance given by 1 1 xd dL − xd 1 κ xd + κ 2 dL − κ 2 xd = + = + = 1 e (dx)W e (dx)W κ κ dC  2 0 κ 1κ 2 e0WL( dx )   1 0  κ 2 e0 W L(dx) κ 1 e0 W L(dx)      xd / L   d − xd / L  dC = κ 1 κ 2 e0 W L(dx) κ 2 dL + (κ 1 − κ 2 )xd The whole capacitor is all the slices in parallel: C = ∫ dC = ∫ L x=0 C= (b) L κ 1 κ 2 e0W L(dx) κ κ e WL −1 = 1 2 0 ∫ (κ 2Ld + (κ 1 − κ 2 )xd) (κ 1 − κ 2 )d(dx) κ 2 Ld + (κ 1 − κ 2 )xd (κ 1 − κ 2 )d x=0 L κ 1 κ 2 e0 W L κ κ e WL ln [κ 2 Ld + (κ 1 − κ 2 )xd] = 1 2 0 [ln κ 1Ld − ln κ 2Ld] = 0 (κ 1 − κ 2 )d (κ 1 − κ 2 )d κ 1κ 2 e0 W L κ ln 1 κ2 (κ 1 − κ 2 )d To take the limit κ 1 → κ 2 , write κ 1 = κ 2 (1 + x) and let x → 0. Then C= κ 22 (1 + x)e0 W L ln (1 + x) (κ 2 + κ 2 x − κ 2 )d Use the expansion of ln(1 + x) from Appendix B.5. C= κ 22 (1 + x) ∈0 W L κ (1 + x) ∈0 W L (x − 21 x 2 + 31 x 3 . . . ) = 2 (1 − 21 x + . . . ) κ 2 xd d lim C = x→0 26.71 κ 2 ∈0 W L κ ∈0 A = d d The vertical orientation sets up two capacitors in parallel, with equivalent capacitance Cp = ∈0 ( A 2) κ ∈0 ( A 2)  κ + 1 ∈0 A + =  2  d d d where A is the area of either plate and d is the separation of the plates. The horizontal orientation produces two capacitors in series. If f is the fraction of the horizontal capacitor filled with dielectric, the equivalent capacitance is fd (1 − f )d =  f + κ (1 − f )  d , or 1 = +   Cs κ ∈0 A ∈0 A κ   ∈0 A Requiring that Cp = Cs gives   ∈0 A κ Cs =    f + κ (1 − f )  d κ +1 κ = , or (κ + 1)[ f + κ (1 − f )] = 2κ 2 f + κ (1 − f ) For κ = 2.00, this yields 3.00[2.00 − (1.00) f ] = 4.00 , with the solution f = 2 / 3 . Chapter 26 Solutions 111 26.72 Initially (capacitors charged in parallel), q1 = C1(∆V) = (6.00 µF)(250 V) = 1500 µC q2 = C2(∆V) = (2.00 µF)(250 V) = 500 µC After reconnection (positive plate to negative plate), qtotal ′ = q1 – q2 = 1000 µC ∆V ′ = and qtotal ′ 1000 µ C = = 125 V Ctotal 8.00 µ F Therefore, q1′ = C1(∆V ′ ) = (6.00 µ F)(125 V) = 750 µC q2′ = C2 (∆V ′ ) = (2.00 µ F)(125 V) = 250 µC 26.73 Emax occurs at the inner conductor's surface. Emax = 2ke λ from Equation 24.7. a ∆V = 2ke λ ln Emax =  b from Example 26.2  a ∆V a ln(b / a) ∆Vmax = Emax a ln 26.74 E= 2κλ ; a  b  3.00  = (18.0 × 106 V / m)(0.800 × 10 −3 m) ln = 19.0 kV  0.800   a ∆V = 2κλ ln ∆Vmax = Emax a ln  b  a  b  a   b dVmax  1   b  + a = Emax ln =0  −    b / a   a 2   a da  ln  b =1  a or b = e1 a so a = b e © 2000 by Harcourt, Inc. All rights reserved. 112 Chapter 26 Solutions 26.75 Assume a potential difference across a and b, and notice that the potential difference across the 8.00 µ F capacitor must be zero by symmetry. Then the equivalent capacitance can be determined from the following circuit: ⇒ ⇒ Cab = 3.00 µ F 26.76 By symmetry, the potential difference across 3C is zero, so the circuit reduces to Ceq = 4 (2C)(4C) 8 = C= C 3 2C + 4C 6 ⇒ ⇒ Chapter 27 Solutions ∆Q ∆t I= 27.1 N= ∆Q = I ∆t = (30.0 × 10–6 A)(40.0 s) = 1.20 × 10–3 C Q 1.20 × 10–3 C = = 7.50 × 1015 electrons e 1.60 × 10–19 C/electron The atomic weight of silver = 107.9, and the volume V is *27.2 V = (area)(thickness) = (700 × 10-4 m2)(0.133 × 10-3 m) = 9.31 × 10-6 m3 ( )( ) The mass of silver deposited is mAg = ρ V = 10.5 × 10 3 kg / m 3 9.31 × 10 −6 m 3 = 9.78 × 10 −2 kg. and the number of silver atoms deposited is N = (9.78 × 10-2 kg) I= 6.02 × 1026 atoms = 5.45 × 1023 107.9 kg 12.0 V V = = 6.67 A = 6.67 C/s R 1.80 Ω ∆t = ∆Q Ne (5.45 × 1023)(1.60 × 10-19 C) = = = 1.31 × 104 s = 3.64 h 6.67 C/s I I t Q(t) = ∫ 0 Idt = I0τ (1 – e –t/τ ) 27.3 27.4 (a) Q(τ) = I0τ (1 – e–1) = (0.632)I0τ (b) Q(10τ ) = I0τ (1 – e–10) = (0.99995)I0τ (c) Q(∞) = I0τ (1 – e– ∞) = I0τ (a) Using (b) The time for the electron to revolve around the proton once is: t= ke e 2 mv 2 = , we get: r r2 v= ke e 2 = 2.19 × 106 m / s . mr 2 π r 2 π (5.29 × 10 −11 m) = = 1.52 × 10 −16 s v (2.19 × 106 m / s) The total charge flow in this time is 1.60 × 10 −19 C, so the current is © 2000 by Harcourt, Inc. All rights reserved. 112 Chapter 27 Solutions I= 1.60 × 10 −19 C = 1.05 × 10 −3 A = 1.05 mA 1.52 × 10 −16 s Chapter 27 Solutions 113 2π T ω= 27.5 I= where T is the period. q ω (8.00 × 10 − 9 C)(100 π rad / s) q = = = 4.00 × 10 − 7 A = 400 nA T 2π 2π The period of revolution for the sphere is T = 27.6 revolving charge is I = 2π , and the average current represented by this ω qω q = . T 2π 2  1.00 m  −4 2 A = (2.00 cm )   = 2.00 × 10 m  100 cm  2 3 q = 4t + 5t + 6 27.7 (a) (b) 27.8 I (1.00 s) = dq dt ( = 12t 2 + 5 t=1.00 s t=1.00 s = 17.0 A J= I 17.0 A = = 85.0 kA / m 2 A 2.00 × 10 − 4 m 2 I= dq dt q = ∫ dq = ∫ I dt = ∫ 1/240 s 0 27.9 ) (100 A) sin (120 π t / s)dt q= −100 C +100 C [cos(π / 2) − cos 0] = 120π = 0.265 C 120 π (a) J= I 5.00 A = = 99.5 kA/m2 A π (4.00 × 10 −3 m)2 (b) J2 = 1 J1 ; 4 I 1 I = A2 4 A1 A1 = 1 A2 4 so π (4.00 × 10 −3 )2 = 1 2 π r2 4 r2 = 2(4.00 × 10 −3 ) = 8.00 × 10 −3 m = 8.00 mm © 2000 by Harcourt, Inc. All rights reserved. 114 Chapter 27 Solutions 27.10 (a) 1 K = 2 mv2 The speed of each deuteron is given by 1 (2.00 × 106)(1.60 × 10–19 J) = 2 (2 × 1.67 × 10–27 kg) v2 v = 1.38 × 107 m/s and The time between deuterons passing a stationary point is t in I = q /t 10.0 × 10–6 C/s = 1.60 × 10–19 C/t t = 1.60 × 10–14 s or ( )( ) So the distance between them is vt = 1.38 × 107 m / s 1.60 × 10 −14 s = 2.21 × 10–7 m (b) One nucleus will put its nearest neighbor at potential V= k eq (8.99 × 109 N · m2/C 2)(1.60 × 10-19 C) = = 6.49 × 10–3 V r 2 .21 × 10–7 m This is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect. 27.11 8.00 × 10 −6 A I = A π 1.00 × 10 −3 m (a) J= (b) From J = nevd , we have n= From I = ∆Q / ∆t , we have 6.02 × 10 23 1.60 × 10 −19 C ∆Q N A e ∆t = = = = 1.20 × 1010 s I I 8.00 × 10 −6 A (c) ( ) 2 = 2.55 A / m 2 2.55 A / m 2 J = = evd 1.60 × 10 -19 C 3.00 × 108 m / s ( )( ( 5.31 × 1010 m − 3 ) )( ) (This is about 381 years!) *27.12 We use I = nqAv d where n is the number of charge carriers per unit volume, and is identical to the number of atoms per unit volume). We assume a contribution of 1 free electron per atom in the relationship above. For aluminum, which has a molecular weight of 27, we know that Avogadro's number of atoms, N A , has a mass of 27.0 g. Thus, the mass per atom is 27.0 g 27.0 g = = 4.49 × 10–23 g/atom NA 6.02 × 1023 Thus, n = density of aluminum 2.70 g/cm3 = mass per atom 4.49 × 10–23 g/atom n = 6.02 × 1022 Therefore, or, vd = atoms atoms = 6.02 × 1028 m3 cm 3 I 5.00 A = 28 –3 nqA (6.02 × 10 m )(1.60 × 10–19 C)(4.00 × 10–6 m2) vd = 0.130 mm/s = 1.30 × 10–4 m/s Chapter 27 Solutions 115 I= *27.13 27.14 (a) Applying its definition, we find the resistance of the rod, R= (b) ∆V 120 V = = 0.500 A = 500 mA R 240 Ω ∆V 15.0 V = = 3750 Ω = 3.75 kΩ I 4.00 × 10 −3 A The length of the rod is determined from Equation 27.11: R = ρ / A. Solving for and substituting numerical values for R, A, and the values of ρ given for carbon in Table 27.1, we obtain = RA (3.75 × 10 3 Ω)(5.00 × 10 −6 m 2 ) = = 536 m ρ (3.50 × 10 −5 Ω ⋅ m) ∆V = IR 27.15 ∆V = R= and ρl : A 1.00 m  A = 0.600 mm2  1000 mm  Iρ l : A I= 2 = 6.00 × 10–7 m2 (0.900 V)(6.00 × 10–7 m2) ∆VA = (5.60 × 10–8 Ω · m)(1.50 m) ρl I = 6.43 A J= 27.16 3.00 A I = σE = = σ (120 N/C) 2 π (0.0120 m)2 πr σ = 55.3(Ω · m)-1 ρ = 27.17 (a) 1 = 0.0181 Ω · m σ Given M = ρdV = ρd Al where ρd ≡ mass density , Taking ρr ≡ resistivity, R= Thus, l= ρr l A = ρr l  M    ρdl = ( we obtain: A = M ρdl ρr ρdl2 M ) 1.00 × 10 −3 (0.500) MR = = 1.82 m ρr ρ d (1.70 × 10 −8 )(8.92 × 10 3 ) © 2000 by Harcourt, Inc. All rights reserved. 116 Chapter 27 Solutions (b) V= M , or ρd Thus, r = π r 2l = M = πρdl M ρd 1.00 × 10 −3 = 1.40 × 10 −4 m π (8.92 × 10 3 )(1.82) The diameter is twice this distance: *27.18 (a) diameter = 280 µm Suppose the rubber is 10 cm long and 1 mm in diameter. ( )( ) 13 -1 ρ l 4ρ l 4 10 Ω ⋅ m 10 m R= = ~ = ~1018 Ω 2 −3 A π d2 π 10 m ( ) (b) R= 4(1.7 × 10–8 Ω · m)(10–3 m) 4ρ l ~ ~ 10–7 Ω π (2 × 10–2 m)2 π d2 (c) I= 10 2 V ∆V ~ ~ 10–16 A R 1018 Ω I~ 10 2 V ~ 10 9 A 10–7 Ω The distance between opposite faces of the cube is 27.19  90.0 g  l=  3  10.5 g cm  (a) R= ρ l ρ l ρ 1.59 × 10 −8 Ω ⋅ m = = = = 7.77 × 10 −7 Ω = 777 nΩ A l2 l 2.05 × 10 -2 m (b) I= ∆V 1.00 × 10 −5 V = = 12.9 A R 7.77 × 10 −7 Ω n= 10.5 g cm 3  electrons  6.02 × 10 23  mol  107.87 g mol 13 = 2.05 cm electrons   1.00 × 106 cm 3   28 3 n = 5.86 × 10 22   = 5.86 × 10 m  1.00 m 3 cm 3    I = nqvA and v= 12.9 C s I = = 3.28 µm/s 2 28 3 nqA 5.86 × 10 m 1.60 × 10 −19 C (0.0205 m ) ( )( ) Chapter 27 Solutions 117 27.20 27.21 ρl A Originally, R= Finally, Rf = ρ (l/ 3) ρ l R = = 3A 9A 9 The total volume of material present does not change, only its shape. Thus, A f lf = A f (1.25li ) = Aili giving A f = Ai 1.25 The final resistance is then: R f = 27.22 ρ All π (r Al ) 2 = J = σE 27.24 R= 27.25 ρ1l1 A1 = ρ (1.25li )  ρl  = 1.56  i  = 1.56R Ai 1.25  Ai  2 2.82 × 10 −8 = 1.29 1.70 × 10 −8 σ= so + Af ρCul π (rCu ) rAl ρAl = = rCu ρCu 27.23 ρ lf ρ2l 2 A2 J 6.00 × 10 −13 A / m 2 = = 6.00 × 10-15 (Ω · m)-1 100 V / m E = (ρ1l1 + ρ2l 2 ) / d 2 R= (4.00 × 10 − 3 Ω ⋅ m)( 0.250 m) + ( 6.00 × 10 − 3 Ω ⋅ m)( 0.400 m) = 378 Ω (3.00 × 10 − 3 m)2 ρ= m nq 2 τ vd = qE m τ Therefore m 9.11 × 10–31 = = 2.47 × 10–14 s (1.70 × 10–8)(8.49 × 1028)(1.60 × 1019)2 ρ nq 2 so τ= so 7.84 × 10–4 = (1.60 × 10–19)E(2.47 × 10–14) 9.11 × 10–31 E = 0.181 V/m © 2000 by Harcourt, Inc. All rights reserved. 118 Chapter 27 Solutions Goal Solution If the drift velocity of free electrons in a copper wire is 7.84 × 10 −4 m / s , what is the electric field in the conductor? G: For electrostatic cases, we learned that the electric field inside a conductor is always zero. On the other hand, if there is a current, a non-zero electric field must be maintained by a battery or other source to make the charges flow. Therefore, we might expect the electric field to be small, but definitely not zero. O: The drift velocity of the electrons can be used to find the current density, which can be used with Ohm’s law to find the electric field inside the conductor. A : We first need the electron density in copper, which from Example 27.1 is n = 8.49 × 10 28 e - / m 3 . current density in this wire is then The J = nqvd = (8.49 × 10 28 e − / m 3 )(1.60 × 10 −19 C / e - )(7.84 × 10 −4 m / s) = 1.06 × 107 A / m 2 Ohm’s law can be stated as J = σE = E / ρ where ρ = 1.7 × 10 −8 Ω ⋅ m for copper, so then E = ρ J = (1.70 × 10 −8 Ω ⋅ m)(1.06 × 107 A / m 2 ) = 0.181 V / m L : This electric field is certainly smaller than typical static values outside charged objects. The direction of the electric field should be along the length of the conductor, otherwise the electrons would be forced to leave the wire! The reality is that excess charges arrange themselves on the surface of the wire to create an electric field that “steers” the free electrons to flow along the length of the wire from low to high potential (opposite the direction of a positive test charge). It is also interesting to note that when the electric field is being established it travels at the speed of light; but the drift velocity of the electrons is literally at a “snail’s pace”! 27.26 (a) n is unaffected J = (c) J = nevd so v d doubles (d) τ = 27.27 I ∝ I so it doubles A (b) mσ is unchanged nq 2 as long as σ does not change due to heating. From Equation 27.17, τ= me 9.11 × 10 −31 = = 2.47 × 10 −14 s −19 2 28 −8 nq 2ρ 1.60 × 10 8.49 × 10 1.70 × 10 ( ( )( )( )( ) ) l = vτ = 8.60 × 10 5 m / s 2.47 × 10 −14 s = 2.12 × 10 − 8 m = 21.2 nm Chapter 27 Solutions 119 27.28 At the low temperature TC we write RC = ∆V = R0 [1 + α (TC − T0 )] where T0 = 20.0°C IC At the high temperature T h , Rh = ∆V ∆V Ih = 1 A = R0 [1 + α (T h − T0 )] (∆V)/(1.00 A) Then (∆V)/IC R = R0 [1 + α (∆T)] 1 + (3.90 × 10–3)(–108) [ 140 Ω = (19.0 Ω) 1 + (4.50 × 10 −3 /°C)∆T gives T = 1.44 × 103 °C And, the final temperature is R = R c + R n = R c [1 + α c (T − T0 )] + R n [1 + α n (T − T0 )] 27.30 0 = Rcα c (T – T 0) + R nα n (T – T 0) R = –R n ] ∆T = 1.42 × 103 °C = T – 20.0 °C Solving, so R c = –R n αn αc αn + Rn αc R n = R(1 – α n /α c)–1 R c = R(1 – α c /α n)–1 (0.400 × 10–3/C°)   Rn = 10.0 k Ω  1 –  (– 0.500 × 10–3/C°)  Rn = 5.56 k Ω 27.31 1 + (3.90 × 10–3)(38.0) IC = (1.00 A)(1.15/0.579) = 1.98 A and *27.29 = and –1 Rc = 4.44 k Ω [ ] (a) ρ = ρ0 [1+ α (T − T0 )] = (2.82 × 10 −8 Ω ⋅ m) 1+ 3.90 × 10 −3 (30.0°) = 3.15 × 10-8 Ω · m (b) J= (c) I = JA = (d) n= 0.200 V / m E = = 6.35 × 106 A / m 2 ρ 3.15 × 10 − 8 Ω ⋅ m 6.02 × 10 23 electrons = 6.02 × 10 28 electrons / m 3   26.98 g  2.70 × 106 g / m 3    vd = (e) π d2 π (1.00 × 10 −4 m)2 J= (6.35 × 106 A / m 2 ) = 49.9 mA 4 4 (6.35 × 106 A / m 2 ) J = = 659 µm/s 28 ne (6.02 × 10 electrons / m 3 )(1.60 × 10 −19 C) ∆V = E = (0.200 V / m)(2.00 m) = 0.400 V © 2000 by Harcourt, Inc. All rights reserved. 120 Chapter 27 Solutions *27.32 For aluminum, R= 27.33 αE = 3.90 × 10–3/°C (Table 27.1) α = 24.0 × 10–6/°C (Table 19.2) (1 + αE ∆T) (1.39) ρ l ρ0 (1 + α E ∆T )(1 + α∆T ) = = R0 = (1.234 Ω) (1.0024) = 1.71 Ω 2 (1 + α ∆T) A A(1 + α ∆T ) R = R0[1 + α ∆T] R – R 0 = R 0α ∆T R – R0 -3)25.0 = 0.125 R 0 = α ∆T = (5.00 × 10 27.34 Assuming linear change of resistance with temperature, R = R 0(1 + α ∆T) R 27.35 27.36 77 K = ρ = ρ0 (1 + α ∆T ) or ∆TW = 0.153 Ω 1 αW  ρW  − 1  ρ  0W  ( ) Require that ρW = 4ρ0 Cu so that −8     4 1.70 × 10 1 ∆TW =  − 1  = 47.6 °C   −8 −3   4.50 × 10 /°C   5.60 × 10  Therefore, TW = 47.6 °C + T0 = 67.6°C α= so, *27.37 (1.00 Ω)[1 + (3.92 × 10−3 )(−216°C)] = I= 1  ∆R   1  2R0 − R0 1 =  = R0  ∆T   R0  T − T0 T − T0 T=  1 + T0 α and 600 W P = = 5.00 A 120 V ∆V and R = 120 V ∆V = 5.00 A = 24.0 Ω I   1 T= +20.0 °C −3 −1   0.400 × 10 C°  so T = 2.52 × 103 °C Chapter 27 Solutions 121 27.38 P = 0.800(1500 hp)(746 W/hp) = 8.95 × 105 W P = I (∆V) 8.95 × 105 = I(2000) I = 448 A 27.39 The heat that must be added to the water is Q = mc ∆T = (1.50 kg)(4186 J/kg°C)(40.0°C) = 2.51 × 105 J Thus, the power supplied by the heater is P = W Q 2.51 × 105 J = = = 419 W t t 600 s and the resistance is 27.40 R= ( ∆V )2 P (110 V)2 = 419 W = 28.9 Ω The heat that must be added to the water is Q = mc(T 2 – T 1) Thus, the power supplied by the heat is P= and the resistance is R= W Q mc(T 2 − T1 ) = = ∆t ∆t t ( ∆V )2 P = 2 27.41 2 P (∆V)2 / R  ∆V   140  = 1.361 = =  = 2 P 0 (∆V0 ) / R  ∆V0  120   P − P0  P  ∆% =  − 1 (100%) = (1.361 − 1)100 = 36.1%  (100%) =   P0   P0  © 2000 by Harcourt, Inc. All rights reserved. ( ∆V )2 t mc(T 2 − T1 ) 122 Chapter 27 Solutions Goal Solution Suppose that a voltage surge produces 140 V for a moment. By what percentage does the power output of a 120-V, 100-W light bulb increase? (Assume that its resistance does not change.) G: , the power will increase as the square of The voltage increases by about 20%, but since the voltage: or a 36.1% increase. O: We have already found an answer to this problem by reasoning in terms of ratios, but we can also calculate the power explicitly for the bulb and compare with the original power by using Ohm’s law and the equation for electrical power. To find the power, we must first find the resistance of the bulb, which should remain relatively constant during the power surge (we can check the validity of this assumption later). A : From , we find that If = The final current is, ∆V f R = 140 V = 0.972 A 144 Ω The power during the surge is So the percentage increase is L: 136 W − 100 W = 0.361 = 36.1% 100 W Our result tells us that this 100 - W light bulb momentarily acts like a 136 - W light bulb, which explains why it would suddenly get brighter. Some electronic devices (like computers) are sensitive to voltage surges like this, which is the reason that surge protectors are recommended to protect these devices from being damaged. In solving this problem, we assumed that the resistance of the bulb did not change during the voltage surge, but we should check this assumption. Let us assume that the filament is made of tungsten and that its resistance will change linearly with temperature according to equation 27.21. Let us further assume that the increased voltage lasts for a time long enough so that the filament comes to a new equilibrium temperature. The temperature change can be estimated from the power surge according to Stefan’s law (equation 20.18), assuming that all the power loss is due to radiation. so that a 36% change in power should correspond to only about a 8% increase i n By this law, temperature. A typical operating temperature of a white light bulb is about 3000 °C, so ∆T ≈ 0.08( 3273 °C) = 260 °C. Then the increased resistance would be roughly ( ) R = R0 (1 + α (T − T0 )) = (144 Ω) 1 + 4.5 × 10 −3 ( 260) ≅ 310 Ω It appears that the resistance could change double from 144 Ω. On the other hand, if the voltage surge lasts only a very short time, the 136 W we calculated originally accurately describes the conversion of electrical into internal energy in the filament. Chapter 27 Solutions 123 P = I (∆V) = 27.42 (∆V)2 = 500 W R R= (110 V)2 = 24.2 Ω (500 W) (a) R= ρ l A (b) R = R 0 [1 + α ∆T] = 24.2 Ω 1 + (0.400 × 10 −3 )(1180) = 35.6 Ω 27.43 l= so RA (24.2 Ω)π (2.50 × 10 −4 m)2 = = 3.17 m ρ 1.50 × 10 −6 Ω ⋅ m [ ] P= (∆V)2 (110)2 = = 340 W R 35.6 R= –6 ρ l (1.50 × 10 Ω · m)25.0 m = = 298 Ω π (0.200 × 10–3 m)2 A ∆V = IR = (0.500 A)(298 Ω) = 149 V 149 V ∆V = 25.0 m = 5.97 V/m l (a) E= (b) P = (∆V)I = (149 V)(0.500 A) = 74.6 W (c) R = R0 [1 + α (T − T0 )] = 298 Ω 1 + (0.400 × 10 −3 / C°)320 C° = 337 Ω [ I= ] (149 V) ∆V R = (337 Ω) = 0.443 A P = (∆V)I = (149 V)(0.443 A) = 66.1 W 27.44 27.45 (a) 1J 1W·s 1C ∆U = q (∆V) = It (∆V) = (55.0 A · h)(12.0 V)  1 A · s  1 V · C  1 J  = 660 W · h = 0.660 kWh     (b) $0.0600 Cost = 0.660 kWh  = 3.96¢ 1 kWh  P = I (∆V) P = ∆V = IR (∆V)2 (10.0)2 = = 0.833 W R 120 © 2000 by Harcourt, Inc. All rights reserved. 124 Chapter 27 Solutions The total clock power is 27.46 From e = (270 × 10 6 ) J s   3600 s   = 2.43 × 1012 J h clocks 2.50  clock   1 h  W out , the power input to the generating plants must be: Qin Qin W out t 2.43 × 1012 J h = = = 9.72 × 1012 J h t e 0.250 and the rate of coal consumption is  1.00 kg coal  metric ton 5 kg coal Rate = 9.72 × 1012 J h  = 295  = 2.95 × 10 h h  33.0 × 106 J  ( ) P = I ( ∆V ) = (1.70 A )(110 V ) = 187 W 27.47 Energy used in a 24-hour day = (0.187 kW)(24.0 h) = 4.49 kWh $0.0600 = $0.269 = 26.9¢ ∴ cost = 4.49 kWh   kWh  P = I (∆V) = (2.00 A)(120 V) = 240 W 27.48 ∆U = (0.500 kg)(4186 J/kg°C)(77.0°C) = 161 kJ = 1.61 × 105 J = 672 s 240 W At operating temperature, 27.49 (a) P = I (∆V) = (1.53 A)(120 V) = 184 W (b) Use the change in resistance to find the final operating temperature of the toaster. R = R 0(1 + α ∆T) [ 120 120 −3 1.53 = 1.80 1 + (0.400 × 10 )∆T ∆T = 441°C T = 20.0°C + 441°C = 461°C ] Chapter 27 Solutions 125 Goal Solution A certain toaster has a heating element made of Nichrome resistance wire. When the toaster is first connected to a 120-V source of potential difference (and the wire is at a temperature of 20.0 °C), the initial current is 1.80 A. However, the current begins to decrease as the resistive element warms up. When the toaster has reached its final operating temperature, the current has dropped to 1.53 A. (a) Find the power the toaster consumes when it is at its operating temperature. (b) What is the final temperature of the heating element? G: Most toasters are rated at about 1000 W (usually stamped on the bottom of the unit), so we might expect this one to have a similar power rating. The temperature of the heating element should be hot enough to toast bread but low enough that the nickel-chromium alloy element does not melt. (The melting point of nickel is 1455 °C , and chromium melts at 1907 °C.) O: The power can be calculated directly by multiplying the current and the voltage. The temperature can be found from the linear conductivity equation for Nichrome, with α = 0.4 × 10 −3 °C -1 from Table 27.1. A : (a) (b) P = ( ∆V ) I = (120 V )(1.53 A ) = 184 W ∆V 120 V = = 66.7 Ω I 1.80 A The resistance at 20.0 °C is R0 = At operating temperature, R= 120 V = 78.4 Ω 1.53 A Neglecting thermal expansion, R= ρ l ρ0 (1 + α (T − T0 ))l = = R0 (1 + α (T − T0 )) A A T = T0 + R R0 − 1 78.4 Ω 66.7 Ω − 1 = 20.0 °C + = 461 °C α 0.4 × 10 −3 °C -1 L : Although this toaster appears to use significantly less power than most, the temperature seems high enough to toast a piece of bread in a reasonable amount of time. In fact, the temperature of a typical 1000-W toaster would only be slightly higher because Stefan’s radiation law (Eq. 20.18) tells us that , so that the temperature might be about 700 °C. (assuming all power is lost through radiation) In either case, the operating temperature is well below the melting point of the heating element. 27.50 P = (10.0 W / ft 2 )(10.0 ft)(15.0 ft) = 1.50 kW Energy = P t = (1.50 kW)(24.0 h) = 36.0 kWh Cost = (36.0 kWh)($ 0.0800 / kWh) = $2.88 *27.51 Consider a 400-W blow dryer used for ten minutes daily for a year. The energy converted is  1 kWh  P t = ( 400 J s)(600 s d)( 365 d) ≅ 9 × 107 J  ≅ 20 kWh  3.6 × 106 J  We suppose that electrical energy costs on the order of ten cents per kilowatt-hour. Then the cost of using the dryer for a year is on the order of Cost ≅ ( 20 kWh )($ 0.100 kWh ) = $ 2 ~$1 © 2000 by Harcourt, Inc. All rights reserved. 126 Chapter 27 Solutions *27.52 (a) I= ∆V R so (120 V)2 = 576 Ω 25.0 W = (b) I= t= (c) and = (120 V)2 = 144 Ω 100 W 25.0 W Q 1.00 C = 0.208 A = = 120 V t t 1.00 C 0.208 A = 4.80 s P = 25.0 W = t= (d) = (∆V)2 R P = (∆V)I = The charge has lower potential energy . ∆U 1.00 J = t t 1.00 J 25.0 W = 0.0400 s The energy changes from electrical to heat and light . ∆U = P t = (25.0 J/s)(86400 s/d)(30.0 d) = 64.8 × 106 J The energy company sells energy . k W·s h $0.0700 Cost = 64.8 × 106 J  kWh   1000  J   3600 s = $1.26      Cost per joule = *27.53 I = nqv d A = nqv d π r 2 We find the drift velocity from vd = v= *27.54 $0.0700  k W h  –8 k W h 3.60 × 106 J = $1.94 × 10 /J I n qπ r x t 2 = 1000 A 8.00 × 10 28 m –3 (1.60 × 10–19 C)π (10–2 m) 2 = 2.49 × 10–4 m/s x 200 × 103 m t= v = = 8.04 × 108 s = 25.5 yr 2.49 × 10–4 m/s  0.500 Ω  (100 mi) = 50.0 Ω The resistance of one wire is   mi  The whole wire is at nominal 700 kV away from ground potential, but the potential difference between its two ends is IR = (1000 A)(50.0 Ω) = 50.0 kV Then it radiates as heat power P = (∆V)I = (50.0 × 103 V)(1000 A) = 50.0 MW Chapter 27 Solutions 127 27.55 (a) α= Separating variables, ⌠ρ d ρ ⌠T ⌡ρ0 ρ = ⌡T0 α d T ρ  ln   = α (T – T 0)  ρ 0 (b) 1 dρ ρ dT We begin with the differential equation ρ = ρ0eα(T – T0) and From the series expansion e x ≅ 1 + x, (x << 1), ρ ≅ ρ0 [1 + α (T − T0 )] *27.56 Consider a 1.00-m length of cable. The potential difference between its ends is = 6.67 mV R= The resistance is Then 6.67 × 10–3 V ∆V = = 22.2 µ Ω 300 A I gives 1.56 cm 27.57 (m) 0.540 1.028 1.543 ρ (Ω · m) 1.41 × 10–6 1.50 × 10–6 1.50 × 10–6 R(Ω) 10.4 21.1 31.8 ρ– = 1.47 × 10–6 Ω · m (in agreement with tabulated value) 1.50 × 10–6 Ω · m (Table 27.1) © 2000 by Harcourt, Inc. All rights reserved. 128 Chapter 27 Solutions 2 wires → 27.58 R= = 100 m 0.108 Ω 300 m (100 m) = 0.0360 Ω (a) (∆V)home = (∆V)line – IR = 120 – (110)(0.0360) = 116 V (b) P = I (∆V) = (110 A)(116 V) = 12.8 kW (c) P wires = I 2R = (110 A)2(0.0360 Ω) = 436 W *27.59 (a) E=− dV (0 − 4.00) V = 8.00i V m i=− dx (0.500 − 0) m ( ) (b) R= −8 ρ l 4.00 × 10 Ω ⋅ m (0.500 m ) = = 0.637 Ω 2 A π 1.00 × 10 −4 m (c) I= ∆V 4.00 V = = 6.28 A R 0.637 Ω (d) J= I 6.28 A i= A π 1.00 × 10 −4 m (e) ρ J = 4.00 × 10 −8 Ω ⋅ m 2.00 × 108 i A m 2 = 8.00i V m = E *27.60 (a) ( ( ( E=– )( V dV(x) i= L i dx 4ρL ρl = πd2 A (b) R= (c) ∆V Vπ d 2 I= R = 4ρL (d) I V i J = A i= ρL (e) V ρJ = L i= E ) 2 ) = 2.00 × 108 i A m 2 = 200i MA m 2 ) Chapter 27 Solutions 129 R = R0 [1 + α (T − T0 )] 27.61 In this case, R= 27.62 I0 I = 10 , 27.63 (a) so 12.0 6.00 ∆V = = I (I – 3.00) I Therefore, R = T = T0 + so T = T0 + 1 9 (9) = 20° + = 2020 °C 0.00450/°C α thus (a) (b) 12.0I – 36.0 = 6.00I and I = 6.00 A 12.0 V 6.00 A = 2.00 Ω = I ( ∆V ) so (b) 27.64  1R 1 I  − 1 = T0 +  0 − 1  α  R0 αI   667 A ( ρ l ρ0 [1 + α (T − T0 )] l0 [1 + α ′ (T − T0 )] = , A0 [1 + 2α ′ (T − T0 )] A We begin with R= which reduces to R= For copper: ρ0 = 1.70 × 10 − 8 Ω ⋅ m , α = 3.90 × 10 R0 = R0 [1 + α (T − T0 )] [1 + α ′ (T − T0 )] [1 + 2α ′ (T − T0 )] −3 °C −1 , and α ′ = 17.0 × 10 ρ0l0 (1.70 × 10 − 8 )(2.00) = = 1.08 Ω A0 π (0.100 × 10 − 3 )2 The simple formula for R gives: [ ] R = (1.08 Ω) 1 + (3.90 × 10 −3 °C −1 )(100°C − 20.0°C) = 1.420 Ω while the more complicated formula gives: R= ) d = vt = ( 20.0 m / s) 2.50 × 10 3 s = 50.0 km and (1.08 Ω)[1+ (3.90 × 10 −3 °C −1 )(80.0°C)][1+ (17.0 × 10 −6 °C −1 )(80.0°C)] [1 + 2(17.0 × 10 −6 ] °C −1 )(80.0°C) © 2000 by Harcourt, Inc. All rights reserved. = 1.418 Ω −6 °C −1 130 Chapter 27 Solutions Let α be the temperature coefficient at 20.0°C, and α′ be the temperature coefficient at 0 °C. Then ρ = ρ0 [1 + α (T − 20.0°C)], and ρ = ρ ′ [1 + α ′(T − 0°C)] must both give the correct resistivity at any temperature T. That is, we must have: ρ0 [1 + α (T − 20.0°C)] = ρ ′ [1 + α ′ (T − 0°C)] (1) 27.65 Setting T = 0 in equation (1) yields: ρ ′ = ρ0 [1 − α (20.0°C)] , and setting T = 20.0°C in equation (1) gives: ρ0 = ρ ′ [1 + α ′ (20.0°C)] Put ρ ′ from the first of these results into the second to obtain: ρ0 = ρ0 [1 − α (20.0°C)][1 + α ′(20.0°C)] Therefore 1 + α ′ ( 20.0°C) = which simplifies to α′= 1 1 − α ( 20.0°C) α [1 − α (20.0 °C)] From this, the temperature coefficient, based on a reference temperature of 0°C, may be computed for any material. For example, using this, Table 27.1 becomes at 0°C : Material Silver Copper Gold Aluminum Tungsten Iron Platinum Lead Nichrome Carbon Germanium Silicon 27.66 (a) R= (b) R= Temp Coefficients at 0°C 4.1 × 10− 3/˚C 4.2 × 10− 3/˚C 3.6 × 10− 3/˚C 4.2 × 10− 3/˚C 4.9 × 10− 3/˚C 5.6 × 10− 3/˚C 4.25 × 10− 3/˚C 4.2 × 10− 3/˚C 0.4 × 10− 3/˚C −0.5 × 10− 3/˚C −24 × 10− 3/˚C −30 × 10− 3/˚C ρl ρL = 2 A π rb − r a2 ( (3.50 × 10 [ 5 ) ) Ω ⋅ m (0.0400 m ) π (0.0120 m ) − (0.00500 m ) 2 ] = 3.74 × 107 Ω = 37.4 MΩ , so R = (c) (d) 2 R= (3.50 × 10 5 Ω⋅m 2 π (0.0400 m ) ρ rb dr = 2π L ∫ ra r ) ln 1.20  = 1.22 × 10  0.500  6 r  ρ ln b  2π L  ra  Ω = 1.22 MΩ Chapter 27 Solutions 131 27.67 Each speaker receives 60.0 W of power. Using I= = 60.0 W 4.00 Ω = I 2 R, we then have = 3.87 A The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less . 27.68 ∆V = –E · l or dV = –E · dx ∆V = –IR = –E · l I= dq E ⋅l A A dV = = E ⋅l = E = − σA = σA R dx dt ρl ρ dV dx Current flows in the direction of decreasing voltage. Energy flows as heat in the direction of decreasing temperature. 27.69 R=∫ R= R = 27.70 ρ dx ρ dx =∫ A wy where y = y1 + y2 − y1 x L L y − y1  ρ dx ρL  = ln  y1 + 2 x ∫ − y y w0y + 2 L 1  x w(y2 − y1 )  1 L L 0 y  ρL ln 2  w(y2 − y1 )  y1  From the geometry of the longitudinal section of the resistor shown in the figure, we see that (b − r) (b − a) = h y From this, the radius at a distance y from the base is r = (a − b) For a disk-shaped element of volume dR = Using the integral formula du ρ dy : πr 2 1 ∫ (au + b)2 = − a(au + b) , y +b h R= R= © 2000 by Harcourt, Inc. All rights reserved. h dy ρ . ∫ π 0 [(a − b)(y / h) + b]2 ρ h π ab 132 Chapter 27 Solutions [ ] I = I 0 exp( e ∆V / kBT ) − 1 27.71 with and R= ∆V I I 0 = 1.00 × 10 −9 A, e = 1.60 × 10 −19 C, and kB = 1.38 × 10 −23 J K . The following includes a partial table of calculated values and a graph for each of the specified temperatures. (i) For T = 280 K : ∆V ( V ) I (A) R (Ω ) 0.400 0.440 0.480 0.520 0.560 0.600 0.0156 0.0818 0.429 2.25 11.8 61.6 25.6 5.38 1.12 0.232 0.0476 0.0097 (ii) For T = 300 K : ∆V ( V ) I (A) R (Ω ) 0.400 0.440 0.480 0.520 0.560 0.600 0.005 0.024 0.114 0.534 2.51 11.8 77.3 18.1 4.22 0.973 0.223 0.051 (iii) For T = 320 K : ∆V ( V ) I (A) R (Ω ) 0.400 0.440 0.480 0.520 0.560 0.600 0.0020 0.0084 0.0357 0.152 0.648 2.76 203 52.5 13.4 3.42 0.864 0.217 Chapter 28 Solutions (a) P = 28.1 ( ∆V )2 becomes R (b) ∆V = IR ε = IR + Ir 20.0 W = (11.6 V)2 R so 11.6 V = I (6.73 Ω) so 15.0 V = 11.6 V + (1.72 A)r and so R = 6.73 Ω I = 1.72 A Figure for Goal Solution r = 1.97 Ω Goal Solution A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.6 V when it is delivering 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance of the battery? G: The internal resistance of a battery usually is less than 1 Ω, with physically larger batteries having less resistance due to the larger anode and cathode areas. The voltage of this battery drops significantly (23%), when the load resistance is added, so a sizable amount of current must be drawn from the battery. If we assume that the internal resistance is about 1 Ω, then the current must be about 3 A to give the 3.4 V drop across the battery’s internal resistance. If this is true, then the load resistance must be about R ≈ 12 V / 3 A = 4 Ω. O: We can find R exactly by using Joule’s law for the power delivered to the load resistor when the voltage is 11.6 V. Then we can find the internal resistance of the battery by summing the electric potential differences around the circuit. A : (a) R= Combining Joule's law, P = ∆VI , and the definition of resistance, ∆V = IR , gives ∆V 2 (11.6 V ) = = 6.73 Ω P 20.0 W 2 (b) The electromotive force of the battery must equal the voltage drops across the resistances: ε = IR + Ir , where I = ∆V R. r= ε − IR = (ε − ∆V )R = (15.0 V − 11.6 V)(6.73 Ω) = 1.97 Ω I ∆V 11.6 V L : The resistance of the battery is larger than 1 Ω, but it is reasonable for an old battery or for a battery consisting of several small electric cells in series. The load resistance agrees reasonably well with our prediction, despite the fact that the battery’s internal resistance was about twice as large as we assumed. Note that in our initial guess we did not consider the power of the load resistance; however, there is not sufficient information to accurately solve this problem without this data. © 2000 by Harcourt, Inc. All rights reserved. Chapter 28 Solutions 135 28.2 (a) ∆Vterm = IR becomes 10.0 V = I (5.60 Ω) so I = 1.79 A (b) ∆Vterm = ε – Ir becomes so ε= 10.0 V = ε – (1.79 A)(0.200 Ω) 10.4 V The total resistance is R = 28.3 3.00 V 0.600 A = 5.00 Ω (a) Rlamp = R – rbatteries = 5.00 Ω – 0.408 Ω = 4.59 Ω (b) 28.4 2 P batteries (0.408 Ω)I = 2 = 0.0816 = 8.16% (5.00 Ω)I P total (a) Here Then, ε = I(R + r), so I = ε R + r = 12.6 V = 2.48 A (5.00 Ω + 0.0800 Ω) ∆V = IR = ( 2.48 A )( 5.00 Ω) = 12.4 V (b) Let I1 and I 2 be the currents flowing through the battery and the headlights, respectively. 28.5 ε − I1r − I 2 R = 0 Then, I1 = I 2 + 35.0 A , and so ε = (I 2 + 35.0 A)(0.0800 Ω) + I 2 (5.00 Ω) = 12.6 V giving I 2 = 1.93 A Thus, ∆V 2 = (1.93 A)(5.00 Ω) = 9.65 V ∆V = I 1R1 = (2.00 A)R 1 and ∆V = I 2(R 1 + R 2) = (1.60 A)(R 1 + 3.00 Ω) Therefore, (2.00 A)R 1 = (1.60 A)(R 1 + 3.00 Ω) or R 1 = 12.0 Ω © 2000 by Harcourt, Inc. All rights reserved. 136 Chapter 28 Solutions 28.6 (a) Rp = 1 (1 7.00 Ω) + (1 10.0 Ω) = 4.12 Ω Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω (b) ∆V = IR 34.0 V = I (17.1 Ω) I = 1.99 A for 4.00 Ω, 9.00 Ω resistors Applying ∆V = IR, *28.7 (1.99 A)(4.12 Ω) = 8.18 V 8.18 V = I (7.00 Ω) so I = 1.17 A for 7.00 Ω resistor 8.18 V = I (10.0 Ω) so I = 0.818 A for 10.0 Ω resistor If all 3 resistors are placed in parallel , 1 2 5 1 = + = 500 250 500 R *28.8 and R = 100 Ω For the bulb in use as intended, I= P 75.0 W = = 0.625 A ∆V 120 V and ∆V 120 V R= I = = 192 Ω 0.625 A Now, presuming the bulb resistance is unchanged, I= 120 V = 0.620 A 193.6 Ω Across the bulb is ∆V = IR = 192 Ω(0.620 A) = 119 V so its power is P = (∆V)I = 119 V(0.620 A) = 73.8 W Chapter 28 Solutions 137 28.9 If we turn the given diagram on its side, we find that it is the same as Figure (a). The 20.0-Ω and 5.00-Ω resistors are in series, so the first reduction is as shown in (b). In addition, since the 10.0-Ω, 5.00-Ω, and 25.0-Ω resistors are then in parallel, we can solve for their equivalent resistance as: Req = ( 1 1 10.0 Ω + 1 5.00 Ω + 1 25.0 Ω ) = 2.94 Ω This is shown in Figure (c), which in turn reduces to the circuit shown in (d). Next, we work backwards through the diagrams applying I = ∆V/R and ∆V = IR. The 12.94-Ω resistor is connected across 25.0-V, so the current through the battery in every diagram is I= 25.0 V ∆V = = 1.93 A R 12.94 Ω (a) In Figure (c), this 1.93 A goes through the 2.94-Ω equivalent resistor to give a potential difference of: ∆V = IR = (1.93 A)(2.94 Ω) = 5.68 V From Figure (b), we see that this potential difference is the same across Vab, the 10-Ω resistor, and the 5.00-Ω resistor. (b) (b) Therefore, V ab = 5.68 V (a) Since the current through the 20.0-Ω resistor is also the current through the 25.0-Ω line ab, 5.68 V Vab = = 0.227 A = 227 mA I =R ab 25.0 Ω 28.10  ρl ρl ρl ρl  (120 V ) 120 V = IReq = I  + + +  , or Iρ l =  1  A1 A2 A3 A4  1 1 1  + + +    A1 A2 A3 A4  ∆V 2 = Iρ l = A2 (120 V ) = 29.5 V  1 1 1 1  A2  + + +   A1 A2 A3 A4  © 2000 by Harcourt, Inc. All rights reserved. (c) (d) 138 Chapter 28 Solutions 28.11 (a) Since all the current flowing in the circuit must pass through the series 100-Ω resistor, P = RI 2 P max = RI max 2 so P = R I max = 1 1  R eq = 100 Ω +  + 100 100   –1 25.0 W = 0.500 A 100 Ω Ω = 150 Ω ∆Vmax = R eq I max = 75.0 V (b) P = (∆V)I = (75.0 V)(0.500 A) = 37.5 W P 2 = P 3 = RI 2 = (100 Ω)(0.250 A)2 = 6.25 W P 1 = 25.0 W 28.12 Using 2.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series 2.00 Ω 3.00 Ω 4.00 Ω 5.00 Ω 28.13 6.00 Ω 7.00 Ω 9.00 Ω Parallel Mixed 0.923 Ω1.56 Ω 1.20 Ω 2.00 Ω 1.33 Ω 2.22 Ω 1.71 Ω 3.71 Ω 4.33 Ω 5.20 Ω The resistors may be arranged in patterns: The potential difference is the same across either combination. ∆V = IR = 3I ( 1 R 1 1 + 500 ) R 1 + 500 = 3 28.14 total power 1  1 + =3  R 500  so R and R = 1000 Ω = 1.00 kΩ If the switch is open, I = ε / ( R′ + R) and P = ε 2 R′ / ( R′ + R)2 If the switch is closed, I = ε / (R + R′ / 2) and P = ε 2 ( R′ / 2) / (R + R′ / 2)2 Then, ε 2 R′ ( R′ + R) 2 = ε 2 R′ 2(R + R′ / 2)2 2R 2 + 2RR′ + R′ 2 / 2 = R′ 2 + 2RR′ + R 2 The condition becomes R 2 = R′ 2 / 2 so R′ = 2 R = 2 (1.00 Ω) = 1.41Ω Chapter 28 Solutions 139 28.15 Rp = 1   1 +  3.00 1.00  −1 = 0.750 Ω Rs = ( 2.00 + 0.750 + 4.00) Ω = 6.75 Ω I battery = ∆V 18.0 V = = 2.67 A Rs 6.75 Ω P = I 2 R: P 2 = ( 2.67 A ) ( 2.00 Ω) 2 P 2 = 14.2 W in 2.00 Ω P 4 = ( 2.67 A ) ( 4.00 Ω) = 28.4 W 2 ∆V 2 = ( 2.67 A )( 2.00 Ω) = 5.33 V, ∆V 4 = ( 2.67 A )( 4.00 Ω) = 10.67 V ∆V p = 18.0 V − ∆V 2 − ∆V 4 = 2.00 V 28.16 (= ∆V3 = ∆V1 ) P3 = ( ∆V3 )2 = (2.00 V)2 = 1.33 W in 3.00 Ω P1 = 2 ( ∆V1 )2 = (2.00 V) = 4.00 W in 1.00 Ω R3 R1 3.00 Ω 1.00 Ω in 4.00 Ω Denoting the two resistors as x and y, x + y = 690, and 1 1 1 150 = x + y 1 1 (690 – x) + x 1 150 = x + 690 – x = x(690 – x) x 2 – 690x + 103,500 = 0 x= 690 ± (690)2 – 414,000 2 x = 470 Ω y = 220 Ω © 2000 by Harcourt, Inc. All rights reserved. 140 Chapter 28 Solutions 28.17 (a) ∆V = IR : 33.0 V = I1 (11.0 Ω) 33.0 V = I 2 ( 22.0 Ω) I1 = 3.00 A P = I 2 R: I 2 = 1.50 A P 1 = ( 3.00 A ) (11.0 Ω) P 2 = (1.50 A ) ( 22.0 Ω) P 1 = 99.0 W P 2 = 49.5 W 2 2 The 11.0-Ω resistor uses more power. P = I ( ∆V ) = ( 4.50)( 33.0) = 148 W (b) P 1 + P 2 = 148 W (c) Rs = R1 + R2 = 11.0 Ω + 22.0 Ω = 33.0 Ω ∆V = IR : 33.0 V = I ( 33.0 Ω), so I = 1.00 A P = I 2 R: P 1 = (1.00 A ) (11.0 Ω) P 2 = (1.00 A ) ( 22.0 Ω) P 1 = 11.0 W P 2 = 22.0 W 2 2 The 22.0-Ω resistor uses more power. (d) P 1 + P 2 = I 2 ( R1 + R2 ) = (1.00 A ) ( 33.0 Ω) = 33.0 W 2 P = I ( ∆V ) = (1.00 A )( 33.0 V ) = 33.0 W (e) 28.18 The parallel configuration uses more power. +15.0 – (7.00)I1 – (2.00)(5.00) = 0 5.00 = 7.00I1 so I1 = 0.714 A so I2 = 1.29 A I3 = I1 + I2 = 2.00 A 0.714 + I2 = 2.00 +ε – 2.00(1.29) – (5.00)(2.00) = 0 ε = 12.6 V Chapter 28 Solutions 141 28.19 We name the currents I1 , I 2 , and I 3 as shown. From Kirchhoff's current rule, I3 = I1 + I2 Applying Kirchhoff's voltage rule to the loop containing I2 and I3 , 12.0 V – (4.00)I3 – (6.00)I2 – 4.00 V = 0 8.00 = (4.00)I3 + (6.00)I2 Applying Kirchhoff's voltage rule to the loop containing I1 and I2 , – (6.00)I2 – 4.00 V + (8.00)I1 = 0 (8.00)I1 = 4.00 + (6.00)I2 Solving the above linear systems, I1 = 846 mA, I2 = 462 mA, I3 = 1.31 A All currents flow in the directions indicated by the arrows in the circuit diagram. *28.20 The solution figure is shown to the right. *28.21 We use the results of Problem 19. (a) By the 4.00-V battery: By the 12.0-V battery: (b) By the 8.00 Ω resistor: ∆U = (∆V)It = 4.00 V(– 0.462 A)120 s = – 222 J 12.0 V (1.31 A) 120 s = 1.88 kJ I 2 Rt = (0.846 A)2(8.00 Ω ) 120 s = 687 J By the 5.00 Ω resistor: (0.462 A)2(5.00 Ω ) 120 s = 128 J By the 1.00 Ω resistor: (0.462 A)2(1.00 Ω ) 120 s = 25.6 J By the 3.00 Ω resistor: (1.31 A)2(3.00 Ω ) 120 s = 616 J By the 1.00 Ω r e s i s t o r : (1.31 A)2(1.00 Ω ) 120 s = 205 J (c) –222 J + 1.88 kJ = 1.66 kJ from chemical to electrical. 687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrical to heat. © 2000 by Harcourt, Inc. All rights reserved. 142 Chapter 28 Solutions 28.22 We name the currents I1 , I 2 , and I 3 as shown. [1] 70.0 – 60.0 – I2 (3.00 kΩ) – I1 (2.00 kΩ) = 0 [2] 80.0 – I3 (4.00 kΩ) – 60.0 – I2 (3.00 kΩ) = 0 [3] I2 = I1 + I3 (a) Substituting for I2 and solving the resulting simultaneous equations yields I1 = 0.385 mA (through R 1) I3 = 2.69 mA (through R 3) I2 = 3.08 mA (through R 2) (b) ∆V cf = – 60.0 V – (3.08 mA)(3.00 kΩ) = – 69.2 V Point c is at higher potential. 28.23 Label the currents in the branches as shown in the first figure. Reduce the circuit by combining the two parallel resistors as shown in the second figure. Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain: (2.71R)I1 + (1.71R)I 2 = 250 and (1.71R)I1 + (3.71R)I 2 = 500 (a) With R = 1000 Ω, simultaneous solution of these equations yields: I1 = 10.0 mA and I 2 = 130.0 mA From Figure (b), Vc − V a = ( I1 + I 2 )(1.71R) = 240 V Thus, from Figure (a), I 4 = Vc − V a 240 V = = 60.0 mA 4000 Ω 4R Finally, applying Kirchhoff’s point rule at point a in Figure (a) gives: I = I 4 − I1 = 60.0 mA − 10.0 mA = + 50.0 mA , or I = 50.0 mA flowing from point a to point e . (b) Chapter 28 Solutions 143 28.24 Name the currents as shown in the figure to the right. w + x + z = y. Loop equations are Then – 200w – 40.0 + 80.0x = 0 – 80.0x + 40.0 + 360 – 20.0y = 0 + 360 – 20.0y – 70.0z + 80.0 = 0 Eliminate y by substitution. x = 2.50 w + 0.500  400 − 100 x − 20.0 w − 20.0 z = 0 440 − 20.0 w − 20.0 x − 90.0 z = 0  Eliminate x : 350 − 270 w − 20.0 z = 0  430 − 70.0 w − 90.0 z = 0 Eliminate z = 17.5 – 13.5w to obtain 430 − 70.0 w − 1575 + 1215 w = 0 w = 70.0/70.0 = 1.00 A upward in 200 Ω z = 4.00 A upward in 70.0 Ω Now x = 3.00 A upward in 80.0 Ω y = 8.00 A downward in 20.0 Ω and for the 200 Ω, 28.25 ∆V = IR = (1.00 A)(200 Ω) = 200 V Using Kirchhoff’s rules, 12.0 − (0.0100) I1 − (0.0600) I 3 = 0 10.0 + (1.00) I 2 − (0.0600) I 3 = 0 and I1 = I 2 + I 3 12.0 − (0.0100) I 2 − (0.0700) I 3 = 0 10.0 + (1.00) I 2 − (0.0600) I 3 = 0 Solving simultaneously, dead battery, I 2 = 0.283 A downward in the and I 3 = 171 A downward in the starter. © 2000 by Harcourt, Inc. All rights reserved. 144 Chapter 28 Solutions 28.26 V ab = (1.00) I1 + (1.00)( I1 − I 2 ) V ab = (1.00) I1 + (1.00) I 2 + ( 5.00)( I − I1 + I 2 ) V ab = ( 3.00)( I − I1 ) + ( 5.00)( I − I1 + I 2 ) Let I = 1.00 A , I1 = x, and I 2 = y Then, the three equations become: V ab = 2.00 x − y , or y = 2.00 x − V ab V ab = − 4.00 x + 6.00 y + 5.00 and V ab = 8.00 − 8.00 x + 5.00 y Substituting the first into the last two gives: 7.00V ab = 8.00 x + 5.00 and 6.00V ab = 2.00 x + 8.00 Solving these simultaneously yields V ab = Then, Rab = 28.27 V ab 27 17 V = I 1.00 A 27 V 17 Rab = or We name the currents I1 , I 2 , and I 3 as shown. (a) I1 = I2 + I3 Counterclockwise around the top loop, 12.0 V – (2.00 Ω)I3 – (4.00 Ω)I1 = 0 Traversing the bottom loop, 8.00 V – (6.00 Ω)I2 + (2.00 Ω)I3 = 0 1 I1 = 3.00 – 2 I3 4 1 I2 = 3 + 3 I 3 (b) Va – (0.909 A)(2.00 Ω) = Vb V b – V a = –1.82 V and I3 = 909 mA 27 Ω 17 Chapter 28 Solutions 145 We apply Kirchhoff's rules to the second diagram. 28.28 50.0 – 2.00I1 – 2.00I2 = 0 (1) 20.0 – 2.00I3 + 2.00I2 = 0 (2) I1 = I2 + I3 (3) Substitute (3) into (1), and solve for I1, I2, and I3 I1 = 20.0 A; I2 = 5.00 A; I3 = 15.0 A Then apply P = I 2R to each resistor: (2.00 Ω)1 : (4.00 Ω) : (2.00 Ω)3 : 28.29 P = I12 (2.00 Ω) = (20.0 A)2 (2.00 Ω) = 800 W 5.00  2 P = A (4.00 Ω) = 25.0 W  2  (Half of I2 goes through each) P = I 3 2 (2.00 Ω) = (15.0 A)2(2.00 Ω) = 450 W (a) RC = (1.00 × 106 Ω)(5.00 × 10–6 F) = 5.00 s (b) Q = Cε = (5.00 × 10–6 C)(30.0 V) = 150 µC (c) 28.30 I(t) = ε e −t/RC = R 30.0 −10.0   exp  = 4.06 µA 6 6 −6  1.00 × 10  (1.00 × 10 )(5.00 × 10 )  (a) I(t) = –I0e–t/RC I0 = Q 5.10 × 10–6 C = = 1.96 A RC (1300 Ω)(2.00 × 10–9 F) –9.00 × 10–6 s   I(t) = – (1.96 A) exp   = – 61.6 mA –9 (1300 Ω)(2.00 × 10 F) – 8.00 × 10–6 s   (b) q(t) = Qe–t/RC = (5.10 µC) exp   = 0.235 µC –9 (1300 Ω)(2.00 × 10 F) (c) The magnitude of the current is I0 = 1.96 A © 2000 by Harcourt, Inc. All rights reserved. 146 Chapter 28 Solutions 28.31 2 U = 1 C ( ∆V ) 2 and ∆V = Q C Therefore, U = Q 2 2C and when the charge decreases to half its original value, the stored energy is one-quarter its original value: 28.32 U f = 1 U0 4 (a) τ = RC = (1.50 × 105 Ω)(10.0 × 10–6 F) = 1.50 s (b) τ = (1.00 × 105 Ω)(10.0 × 10–6 F) = 1.00 s 10.0 V = 200 µA 50.0 × 103 Ω (c) The battery carries current  10.0 V  –t/ 1.00 s I = I0e–t/RC =  3  e 100 × 10 Ω The 100 kΩ carries current of magnitude 200 µA + (100 µA)e–t / 1.00 s So the switch carries downward current 28.33 (a) Call the potential at the left junction V L and at the right V R . "long" time, the capacitor is fully charged. VL = 8.00 V because of voltage divider: IL = 10.0 V 5.00 Ω After a = 2.00 A VL = 10.0 V – (2.00 A)(1.00 Ω) = 8.00 V Likewise, 2.00 Ω   VR =   10.0 V = 2.00 V 2.00 Ω + 8.00 Ω or IR = 10.0 V = 1.00 A 10.0 Ω VR = (10.0 V) – (8.00 Ω)(1.00 A) = 2.00 V ∆V = VL – VR = 8.00 – 2.00 = 6.00 V Therefore, R= (b) Redraw the circuit 1 = 3.60 Ω (1/ 9.00 Ω) + (1/ 6.00 Ω) RC = 3.60 × 10–6 s and e–t/RC = 1 10 so t = RC ln 10 = 8.29 µs Chapter 28 Solutions 147 28.34 ( )( ) (a) τ = RC = 4.00 × 106 Ω 3.00 × 10 −6 F = 12.0 s (b) I = ε e − t/RC = R [ 12.0 e − t/12.0 s 6 4.00 × 10 ] [ q = C ε 1 − e − t/RC = 3.00 × 10 − 6 (12.0) 1 − e − t/12.0 [ q = 36.0 µ C 1 − e − t/12.0 28.35 ∆V0 = ] ) I = 3.00 µ Ae − t/12.0 Q C Then, if q(t) = Qe − t/RC ∆V(t) = ∆V0 e − t/RC ∆V (t ) = e − t/RC ∆V0  1 4.00 = exp  −  R 3.60 × 10 − 6 2  Therefore ln ( ) 4.00  1 =−  2 R 3.60 × 10 − 6 ) (     R = 1.60 MΩ 28.36 ∆V0 = Q C Then, if q(t) = Q e −t RC ∆V(t) = ( ∆V0 ) e −t RC and ( ∆V0 ) When ∆V(t) = ∆V(t) 1 ( ∆V0 ) , then 2 e −t RC = − Thus, = e −t RC 1 2 t  1 = − ln 2 = ln  2 RC R= t C(ln 2) © 2000 by Harcourt, Inc. All rights reserved. 148 Chapter 28 Solutions 28.37 28.38 [ ] so q(t) = 1 − e −t/RC Q 0.600 = 1 − e − 0.900/RC or e − 0.900/RC = 1 − 0.600 = 0.400 − 0.900 = ln(0.400) RC thus RC = Applying Kirchhoff’s loop rule, −I g (75.0 Ω) + I − I g Rp = 0 q(t) = Q 1 − e −t/RC ( − 0.900 = 0.982 s ln(0.400) ) Therefore, if I = 1.00 A when I g = 1.50 mA , Rp = 28.39 I g (75.0 Ω) (I − Ig ) = (1.50 × 10 −3 ) A (75.0 Ω) 1.00 A − 1.50 × 10 −3 A = 0.113 Ω Series Resistor → Voltmeter 25.0 = 1.50 × 10-3(Rs + 75.0) ∆V = IR: Rs = 16.6 kΩ Solving, Figure for Goal Solution Goal Solution The galvanometer described in the preceding problem can be used to measure voltages. In this case a large resistor is wired in series with the galvanometer in a way similar to that shown in Figure P28.24b This arrangement, in effect, limits the current that flows through the galvanometer when large voltages are applied. Most of the potential drop occurs across the resistor placed in series. Calculate the value of the resistor that enables the galvanometer to measure an applied voltage of 25.0 V at full-scale deflection. G: The problem states that the value of the resistor must be “large” in order to limit the current through the galvanometer, so we should expect a resistance of kΩ to MΩ. O: The unknown resistance can be found by applying the definition of resistance to the portion of the circuit shown in Figure 28.24b. Rg = 75.0 Ω . For the two resistors in series, A : ∆V ab = 25.0 V; From Problem 38, I = 1.50 mA and Req = Rs + Rg so the definition of resistance gives us: ∆V ab = I(Rs + Rg ) Therefore, Rs = ∆V ab 25.0 V − Rg = − 75.0 Ω = 16.6 kΩ I 1.50 × 10 −3 A L : The resistance is relatively large, as expected. It is important to note that some caution would be necessary if this arrangement were used to measure the voltage across a circuit with a comparable resistance. For example, if the circuit resistance was 17 kΩ, the voltmeter in this problem would cause a measurement inaccuracy of about 50%, because the meter would divert about half the current that normally would go through the resistor being measured. Problems 46 and 59 address a similar concern about measurement error when using electrical meters. Chapter 28 Solutions 149 28.40 We will use the values required for the 1.00-V voltmeter to obtain the internal resistance of the galvanometer. ∆V = Ig (R + rg) rg = Solve for rg : 1.00 V ∆V –R= – 900 Ω = 100 Ω Ig 1.00 × 10-3 A We then obtain the series resistance required for the 50.0-V voltmeter: R= 28.41 V 50.0 V Ig – rg = 1.00 × 10-3 A – 100 Ω = 49.9 kΩ ( ) ∆V = I g r g = I − I g Rp , or Rp = I grg (I − Ig ) = I g (60.0 Ω) (I − Ig ) Therefore, to have I = 0.100 A = 100 mA when I g = 0.500 mA : Rp = (0.500 mA)(60.0 Ω) = 99.5 mA 0.302 Ω Figure for Goal Solution Goal Solution Assume that a galvanometer has an internal resistance of 60.0 Ω and requires a current of 0.500 mA to produce full-scale deflection. What resistance must be connected in parallel with the galvanometer if the combination is to serve as an ammeter that has a full-scale deflection for a current of 0.100 A? G: An ammeter reads the flow of current in a portion of a circuit; therefore it must have a low resistance so that it does not significantly alter the current that would exist without the meter. Therefore, the resistance required is probably less than 1 Ω. O: From the values given for a full-scale reading, we can find the voltage across and the current through the shunt (parallel) resistor, and the resistance value can then be found from the definition of resistance. A : The voltage across the galvanometer must be the same as the voltage across the shunt resistor i n parallel, so when the ammeter reads full scale, ∆V = (0.500 mA )(60.0 Ω) = 30.0 mV Through the shunt resistor, I = 100 mA − 0.500 mA = 99.5 mA Therefore, R= ∆V 30.0 mV = = 0.302 Ω I 99.5 mA L : The shunt resistance is less than 1 Ω as expected. It is important to note that some caution would be necessary if this meter were used in a circuit that had a low resistance. For example, if the circuit resistance was 3 Ω , adding the ammeter to the circuit would reduce the current by about 10%, so the current displayed by the meter would be lower than without the meter. Problems 46 and 59 address a similar concern about measurement error when using electrical meters. © 2000 by Harcourt, Inc. All rights reserved. 150 Chapter 28 Solutions R 2R 3 R 2R 3 1000 Ω = 400 Ω 2.50 28.42 Rx = 28.43 Using Kirchhoff’s rules with Rg << 1, R1 = 2.50 R 2 = − ( 21.0 Ω) I1 + (14.0 Ω) I 2 = 0, so I1 = ( 21.0 Ω 2 I2 3 G ) I2 70.0 − 21.0I1 − 7.00 I1 + I g = 0 , and ( 3 ) + 70.0 V and 10.0 − 3.00I 2 = −I g Solving simultaneously yields: Ig = 0.588 A 28.44 R= ρL ρL and Ri = i A Ai But, V = AL = Ai Li , so R = ρ L2 V and Ri = [ ] 2 ρ Li 1 + ( ∆L Li ) ρ ( Li + ∆L) Therefore, R = = V V This may be written as: 28.45 ε x = ε s ; ε = ε s Rx x Rs Rs Rs 7.00 Ω I2 - Ig The last two equations simplify to ( ) Ig 14.0 Ω 70.0 − 14.0I 2 − 7.00 I 2 − I g = 0 10.0 − 4.00 2 I 2 = I g , 7.00 Ω I1 + Ig I1 ρ L2i V 2 = Ri [1 + α ] R = R i (1 + 2α + α 2)  48.0 Ω  =  (1.0186 V) = 1.36 V  36.0 Ω  2 where α ≡ ∆L L Chapter 28 Solutions 151 *28.46 (a) In Figure (a), the emf sees an equivalent resistance of 200.00 Ω. 6.0000 V A V 6.000 0 V I= = 0.030 000 A 200.00 Ω 20.000 Ω 20.000 Ω 20.000 Ω A V 180.00 Ω 180.00 Ω 180.00 Ω (a) (b) (c) ∆V = IR = (0.030 000 A )(180.00 Ω) = 5.400 0 V The terminal potential difference is   1 1 Req =  +   180.00 Ω 20 000 Ω  (b) In Figure (b), −1 = 178.39 Ω The equivalent resistance across the emf is 178.39 Ω + 0.500 00 Ω + 20.000 Ω = 198.89 Ω The ammeter reads I= and the voltmeter reads ∆V = IR = (0.030 167 A )(178.39 Ω) = 5.381 6 V ε R = 6.000 0 V = 0.030 167 A 198.89 Ω   1 1  180.50 Ω + 20 000 Ω    (c) In Figure (c), −1 = 178.89 Ω Therefore, the emf sends current through Rtot = 178.89 Ω + 20.000 Ω = 198.89 Ω The current through the battery is I= 6.000 0 V = 0.030 168 A 198.89 Ω but not all of this goes through the ammeter. The voltmeter reads ∆V = IR = (0.030 168 A )(178.89 Ω) = 5.396 6 V The ammeter measures current I= ∆V 5.396 6 V = = 0.029 898 A R 180.50 Ω The connection shown in Figure (c) is better than that shown in Figure (b) for accurate readings. 28.47 I= P 1500 W = = 12.5 A ∆V 120 V For the Toaster, I= 750 W 120 W = 6.25 A And for the Grill, I= 1000 W = 8.33 A (Grill) 120 V (a) P = I(∆V) So for the Heater, (b) 12.5 + 6.25 + 8.33 = 27.1 A sufficient. The current draw is greater than 25.0 amps, so this would not be © 2000 by Harcourt, Inc. All rights reserved. 152 Chapter 28 Solutions 28.48 (a) P = I 2 R = I 2 2 −8  ρ l (1.00 A) (1.70 × 10 Ω ⋅ m)(16.0 ft)(0.3048 m / ft) = = 0.101 W  A π (0.512 × 10 −3 m)2 (b) P = I 2 R = 100(0.101 Ω) = 10.1 W 28.49 2 2 I Al RAl = I Cu RCu I Al = so RCu ρCu I Cu = I Cu = RAl ρAl 1.70 (20.0) = 0.776(20.0) = 15.5 A 2.82 *28.50 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is a chunk of rubber with contact area 4 mm 2 and thickness 1 mm. Its resistance is R= ( )( ) −3 13 ρ l 10 Ω ⋅ m 10 m ≅ ≅ 2 × 1015 Ω A 4 × 10 −6 m 2 The current will be driven by 120 V through total resistance (series) 2 × 1015 Ω + 10 4 Ω + 2 × 1015 Ω ≅ 5 × 1015 Ω It is: I = ∆V 120 V ~ R 5 × 1015 Ω ~ 10 −14 A (b) The resistors form a voltage divider, with the center of your hand at potential V h 2 , where V h is the potential of the "hot" wire. The potential difference between your finger and thumb is ∆V = IR ~ 10 −14 A 10 4 Ω ~ 10 −10 V . So the points where the rubber meets your fingers are at ( )( ) potentials of ~ *28.51 Vh + 10 −10 V 2 and ~ Vh − 10 −10 V 2 The set of four batteries boosts the electric potential of each bit of charge that goes through them by 4 × 1.50 V = 6.00 V. The chemical energy they store is ∆U = q∆V = (240 C)(6.00 J/C) = 1440 J The radio draws current 6.00 V ∆V = 0.0300 A I= R = 200 Ω So, its power is P = (∆V)I = (6.00 V)(0.0300 A) = 0.180 W = 0.180 J/s Then for the time the energy lasts, we have P = E t: t= We could also compute this from I = Q/t: t= E 1440 J = = 8.00 × 10 3 s P 0.180 J / s 240 C Q = = 8.00 × 103 s = 2.22 h 0.0300 A I Chapter 28 Solutions 153 *28.52 I= ε R+r Let x ≡ ε 2R or ( R + r )2 =  ( R + r )2 = xR or R 2 + ( 2r − x )R − r 2 = 0 , so P = I 2 R = ε2 , P then ( R + r )2 ε2 R  P  With r = 1.20 Ω , this becomes R 2 + ( 2.40 − x )R − 1.44 = 0, which has solutions of R= (a) With ε = 9.20 V and P = 12.8 W , x = 6.61: R= −( 2.40 − x ) ± (2.40 − x)2 − 5.76 2 ( 4.21)2 − 5.76 + 4.21 ± 2 = 3.84 Ω = 1.59 ± −3.22 2 or 0.375 Ω (b) For ε = 9.20 V and P = 21.2 W, x ≡ ε 2 = 3.99 R= P (1.59)2 − 5.76 +1.59 ± 2 The equation for the load resistance yields a complex number, so there is no resistance that will extract 21.2 W from this battery. The maximum power output occurs when R = r = 1.20 Ω , and that maximum is: P max = ε 2 4r = 17.6 W 28.53 +12.0 − 2.00 I − 4.00 I = 0, Using Kirchhoff’s loop rule for the closed loop, so I = 2.00 A Vb − V a = + 4.00 V − ( 2.00 A )( 4.00 Ω) − (0)(10.0 Ω) = − 4.00 V Thus, 28.54 ∆V ab = 4.00 V and point a is at the higher potential . The potential difference across the capacitor ( ) − 3.00 × 10 5 Ω R e Taking the natural logarithm of both sides, − R=− ( ) − 3.00 × 10 5 Ω R or and ] − ( 3.00 s )  4.00 V = (10.0 V )1 − e  Using 1 Farad = 1 s Ω, Therefore, 0.400 = 1.00 − e [ ∆V (t ) = ∆Vmax 1 − e −t RC = 0.600 3.00 × 10 5 Ω = ln(0.600) R 3.00 × 10 5 Ω = + 5.87 × 10 5 Ω = 587 kΩ ln(0.600) © 2000 by Harcourt, Inc. All rights reserved. ( R 10.0 × 10 −6 s Ω )   154 Chapter 28 Solutions 28.55 Let the two resistances be x and y. y x Ps 225 W = = 9.00 Ω 2 I ( 5.00 A)2 Then, Rs = x + y = and Pp xy 50.0 W Rp = = 2 = = 2.00 Ω x+y I ( 5.00 A)2 so x(9.00 Ω − x ) = 2.00 Ω x + (9.00 Ω − x ) y = 9.00 Ω – x x y x 2 − 9.00 x + 18.0 = 0 Factoring the second equation, ( x − 6.00)( x − 3.00) = 0 so x = 6.00 Ω or x = 3.00 Ω Then, y = 9.00 Ω − x gives y = 3.00 Ω or The two resistances are found to be 6.00 Ω and 3.00 Ω . 28.56 Let the two resistances be x and y. Then, Rs = x + y = Pp xy Ps and Rp = = 2. 2 x+y I I From the first equation, y = becomes ( x Ps I 2 − x ( ) x + Ps I − x 2 ) = Pp I 2 Ps − x , and the second I2 Ps Pp P  or x 2 −  2s  x + 4 = 0. I  I Using the quadratic formula, x = Then, y = P s ± P s2 − 4P s P p 2I 2 . P s > P s2 − 4P s P p Ps − x gives y = . I2 2I 2 The two resistances are P s + P s2 − 4P s P p 2I 2 and P s − P s2 − 4P s P p 2I 2 y = 6.00 Ω Chapter 28 Solutions 155 28.57 The current in the simple loop circuit will be I = (a) ∆Vter = ε – Ir = (b) I= (c) P = I 2R = ε 2 εR R+r ε R+r ε R+r and ∆Vter → ε as R → ∞ and I→ ε r as R → 0 R (R + r)2 dP ε2 = 0 −2ε 2 R = ε 2 R(−2)(R + r)−3 + ε 2 (R + r)−2 = + dR (R + r)3 (R + r)2 Then 2R = R + r Figure for Goal Solution R=r and Goal Solution A battery has an emf ε and internal resistance r. A variable resistor R is connected across the terminals of the battery. Determine the value of R such that (a) the potential difference across the terminals is a maximum, (b) the current in the circuit is a maximum, (c) the power delivered to the resistor is a maximum. G: If we consider the limiting cases, we can imagine that the potential across the battery will be a maximum when R = ∞ (open circuit), the current will be a maximum when R = 0 (short circuit), and the power will be a maximum when R is somewhere between these two extremes, perhaps when R = r. O: We can use the definition of resistance to find the voltage and current as functions of R, and the power equation can be differentiated with respect to R. A : (a) The battery has a voltage ∆Vterminal = ε − Ir = ε (b) The circuit's current is I= (c) The power delivered is P = I 2R = εR R+r or as R → ∞ , ∆Vterminal → ε or as R → 0, R+r I→ ε r ε 2R (R + r)2 To maximize the power P as a function of R, we differentiate with respect to R, and require that dP /dR = 0 −2ε 2 R dP ε2 = 0 = ε 2 R(−2)(R + r)−3 + ε 2 (R + r)−2 = + 3 dR (R + r) (R + r)2 Then 2R = R + r and R=r L : The results agree with our predictions. Making load resistance equal to the source resistance to maximize power transfer is called impedance matching. © 2000 by Harcourt, Inc. All rights reserved. 156 Chapter 28 Solutions 28.58 (a) ε − I(ΣR) − (ε1 + ε 2 ) = 0 40.0 V − (4.00 A)[(2.00 + 0.300 + 0.300 + R)Ω] − (6.00 + 6.00) V = 0; (b) Inside the supply, P = I 2 R = ( 4.00 A ) ( 2.00 Ω) = 2 so R = 4.40 Ω 32.0 W Inside both batteries together, P = I 2 R = ( 4.00 A ) (0.600 Ω) = 9.60 W For the limiting resistor, P = ( 4.00 A ) ( 4.40 Ω) = 70.4 W 2 2 (c) P = I(ε 1 + ε 2 ) = (4.00 A)[(6.00 + 6.00)V ] = 48.0 W 28.59 Let R m = measured value, R = actual value, (a) IR = current through the resistor R I = current measured by the ammeter. (a) When using circuit (a), IRR = ∆V = 20 000(I – IR) or Figure for Goal solution ∆V ∆V and IR = R , we have But since I = Rm IR and R = 20 000 When R > R m , we require (R – R m ) ≤ 0.0500 R Therefore, R m ≥ R (1 – 0.0500) and from (1) we find (b) When using circuit (b), But since IR = ∆V , Rm (b)   I R = 20 000  − 1   IR I R = R m (R – R m) Rm (1) R ≤ 1050 Ω IRR = ∆V – IR (0.5 Ω). Rm = (0.500 + R) When R m > R, we require (R m – R) ≤ 0.0500 R From (2) we find R ≥ 10.0 Ω (2) Chapter 28 Solutions 157 Goal Solution The value of a resistor R is to be determined using the ammeter-voltmeter setup shown in Figure P28.59. The ammeter has a resistance of 0.500 Ω, and the voltmeter has a resistance of 20 000 Ω. Within what range of actual values of R will the measured values be correct to within 5.00% if the measurement is made using (a) the circuit shown in Figure P28.59a (b) the circuit shown in Figure P28.59b? G: An ideal ammeter has zero resistance, and an ideal voltmeter has infinite resistance, so that adding the meter does not alter the current or voltage of the existing circuit. For the non-ideal meters in this problem, a low values of R will give a large voltage measurement error in circuit (b), while a large value of R will give significant current measurement error in circuit (a). We could hope that these meters yield accurate measurements in either circuit for typical resistance values of 1 Ω to 1 MΩ . O: The definition of resistance can be applied to each circuit to find the minimum and maximum current and voltage allowed within the 5.00% tolerance range. A : (a) In Figure P28.59a, at least a little current goes through the voltmeter, so less current flows through the resistor than the ammeter reports, and the resistance computed by dividing the voltage by the inflated ammeter reading will be too small. Thus, we require that ∆V/ I = 0.950R where I is the current through the ammeter. Call I R the current through the resistor; then I − I R is the current i n the voltmeter. Since the resistor and the voltmeter are in parallel, the voltage across the meter equals the voltage across the resistor. Applying the definition of resistance: ∆V = I R R = ( I − I R )( 20 000 Ω) Our requirement is so I= I R (R + 20 000 Ω) 20 000 Ω IRR ≥ 0.95R  I R (R + 20 000 Ω)    20 000 Ω   Solving, 20 000 Ω ≥ 0.95(R + 20 000 Ω) = 0.95R + 19000 Ω and R≤ 1000 Ω 0.95 or R ≤ 1.05 kΩ (b) If R is too small, the resistance of an ammeter in series will significantly reduce the current that would otherwise flow through R. In Figure 28.59b, the voltmeter reading is I (0.500 Ω) + IR , at least a little larger than the voltage across the resistor. So the resistance computed by dividing the inflated voltmeter reading by the ammeter reading will be too large. We require V ≤ 1.05R I Thus, 0.500 Ω ≤ 0.0500R so that and I (0.500 Ω) + IR ≤ 1.05R I R ≥ 10.0 Ω L : The range of R values seems correct since the ammeter’s resistance should be less than 5% of the smallest R value ( 0.500 Ω ≤ 0.05R means that R should be greater than 10 Ω), and R should be less than 5% of the voltmeter’s internal resistance ( R ≤ 0.05 × 20 kΩ = 1 kΩ ). Only for the restricted range between 10 ohms and 1000 ohms can we indifferently use either of the connections (a) and (b) for a reasonably accurate resistance measurement. For low values of the resistance R, circuit (a) must be used. Only circuit (b) can accurately measure a large value of R. © 2000 by Harcourt, Inc. All rights reserved. 158 Chapter 28 Solutions 28.60 The battery supplies energy at a changing rate dE E  = P = E I = E e −1/RC R  dt Then the total energy put out by the battery is ∫ dE = ∫ ∫ dE = ε 2 (−RC) R t=0  t   dt   t  − = − ε 2C exp − exp −  RC   RC   RC  ∞ ∫0 ∞ ∞ ε 2 exp  − R t  dt  RC  = − ε 2C[0 − 1] = ε 2 C 0 The heating power of the resistor is dE ε 2  2t  = P = ∆V R I = I 2 R = R 2 exp −  RC  dt R So the total heat is ∫ dE = ∫0 ∫ ε 2  − RC  dE = R  2 ∞ ∞  ∫0 ε 2C exp − 2t   2t   2dt  exp − − =−  RC   RC   RC  2 ∞ =− ε 2 exp  − R 2t  dt  RC  ε 2C [0 − 1] = ε 2C 0 2 2 The energy finally stored in the capacitor is U = 2 C (∆V)2 = 2 C ε 2. Thus, energy is conserved: 1 ε 2C = 12 ε 2C + 12 ε 2C 28.61 [ (a) q = C( ∆V ) 1 − e −t RC ( and resistor and capacitor share equally in the energy from the battery. ] ) q = 1.00 × 10 −6 F (10.0 V ) (b) I = I= (c) 1 = 9.93 µC dq  ∆V  −t RC e = dt  R   10.0 V  −5.00 = 3.37 × 10 −8 A = 33.7 nA e  2.00 × 106 Ω  dU d  1 q 2  q dq  q  I = = = dt dt  2 C  C dt  C  dU  9.93 × 10 −6 C  = 3.37 × 10 −8 A = 3.34 × 10 −7 W = 334 nW dt  1.00 × 10 −6 C V  ( ( ) ) (d) P battery = IE = 3.37 × 10 −8 A (10.0 V ) = 3.37 × 10 −7 W = 337 nW Chapter 28 Solutions 159 28.62 Start at the point when the voltage has just reached 2 2 V and the switch has just closed. The voltage is V 3 3 and is decaying towards 0 V with a time constant RBC. 2  VC (t) =  V  e −t/RBC 3  We want to know when V C(t) will reach 1 2   1 V =  V  e −t/RBC or e −t/RBC =  3 2 3  Therefore, or 1 V. 3 t1 = R BC ln 2 After the switch opens, the voltage is 1 V, 3 increasing toward V with time constant ( RA + RB )C : 2  V C (t) = V –  V  e −t/(RA +RB )C 3  When V C (t) = so 28.63 2 V, 3 2 2 V = V − Ve −t/(RA +RB )C 3 3 or e −t/(RA +RB )C = t2 = (R A + R B)C ln 2 and T = t1 + t2 = (R A + 2R B)C ln 2 (a) First determine the resistance of each light bulb: R= 1 2 P = (∆V)2 R (∆V)2 (120 V)2 = = 240 Ω P 60.0 W We obtain the equivalent resistance Req of the network of light bulbs by applying Equations 28.6 and 28.7: Req = R1 + 1 = 240 Ω + 120 Ω = 360 Ω (1/ R2 ) + (1/ R3 ) The total power dissipated in the 360 Ω is (b) The current through the network is given by P = I 2 Req : The potential difference across R 1 is P= I= (∆V)2 (120 V)2 = = 40.0 W Req 360 Ω P = Req ∆V1 = IR1 = 40.0 W 1 = A 3 360 Ω 1  A (240 Ω) = 80.0 V 3  The potential difference ∆V 23 across the parallel combination of R 2 and R 3 is ∆V 23 = IR23 =  1  1  = 40.0 V A  3   (1/ 240 Ω) + (1/ 240 Ω)  © 2000 by Harcourt, Inc. All rights reserved. 160 Chapter 28 Solutions 28.64 ∆V = IR (a) 20.0 V = (1.00 × 10-3 A)(R1 + 60.0 Ω) R1 = 1.994 × 104 Ω = 19.94 kΩ 28.65 (b) 50.0 V = (1.00 × 10-3 A)(R2 + R1 + 60.0 Ω) R 2 = 30.0 kΩ (c) 100 V = (1.00 × 10-3 A)(R3 + R1 + 60.0 Ω) R 3 = 50.0 kΩ Consider the circuit diagram shown, realizing that I g = 1.00 mA. For the 25.0 mA scale: (24.0 mA)( R1 + R2 + R3 ) = (1.00 mA)(25.0 Ω)  25.0  Ω  24.0  or R1 + R2 + R3 = For the 50.0 mA scale: ( 49.0 mA)( R1 + R2 ) = (1.00 mA)(25.0 Ω + R3 ) or 49.0( R1 + R2 ) = 25.0 Ω + R3 For the 100 mA scale: (99.0 mA)R1 = (1.00 mA)(25.0 Ω + R2 + R3 ) or 99.0R1 = 25.0 Ω + R2 + R3 (1) (2) (3) Solving (1), (2), and (3) simultaneously yields R1 = 0.260 Ω, 28.66 ( R2 = 0.261 Ω, ) Ammeter: I g r = 0.500 A − I g (0.220 Ω) or I g (r + 0.220 Ω) = 0.110 V (1) Voltmeter: 2.00 V = I g (r + 2500 Ω) (2) Solve (1) and (2) simultaneously to find: Ig = 0.756 mA and r = 145 Ω R3 = 0.521 Ω Chapter 28 Solutions 161 28.67 (a) After steady-state conditions have been reached, there is no DC current through the capacitor. Thus, for R 3: I R3 = 0 (steady-state) For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the 12-k Ω and 15-k Ω resistors in series: For R 1 and R 2: I(R1 +R2 ) = ε R1 + R2 = 9.00 V = 333 µA (steady-state) (12.0 kΩ + 15.0 kΩ) (b) After the transient currents have ceased, the potential difference across C is the same as the potential difference across R 2(= IR 2) because there is no voltage drop across R 3 . Therefore, the charge Q on C is Q = C (∆V)R2 = C (IR2) = (10.0 µF)(333 µA)(15.0 k Ω) = 50.0 µC (c) When the switch is opened, the branch containing R 1 is no longer part of the circuit. The capacitor discharges through (R 2 + R 3) with a time constant of (R 2 + R 3)C = (15.0 k Ω + 3.00 k Ω)(10.0 µF) = 0.180 s. The initial current Ii in this discharge circuit is determined by the initial potential difference across the capacitor applied to (R2 + R3) in series: Ii = (333 µ A)(15.0 kΩ) (∆V)C IR2 = = = 278 µA (R2 + R3 ) (R2 + R3 ) (15.0 kΩ + 3.00 kΩ) Thus, when the switch is opened, the current through R 2 changes instantaneously from 333 µA (downward) to 278 µA (downward) as shown in the graph. Thereafter, it decays according to I R2 = Ii e −t/(R2 +R3 )C = (278 µ A)e −t/(0.180 s) (for t > 0) (d) The charge q on the capacitor decays from Q i to Q i/5 according to q = Qi e −t/(R2 +R3 )C Qi = Qi e(−t/0.180 s) 5 5 = e t/0.180 s ln 5 = t 180 ms t = (0.180 s)(ln 5) = 290 ms © 2000 by Harcourt, Inc. All rights reserved. (a) 162 Chapter 28 Solutions 28.68 ∆V = ε e −t RC so ln A plot of ln  ε   1  = t  ∆V   RC  1  ε  versus t should be a straight line with slope = .  ∆V  RC t (s) 0 4.87 11.1 19.4 30.8 46.6 67.3 102.2 Using the given data values: ln(ε ∆V ) 0 0.109 0.228 0.355 0.509 0.695 0.919 1.219 ∆V (V) 6.19 5.55 4.93 4.34 3.72 3.09 2.47 1.83 (a) A least-square fit to this data yields the graph to the right. Σxi = 282 , Slope = Σxi2 = 1.86 × 10 4 , N ( Σxi yi ) − ( Σxi )( Σyi ) N ( ) − (Σx ) Σxi2 2 Σxi yi = 244, 2 i and the capacitance is C = i i 2 i i (b) Thus, the time constant is τ = RC = N=8 (Σx )(Σy ) − (Σx )(Σx y ) = 0.0882 Intercept = N ( Σx ) − ( Σx ) = 0.0118 The equation of the best fit line is: Σyi = 4.03 , ln  ε  = (0.0118) t + 0.0882  ∆V  1 1 = = 84.7 s slope 0.0118 84.7 s τ = = 8.47 µF R 10.0 × 106 Ω i i 2 i Chapter 28 Solutions 163 r 28.69 r i/6 r r r r a r b r i/6 i/3 i/6 i/3 a r 3 junctions at the same potential 28.70 i/3 b i/6 r r i/3 i/6 i/3 r i/6 i/3 another set of 3 junctions at the same potential (a) For the first measurement, the equivalent circuit is as shown in Figure 1. Rab = R1 = Ry + Ry = 2Ry Ry = so 1 R1 2 R1 a Ry Rac = R2 = (1) 1 Ry + Rx 2 Ry Rx Figure 1 For the second measurement, the equivalent circuit is shown in Figure 2. Thus, b c R2 a Ry (2) Ry c Rx Figure 2 Substitute (1) into (2) to obtain: R2 = (b) If R1 = 13.0 Ω and R2 = 6.00 Ω , then 1 1  R1 + Rx , 22  or Rx = R2 − 1 R1 4 Rx = 2.75 Ω The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω. 28.71 Since the total current passes through R3 , that resistor will dissipate the most power. When that resistor is operating at its power limit of 32.0 W, the current through it is 2 I total = P 32.0 W = = 16.0 A 2 , or I total = 4.00 A R 2.00 Ω Half of this total current (2.00 A) flows through each of the other two resistors, so the power dissipated in each of them is: P= ( ) 2 1 R I 2 total = (2.00 A)2 (2.00 Ω) = 8.00 W Thus, the total power dissipated in the entire circuit is: © 2000 by Harcourt, Inc. All rights reserved. R1 R3 R2 R 1 = R2 = R3 = 2.00 Ω 164 Chapter 28 Solutions 28.72 P total = 32.0 W + 8.00 W + 8.00 W = 48.0 W The total resistance between points b and c is: R= .00 kΩ 1 = 2.00 µF (2 .00 kΩ)(3.00 kΩ) = 1.20 kΩ 2 .00 kΩ + 3.00 k Ω .00 kΩ = 3.00 µF 2 20 V The total capacitance between points d and e is: C = 2.00 µ F + 3.00 µ F = 5.00 µ F The potential difference between point d and e in this series RC circuit at any time is: [ ] [ ∆V = ε 1 − e −t RC = (120.0 V ) 1 − e −1000t 6 ] Therefore, the charge on each capacitor between points d and e is: [ q1 = C1 ( ∆V ) = ( 2.00 µ F )(120.0 V ) 1 − e −1000t [ 6 ]= and q2 = C2 ( ∆V ) = ( 3.00 µ F )(120.0 V ) 1 − e −1000t *28.73 (a) Req = 3R (b) Req = 1 R = (1/ R) + (1/ R) + (1/ R) 3 (240 µC)[1 − e −1000t 6 6 ] ]= (360 µC)[1 − e −1000t I= ε I= 3R 3ε R (c) Nine times more power is converted in the parallel connection. 6 ] P series = ε I = ε2 P parallel = ε I = 3ε 2 R 3R Chapter 29 Solutions 29.1 (a) up (b) out of the page, since the charge is negative. (c) no deflection (d) into the page At the equator, the Earth's magnetic field is horizontally north. Because an electron has negative charge, F = q v × B is opposite in direction to v × B. Figures are drawn looking down. 29.2 (a) Down × North = East, so the force is directed West (b) North × North = sin 0° = 0: Zero deflection (c) West × North = Down, so the force is directed Up (d) Southeast × North = Up, so the force is Down FB = q v × B; 29.3 (a) F B (–j) = –e v i × B Therefore, B = B (–k) which indicates the negative z direction *29.4 (a) FB = qvB sin θ = (1.60 × 10–19 C)(3.00 × 106 m/s)(3.00 × 10–1 T) sin 37.0° FB = 8.67 × 10–14 N (b) 29.5 F 8.67 × 10–14 N = 5.19 × 1013 m/s2 a= m = 1.67 × 10–27 kg F = ma = (1.67 × 10–27 kg)(2.00 × 1013 m/s2) = 3.34 × 10–14 N = qvB sin 90° B= F 3.34 × 10–14 N = = 2.09 × 10–2 T qv (1.60 × 10–19 C)(1.00 × 107 m/s) The right-hand rule shows that B must be in the –y direction to yield a force in the +x direction when v is in the z direction. © 2000 by Harcourt, Inc. All rights reserved. (c) (d) 2 Chapter 29 Solutions *29.6 ∆K = 2 mv 2 = e(∆V) = ∆U ( = 2.90 × 107 m / s v= 29.7 1 First find the speed of the electron: ) 2 1.60 × 10 −19 C ( 2400 J / C) 2e( ∆V ) = m ( 9.11 × 10 -31 kg ) (a) FB, max = qvB = (1.60 × 10–19 C)(2.90 × 107 m/s)(1.70 T) = 7.90 × 10–12 N (b) FB, min = 0 occurs when v is either parallel to or anti-parallel to B Gravitational force: Fg = mg = (9.11 × 10–31 kg)(9.80 m/s2) = 8.93 × 10–30 N down Electric force: Fe = qE = (–1.60 × 10–19 C)100 N/C down = 1.60 × 10–17 N up Magnetic force: N⋅s  m    F B = qv × B = −1.60 × 10 −19 C 6.00 × 106 E × 50.0 × 10 −6 N    C⋅m  s ( ) FB = – 4.80 × 10–17 N up = 4.80 × 10–17 N down 29.8 We suppose the magnetic force is small compared to gravity. Then its horizontal velocity component stays nearly constant. We call it v i. From vy 2 = vyi 2 + 2ay ( y − yi ) , the vertical component at impact is − 2gh j. Then, ( ) F B = qv × B = Q vi − 2gh j × Bk = QvB( −j) − Q 2gh Bi FB = QvB vertical + Q 2gh B horizontal FB = 5.00 × 10–6 C(20.0 m/s)(0.0100 T) j + 5.00 × 10–6 C 2(9.80 m/s2)(20.0 m) (0.0100 T) i FB = (1.00 × 10–6 N) vertical + (0.990 × 10–6 N) horizontal 29.9 F B = qvB sin θ so 8.20 × 10-13 N = (1.60 × 10-19 C)(4.00 × 106 m/s)(1.70 T) sin θ sin θ = 0.754 and θ = sin-1(0.754) = 48.9° or 131° Chapter 29 Solutions 29.10 ( ) ( ) qE = −1.60 × 10 −19 C ( 20.0 N / C)k = −3.20 × 10 −18 N k Σ F = qE + qv × B = ma (−3.20 × 10 ) −18 N k – 1.60 × 10-19 C (1.20 × 104 m/s i) × B = (9.11 × 10-31)(2.00 × 1012 m/s2)k – (3.20 × 10-18 N)k – (1.92 × 10-15 C · m/s)i × B = (1.82 × 10-18 N)k (1.92 × 10-15 C · m/s)i × B = – (5.02 × 10-18 N)k The magnetic field may have any x-component . Bz = 29.11 0 and By = –2.62 mT F B = qv × B i j k v × B = +2 −4 +1 = (12 − 2) i + (1 + 6) j + ( 4 + 4) k = 10i + 7 j + 8 k +1 +2 −3 v × B = 10 2 + 7 2 + 8 2 = 14.6 T ⋅ m / s ( ) F B = q v × B = 1.60 × 10 −19 C (14.6 T ⋅ m s) = 2.34 × 10 −18 N 29.12 ( F B = qv × B = −1.60 × 10 ( −19 ) i j k 0 3.70 × 10 1.40 2.10 )[ ( 5 0 0 ( )) ] F B = −1.60 × 10 −19 C (0 − 0) i + (0 − 0) j + 0 − (1.40 T ) 3.70 × 10 5 m s k = 29.13 (8.29 × 10 −14 ) k N FB = ILB sin θ with F B = F g = mg mg = ILB sin θ so m g = IB sin θ L I = 2.00 A and  100 cm / m  m −2 = (0.500 g / cm )  = 5.00 × 10 kg / m L  1000 g / kg  Thus (5.00 × 10–2)(9.80) = (2.00)B sin 90.0° B = 0.245 Tesla with the direction given by right-hand rule: eastward © 2000 by Harcourt, Inc. All rights reserved. 3 Chapter 29 Solutions 4 Goal Solution A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward? G: Since I = 2.00 A south, B must be to the right-hand rule for currents in a magnetic field. east to make F upward according to the The magnitude of B should be significantly greater than the earth’s magnetic field (~ 50 µ T), since we do not typically see wires levitating when current flows through them. O: The force on a current-carrying wire in a magnetic field is F B = I1 × B, from which we can find B. A : With I to the south and B to the east, the force on the wire is simply FB = I lBsin 90°, which must oppose the weight of the wire, mg. So, B= g   10 2 cm / m  FB mg g  m   9.80 m / s 2   0.500 = = = = 0.245 T  cm   10 3 g / kg  Il I  l   2.00 A   Il L : The required magnetic field is about 5000 times stronger than the earth’s magnetic field. Thus it was reasonable to ignore the earth’s magnetic field in this problem. In other situations the earth’s field can have a significant effect. FB = I L × B = (2.40 A)(0.750 m)i × (1.60 T)k = (–2.88 j) N 29.14 29.15 29.16 (a) FB = ILB sin θ = (5.00 A)(2.80 m)(0.390 T) sin 60.0° = 4.73 N (b) FB = (5.00 A)(2.80 m)(0.390 T) sin 90.0° = 5.46 N (c) FB = (5.00 A)(2.80 m)(0.390 T) sin 120° = 4.73 N F B mg I L × B = = L L L I= mg (0.0400 kg/m)(9.80 m/s2) = = 0.109 A BL 3.60 T The direction of I in the bar is to the right . Chapter 29 Solutions 29.17 The magnetic and gravitational forces must balance. Therefore, it is necessary to have F B = BIL = mg, or I = (mg/BL) = (λ g/B) [λ is the mass per unit length of the wire]. Thus, I = (1.00 × 10-3 kg/m)(9.80 m/s2) = 196 A (5.00 × 10-5 T) (if B = 50.0 µT) The required direction of the current is eastward , since East × North = Up. 29.18 For each segment, I = 5.00 A and B = 0.0200 N / A ⋅ m j L F B = I (L × B) ab −0.400 m j 0 bc 0.400 m k (40.0 mN)(– i) cd – 0.400 m i + 0.400 m j (40.0 mN)(– k) da 0.400 m i – 0.400 m k (40.0 mN)(k + i) Segment 29.19 F B = I (d × B) = Id(k ) × B( −j) = IdB(i ) The rod feels force (Ktrans + Krot )i + ∆E = ( Ktrans + Krot ) f The work-energy theorem is 0 + 0 + Fscos θ = 21 mv 2 + 21 Iω 2 ( )  v IdBLcos 0˚ = 21 mv 2 + 21 21 mR 2  R v= 29.20 5 4IdBL = 3m 2 and IdBL = 43 mv 2 4(48.0 A)(0.120 m)(0.240 T)(0.450 m) = 1.07 m/s 3(0.720 kg) The rod feels force F B = I (d × B) = Id(k ) × B( −j) = IdB(i ) The work-energy theorem is (Ktrans + Krot )i + ∆E = (Ktrans + Krot ) f 0 + 0 + Fscos θ = 21 mv 2 + 21 Iω 2 ( )  v IdBLcos 0˚ = 21 mv 2 + 21 21 mR 2  R 2 and v= 4IdBL 3m © 2000 by Harcourt, Inc. All rights reserved. 6 Chapter 29 Solutions The magnetic force on each bit of ring is I ds × B = I ds B radially inward and upward, at angle θ above the radial line. The radially inward components tend to squeeze the ring but all cancel out as forces. The upward components I ds B sin θ all add to 29.21 I 2π rB sin θ up . Take the x-axis east, the y-axis up, and the z-axis south. The field is *29.22 B = ( 52.0 µ T ) cos 60.0˚ ( −k ) + ( 52.0 µ T ) sin 60.0˚ ( −j) The current then has equivalent length: L′ = 1.40 m ( −k ) + 0.850 m ( j) F B = I L′ × B = (0.0350 A )(0.850 j − 1.40 k ) m × ( − 45.0 j − 26.0 k ) 10 −6 T F B = 3.50 × 10 −8 N ( −22.1i − 63.0i ) = 2.98 × 10 −6 N ( −i ) = 2.98 µ N west 29.23 (a) 2π r = 2.00 m so r = 0.318 m [ ] µ = IA = (17.0 × 10-3 A) π (0.318)2 m 2 = 5.41 mA · m2 (b) *29.24 τ=µ×B so τ = (5.41 × 10-3 A · m2)(0.800 T) = 4.33 mN · m τ = µB sin θ so 4.60 × 10-3 N · m = µ(0.250) sin 90.0° µ = 1.84 × 10–2 A · m2 = 18.4 mA · m2 29.25 τ = NBAI sin θ ( ) τ = 100(0.800 T ) 0.400 × 0.300 m 2 (1.20 A ) sin 60° τ = 9.98 N · m Note that θ is the angle between the magnetic moment and the B field. The loop will rotate so as to align the magnetic moment with the B field. Looking down along the y-axis, the loop will rotate in a clockwise direction. (a) (b) Chapter 29 Solutions 29.26 (a) 7 Let θ represent the unknown angle; L , the total length of the wire; and d, the length of one side of the square coil. Then, use the right-hand rule to find L µ = NAI =   d 2 I  4d  At equilibrium, at angle θ with the horizontal. Στ = (µ × B) – (r × mg) = 0  mgd   ILBd  sin(90.0° − θ ) − sin θ = 0  4   2  and  mgd   ILBd  cos θ sin θ =  4   2   ILB  −1  (3.40 A)(4.00 m)(0.0100 T)  θ = tan −1   = tan   = 3.97° 2  2mg   2(0.100 kg)(9.80 m / s )  (b) τm = 1  ILBd  cos θ = (3.40 A)(4.00 m)(0.0100 T)(0.100 m) cos 3.97° = 3.39 mN · m  4  4 From τ = µ × B = IA × B, the magnitude of the torque is IAB sin 90.0° 29.27 (a) Each side of the triangle is 40.0 cm/3. Its altitude is 13.3 2 – 6.67 2 cm = 11.5 cm and its area is 1 A = 2 (11.5 cm)(13.3 cm) = 7.70 × 10-3 m2 Then τ = (20.0 A)(7.70 × 10-3 m2)(0.520 N · s/C · m) = 80.1 mN · m (b) Each side of the square is 10.0 cm and its area is 100 cm2 = 10-2 m2. τ = (20.0 A)(10-2 m2)(0.520 T) = 0.104 N· m (c) r = 0.400 m/2π = 0.0637 m A = π r 2 = 1.27 × 10-2 m2 τ = (20.0 A)(1.27 × 10-2 m2)(0.520) = 0.132 N · m (d) The circular loop experiences the largest torque. *29.28 Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of magnitude µ Bsin θ on the dipole, tending to turn the dipole moment in the direction of decreasing θ . Its energy is given by U −0= ∫ θ 90.0° θ µ Bsin θ dθ = µ B( − cos θ ) 90.0° = − µ Bcos θ + 0 © 2000 by Harcourt, Inc. All rights reserved. or U=–µ·B 8 Chapter 29 Solutions *29.29 (a) The field exerts torque on the needle tending to align it with the field, so the minimum energy orientation of the needle is: pointing north at 48.0° below the horizontal where its energy is ( )( ) U min = − µ Bcos 0˚ = − 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = − 5.34 × 10 −7 J It has maximum energy when pointing in the opposite direction, south at 48.0° above the horizontal 29.30 ( )( ) where its energy is U max = − µ B cos 180˚ = + 9.70 × 10 −3 A ⋅ m 2 55.0 × 10 −6 T = + 5.34 × 10 −7 J (b) U min + W = U max : W = U max − U min = + 5.34 × 10 −7 J − −5.34 × 10 −7 J = 1.07 µJ (a) τ = µ × B, so ( ) τ = µ × B= µB sin θ = NIAB sin θ [ ]( ) τ max = NIABsin 90.0˚ = 1( 5.00 A ) π (0.0500 m )2 3.00 × 10 −3 T = 118 µN · m (b) U = − µ ⋅ B, so − µ B ≤ U ≤ + µ B [ Since µ B = ( NIA)B = 1( 5.00 A ) π (0.0500 m ) the range of the potential energy is: 29.31 (a) B = 50.0 × 10-6 T; 2 ](3.00 × 10 T) = 118 µJ, −3 −118 µ J ≤ U ≤ +118 µ J v = 6.20 × 106 m/s Direction is given by the right-hand-rule: southward F B = qvB sin θ FB = (1.60 × 10-19 C)(6.20 × 106 m/s)(50.0 × 10-6 T) sin 90.0° = 4.96 × 10-17 N 29.32 (b) F= (a) 1 2 mv 2 r so m v 2 = q(∆V) r= mv 2 (1.67 × 10-27 kg)(6.20 × 106 m/s)2 = = 1.29 km F 4.96 × 10-17 N 1 2 (3.20 × 10-26 kg) v2 = (1.60 × 10-19 C)(833 V) The magnetic force provides the centripetal force: qvB sin θ = mv 2 r v = 91.3 km/s Chapter 29 Solutions r= (3.20 × 10-26 kg)(9.13 × 104 m/s) mv = = 1.98 cm qB sin 90.0° (1.60 × 10-19 C)(0.920 N · s/C · m) © 2000 by Harcourt, Inc. All rights reserved. 9 10 29.33 Chapter 29 Solutions For each electron, mv 2 r q vB sin 90.0° = and v= eBr m The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic: 2 K = 21 mv1i2 + 0 = 21 mv1f + 21 mv22 f  e 2B2 R12  1  e 2B2 R22  e 2B2 2 2 K = 21 m  + m  = 2m R1 + R2  m2  2  m2  ( K= 29.34 ) e(1.60 × 10 −19 C)(0.0440 N ⋅ s / C ⋅ m)2 (0.0100 m)2 + (0.0240 m)2 = 115 keV −31 2(9.11 × 10 kg) [ We begin with qvB = ] mv 2 qRB R , so v = m T= The time to complete one revolution is 2π R 2π R 2π m = = v qB  qRB   m  2π m Solving for B, B = qT = 6.56 × 10-2 T 29.35 q( ∆V ) = 21 mv 2 Also, qvB = mv 2 r Therefore, or v= so r= 2q( ∆V ) m mv m 2q( ∆V ) = = qB qB m r p2 = rd2 = and The conclusion is: 2m( ∆V ) qB2 2mp ( ∆V ) eB2 ( )  2mp ( ∆V )  2md ( ∆V ) 2 2mp ( ∆V ) 2 = = 2  = 2 rp 2 2 2 qd B eB  eB  rα 2 = ( )  2mp ( ∆V )  2mα ( ∆V ) 2 4mp ( ∆V ) 2 = = 2  = 2 rp 2 2 2 qα B (2e )B  eB  rα = r d = 2 r p Chapter 29 Solutions 11 Goal Solution 29.35 A proton (charge +e, mass mp ), a deuteron (charge +e, mass 2mp ), and an alpha particle, (charge + 2e , mass 4mp ) are accelerated through a common potential difference ∆V . The particles enter a uniform magnetic field B with a velocity in a direction perpendicular to B. The proton moves in a circular path of radius r p . Determine the values of the radii of the circular orbits for the deuteron rd and the alpha particle rα in terms of r p . G: In general, particles with greater speed, more mass, and less charge will have larger radii as they move in a circular path due to a constant magnetic force. Since the effects of mass and charge have opposite influences on the path radius, it is somewhat difficult to predict which particle will have the larger radius. However, since the mass and charge ratios of the three particles are all similar i n magnitude within a factor of four, we should expect that the radii also fall within a similar range. O: The radius of each particle’s path can be found by applying Newton’s second law, where the force causing the centripetal acceleration is the magnetic force: F = qv × B. The speed of the particles can be found from the kinetic energy resulting from the change in electric potential given. A : An electric field changes the speed of each particle according to assuming that the particles start from rest, we can write q∆V = The magnetic field changes their direction as described by ΣF = ma: thus L: ( K + U )i = ( K + U ) f . Therefore, 1 mv 2 . 2 mv 2 r mv m 2q∆V 1 2m∆V r= = = m q qB qB B qvBsin 90° = For the protons, rp = 1 2mp ∆V B e For the deuterons, rd = 1 2(2mp )∆V = 2r p B e For the alpha particles, rα = 1 2(4mp )∆V = 2r p B 2e Somewhat surprisingly, the radii of the deuterons and alpha particles are the same and are only 41% greater than for the protons. 29.36 (a) We begin with qvB = L = qB Therefore, R = (b) Thus, v = mv 2 , or qRB = mv . But, L = mvR = qR 2B. R 4.00 × 10 −25 J ⋅ s (1.60 × 10 −19 )( C 1.00 × 10 −3 T ) = 0.0500 m = 5.00 cm L 4.00 × 10 −25 J ⋅ s = = 8.78 × 106 m s −31 mR kg (0.0500 m ) 9.11 × 10 ( ) © 2000 by Harcourt, Inc. All rights reserved. 12 Chapter 29 Solutions 29.37 ω= 29.38 1 2 mv 2 = q(∆V) r= 29.39 29.40 qB (1.60 × 10 −19 C)(5.20 T) = = 4.98 × 108 rad/s m 1.67 × 10 −27 kg mv qB so v= so r= 2q( ∆V ) m m 2q(∆V) / m qB r2 = m 2(∆V) ⋅ q B2 and ( r ′ )2 = m= qB2 r 2 2(∆V) and ( m′ ) = E = 21 mv 2 = e(∆V) m′ 2(∆V) ⋅ q′ B2 (q′)B2 (r ′)2 2(∆V) m′ q ′ ( r ′ )  2e   2R  = ⋅ 2 = = 8  e  R  q r m 2 so evBsin 90° = mv 2 R and B= mv m 2e(∆V) 1 2m(∆V) = = eR eR R m e B = 2(1.67 × 10 −27 kg)(10.0 × 106 V) 1 = 7.88 × 10-12 T 5.80 × 1010 m 1.60 × 10 −19 C mv r= qB so 2 ( )( ) 7.94 × 10 −3 m 1.60 × 10 −19 C (1.80 T ) rqB m= = v 4.60 × 10 5 m s   1u m = 4.97 × 10 −27 kg = 2.99 u −27 kg   1.66 × 10 The particle is singly ionized: either a tritium ion, 29.41 FB = Fe E = vB = so qvB = qE where v = 2K / m . (  2(750) 1.60 × 10 −19 2K B=  m 9.11 × 10 −31  )    3 + 1H , or a helium ion, . K is kinetic energy of the electrons. 1/2 (0.0150) = 3 + 2 He 244 kV/m Chapter 29 Solutions 1 K = 2 mv 2 = q(∆V) 29.42 F B = qv × B = 2q( ∆V ) m mv 2 mv m 2q(∆V) / m 1 2m(∆V) r= = = qB q B r q B 2(238 × 1.66 × 10 −27 )2000  1  = 8.28 × 10 −2 m = 8.28 cm  1.20  1.60 × 10 −19 (a) r238 = (b) r235 = 8.23 cm r238 = r235 v= so m238 = m235 238.05 = 1.0064 235.04 The ratios of the orbit radius for different ions are independent of ∆V and B. 29.43 v= E 2500 V m = = 7.14 × 10 4 m s 0.0350 T B In the deflection chamber: r= 2.18 × 10 −26 kg 7.14 × 10 4 m s mv = = 0.278 m qB 1.60 × 10 −19 C (0.0350 T ) (a) ( )( ( ) ) K = 2 m v 2: (34.0 × 10 v = 8.07 × 107 m/s (1.67 × 10-27 kg)(8.07 × 107 m/s) mv = 0.162 m r = qB = (1.60 × 10-19 C)(5.20 T) 1 29.44 29.45 In the velocity selector: F B = qvB = 6 )( ) (1.67 × 10 eV 1.60 × 10 −19 J / eV = mv 2 R v qBR qB (1.60 × 10-19 C)(0.450 T) ω = R = mR = m = = 4.31 × 107 rad/s 1.67 × 10-27 kg (b) 29.46 (1.60 × 10-19 C)(0.450 T)(1.20 m) qBR = 5.17 × 107 m/s v= m = 1.67 × 10-27 kg F B = qvB = B= mv 2 r 4.80 × 10–16 kg · m/s mv = = 3.00 T qr (1.60 × 10–19 C)(1000 m) © 2000 by Harcourt, Inc. All rights reserved. 1 2 −27 ) kg v 2 13 14 Chapter 29 Solutions 25.0 θ = tan-1 10.0 = 68.2° 29.47 1.00 cm R = sin 68.2° = 1.08 cm and Ignoring relativistic correction, the kinetic energy of the electrons is 1 2q( ∆V ) 2 so v= = 1.33 × 108 m / s 2 mv = q(∆V) m From the centripetal force mv2 R = qvB, we find the magnetic field B= 29.48 mv (9.11 × 10 −31 kg)(1.33 × 108 m / s) = = 70.1 mT qR (1.60 × 10 −19 C)(1.08 × 10 −2 m) 1 nq (a) RH ≡ (b) ∆VH = so n= 1 1 = = 7.44 × 10 28 m −3 −19 qRH C 0.840 × 10 −10 m 3 C 1.60 × 10 ( )( ) IB nqt ( )( )( )( ) 7.44 × 10 28 m −3 1.60 × 10 −19 C 0.200 × 10 −3 m 15.0 × 10 −6 V nqt( ∆VH ) B= = = 1.79 T I 20.0 A 29.49 1 t( ∆VH ) (35.0 × 10 −6 V)(0.400 × 10 -2 m) = = = 3.70 × 10-9 m3/C IB nq (21.0 A)(1.80 T) 29.50 IB , and given that I = 50.0 A, B = 1.30 T, and t = 0.330 mm, the number of nqt charge carriers per unit volume is Since ∆V H = n= IB = 1.28 × 1029 m-3 e(∆V H )t The number density of atoms we compute from the density: n0 = 8.92 g  1 mole   6.02 × 10 23 atoms   106 cm 3  = 8.46 × 10 28 atom / m 3    3  mole 1 m cm 3  63.5 g     So the number of conduction electrons per atom is n 1.28 × 10 29 = = 1.52 n0 8.46 × 10 28 Chapter 29 Solutions ( )( )( )( 8.48 × 10 28 m −3 1.60 × 10 −19 C 5.00 × 10 −3 m 5.10 × 10 −12 V nqt( ∆VH ) B= = I 8.00 A 29.51 15 ) B = 4.32 × 10 − 5 T = 43.2 µT Goal Solution In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.500 cm thick is positioned along an east-west direction. If a current of 8.00 A in the conductor results in a Hall voltage of 5.10 pV, what is the magnitude of the Earth's magnetic field? (Assume that 28 3 n = 8.48 × 10 electrons/m and that the plane of the bar is rotated to be perpendicular to the direction of B.) G: The Earth’s magnetic field is about 50 µ T (see Table 29.1), so we should expect a result of that order of magnitude. O: The magnetic field can be found from the Hall effect voltage: ∆VH = IB nqt B= or nqt∆VH I A : From the Hall voltage, (8.48 × 10 B= 28 )( ) ( e - m 3 1.60 × 10 −19 C e - (0.00500 m ) 5.10 × 10 -12 V 8.00 A ) = 4.32 × 10 −5 T = 43.2 µ T L : The calculated magnetic field is slightly less than we expected but is reasonable considering that the Earth’s local magnetic field varies in both magnitude and direction. 29.52 (a) ∆VH = IB nqt nqt B 0.0800 T T = = = 1.14 × 10 5 I ∆VH 0.700 × 10 −6 V V so B= Then, the unknown field is ( )(  nqt  ( ∆VH )  I  ) B = 1.14 × 10 5 T V 0.330 × 10 −6 V = 0.0377 T = 37.7 mT (b) nqt T = 1.14 × 10 5 I V T I  n = 1.14 × 10 5  V  qt so 0.120 A T  n = 1.14 × 10 5 = 4.29 × 10 25 m −3 −19   V 1.60 × 10 C 2.00 × 10 −3 m ( )( ) © 2000 by Harcourt, Inc. All rights reserved. 16 Chapter 29 Solutions q vB sin 90° = 29.53 (a) mv 2 r θ v eB = = t r m ∴ω= The time it takes the electron to complete π radians is t= θ θm (π rad)(9.11 × 10–31 kg) = = = 1.79 × 10–10 s eB (1.60 × 10–19 C)(0.100 N · s/C · m) ω q Br (b) 29.54 Since v = m Ke = 21 mv 2 = q 2B2 r 2 e(1.60 × 10 −19 C)(0.100 N ⋅ s / Cm)2 (2.00 × 10 −2 m)2 = = 351 keV 2m 2(9.11 × 10 −31 kg) ∑ Fy = 0: +n – mg = 0 ∑ Fx = 0: –µ kn + IBd sin 90.0° = 0 B= 29.55 (a) , µk m g 0.100(0.200 kg)(9.80 m/s2) = = 39.2 mT (10.0 A)(0.500 m) Id The electric current experiences a magnetic force . I(h × B) in the direction of L. (b) The sodium, consisting of ions and electrons, flows along the pipe transporting no net charge. But inside the section of length L, electrons drift upward to constitute downward electric current J × (area) = JLw. The current then feels a magnetic force I h × B = JLwhB sin 90° This force along the pipe axis will make the fluid move, exerting pressure JLwhB F = hw = JLB area 29.56 The magnetic force on each proton, F B = qv × B = qvB sin 90° downward perpendicular to velocity, supplies centripetal force, guiding it into a circular path of radius r, with qvB = mv 2 r r= and mv qB 1 We compute this radius by first finding the proton's speed: K = 2 m v 2 v= 2K = m ( )( 2 5.00 × 106 eV 1.60 × 10 −19 J / eV 1.67 × 10 −27 kg ) = 3.10 × 10 7 m/s Chapter 29 Solutions Now, r= mv (1.67 × 10–27 kg)(3.10 × 107 m/s)(C · m) = = 6.46 m qB (1.60 × 10–19 C)(0.0500 N · s) sin ∝ = 1m 1.00 m = 6.46 m r α = 8.90° (b) From the figure, observe that (a) The magnitude of the proton momentum stays constant, and its final y component is – (1.67 × 10–27 kg)(3.10 × 107 m/s) sin(8.90°) = – 8.00 × 10–21 kg · m/s *29.57 (a) ( ) F B = qv × B = e( vi i ) × Bx i + By j + Bz k = 0 + evi By k − evi Bz j If B = Bx i + By j + Bz k , Since the force actually experienced is F B = Fi j, observe that Bx could have any value , By = 0 , and Bz = − Fi evi F B = qv × B = e( − vi i ) × (Bx i + 0 j − Fi evi k ) = –F i j (b) If v = − vi i , then (c) If q = −e and v = vi i , then F B = qv × B = − e( vi i ) × (Bx i + 0 j − Fi evi k ) = –F i j Reversing either the velocity or the sign of the charge reverses the force. 29.58 A key to solving this problem is that reducing the normal force will reduce the friction force: FB = BIL or B = FB IL When the wire is just able to move, ΣFy = n + FB cos θ − mg = 0 so n = mg − FB cos θ and f = µ ( mg − FB cos θ ) Also, ΣFx = FB sin θ − f = 0 so FB sin θ = f : FB sin θ = µ ( mg − FB cos θ ) We minimize B by minimizing FB : dFB cos θ − µ sin θ = (µ mg ) =0 ⇒ dθ (sin θ + µ cos θ )2 and  1 Thus, θ = tan −1   = tan −1 ( 5.00) = 78.7° for the smallest field, and µ B= ( m L) FB  µ g  = IL  I  sin θ + µ cos θ ( )  (0.200) 9.80 m s 2  0.100 kg m  Bmin =  = 0.128 T 1.50 A   sin 78.7° + (0.200) cos 78.7°   Bmin = 0.128 T pointing north at an angle of 78.7˚ below the horizontal © 2000 by Harcourt, Inc. All rights reserved. FB = µ mg sin θ + µ cos θ µ sin θ = cos θ 17 18 29.59 Chapter 29 Solutions (a) F = qE + qv × B = q(E + v × B) The net force is the Lorentz force given by ( F = 3.20 × 10 −19 )[(4 i − 1j − 2 k) + (2i + 3 j − 1k) × (2i + 4 j + 1k)] N F= Carrying out the indicated operations, we find: (b) 29.60  F θ = cos −1  x  = cos −1   F   r= (3.52i − 1.60 j) × 10−18 N   = 24.4° (3.52)2 + (1.60)2  3.52 mv (1.67 × 10 −27 )(1.50 × 108 ) = m = 3.13 × 104 m = 31.3 km qB (1.60 × 10 −19 )(5.00 × 10 −5 ) No, the proton will not hit the Earth . 29.61 Let ∆x1 be the elongation due to the weight of the wire and let ∆x2 be the additional elongation of the springs when the magnetic field is turned on. Then Fmagnetic = 2k ∆x2 where k is the force constant of the spring and can be determined from k = mg/2∆x1. (The factor 2 is included in the two previous equations since there are 2 springs in parallel.) Combining these two equations, we find  mg  mg ∆x2 Fmagnetic = 2 ;  ∆x2 = ∆x1  2∆x1  I= Therefore, where *29.62 but 24.0 V = 2.00 A, 12.0 Ω F B = I L × B = ILB B= mg ∆x2 (0.0100)(9.80)(3.00 × 10 −3 ) = = 0.588 T IL ∆x1 (2.00)(0.0500)(5.00 × 10 −3 ) Suppose the input power is 120 W = (120 V ) I : I ~ 1 A = 100 A ω = 2000 Suppose rev  1 min   2 π rad  rad ~200 min  60 s   1 rev  s rad   and the output power is 20 W = τω = τ 200  s  τ ~10 −1 N ⋅ m Suppose the area is about ( 3 cm) × ( 4 cm), or A~10 −3 m 2 From Table 29.1, suppose that the field is B~10 −1 T Then, the number of turns in the coil may be found from τ ≅ NIAB: ( )  C  −3 2  −1 N ⋅ s  10 m 10 giving 0.1 N ⋅ m ~ N 1  s  Cm  N ~10 3 Chapter 29 Solutions Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium: 29.63 ΣFx = T sin θ − ILB sin 90.00 = 0 ΣFy = T cos θ − mg = 0 , Therefore, tan θ = B= (0.0100 ILB IB = mg ( m L) g ( kg m ) 9.80 m s 2 5.00 A or or T sin θ = ILB T cos θ = mg or B= ) tan(45.0˚ ) = (m L) g tan θ I 19.6 mT Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium: 29.64 ΣFx = T sin θ − ILBsin 90.00 = 0 ΣFy = T cos θ − mg = 0 , tan θ = ILB IB = mg ( m L) g Σ F = ma or qvB sin 90.0° = 29.65 T sin θ = ILB T cos θ = mg or B= (m L) g tan θ = I mv 2 r ∴ the angular frequency for each ion is ∆f = f 12 − f 14 = or or qB v = ω = m = 2π f and r qB  1 1  (1.60 × 10 −19 C)(2.40 T)  1 1  − −  = 2 π  m12 m14  2 π (1.66 × 10 −27 kg / u)  12.0 u 14.0 u  ∆f = f12 – f14 = 4.38 × 105 s-1 = 438 kHz Let v x and v ⊥ be the components of the velocity of the positron parallel to and perpendicular to the direction of the magnetic field. 29.66 (a) The pitch of trajectory is the distance moved along x by the positron during each period, T (see Equation 29.15).  2π m  p = vxT = (v cos 85.0°)   Bq  p= (b) (5.00 × 106 )(cos 85.0°)(2 π )(9.11 × 10 −31 ) = 1.04 × 10-4 m (0.150)(1.60 × 10 −19 ) From Equation 29.13, r= r= mv ⊥ mv sin 85.0° = Bq Bq (9.11 × 10 −31 )(5.00 × 106 )(sin 85.0°) = 1.89 × 10-4 m −19 (0.150)(1.60 × 10 ) © 2000 by Harcourt, Inc. All rights reserved. µg tan θ I 19 Chapter 29 Solutions 20 τ = IAB where the effective current due to the orbiting electrons is 29.67 and the period of the motion is The electron's speed in its orbit is found by requiring ke q 2 mv 2 = or R R2 Substituting this expression for v into the equation for T, we find ∆q q = T ∆t 2π R T= v ke v=q mR I= T = 2π mR 3 q 2 ke (9.11 × 10 −31 )(5.29 × 10 −11 )3 = 1.52 × 10 −16 s (1.60 × 10 −19 )2 (8.99 × 10 9 ) T = 2π τ = Therefore, 1.60 × 10 −19  q π (5.29 × 10 −11 )2 (0.400) = 3.70 × 10-24 N · m AB = −16 T 1.52 × 10 Goal Solution Consider an electron orbiting a proton and maintained in a fixed circular path of radius R = 5.29 × 10-11 m by the Coulomb force. Treating the orbiting charge as a current loop, calculate the resulting torque when the system is in a magnetic field of 0.400 T directed perpendicular to the magnetic moment of the electron. -30 G: Since the mass of the electron is very small (~10 kg), we should expect that the torque on the -30 orbiting charge will be very small as well, perhaps ~10 N⋅m. O: The torque on a current loop that is perpendicular to a magnetic field can be found from τ = IAB sin θ . The magnetic field is given, θ = 90°, the area of the loop can be found from the radius of the circular path, and the current can be found from the centripetal acceleration that results from the Coulomb force that attracts the electron to proton. ( A : The area of the loop is A = π r 2 = π 5.29 × 10 −11 m ) 2 = 8.79 × 10 −21 m 2 . If v is the speed of the electron, then the period of its circular motion will be T = 2 πR v, and the effective current due to the orbiting electron is I = ∆Q / ∆t = e T . Applying Newton’s second law with the Coulomb force acting as the central force gives ΣF = ke q 2 mv 2 = R R2 T = 2π so that v=q ke mR and (9.10 × 10 −31 kg)(5.29 × 10 -11 m)3 (1.60 × 10 The torque is -19 τ = C ) (8.99 × 10 2 9 N⋅m C 2 2 ) T = 2π mR 3 q 2 ke = 1.52 × 10 −16 s 1.60 × 1019 C  q AB: τ = ( π )(5.29 × 10 −11 m)2 (0.400 T) = 3.70 × 10 −24 N ⋅ m T 1.52 × 10 -16 s L : The torque is certainly small, but a million times larger than we guessed. This torque will cause the atom to precess with a frequency proportional to the applied magnetic field. A similar process on the nuclear, rather than the atomic, level leads to nuclear magnetic resonance (NMR), which is used for magnetic resonance imaging (MRI) scans employed for medical diagnostic testing (see Section 44.2). Chapter 29 Solutions (a) qB m or m= qB qB = ω 2π f (1.60 × 10 −19 C)(5.00 × 10 −2 T) = 3.82 × 10-25 kg (2 π )(5.00 rev / 1.50 × 10 −3 s) m= 29.69 ω= Use the equation for cyclotron frequency 29.68 21 ( ) J   K = 21 mv 2 = 6.00 MeV = 6.00 × 106 eV 1.60 × 10 −19  eV  θ’ x x K = 9.60 × 10-13 J ( 2 9.60 × 10 v= x −13 J ) = 3.39 × 10 1.67 × 10 −27 kg FB = qvB = x 45˚ x mv 2 R R= so 45˚ R 7 ms ( 45.0˚ v )( x x x in = x B x x x x x x x x x x x x x x x x x x x x x x x x x x x x x ) 1.67 × 10 −27 kg 3.39 × 107 m s mv = = 0.354 m qB 1.60 × 10 −19 C (1.00 T ) ( ) Then, from the diagram, x = 2R sin 45.0˚ = 2(0.354 m ) sin 45.0˚ = 0.501 m 29.70 (b) From the diagram, observe that θ' = 45.0° . (a) See graph to the right. The Hall voltage is directly proportional to the magnetic field. A least-square fit to the data gives the equation of the best fitting line as: ( ∆VH = 1.00 × 10 −4 120.00 100.00 80.00 ∆V ) H (µV) 60.00 V T B 40.00 20.00 (b) Comparing the equation of the line which fits the data best to 0.00 0.00 0.20 0.40 0.60 0.80 1.00 1.20 B (T)  I  ∆VH =  B  nqt  I I = 1.00 × 10 −4 V T , or t = nqt nq 1.00 × 10 −4 V T ( observe that: ) Then, if I = 0.200 A, q = 1.60 × 10 −19 C, and n = 1.00 × 10 26 m −3 , the thickness of the sample is t= ( )( 0.200 A )( 1.00 × 10 26 m −3 1.60 × 10 −19 C 1.00 × 10 −4 V T ) = 1.25 × 10 −4 m = 0.125 mm © 2000 by Harcourt, Inc. All rights reserved. x x T x 1.00 22 Chapter 29 Solutions *29.71 (a) (b) The magnetic force acting on ions in the blood stream will deflect positive charges toward point A and negative charges toward point B. This separation of charges produces an electric field directed from A toward B. At equilibrium, the electric force caused by this field must balance the magnetic force, so qvB = qE = q or v=  ∆V   d  ∆V 160 × 10 −6 V = = 1.33 m/s Bd (0.040 0 T ) 3.00 × 10 −3 m ( ) N o . Negative ions moving in the direction of v would be deflected toward point B, giving A a higher potential than B. Positive ions moving in the direction of v would be deflected toward A, again giving A a higher potential than B. Therefore, the sign of the potential difference does not depend on whether the ions in the blood are positively or negatively charged. When in the field, the particles follow a circular path according to qvB = mv 2 r, so the radius of the path is: r = mv / qB *29.72 (a) (b) qBh mv , that is, when v = , the m qB particle will cross the band of field. It will move in a full semicircle of radius h, leaving the field at v f = −v j . (2h, 0, 0) with velocity When r = h = When v < qBh mv , the particle will move in a smaller semicircle of radius r = < h. It will qB m leave the field at ( 2r, 0, 0) with velocity (c) v f = −v j . qBh mv , the particle moves in a circular arc of radius r = > h, centered at (r, 0, 0) . qB m The arc subtends an angle given by θ = sin −1 ( h r ) . It will leave the field at the point with When v > coordinates [r(1 − cos θ ), h, 0] with velocity v f = v sin θ i + v cos θ j . Chapter 30 Solutions µ 0I 2R µ0q(v/2π R) = 12.5 T 2R 30.1 B= *30.2 We use the Biot-Savart law. For bits of wire along the straight-line sections, ds is at 0° or 180° to ~, so ds × ~= 0. Thus, only the curved section of wire contributes to B at P. Hence, ds is = tangent to the arc and ~ is radially inward; so ds × ~= d s l sin 90° = d s along the curve are the same distance r = 0.600 m from the field point, so B= ⌠ dB ⌡ all current ⌠µ =⌡ 0 4π I ds × ~ r 2 = µ0 I ⌠ 2 ⌡ ds 4π r = ⊗. All points µ0 I 2 s 4π r where s is the arclength of the curved wire,  2π   = 0.314 m s = rθ = (0.600 m)30.0° 360° Then, B =  10–7  T · m (3.00 A) (0.314 m) A  (0.600 m)2 B = 261 nT into the page 30.3 (a) B= 4µ0I  3π π l  cos – cos  where a = 2 4 4 4π a  is the distance from any side to the center. B= (b) 4.00 × 10–6  2 2  + 2  = 2 2 × 10–5 T = 28.3 µT into the paper 0.200  2 For a single circular turn with 4 l = 2π R, B= µ 0I µ0 π I (4π 2 × 10–7)(10.0) = 4l = = 24.7 µT into the paper 4(0.400) 2R © 2000 by Harcourt, Inc. All rights reserved. Figure for Goal Solution Chapter 30 Solutions 191 Goal Solution (a) A conductor in the shape of a square of edge length l = 0.400 m carries a current I = 10.0 A (Fig. P30.3). Calculate the magnitude and direction of the magnetic field at the center of the square. (b) If this conductor is formed into a single circular turn and carries the same current, what is the value of the magnetic field at the center? G: As shown in the diagram above, the magnetic field at the center is directed into the page from the clockwise current. If we consider the sides of the square to be sections of four infinite wires, then we could expect the magnetic field at the center of the square to be a little less than four times the strength of the field at a point l/2 away from an infinite wire with current I. B<4 ( )  4 π × 10 −7 T ⋅ m / A (10.0 A )  µ0I = 4  = 40.0 µ T   2 π (0.200 m ) 2π a   Forming the wire into a circle should not significantly change the magnetic field at the center since the average distance of the wire from the center will not be much different. O: Each side of the square is simply a section of a thin, straight conductor, so the solution derived from the Biot-Savart law in Example 30.1 can be applied to part (a) of this problem. For part (b), the BiotSavart law can also be used to derive the equation for the magnetic field at the center of a circular current loop as shown in Example 30.3. A : (a) We use Equation 30.4 for the field created by each side of the square. Each side contributes a field away from you at the center, so together they produce a magnetic field: B= so at the center of the square, ( ) −6 4µ 0 I  3 π  4 4 π × 10 T ⋅ m / A (10.0 A )  2 2 π = cos − cos +   4π a  4 4  4 π (0.200 m ) 2   2 B = 2.00 2 × 10 −5 T = 28.3 µ T perpendicularly into the page (b) As in the first part of the problem, the direction of the magnetic field will be into the page. The new radius is found from the length of wire: 4 = 2 π R, so R = 2 /π = 0.255 m. Equation 30.8 gives the magnetic field at the center of a circular current loop: B= µ 0 I (4 π × 10 −7 T ⋅ m / A)(10.0 A ) = = 2.47 × 10 −5 T = 24.7 µ T 2R 2(0.255 m) Caution! If you use your calculator, it may not understand the keystrokes: . get the right answer, you may need to use To L : The magnetic field in part (a) is less than 40µ T as we predicted. Also, the magnetic fields from the square and circular loops are similar in magnitude, with the field from the circular loop being about 15% less than from the square loop. Quick tip: A simple way to use your right hand to find the magnetic field due to a current loop is to curl the fingers of your right hand in the direction of the current. Your extended thumb will then point in the direction of the magnetic field within the loop or solenoid. © 2000 by Harcourt, Inc. All rights reserved. 192 Chapter 30 Solutions µ 0 I 4 π × 10 − 7 (1.00 A) = = 2.00 × 10-7 T 2π r 2 π (1.00 m) 30.4 B= 30.5 For leg 1, ds × ~= 0, so there is no contribution to the field from this segment. For leg 2, the wire is only semi-infinite; thus, B= µ0I 2R µ 0I 1  µ0I  into the paper =   4π x 2  2π x  R= µ 0 I 20.0 π × 10 − 7 = = 31.4 cm 2B 2.00 × 10 − 5 30.6 B= 30.7 We can think of the total magnetic field as the superposition of the field due to the long straight wire (having magnitude µ 0 I 2 πR and directed into the page) and the field due to the circular loop (having magnitude µ 0 I 2R and directed into the page). The resultant magnetic field is: ( ) −7 1  µ0I  1  4 π × 10 T ⋅ m / A (7.00 A )  B= 1 + = 1 + = 5.80 × 10 − 5 T  π  2R  π 2(0.100 m ) or 30.8 B = 58.0 µ T We can think of the total magnetic field as the superposition of the field due to the long straight wire (having magnitude µ 0 I 2 πR and directed into the page) and the field due to the circular loop (having magnitude µ 0 I 2R and directed into the page). The resultant magnetic field is: 1  µ0I  B= 1 +  π  2R 30.9 (directed into the page) (directed into the page) For the straight sections ds × ~= 0. loop: B= 1 µ0I µ0I = into the paper 4 2R 8R The quarter circle makes one-fourth the field of a full B= (4 π × 10 − 7 T ⋅ m / A)(5.00 A) = 26.2 µT into the paper 8(0.0300 m) Chapter 30 Solutions 193 30.10 Along the axis of a circular loop of radius R, B= ( µ 0 IR 2 2 x +R 2 1.00 ) 2 32 0.80 B/B  1 B   = B0  ( x R)2 + 1    or B Along Axis of Circular Loop 0.60 0 32 0.40 0.20 0.00 0.00 1.00 2.00 where B0 ≡ µ 0 I 2R. xR 0.00 1.00 2.00 3.00 4.00 5.00 30.11 dB = B= B= 30.12 B B0 1.00 0.354 0.0894 0.0316 0.0143 0.00754 µ 0 I d1× ~ 4π r 2 1 1 µ 0 I  6 2π a 6 2π b  −  4 π  a2 b2  µ 0 I  1 1 − directed out of the paper 12  a b  Apply Equation 30.4 three times: B= B= µ 0I   cos 0 − 4 π a    toward you d 2 + a2  d   away from you d 2 + a2  + µ0I 4π d  a +  2  d + a2  + µ0I 4π a   −d − cos 180° toward you  2   d + a2   µ 0 I  a2 + d 2 − d a2 + d 2  2 π a d a2 + d 2 3.00 x/R a away from you © 2000 by Harcourt, Inc. All rights reserved. 4.00 5.00 194 Chapter 30 Solutions 30.13 The picture requires L = 2R B= 1 2  µ0I  µ0I µ0I  2 R  + 4 π R (cos 90.0° − cos 135°) + 4 π R (cos 45.0° − cos 135°)   + B= µ0I (cos 45.0° − cos 90.0°) into the page 4π R µ0I  1 1  µ I = 0.475 0 (into the page) +    R  R  4 π 2 Label the wires 1, 2, and 3 as shown in Figure (a) and let the magnetic field created by the currents in these wires be B1 , B2 , and B3 respectively. 30.14 (a) At Point A : B1 = B2 = µ0I µ0I and B3 = . 2 π ( 3a) 2π a 2 ( ) The directions of these fields are shown in Figure (b). Observe that the horizontal components of B1 and B2 cancel while their vertical components both add to B3 . Figure (a) Therefore, the net field at point A is: BA = B1 cos 45.0˚ + B2 cos 45.0˚ + B3 = µ0I  2 1 cos 45.0˚ +  2 π a  2 3 (4π × 10 T ⋅ m A)(2.00 A)  2 cos 45˚ + 1  = =  2 3  2 π (1.00 × 10 m ) −7 BA (b) At point B : B1 and B2 cancel, leaving BB = B3 = BB = (c) −2 (4π × 10 ) T ⋅ m A ( 2.00 A ) −7 ( 2 π ( 2) 1.00 × 10 −2 m ) 53.3 µT Figure (b) µ0I . 2 π ( 2a) = 20.0 µT Figure (c) µ0I µ I and B3 = 0 with the directions shown in Figure (c). Again, 2π a 2π a 2 the horizontal components of B1 and B2 cancel. The vertical components both oppose B3 giving At point C : B1 = B2 = ( )  µ I µ I  2 cos 45.0˚  µ I  0 BC = 2  cos 45.0˚  − 0 = 0  − 1 = 0 2 2 π a π a 2     2 π a 2 ( ) Chapter 30 Solutions 195 Take the x-direction to the right and the y-direction up in the plane of the paper. Current 1 creates at P a field 30.15 ( ) 2.00 × 10 −7 T ⋅ m ( 3.00 A ) µ0I = 2π a A(0.0500 m ) B1 = B1 = 12.0 µ T downward and leftward, at angle 67.4° below the –x axis. Current 2 contributes B2 = (2.00 × 10 −7 ) T ⋅ m ( 3.00 A ) clockwise perpendicular to 12.0 cm A(0.120 m ) B2 = 5.00 µ T to the right and down, at angle –22.6° Then, B = B1 + B2 = (12.0 µ T ) ( −i cos 67.4° −j sin 67.4°) + ( 5.00 µ T ) (i cos 22.6° −j sin 22.6°) B = ( −11.1 µ T )j − (1.92 µ T )j = (–13.0 µT)j Let both wires carry current in the x direction, the first at y = 0 and the second at y = 10.0 cm . *30.16 (a) B= ( ) 4 π × 10 −7 T ⋅ m A ( 5.00 A ) µ0I k= k 2π r 2 π (0.100 m ) B = 1.00 × 10 −5 T out of the page (b) [ ( )] ( ) F B = I 2L × B = (8.00 A ) (1.00 m ) i × 1.00 × 10 −5 T k = 8.00 × 10 −5 N ( − j ) F B = 8.00 × 10 −5 N toward the first wire (c) B= ( ) 4 π × 10 −7 T ⋅ m A (8.00 A ) µ0I − k) = ( (− k) = (1.60 × 10−5 T) (− k) 2 π (0.100 m ) 2π r B = 1.60 × 10 −5 T into the page (d) [ ( ) ] ( ) F B = I1L × B = ( 5.00 A ) (1.00 m ) i × 1.60 × 10 −5 T ( − k ) = 8.00 × 10 −5 N ( + j ) F B = 8.00 × 10 −5 N toward the second wire 30.17 By symmetry, we note that the magnetic forces on the top and bottom segments of the rectangle cancel. The net force on the vertical segments of the rectangle is (using Equation 30.12) FB = 1 µ 0 I1I2 l  1 –  i c + a c  2π Substituting given values FB = –2.70 × 10–5 i N = – 27.0 µN i © 2000 by Harcourt, Inc. All rights reserved. 196 Chapter 30 Solutions Goal Solution In Figure P30.17, the current in the long, straight wire is I 1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and l = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. G: Even though there are forces in opposite directions on the loop, we must remember that the magnetic field is stronger near the wire than it is farther away. By symmetry the forces exerted o n sides 2 and 4 (the horizontal segments of length a) are equal and opposite, and therefore cancel. The magnetic field in the plane of the loop is directed into the page to the right of I1. By the right-hand rule, F = I1 × B is directed toward the left for side 1 of the loop and a smaller force is directed toward the right for side 3. Therefore, we should expect the net force to be to the left, possibly in the µ N range for the currents and distances given. O: The magnetic force between two parallel wires can be found from Equation 30.11, which can be applied to sides 1 and 3 of the loop to find the net force resulting from these opposing force vectors. A : F = F1 + F 2 = F= (4π × 10 µ 0 I1I 2 l  1 µ I I l  −a  1 i − i= 0 1 2    2π 2 π  c(c + a)  c+a c −7 ) N / A 2 ( 5.00 A )(10.0 A)(0.450 m)  − 0.150 m   i  (0.100 m)(0.250 m)  2π F = ( −2.70 × 10 −5 i) N or F = 2.70 × 10 −5 N toward the left L : The net force is to the left and in the µ N range as we expected. The symbolic representation of the net force on the loop shows that the net force would be zero if either current disappeared, if either dimension of the loop became very small ( a → 0 or l → 0), or if the magnetic field were uniform ( c → ∞) . The separation between the wires is 30.18 a = 2(6.00 cm) sin 8.00° = 1.67 cm. (a) Because the wires repel, the currents are in opposite directions . (b) Because the magnetic force acts horizontally, µ 0I 2 l FB = = tan 8.00° Fg 2π a mg I2= mg 2π a tan 8.00° l µ0 so I = 67.8 A Chapter 30 Solutions 197 30.19 Each wire is distant from P by (0.200 m) cos 45.0° = 0.141 m Each wire produces a field at P of equal magnitude: BA = (2.00 × 10–7 T · m)(5.00 A) µ 0I = A(0.141 m) 2π a = 7.07 µT Carrying currents into the page, A produces at P a field of 7.07 µT to the left and down at –135°, while B creates a field to the right and down at – 45°. Carrying currents toward you, C produces a field downward and to the right at – 45°, while D 's contribution is downward and to the left. The total field is then 4 (7.07 µT) sin 45.0° = 20.0 µT toward the page's bottom 30.20 Let the current I flow to the right. It creates a field B = µ 0 I 2 π d at the proton's location. And we have a balance between the weight of the proton and the magnetic force mg(− j) + qv(− i) × d= 30.21 µ0I (k) = 0 at a distance d from the wire 2π d qv µ 0 I (1.60 × 10 −19 C)(2.30 × 10 4 m / s)(4 π × 10 − 7 T ⋅ m / A)(1.20 × 10 − 6 A) = = 5.40 cm 2 π mg 2 π (1.67 × 10 − 27 kg) (9.80 m / s 2 ) From Ampère's law, the magnetic field at point a is given by Ba = µ 0 I a 2 π r a , where I a is the net current flowing through the area of the circle of radius r a . In this case, I a = 1.00 A out of the page (the current in the inner conductor), so Ba = (4π × 10 −7 ) T ⋅ m / A (1.00 A) 2 π (1.00 × 10 −3 m) Similarly at point b : Bb = = 200 µ T toward top of page µ 0 Ib , where Ib is the net current flowing through the area of the 2 π rb circle having radius rb . Taking out of the page as positive, Ib = 1.00 A − 3.00 A = − 2.00 A , or Ib = 2.00 A into the page. Therefore, Bb = (4 π × 10 − 7 T ⋅ m / A)(2.00 A) = 133 µ T toward bottom of page 2 π (3.00 × 10 − 3 m) © 2000 by Harcourt, Inc. All rights reserved. 198 Chapter 30 Solutions *30.22 (a) In B = µ 0I 2π r , the field will be one-tenth as large at a ten-times larger distance: 400 cm µ 0I µ 0I k+ (–k) 2π r 1 2π r 2 B= so 4π × 10–7 T · m (2.00 A)  1 1  = 7.50 nT – 0.3985 m 0.4015 m  2π A (b) B= (c) Call r the distance from cord center to field point and 2d = 3.00 mm the distance between conductors. B= 1  µ0I 2d µ 0I  1 – = r – d r + d   2π r 2 – d 2 2π (3.00 × 10–3 m) T · m (2.00 A) 2 7.50 × 10–10 T =  2.00 × 10–7 A   r – 2.25 × 10–6 m 2 so r = 1.26 m The field of the two-conductor cord is weak to start with and falls off rapidly with distance. (d) The cable creates zero field at exterior points, since a loop in Ampère's law encloses zero total current. Shall we sell coaxial-cable power cords to people who worry about biological damage from weak magnetic fields? 30.23 (a) B inner = µ0NI = 3.60 T 2π r (b) Bouter = µ0NI = 1.94 T 2π r µ 0I r for r ≤ a 2π a 2 B= (b) r= (a) One wire feels force due to the field of the other ninety-nine. 30.25 µ 0I 2π B = so µ0(2.50 A) = 0.0500 m = 2π (10.0 × 10–6 T) B= µ0(2.50 A) (0.0125 m) = 10.0 µT 2π (0.0250 m)2 *30.24 (a) 2.50 cm beyond the conductor's surface  µ I  Within the bundle, B =  0 2  r = 3.17 × 10 −3 T .  2π R  The force, acting inward, is F B = I lB, and the force per unit length is FB –3 l = 6.34 × 10 N/m inward (b) B ∝ r, so B is greatest at the outside of the bundle. Since each wire carries the same current, F is greatest at the outer surface . Figures for Goal Solution Chapter 30 Solutions 199 Goal Solution A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R = 0.500 cm. (a) If each wire carries 2.00 A, what are the magnitude and direction of the magnetic force per unit length acting on a wire located 0.200 cm from the center of the bundle? (b) Would a wire on the outer edge of the bundle experience a force greater or less than the value calculated in part (a)? G: The force on one wire comes from its interaction with the magnetic field created by the other ninetynine wires. According to Ampere’s law, at a distance r from the center, only the wires enclosed within a radius r contribute to this net magnetic field; the other wires outside the radius produce magnetic field vectors in opposite directions that cancel out at r . Therefore, the magnetic field (and also the force on a given wire at radius r ) will be greater for larger radii within the bundle, and will decrease for distances beyond the radius of the bundle, as shown in the graph to the right. Applying F = I1 × B, the magnetic force on a single wire will be directed toward the center of the bundle, so that all the wires tend to attract each other. O: Using Ampere’s law, we can find the magnetic field at any radius, so that the magnetic force F = I1 × B on a single wire can then be calculated. A : (a) Ampere’s law is used to derive Equation 30.15, which we can use to find the magnetic field at r = 0.200 cm from the center of the cable: ( ) ( ) 4 π × 10 −7 T ⋅ m / A (99)( 2.00 A ) 0.200 × 10 −2 m µ o I or B= = = 3.17 × 10 −3 T 2π R2 2 π (0.500 × 10 −2 m)2 This field points tangent to a circle of radius 0.200 cm and exerts a force F = I1 × B toward the center of the bundle, on the single hundredth wire: ( ) F l = IBsin θ = ( 2.00 A ) 3.17 × 10 −3 T (sin 90°) = 6.34 mN / m (b) As is shown above in Figure 30.12 from the text, the magnetic field increases linearly as a function of r until it reaches a maximum at the outer surface of the cable. Therefore, the force on a single wire at the outer radius r = 5.00 cm would be greater than at r = 2.00 cm by a factor of 5/2. L : We did not estimate the expected magnitude of the force, but 200 amperes is a lot of current. It would be interesting to see if the magnetic force that pulls together the individual wires in the bundle is enough to hold them against their own weight: If we assume that the insulation accounts for about half the volume of the bundle, then a single copper wire in this bundle would have a cross sectional area of about (1 2)(0.01)π (0.500 cm)2 = 4 × 10−7 m 2 with a weight per unit length of ( ) ( ) ρ gA = 8 920 kg / m 3 (9.8 N / kg ) 4 × 10 −7 m 2 = 0.03 N / m Therefore, the outer wires experience an inward magnetic force that is about half the magnitude of their own weight. If placed on a table, this bundle of wires would form a loosely held mound without the outer sheathing to hold them together. 30.26 From ∫ B ⋅ d1 = µ 0 I, I = 2π rB (2π)(1.00 × 10-3)(0.100) = = 500 A µ0 4π × 10–7 © 2000 by Harcourt, Inc. All rights reserved. 200 Chapter 30 Solutions ∫ B ⋅ ds = µ 0 I . Use Ampère’s law, becomes 30.27 For current density J, this ∫ B ⋅ ds = µ 0 ∫ J ⋅ dA (a) For r1 < R , this gives 2 π r1 = µ 0 ∫ r1 0 B= (b) µ 0br12 3 (br )(2 π r dr ) and (for r1 < R or inside the cylinder) When r2 > R , Ampère’s law yields (2π r2 )B = µ 0 ∫0 (br )(2π r dr ) = 2π µ 0bR3 R B= or 30.28 µ 0bR 3 3r2 3, (for r2 > R or outside the cylinder) (a) See Figure (a) to the right. (b) At a point on the z axis, the contribution from each wire has µ0I magnitude B = and is perpendicular to the line from 2 π a2 + z 2 this point to the wire as shown in Figure (b). Combining fields, the vertical components cancel while the horizontal components add, yielding     µ0I z µ0I µ0I z = By = 2  sin θ  =  2 2 2 2 2  2 2 π a + z2  a +z  2π a + z  π a +z ( (Currents are into the paper) Figure (a) ) The condition for a maximum is: dBy dz = − µ 0 I z( 2z) ( π a2 + z ) 2 2 + ( µ0I π a +z 2 2 ) = 0, or ( ( ) ) 2 2 µ0I a − z =0 π a2 + z 2 2 Thus, along the z axis, the field is a maximum at d = a . Figure (b) Chapter 30 Solutions 201 N B = µ0 l I 30.29 30.30 B I = µ n = 31.8 mA 0 so 10.0 (a) I= (b) FB = IB = 39.8 kN/m radially outward l (4π × 10–7)(2000) = 3.98 kA This is the force the windings will have to resist when the magnetic field in the solenoid is 10.0 T. 30.31 The resistance of the wire is Re = ρl ε = ε π r2 . , so it carries current I = Re ρl π r2 If there is a single layer of windings, the number of turns per length is the reciprocal of the wire diameter: n = 1/ 2r . B = nµ 0I = So, *30.32 µ0 επ r 2 ρ l 2r = µ0 ε π r 2ρ l = (4π × 10–7 T · m/A)(20.0 V)π (2.00 × 10–3 m) 2(1.70 × 10–8 Ω · m)(10.0 m) = 464 mT The field produced by the solenoid in its interior is given by T ⋅ m   30.0   B = µ 0nI ( − i ) = 4 π × 10 −7 (15.0 A)( − i)  A   10 -2 m  ( ) B = − 5.65 × 10 − 2 T i The force exerted on side AB of the square current loop is (F B )AB = IL × B = (0.200 A)[(2.00 × 10−2 m) j × (5.65 × 10−2 T)(− i)] (F B )AB = (2.26 × 10− 4 N) k Similarly, each side of the square loop experiences a force, lying in the plane of the loop, of 226 µ N directed away from the center . From the above result, it is seen that the net torque exerted on the square loop by the field of the solenoid should be zero. More formally, the magnetic dipole moment of the square loop is given by ( µ = IA = (0.200 A ) 2.00 × 10 −2 m ) (− i) = − 80.0 µA ⋅ m 2 The torque exerted on the loop is then 2 i ( ) ( ) τ = µ × B = − 80.0 µ A ⋅ m 2 i × − 5.65 × 10 − 2 T i = 0 © 2000 by Harcourt, Inc. All rights reserved. 202 Chapter 30 Solutions 30.33 (a) ( ) Φ B = ∫ B ⋅ dA = B ⋅ A = ( 5i + 4 j + 3 k ) T ⋅ 2.50 × 10 −2 m i 2 Φ B = 3.13 × 10 − 3 T ⋅ m 2 = 3.13 × 10 − 3 Wb = 3.13 mWb 30.34 (b) (ΦB )total = ∫ B ⋅ dA = (a) Φ B = B ⋅ A = BA where A is the cross-sectional area of the solenoid. ΦB = (b) 0 for any closed surface (Gauss’s law for magnetism) ( )  µ 0 NI  π r 2 = 7.40 µ Wb  l  Φ B = B ⋅ A = BA = [(  µ 0 NI  π r22 − r12  l  ( )] )  4 π × 10 −7 T ⋅ m A ( 300)(12.0 A )  2  π (8.00)2 − ( 4.00)2 10 −3 m = 2.27 µ Wb ΦB =    (0.300 m)   30.35 ]( ) (a) (ΦB )flat = B ⋅ A = BπR 2 cos(180 − θ ) = (b) The net flux out of the closed surface is zero: (Φ B )flat + (Φ B )curved = 0 (ΦB )curved = –B π R 2 cos θ B π R 2 cos θ dΦ E d dQ / dt I = (EA) = = dt dt e0 e0 30.36 (a) dE I = = 7.19 × 1011 V/m · s dt e0 A (b) ∫ B ⋅ ds = e0µ 0 B= 30.37 [ ΦE dt so 2 π rB = e0µ 0  d Q ⋅ π r2   dt  e0 A  µ 0 Ir µ 0 (0.200)(5.00 × 10 −2 ) = = 2.00 × 10-7 T 2A 2 π (0.100)2 (a) dΦ E dQ / dt I (0.100 A) = = = = 11.3 × 10 9 V ⋅ m / s e0 e0 8.85 × 10 −12 C 2 / N ⋅ m 2 dt (b) I d = e0 dΦ E = I = 0.100 A dt Chapter 30 Solutions 203 30.38 (a) I= ev 2π r  ev  2 –24 2 µ = IA =   π r = 9.27 × 10 A · m 2 π r   The Bohr model predicts the correct magnetic moment. However, the "planetary model" is seriously deficient in other regards. (b) 30.39 Because the electron is (–), its [conventional] current is clockwise, as seen from above, and µ points downward . Assuming a uniform B inside the toroid is equivalent NI to assuming r << R, then B 0 ≅ µ 0 2π R and a tightly wound solenoid. B0 = µ 0 (630)(3.00) = 0.00189 T 2 π (0.200) With the steel, B = κmB0 = (1 + χ)B0 = (101)(0.00189 T) 30.40  N  B = µ nI = µ  I  2π r  30.41 Φ B = µ nIA so I = (2π r )B = µN ( B = 0.191 T 2 π (0.100 m )(1.30 T ) ) 5000 4 π × 10 −7 Wb A ⋅ m ( 470) = 277 mA 500   B = µ nI = (750 × 4 π × 10 −7 )  (0.500) = 0.188 T  2 π (0.200)  A = 8.00 × 10-4 m2 30.42 and ΦB = (0.188 T)(8.00 × 10-4 m2) = 1.50 × 10-4 T · m2) = 150 µT · m2 The period is T = 2π /ω. The spinning constitutes a current I = µ = IA = µ= Qω Q ω R2 π R2 = 2 2π in the direction of ω (6.00 × 10 − 6 C)(4.00 / s)(0.0200 m)2 = 4.80 × 10-9 A · m2 2 © 2000 by Harcourt, Inc. All rights reserved. Q Qω = . T 2π 204 Chapter 30 Solutions 30.43 B = µ 0 (H + M) 30.44 B = µ 0 ( H + M) so H= B − M = 2.62 × 106 A/m µ0 2.00 T . µ0 But M = xnµ B where µ B is the Bohr magneton, n is the number of atoms per unit volume, and x is the number of electrons that contribute per atom. Thus, If µ 0 M = 2.00 T , then the magnetization of the iron is M = x= *30.45 (a) 2.00 T 2.00 T M = = = 2.02 nµ B nµ Bµ 0 8.50 × 10 28 m −3 9.27 × 10 −24 N ⋅ m T 4 π × 10 −7 T ⋅ m A 30.47 )( )( ) Comparing Equations 30.29 and 30.30, we see that the applied field is described by B0 = µ 0H. B C Then Eq. 30.35 becomes M = C 0 = µ 0 H , and the definition of susceptibility (Eq. 30.32) is T T χ= 30.46 ( M C = µ0 H T ( ) (b) 2.70 × 10 − 4 ( 300 K ) χT K⋅A C= = = 6.45 × 10 4 −7 µ0 T⋅m 4 π × 10 T ⋅ m A (a) Bh = Bcoil = (b) Bh = Bsin φ → B = (a) Number of unpaired electrons = µ 0 NI (4 π × 10 −7 )(5.00)(0.600) = = 12.6 µT 2R 0.300 12.6 µ T Bh = = 56.0 µT sin φ sin 13.0° 8.00 × 1022 A · m2 = 8.63 × 1045 9.27 × 10–24 A · m2 Each iron atom has two unpaired electrons, so the number of iron atoms required is 1 2 (b) (8.63 × 10 ) . Mass = 45 (4.31 × 1045 atoms)(7900 kg/m3) = 4.01 × 1020 kg 8.50 × 1028 atoms/m3 Chapter 30 Solutions 205 Goal Solution The magnetic moment of the Earth is approximately 8.00 × 1022 A·m 2. (a) If this were caused by the complete magnetization of a huge iron deposit, how many unpaired electrons would this correspond to? (b) At two unpaired electrons per iron atom, how many kilograms of iron would this correspond to? (Iron has a density of 7 900 kg/m3, and approximately 8.50 × 1028 atoms/m3.) G: We know that most of the Earth is not iron, so if the situation described provides an accurate model, then the iron deposit must certainly be less than the mass of the Earth ( MEarth = 5.98 × 10 2 4 kg ). One mole of iron has a mass of 55.8 g and contributes 2(6.02 × 10 2 3 ) unpaired electrons, so we should expect the total unpaired electrons to be less than 10 50 . O: The Bohr magneton µ B is the measured value for the magnetic moment of a single unpaired electron. Therefore, we can find the number of unpaired electrons by dividing the magnetic moment of the Earth by µ B . We can then use the density of iron to find the mass of the iron atoms that each contribute two electrons.  1 A  1T J  N ⋅ m  −24 A : (a) µ B = 9.27 × 10 −24  1 A ⋅ m2   = 9.27 × 10    J   N ⋅ s C ⋅ m   C / s  T  N= The number of unpaired electrons is 8.00 × 10 22 A ⋅ m 2 = 8.63 × 10 45 e 9.27 × 10 −24 A ⋅ m 2 (b) Each iron atom has two unpaired electrons, so the number of iron atoms required is 1 N = 1 (8.63 × 10 45 ) = 4.31 × 10 45 iron atoms . 2 2 MFe Thus, (4.31 × 10 = 45 )( atoms 7900 kg / m 3 8.50 × 10 28 atoms / m 3 ) = 4.01 × 10 20 kg L : The calculated answers seem reasonable based on the limits we expected. From the data in this problem, the iron deposit required to produce the magnetic moment would only be about 1/15 000 the mass of the Earth and would form a sphere 500 km in diameter. Although this is certainly a large amount of iron, it is much smaller than the inner core of the Earth, which is estimated to have a diameter of about 3000 km. 30.48 B= µ 0I = 2.00 × 10–5 T = 20.0 µT 2π R 30.49 B= µ 0 IR 2 2(R 2 + R 2 )3/2 30.50 (a) BC = µ0(10.0) µ 0I – =0 2π (0.270) 2π (0.0900) (b) BA = 4µ0(10.0) = 88.9 µT 2π (0.0900) I = 2.00 × 109 A so so flowing west I = 30.0 A out of paper © 2000 by Harcourt, Inc. All rights reserved. 206 Chapter 30 Solutions Suppose you have two 100-W headlights running from a 12-V battery, with the whole 200 W = 17 A current going through the switch 60 cm from the compass. Suppose the 12 V dashboard contains little iron, so µ ≅ µ0. Model the current as straight. Then, *30.51 B= µ 0I 2π r = (4π × 10–7)17 ~ 10– 5 T 2π (0.6) If the local geomagnetic field is 5 × 10–5 T, this is ~10–1 times as large, compass noticeably. enough to affect the A ring of radius r and width dr has area dA = 2π r dr. The current inside radius r is 30.52 r r 0 0 ( I = ∫ 2 π J r dr = 2 π J0 ∫ r dr − 2 π J0 R 2 (a) )∫ r r 3 0 ( )( dr = 2 π J0 r 2 2 − 2 π J0 R 2 r 4 4 ( ) ) Ampère's law says B( 2 π r ) = µ 0 I = µ 0 π J 0 r 2 − r 4 2R 2 , or and or 1  r  1  r 3 − B = µ 0 J0 R   for r ≤ R  2  R  4  R   [ ] B( 2 π r ) = µ 0 I total = µ 0 π J0 R 2 − π J0 R 2 2 = µ 0 π J0 R 2 2 B= µ 0 J0 R 2 µ 0 J0 R = for r ≥ R 4( r R ) 4r (b) 0.300 0.250 0.200 B / µ 0 J 0 R 0.150 0.100 0.050 0.000 0 2 4 6 r/R (c) To locate the maximum in the region r ≤ R, require that This gives the position of the maximum as r = 2 / 3 R .  1  21 2 1  2 3 2  Here B = µ 0 J0 R  −  = 0.272 µ 0 J0 R 4  3    2  3  µ J r2 dB µ 0 J0 = − 3 0 02 = 0 dr 2 4R Chapter 30 Solutions 207 30.53 Consider a longitudinal filament of the strip of width dr as shown in the sketch. The contribution to the field at point P due to the current dI in the element d r is µ 0dI where dI = I ( dr w ) dB = 2π r b+w µ 0I dr µ 0I w k = ln  1 +  k b  2π w r 2π w B = ∫ dB = ⌠ ⌡ b 30.54 We find the total number of turns: N= B= µ 0 NI l (0.0300 T)(0.100 m)A Bl = = 2.39 × 103 µ 0I (4π × 10–7 T · m)(1.00 A) Each layer contains (10.0 cm/0.0500 cm) = 200 closely wound turns (2.39 × 103/200) = 12 layers . so she needs The inner diameter of the innermost layer is 10.0 mm. The outer diameter of the outermost layer is 10.0 mm + 2 × 12 × 0.500 mm = 22.0 mm. The average diameter is 16.0 mm, so the total length of wire is (2.39 × 103)π (16.0 × 10–3 m) = 120 m 30.55 where in this case I = µ 0 IR 2 2(x 2 + R 2 )3/2 B= On the axis of a current loop, the magnetic field is given by q . The magnetic field is directed away from the center, with a (2π /ω) strength of B= 30.56 µ 0ω R 2 q 4π (x + R ) 2 2 3/2 = µ 0 (20.0)(0.100)2 (10.0 × 10 −6 ) [ 4 π (0.0500) + (0.100) 2 ] 2 3/2 = 1.43 × 10–10 T On the axis of a current loop, the magnetic field is given by where in this case I = when x = R , then 2 q (2π /ω) . Therefore, B= B= B= © 2000 by Harcourt, Inc. All rights reserved. µ 0 IR 2 2(x 2 + R 2 )3/2 µ 0ω R 2 q 4π (x 2 + R 2 )3/2 µ 0ω qR 2 ( 4 π 45 R ) 2 3/2 = µ 0qω 2.5 5 π R 208 Chapter 30 Solutions 30.57 (a) Use Equation 30.7 twice: Bx = µ 0IR 2  1 1  +  2 2 3/2 2 2 (x + R ) ((R – x) + R 2)3/2 B = B x1 + B x2 = B= (b) µ 0 IR 2 2(x + R 2 )3/2 2 µ 0IR 2  1 1  2 (x 2 + R 2 )3/2 + (2R 2 + x 2 – 2xR)3/2 dB µ 0 IR 2 = dx 2 (  3 2 2 − 2 ( 2x ) x + R Substituting x = ) −5 2 − ( 3 2R 2 + x 2 − 2xR 2 R and cancelling terms, 2 ) −5 2 (2x − 2R)  dB =0 dx [ 3µ 0 IR 2 2 d 2B = − (x + R 2 )−5 2 − 5x 2 (x 2 + R 2 )−7 2 + (2R 2 + x 2 − 2xR)−5 2 − 5(x − R)2 (2R 2 + x 2 − 2xR)−7 2 2 dx 2 Again substituting x = 30.58 d 2B =0 dx 2 R and cancelling terms, 2 "Helmholtz pair" → separation distance = radius B= [ 2µ 0 IR 2 2 ( R / 2) + R 2 2 ] 3/2 = µ 0 IR 2 1  +1  4  3/2 = R3 ( µ0I for 1 turn 1.40R ) 4 π × 10 −7 100(10.0) µ 0 NI For N turns in each coil, B = = = 1.80 × 10- 3 T 1.40R 1.40(0.500) ] Chapter 30 Solutions 209 Model the two wires as straight parallel wires (!) 30.59 (a) FB = FB = (b) *30.60 (a) µ0 I 2L 2π a (Equation 30.12) (4π × 10–7)(140)22π(0.100) = 2.46 N 2π (1.00 × 10–3) aloop = 2.46 N – m loop g = 107 m/s2 m loop upward upward µ0 Ids × ~, the moving charge constitutes a bit of current as in I = nqvA. For a 4π r 2 µ positive charge the direction of ds is the direction of v , so dB = 0 2 nqA( ds) v × ~. Next, A( ds) 4π r is the volume occupied by the moving charge, and nA( ds) = 1 for just one charge. Then, In dB = B= µ0 qv × ~ 4π r 2 (4π × 10 −7 )( )( T ⋅ m A 1.60 × 10 −19 C 2.00 × 107 m s ) sin 90.0˚ = 3.20 × 10 −13 T (b) B= (c) FB = q v × B = 1.60 × 10 −19 C 2.00 × 107 m s 3.20 × 10 −13 T sin 90.0˚ ( 4 π 1.00 × 10 ( ) −3 2 )( )( ) FB = 1.02 × 10 −24 N directed away from the first proton (d) ( )( 8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C kqq Fe = qE = e 12 2 = 2 r 1.00 × 10 −3 ( ) ) 2 Fe = 2.30 × 10 −22 N directed away from the first proton Both forces act together. The electrical force is stronger by two orders of magnitude. It is productive to think about how it would look to an observer in a reference frame moving along with one proton or the other. *30.61 (a) (b) ( ) 4 π × 10 −7 T ⋅ m A ( 24.0 A ) µ0I B= = = 2.74 × 10 − 4 T 2π r 2 π (0.0175 m ) ( ) conductor DE produces a At point C, conductor AB produces a field 21 2.74 × 10 − 4 T ( − j), field of 21 2.74 × 10 −4 T ( − j) , BD produces no field, and AE produces negligible field. The ( total field at C is ) 2.74 × 10 − 4 T ( − j) . © 2000 by Harcourt, Inc. All rights reserved. 210 Chapter 30 Solutions [( ) ] (1.15 × 10 (c) F B = IL × B = ( 24.0 A )(0.0350 m k ) × 5 2.74 × 10 − 4 T ( − j) = (d) 1.15 × 10 −3 N i  ΣF m a= = = 0.384 2 i −3  m s  3.00 × 10 kg (e) (f) ( −3 ) N i ) The bar is already so far from AE that it moves through nearly constant magnetic field. The force acting on the bar is constant, and therefore the bar’s acceleration is constant . ( ) v 2f = vi2 + 2ax = 0 + 2 0.384 m s 2 (1.30 m ) , so v f = (0.999 m s) i Chapter 30 Solutions 211 At equilibrium, 30.62 IB = 30.63 (a) 2 π a( m l) g FB µ 0 I A I B mg = = or I B = µ0IA 2π a l l (4π × 10 −7 ) T ⋅ m A (150 A ) )= 81.7 A The magnetic field due to an infinite sheet of charge (or the magnetic field at points near a large sheet of charge) is given by B = µ 0 Js 2 . The current density Js = I l and in this case the equivalent current of the moving charged belt is I= dq d = (σ lx) = σ lv; dt dt Therefore, Js = σ v (b) ( 2 π (0.0250 m )(0.0100 kg m ) 9.80 m s 2 v= and dx dt B= µ0σ v 2 If the sheet is positively charged and moving in the direction shown, the magnetic field is out of the page, parallel to the roller axes. TM (4.00 K)(10.0%)(8.00 × 10 27 atoms / m 3 )(5.00)(9.27 × 10 − 24 J / T 2 ) K⋅J = = 2.97 × 10 4 2 B 5.00 T T ⋅ m3 30.64 C= 30.65 At equilibrium, Στ = + µ × B − mg or µ B sin 5.00˚ = Therefore, B= L  cos 5.00˚ = 0, 2  mgL cos 5.00˚ 2 ( ) (0.0394 kg) 9.80 m s2 (0.100 m) mgL = 2µ tan 5.00˚ 2 (7.65 J T ) tan 5.00˚ B = 28.8 mT 30.66 The central wire creates field B = µ 0 I1 2 π R counterclockwise. The curved portions of the loop feels no force since 1 × B = 0 there. The straight portions both feel I 1 × B forces to the right, amounting to F B = I 2 2L µ 0 I1 µ 0 I1 I 2 L = to the right 2π R πR © 2000 by Harcourt, Inc. All rights reserved. 212 Chapter 30 Solutions 30.67 When the conductor is in the rectangular shape shown in figure (a), the segments carrying current straight toward or away from point P1 do not contribute to the magnetic field at P1. Each of the other four setions of length l makes an equal contribution to the total field into the page at P1. To find the contribution of the horizontal section of current in the upper right, we use B= So µ0I (cos θ1 – cos θ 2 ) with a = l, θ1 = 90°, and θ 2 = 135° 4π a B1 = 4 µ0I  µ0I 1  = 0 – 4πl  2 2 πl When the conductor is in the shape of a circular arc, the magnitude or the field at the center is µ I 4l given by Equation 30.6, B = 0 θ . From the geometry in this case, we find R = and θ = π. π 4π R Therefore, B2 = 30.68 I= µ 0 Iπ µ Iπ = 0 ; 4 π (4l/ π ) 16l ( )( so that B1 8 2 = 2 B2 π ) 3 −8 2 π rB 2 π 9.00 × 10 1.50 × 10 = = 675 A µ0 4 π × 10 −7 Flow of positive current is downward or negative charge flows upward . 30.69 By symmetry of the arrangement, the magnitude of the net magnetic field at point P is B = 8B 0x where B 0 is the contribution to the field due to current in an edge length equal to L/2. In order to calculate B0, we use the Biot-Savart law and consider the plane of the square to be the yz-plane with point P on the x-axis. The contribution to the magnetic field at point P due to a current element of length dz and located a distance z along the axis is given by Equation 30.3. B0 = µ 0 I d1× ~ 4π ∫ r 2 From the figure we see that r = x 2 + (L2 / 4) + z 2 and d1× ~ = dz sin θ = dz L2 / 4 + x 2 L / 4 + x2 + z2 2 By symmetry all components of the field B at P cancel except the components along x (perpendicular to the plane of the square); and Chapter 30 Solutions 213 B0x = B0 cos φ Therefore, B0x µ I = 0 4π L/2 cos φ = where L / 4 + x2 2 . L/2 ∫ 0 sin θ cos φ dz and B = 8B0x . r2 Using the expressions given above for sin θ cos φ, and r, we find B= 30.70 (a) µ 0 IL2  L2 L2  2π  x 2 +  x 2 + 2 4  From Equation 30.10, the magnetic field produced by one loop at the center of the second loop is given by B = ( ) ( ) 2 µ 0 IR 2 µ 0 I π R µ µ = = 0 3 where the magnetic moment of either loop is 3 3 2π x 2π x 2x µ = I π R 2 . Therefore, ( 2 dB  µ µ   3  3µ 0 π R I = Fx = µ =µ 0  2π   x 4  dx 2π x 4 (b) 30.71 ( ) 2 = 3π µ 0 I 2 R 4 2 x4 ) ( ) −3 −7 3 π µ 0 I 2 R 4 3 π 4 π × 10 T ⋅ m A (10.0 A ) 5.00 × 10 m Fx = = 4 2 2 x4 5.00 × 10 −2 m ( 2 ) 4 = 5.92 × 10 − 8 N There is no contribution from the straight portion of the wire since ds × ~= 0. For the field of the spiral, dB = B= µ 0 I (ds × ~) (4 π ) r 2 µ0I 4π µ I B= 0 4π 2π ds sin θ ~ ∫ θ =0 r2 2π ∫ r −2 dr = − θ =0 Substitute r = eθ: = µ0I 4π 2π ∫( θ =0 B=− ) 2π ( )θ µ 0 I −1 r 4π   3π   1 2 dr sin  2   4  r =0 [ ] µ 0 I −θ e 4π 2π 0 =− [ ] ( µ 0 I −2 π µ0I e − e0 = 1 − e −2 π 4π 4π © 2000 by Harcourt, Inc. All rights reserved. ) (out of the page) 214 Chapter 30 Solutions 30.72 (a) B = B0 + µ 0 M B − B0 µ0 M= M= and B − B0 µ0 Assuming that B and B0 becomes M = (B − B0 ) µ 0 are parallel, this The magnetization curve gives a plot of M versus B 0. (b) 30.73 The second graph is a plot of the relative permeability (B B0 ) as a function of the applied field B0 . Consider the sphere as being built up of little rings of radius r , centered on the rotation axis. The contribution to the field from each ring is dB = ( µ 0 r 2 dI 2 x +r 2 dI = where ) 2 32 ω dr dQ ω dQ = 2π t r dx x dQ = ρ dV = ρ ( 2 π r dr )( dx ) µ 0ρω r 3 dr dx dB = ( 2 32 +R R 2 −x 2 2 x2 + r where ρ = ) ( Let v = r 2 + x 2 , B=∫ +R x=−R ( Q 4 π R3 3 ) µ 0ρω r 3 dr dx 32 2 x2 + r2 x=−R ∫r=0 B=∫ R ) dv = 2r dr , and ( r2 = v − x2 ) 2 µ 0ρω v − x dv µ ρω ∫v=x 2 2 2 v 3 2 dx = 04 R2 2  R 2 −1 2  2 R −3 2 v dv − x 2  ∫x=−R ∫v=x ∫v=x 2 v dv dx R ( ) 2  1 2 R2 µ ρω 2 −1 2 R  dx = 0 2v v + 2x  ∫x=−R  x2 x2  4  B= µ 0ρω 4 B= µ 0ρω 4 B= 2µ 0ρω  2R 3 4R 2 µ 0ρωR 2 2 = − + 2R  2 3 4  3R  R R  x 2 ∫−R 2 R  2µ 0ρω − 4 x + 2R  dx = 4  R x 2 ∫0 2 R  ∫x=−R 2(R − x ) + 2x R   − 4x + 2R  dx  1 1   R − x   dx    2 Chapter 30 Solutions 215 Consider the sphere as being built up of little rings of radius r , centered on the rotation axis. The current associated with each rotating ring of charge is 30.74 dI = ω dr dQ ω = ρ ( 2 πr dr )( dx ) t 2π [ ] r dx x R The magnetic moment contributed by this ring is dµ = A( dI ) = π r 2 ω ρ ( 2 πr dr )( dx ) = πωρ r 3 dr dx 2π [ ] 4 µ = πωρ ∫ +R x=−R (  R2 − x2  R2 − x2  R 2 −x 2 3  +R  +R  dx dx = r dr = πωρ πωρ   ∫r=0 ∫x=−R ∫x=−R 4 4   µ=  2R 2  2R 5  πωρ +R πωρ  4 R 4 − 2R 2 x 2 + x 4 dx = R ( 2R) − 2R 2   + 5  ∫ 4 x=−R 4   3   µ= 4 2  πωρ R 5  16  πωρ 5  4 πωρ R 5 = R 2− + =  3 5 4  15  15 4 ( [ ] P1 -B2 r To find the field at either point P1 or P2 , find Bs which would exist if the conductor were solid, using Ampère’s law. Next, find B1 and B2 that would be due to the conductors of radius a 2 that could occupy the void where the holes exist. Then use the superposition principle and subtract the field that would be due to the part of the conductor where the holes exist from the field of the solid conductor. At point P1 , Bs = B= ( ), µ 0 J π a2 2π r µ 0 Jπ a2 2π Bs r 2 + ( a 2) a/2 r a/2 2  1 1 1 −   −  r 4(r − a 2) 4(r + a 2)  ) © 2000 by Harcourt, Inc. All rights reserved. 2 θ P2 − B1′ µ J π ( a 2) µ J π ( a 2) B1 = 0 , and B2 = 0 . 2 π (r − a 2) 2 π (r + a 2) µ 0 ( 2I )  4r 2 − a2 − 2r 2  µ 0 I  2r 2 − a2  =   directed to the left 2 π  4r r 2 − a2 4  π r  4r 2 − a2    ( -B1 Bs Therefore, J = 2I π a2 . B = Bs − B1 − B2 = dx up A = π a2 − a2 4 − a2 4 = π a2 2 (a) 2 ) Note that the current I exists in the conductor with a current density J = I A , where 30.75 ) 2 θ θ − B2′ 216 Chapter 30 Solutions (b) At point P2 , Bs = ( ) µ 0 J π a2 2π r and B1′ = B2′ = µ 0 Jπ ( a 2) 2 2 π r 2 + ( a 2) 2 . The horizontal components of B1′ and B2′ cancel while their vertical components add. B = Bs − B1′ cos θ − B2′ cos θ = B= ( ) ( ) − 2  µ 0 J π a2 2π r µ 0 Jπ a2 4  r   2 π r 2 + a2 4  r 2 + a2 4    µ 2I  2 µ 0 J π a2   r2 1 −  = 0 ( ) 1 − 2r = 2 2 π r  2 r 2 + a2 4  2 π r  4r + a2    ( ) µ 0 I  2r 2 + a2    π r  4r 2 + a2  directed toward the top of the page Chapter 31 Solutions 31.1 ε= ∆Φ B ∆( NBA) = = 500 mV ∆t ∆t 31.2 ε= ∆Φ B ∆(B ⋅ A ) = = 1.60 mV ∆t ∆t 31.3 ε = –N ( ∆BA cos θ ∆t ) = −25.0 50.0 × 10 −6 T π (0.500 m ) I loop = and = 2  cos  –NB ε = R 1.60 mV = 0.800 mA 2.00 Ω πr2 180° − cos 0   0.200 s  cos θf – cos θi   ∆t   E = + 9.82 mV 31.4 31.5 (a) ε= dΦB dB ABmax –t/ τ – dt = –A dt = e τ (b) ε= (0.160 m2)(0.350 T) – 4.00/2.00 e = 3.79 mV 2.00 s (c) At t = 0, ε =N ε= 28.0 mV dΦ B ∆(NBA) = = 3.20 kV dt ∆t so I= ε R © 2000 by Harcourt, Inc. All rights reserved. = 160 A Chapter 31 Solutions 219 Goal Solution A strong electromagnet produces a uniform field of 1.60 T over a cross-sectional area of 0.200 m 2 . A coil having 200 turns and a total resistance of 20.0 Ω is placed around the electromagnet. The current in the electromagnet is then smoothly decreased until it reaches zero in 20.0 ms. What is the current induced in the coil? G: A strong magnetic field turned off in a short time ( 20.0 ms ) will produce a large emf, maybe on the order of 1 kV . With only 20.0 Ω of resistance in the coil, the induced current produced by this emf will probably be larger than 10 A but less than 1000 A. O: According to Faraday’s law, if the magnetic field is reduced uniformly, then a constant emf will be produced. The definition of resistance can be applied to find the induced current from the emf. A : Noting unit conversions from F = qv × B and U = qV , the induced voltage is ε = −N d(B ⋅ A) = −N  0 − Bi A cos θ  = ∆t dt I= ε R = ( ) + 200(1.60 T ) 0.200 m 2 (cos 0°)  1 N ⋅ s / C ⋅ m   1 V ⋅ C  20.0 × 10 −3 s  T  N ⋅ m  = 3200 V 3200 V = 160 A 20.0 Ω L : This is a large current, as we expected. The positive sign is indicative that the induced electric field is in the positive direction around the loop (as defined by the area vector for the loop). ε = –N 31.6 ∆t = ε 31.7 dΦ B N(BA – 0) = – dt ∆t NBA ε = NB( π r 2) ε = 500(0.200)π (5.00 × 10-2)2 10.0 × 103 = 7.85 × 10–5 s d(BA) dI = 0.500 µ 0nA dt = 0.480 × 10–3 V dt = ε 4.80 × 10-4 = 1.60 A 3.00 × 10-4 (a) Iring = (b) Bring = (c) Coil's field points downward, and is increasing, so R = µ 0I 2rring = 20.1 µT Bring points upward 31.8 ε = d(BA) dI 2 ∆I = 0.500 µ 0nA = 0.500 µ0n π r 2 dt dt ∆t © 2000 by Harcourt, Inc. All rights reserved. 220 Chapter 31 Solutions ε 2 µ 0 n π r 2 ∆I 2R ∆t (a) Iring = (b) µ 0 n π r2 ∆I µ 0I B = 2r = 4r R 1 1 ∆t (c) The coil's field points downward, and is increasing, so (a) dΦ B = B ⋅ dA = (b) ε= R = 2 2 31.9 − µ0I Ldx : 2π x ΦB = ∫ h+w µ 0 IL x=h 2π Bring points upward . µ IL  h + w  dx = 0 ln  h  x 2π dΦ B d  µ IL  h + w    µ L  h + w   dI = −  0 ln = −  0 ln     dt  2 π h  dt  2 π  h   dt 4 π × 10 ε = −( −7 ) T ⋅ m A (1.00 m ) 2π ln A  1.00 + 10.0   10.0 =    s 1.00 − 4.80 µ V The long wire produces magnetic flux into the page through the rectangle (first figure, above). As it increases, the rectangle wants to produce its own magnetic field out of the page, which it does by carrying counterclockwise current (second figure, above). Chapter 31 Solutions 221 31. 10 Φ B = (µ 0nI)Asolenoid 2 2 ε = −N dΦB = −Nµ 0n(π rsolenoid ) dI = −Nµ 0n(π rsolenoid )(600 dt dt ε = −15.0( 4 π × 10−7 )( ) A s) cos(120t) T ⋅ m A 1.00 × 10 3 m π (0.0200 m ) (600 A s) cos(120t) 2 E = −14.2 cos(120t) mV 31.11 For a counterclockwise trip around the left-hand loop, with B = At [ ] d At(2a2 )cos 0° − I1(5R) − I PQ R = 0 dt and for the right-hand loop, [ ] d At a2 + I PQ R − I 2 (3R) = 0 dt where I PQ = I1 − I 2 is the upward current in QP Thus, 2Aa2 − 5R(I PQ + I 2 ) − I PQ R = 0 Aa2 + I PQ R = I 2 (3R) and 2Aa2 − 6RI PQ − 53 (Aa2 + I PQ R) = 0 31.12 ε= I PQ = Aa2 upward, and since R = (0.100 Ω/m)(0.650 m) = 0.0650 Ω 23R I PQ = (1.00 × 10 − 3 T / s)(0.650 m)2 = 283 µA upward 23(0.0650 Ω) ∆Φ B  dB  A = N (0.0100 + 0.0800t )A =N  dt  ∆t At t = 5.00 s, ε = 30.0(0.410 T)[π (0.0400 m)2 ] = 61.8 mV © 2000 by Harcourt, Inc. All rights reserved. 222 Chapter 31 Solutions 31.13 ( B = µ 0nI = µ 0n( 30.0 A ) 1 − e −1.60t ) ( Φ B = ∫ B dA = µ 0n( 30.0 A ) 1 − e −1.60t ( )∫ dA ) Φ B = µ 0n( 30.0 A ) 1 − e −1.60t π R 2 ε = − N d ΦB = − Nµ 0n(30.0 A)π R 2 (1.60)e − 1.60t dt ε = −(250)(4 π × 10−7 N A 2 )(400 m −1 )(30.0 A)[π (0.0600 m)2 ]1.60 s−1 e −1.60t ε 31.14 = (68.2 mV )e −1.60t counterclockwise B = µ 0nI = µ 0nI max (1 − e − α t ) Φ B = ∫ B dA = µ 0nI max (1 − e − α t )∫ dA Φ B = µ 0nI max (1 − e − α t )π R 2 ε = − N d ΦB = − Nµ 0nImax π R 2α e − α t dt ε 31.15 = Nµ 0nI max π R 2α e − α t counterclockwise ε= l= Nl2 ∆B cos θ d (NBl2 cos θ ) = dt ∆t ε ∆t N ∆B cos θ = (80.0 × 10 −3 V)(0.400 s) = 1.36 m (50)(600 × 10 −6 T − 200 × 10 −6 T)cos(30.0°) Length = 4 lN = 4(1.36 m)(50) = 272 m Chapter 31 Solutions 223 Goal Solution A coil formed by wrapping 50.0 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased uniformly from 200 µ T to 600 µ T in 0.400 s, an emf of 80.0 mV is induced in the coil. What is the total length of the wire? G: If we assume that this square coil is some reasonable size between 1 cm and 1 m across, then the total length of wire would be between 2 m and 200 m. O: The changing magnetic field will produce an emf in the coil according to Faraday’s law of induction. The constant area of the coil can be found from the change in flux required to produce the emf. A : By Faraday’s law, ε = −N For magnitudes, ε and the area is A= dΦ B d dB = −N (BA cos θ ) = −NA cos θ dt dt dt = NA cos θ  ∆B   ∆t  ε  ∆B  N cos θ  ∆t  = 80.0 × 10 −3 V  600 × 10 −6 T − 200 × 10 −6 T  50(cos 30.0°)  0.400 s   = 1.85 m 2 Each side of the coil has length d = A , so the total length of the wire is L = N ( 4d) = 4N A = (4)(50) 1.85 m 2 = 272 m L : The total length of wire is slightly longer than we predicted. With d = 1.36 m , a normal person could easily step through this large coil! As a bit of foreshadowing to a future chapter on AC circuits, an even bigger coil with more turns could be hidden in the ground below high-power transmission lines so that a significant amount of power could be “stolen” from the electric utility. There is a story of one man who did this and was arrested when investigators finally found the reason for a large power loss in the transmission lines! 31.16  ∆Φ B   ∆t  The average induced emf is given by ε Here N = 1, and ∆Φ B = B(Asquare with Acircle = π r 2 = π (0.500 m)2 = 0.785 m 2 = – N − Acircle ) Also, the circumference of the circle is 2π r = 2π (0.500 m) = 3.14 m Thus, each side of the square has a length and So 3.14 m = 0.785 m, 4 Asquare = L2 = 0.617 m 2 L= ∆Φ B = (0.400 T)(0.617 m 2 − 0.785 m 2 ) = − 0.0672 T ⋅ m 2 The average induced emf is therefore: ε=– − 0.0672 T ⋅ m 2 = 0.672 V 0.100 s © 2000 by Harcourt, Inc. All rights reserved. 224 Chapter 31 Solutions In a toroid, all the flux is confined to the inside of the toroid. 31.17 B= µ 0 NI 500 µ 0 I = 2π r 2π r Φ B = ∫ B dA = ΦB = 500 µ 0 I max dz dr sin ω t ∫ 2π r 500 µ 0 I max  b + R asin ω t ln  R  2π ε = N ′ dΦB = 20  500 µ 0 Imax  ω a ln b + R  cos ω t dt 2π R N rad   ( 3.00 + 4.00) cm    4 π × 10 −7 2 ( 50.0 A ) 377 (0.0200 m)ln  cos ω t = (0.422 V) cos ω t       s 2π 4.00 cm A ε = 10 31.18 4  N I  l The field inside the solenoid is: B = µ 0nI = µ 0 Thus, through the single-turn loop Φ B = BAsolenoid = µ 0 and the induced emf in the loop is ε = − ∆ΦB = −µ 0  N  (π r 2 )  ∆I  = ε = −N dΦB 31.19 IR = −N dt Idt = − Q=− N dΦ B R ∆t dΦ B dt N ∫ Idt = − R ∫ dΦB ( N N ∆Φ B = − A Bf − Bi R R )  200  −4 2 Q = −  (100 × 10 m )(−1.10 − 1.10) T = 0.880 C  5.00 Ω  31.20 I= ε R = Blv R v = 1.00 m/s l ( )  N π r2 I  l ∆t − µ 0 Nπ r 2  I 2 – I1   ∆t  l Chapter 31 Solutions 225 31.21 (a) F B = I 1 × B = I lB. FB = When I = E / R and ε = Blv , we get Blv B2l2 v (2.50)2 (1.20)2 (2.00) (lB) = = = 3.00 N R 6.00 R The applied force is 3.00 N to the right (b) P = I 2R = FB = IlB *31.22 I= 31.23 B2l2 v 2 = 6.00 W R E Blv = R R I 2lR lv (a) FB = (b) I 2R = 2.00 W (c) For constant force, or P = Fv = 6.00 W and E = Blv so B= and I= IR lv FB v = 0.500 A R P = F ⋅ v = (1.00 N )( 2.00 m / s) = 2.00 W The downward component of B, perpendicular to v, is (50.0 × 10–6 T) sin 58.0° = 4.24 × 10–5 T ( ) E = Blv = 4.24 × 10 −5 T (60.0 m )( 300 m / s) = 0.763 V The left wing tip is positive relative to the right. 31.24 d ε = –N dt  ∆A  BA cos θ = –NB cos θ    ∆t  ε = –1(0.100 T) cos 0° I= (3.00 m × 3.00 m sin 60.0°) – (3.00 m)2 = 1.21 V 0.100 s 1.21 V = 0.121 A 10.0 Ω The flux is into the page and decreasing. The loop makes its own magnetic field into the page by carrying clockwise current. 31.25 ω = (2.00 rev/s)(2π rad/rev) = (4.00)π rad/s 1 2 E = Bω l2 = 2.83 mV © 2000 by Harcourt, Inc. All rights reserved. 226 Chapter 31 Solutions 31.26 (a) Bext = Bext i and Bext decreases; therefore, the induced field is B0 = B 0 i (to the right). Therefore, the current is to the right in the resistor. (b) Bext = Bext (–i) increases; therefore, the induced field B0 = B 0 (+ i) is to the right, and the current is to the right in the resistor. (c) (a) (b) (c) (d) Bext = Bext (–k) into the paper and Bext decreases; therefore, the induced field is B0 = B 0 (–k) into the paper. Therefore, the current is to the right in the resistor. (d) By the Lorentz force law, F B = q (v × B). Therefore, a positive charge will move to the top of the bar if B is into the paper . 31.27 (a) The force on the side of the coil entering the field (consisting of N wires) is F = N ( ILB) = N ( IwB) The induced emf in the coil is ε =N d(Bwx ) dΦ B =N = NBwv, dt dt so the current is I = ε R = NBwv R counterclockwise. The force on the leading side of the coil is then: F=N N 2B2 w 2 v  NBwv  wB = to the left  R  R Φ B = NBA = constant , so ε = 0, I = 0, and F = 0 (b) Once the coil is entirely inside the field, . (c) As the coil starts to leave the field, the flux decreases at the rate Bwv , so the magnitude of the current is the same as in part (a), but now the current flows clockwise. Thus, the force exerted on the trailing side of the coil is: Chapter 31 Solutions 227 N 2B2 w 2 v to the left again R F= 31.28 (a) Motional emf ε = Bwv appears in the conducting water. Its resistance, if the plates are submerged, is ρL ρw = A ab Kirchhoff's loop theorem says Bwv – IR – I= Bwv R+ (b) 31.29 Iρ w =0 ab ρw ab = abvB abR ρ+ w (100 m)(5.00 m)(3.00 m/s)(50.0 × 10– 6 T) = 0.750 mA 100 Ω · m Isc = Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. The loop will create a field out of the page by carrying a counterclockwise current. Therefore, current must flow from b to a through the resistor. Hence, V a – V b will be negative . 1 2 31.30 E = Bω l2 = 0.259 mV 31.31 Name the currents as shown in the diagram: Left loop: + Bdv2 − I 2 R 2 − I1R 1 = 0 Right loop: + Bdv3 − I 3 R 3 + I1R 1 = 0 At the junction: I 2 = I1 + I 3 Then, Bdv2 − I1R 2 − I 3 R 2 − I1R 1 = 0 I3 = So, Bdv3 I1R 1 + R3 R3 B dv2 − I1 (R 1 + R 2 ) − B dv3 R 2 R3 − I1R 1R 2 R3 =0 © 2000 by Harcourt, Inc. All rights reserved. 228 Chapter 31 Solutions   v2 R 3 − v3 R 2 I1 = Bd   upward  R 1R 2 + R 1R 3 + R 2 R 3    (4.00 m / s)(15.0 Ω) − (2.00 m / s)(10.0 Ω) I1 = (0.0100 T)(0.100 m)    (5.00 Ω)(10.0 Ω) + (5.00 Ω)(15.0 Ω) + (10.0 Ω)(15.0 Ω)  upward 31.32 (a) dB 2 dt = 6.00t – 8.00t ε = F = qE = 8.00 × 10–21 N 31.33 When 6.00t2 – 8.00t = 0, dB = 0.0600t dt At t = 3.00 s, *31.34 145 µA dΦB dt π R 2(dB/dt) 8.00π (0.0250)2 = 2π r 2 2π (0.0500) At t = 2.00 s, E = (b) = ε = clockwise for electron t = 1.33 s dΦB dt 2  dB  –3 E = π r1   = 1.80 × 10 N/C perpendicular to r1 and counterclockwise 2π r 1 dt ε = dΦB = π r 2  dB  = ∫ E ⋅ d1 dt E(2 π R) = π r 2 dt dB , dt or  π r 2  dB E=   2 π R  dt B = µ 0nI dI dB = µ 0n dt dt I = 3.00 e 0.200t dI = 0.600 e 0.200t dt At t = 10.0 s, becomes E= E= π r2 (µ 0n)(0.600 e 0.200t ) 2π R (0.0200 m)2 (4 π × 10 −7 N / A 2 )(1000 turns / m)(0.600)e 2.00 = 2.23 × 10 −5 N / C 2(0.0500 m) Chapter 31 Solutions 229 31.35 (a) ∫ E ⋅ d1 = dΦ B dt 2 π rE = ( π r 2 ) (b) dB dt so E = (9.87 mV/m) cos (100 π t) The E field is always opposite to increasing B. ∴ clockwise © 2000 by Harcourt, Inc. All rights reserved. 230 Chapter 31 Solutions rev  2π rad 1 min For the alternator, ω = 3000 m i n  1 rev   60 s  = 314 rad/s    31.36 ε = −N dΦB = −250 d [(2.50 × 10−4 T ⋅ m 2 )cos(314 t / s)] = +250(2.50 × 10–4 T · m2)(314/s) sin(314t) dt dt 31.37 (a) ε = (19.6 V) sin(314t) (b) ε max = 19.6 V (a) εmax = NABω = (1000)(0.100)(0.200)(120π) = (b) ε(t) = –NBAω · sin ωt = –NBAω sin θ ε is maximal when sin θ = 1, 7.54 kV π or θ = ± 2 , so the plane of coil is parallel to B 31.38 Let θ represent the angle through which the coil turns, starting from θ = 0 at an instant when the horizontal component of the Earth's field is perpendicular to the area. Then, ε =− N d d BA cos θ = − NBA cos ω t = + NBAω sin ω t dt dt Here sin ω t oscillates between +1 and –1, so the spinning coil generates an alternating voltage with amplitude ε max = NBAω = NBA2π f 31.39 ( = 100(2.00 × 10 −5 T)(0.200 m)2 (1500) )( 2 π rad = 12.6 mV 60.0 s ) B = µ 0nI = 4 π × 10 −7 T ⋅ m A 200 m −1 (15.0 A ) = 3.77 × 10 −3 T For the small coil, Thus, ( ) Φ B = NB ⋅ A = NBA cos ω t = NB π r 2 cos ω t ε = − dΦB = NBπ r 2ω sin ω t dt ε = (30.0)(3.77 × 10−3 T)π (0.0800 m)2 ( 4.00π s−1 ) sin( 4.00π t) = (28.6 mV) sin(4.00π t) Chapter 31 Solutions 231 As the magnet rotates, the flux through the coil varies sinusoidally in time with Φ B = 0 at t = 0. Choosing the flux as positive when the field passes from left to right through the area of the coil, the flux at any time may be written as Φ B = −Φ max sin ω t so the induced emf is given by 31.40 1 0.5 I/I max 0 0 0.5 -1 t/T = ( ω t / 2 π ) dt The current in the coil is then (a) 1.5 -0.5 ε = − dΦB = ω Φ max cos ω t . 31.41 1 I= ε R = ω Φ max cos ω t = I max cos ω t R F = NI lB τmax = 2Fr = NI lwB = 0.640 N · m (b) P = τω = (0.640 N · m)(120π rad/s) 1 P max = 241 W (about 3 hp) 31.42 (a) ε max = BAω = B( 21 π R 2 )ω ε ε max = (1.30 T) π (0.250 m)2  4.00π 2 ε max = (b) (c) rad  s  Figure 1 1.60 V 2π ε = ∫0 ε 2π dθ = ε BAω 2π 2π ∫0 sin θ dθ = 0 The maximum and average ε would remain unchanged. Figure 2 (d) See Figure 1 at the right. 31.43 (e) See Figure 2 at the right. (a) ΦB = BA cos θ = BA cos ωt = (0.800 T)(0.0100 m2) cos 2π (60.0)t = (8.00 mT · m2) cos (377t) © 2000 by Harcourt, Inc. All rights reserved. 2 232 Chapter 31 Solutions dΦB = (3.02 V) sin (377t) dt (b) ε=– (c) I = ε R = (3.02 A) sin (377t) (d) P = I 2R = (9.10 W) sin2 (377t) (e) P = Fv = τ ω τ= so P = (24.1 mN · m) sin2 (377t) ω At terminal speed, the upward magnetic force exerted o n the lower edge of the loop must equal the weight of the loop. That is, 31.44 Mg = FB = IwB = B2 w 2 vt ε  Bwvt  wB = wB =  R  R  R Thus, B= (0.150 kg)(9.80 (1.00 m) (2.00 2 ) m s 2 (0.750 Ω) m s) = 0.742 T See the figure above with Problem 31.44. 31.45 (a) At terminal speed, or *31.46 MgR = w 2 vt vt = Mg = FB = IwB = B2 w 2 vt ε  Bwvt  wB = wB =  R  R  R MgR B2 w 2 (b) The emf is directly proportional to vt , but the current is inversely proportional to R. A large R means a small current at a given speed, so the loop must travel faster to get Fm = mg. (c) At given speed, the current is directly proportional to the magnetic field. But the force is proportional to the product of the current and the field. For a small B, the speed must increase to compensate for both the small B and also the current, so vt ∝ B2 . The current in the magnet creates an upward magnetic field, so the N and S poles on the solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B increases as the coil approaches, so a current is induced here to create a downward magnetic field. This is clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic field by being counterclockwise as the picture correctly shows. Chapter 31 Solutions 233 F = ma = qE + qv × B 31.47 e a = [E + v × B] where m a= i v × B = 200 k 0 = − 200(0.400)j + 200(0.300)k 0.200 0.300 0.400 1.60 × 10 −19 [50.0j − 80.0j + 60.0k] = 9.58 × 107 [−30.0j + 60.0k] 1.67 × 10 −27 a = 2.87 × 10 9 [−j + 2k] m s 2 = F = ma = qE + qv × B so 31.48 j 0 (−2.87 × 10 9 j + 5.75 × 10 9 k) m s 2 −e a = [E + v × B] m where i j v × B = 10.0 0 0 a= (−1.60 × 10 ) [2.50i + 5.00 j − 4.00 j] = (−1.76 × 10 )[2.50i + 1.00 j] 9.11 × 10 a= (− 4.39 × 10 k 0 0 0.400 −19 11 −31 11 ) i − 1.76 × 1011 j m s 2 ε = −N d (BA cos θ ) = −N (π r 2 ) cos 0˚ dB dt dt *31.49 ε = − (30.0)π (2.70 × 10−3 m) (1) d [50.0 mT + (3.20 mT) sin (2π dt 2 ε = − (30.0)π (2.70 × 10−3 m) ε= *31.50 (a) ( 2 (3.20 × 10 T)(2π )(523 s) cos (2π 523t −3 ) − 7.22 × 10 −3 V cos ( 2 π 523t s) Doubling the number of turns. Amplitude doubles: period unchanged (b) Doubling the angular velocity. doubles the amplitude: cuts the period in half (c) Doubling the angular velocity while reducing the number of turns to one half the original value. Amplitude unchanged: cuts the period in half ] 523t s) © 2000 by Harcourt, Inc. All rights reserved. s) = − 4.00j 234 Chapter 31 Solutions T ε = −N ∆ (BA cos θ ) = −N (π r 2 ) cos 0˚ ∆B = −1(0.00500 m 2 )(1)  1.50 T − 5.00 = 0.875 V −3  ∆t ∆t *31.51 31.52 20.0 × 10 ε 0.875 V = 43.8 A 0.020 0 Ω (a) I= (b) P = EI = (0.875 V )( 43.8 A ) = 38.3 W R = s In the loop on the left, the induced emf is ε = dΦ B dB 2 =A = π (0.100 m ) (100 T s) = π V dt dt and it attempts to produce a counterclockwise current in this loop. In the loop on the right, the induced emf is ε = dΦ B 2 = π (0.150 m ) (100 T s) = 2.25 π V dt and it attempts to produce a clockwise current. Assume that I1 flows down through the 6.00-Ω resistor, I 2 flows down through the 5.00-Ω resistor, and that I 3 flows up through the 3.00-Ω resistor. From Kirchhoff’s point rule: I 3 = I1 + I 2 (1) Using the loop rule on the left loop: 6.00 I1 + 3.00I 3 = π (2) Using the loop rule on the right loop: 5.00 I 2 + 3.00I 3 = 2.25 π (3) Solving these three equations simultaneously, I1 = 0.0623 A , *31.53 I 2 = 0.860 A , and I 3 = 0.923 A The emf induced between the ends of the moving bar is ε = Blv = (2.50 T)(0.350 m)(8.00 m s) = 7.00 V The left-hand loop contains decreasing flux away from you, so the induced current in it will be clockwise, to produce its own field directed away from you. Let I1 represent the current flowing upward through the 2.00-Ω resistor. The right-hand loop will carry counterclockwise current. Let I 3 be the upward current in the 5.00-Ω resistor. Chapter 31 Solutions 235 (a) (b) Kirchhoff’s loop rule then gives: +7.00 V − I1 ( 2.00 Ω) = 0 I1 = 3.50 A and +7.00 V − I 3 ( 5.00 Ω) = 0 I 3 = 1.40 A The total power dissipated in the resistors of the circuit is P = EI1 + EI 3 = E( I1 + I 3 ) = (7.00 V )( 3.50 A + 1.40 A ) = 34.3 W (c) Method 1: The current in the sliding conductor is downward with value I 2 = 3.50 A + 1.40 A = 4.90 A . The magnetic field exerts a force of Fm = IlB = ( 4.90 A )(0.350 m )( 2.50 T ) = 4.29 N directed toward the right on this conductor. A n outside agent must then exert a force of 4.29 N to the left to keep the bar moving. Method 2: The agent moving the bar P = F ⋅ v = Fv cos 0˚ . The force required is then: F= *31.54 must supply the power according to P 34.3 W = = 4.29 N v 8.00 m s Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose we use a bar magnet to produce field 10 −3 T through the coil in one direction along its axis. Suppose we then flip the magnet to reverse the flux in 10 −1 s. The average induced emf is then ε = − N ∆ΦB = − N ∆[BA cos θ ] = −NB(π r 2 ) cos180˚ − cos 0˚  ∆t ∆t ε = − (20)(10−3 T)π (0.0150 m)2  31.55 I= ε + ε Induced R and ∆t −2  10 −1 s  ~10 − 4 V d ε Induced = – dt (BA) dv F = m dt = IBd dv Bd IBd Bd ) = dt = m = m R (ε + ε Induced m R (ε – Bvd) To solve the differential equation, let u = (ε – Bvd), 1 du Bd – Bd = u dt mR ∫u so Integrating from t = 0 to t = t, e −B u ln 0 du dv = –Bd . dt dt t (Bd)2 du = −∫ dt t=0 mR u u u0 = – (Bd)2 mR t 2 2 d t/mR Since v = 0 when t = 0, u0 = ε and u = ε – Bvd © 2000 by Harcourt, Inc. All rights reserved. or u u0 = 236 Chapter 31 Solutions ε – Bvd = ε e −B d t/mR 2 2 and v= ε Bd (1 − e −B 2 2 d t/mR ) Chapter 31 Solutions 237 31.56 (a) For maximum induced emf, with positive charge at the top of the antenna, F+ = q+ (v × B), so the auto must move east (b) 31.57  65.0 × 103 m  cos 65.0° = 4.58 × 10– 4 V = (5.00 × 10–5 T)(1.20 m)   3600 s  ε = Blv I= ε = R B ∆A R ∆t q = I ∆t = so (15.0 µ T)(0.200 m)2 = 1.20 µC 0.500 Ω Goal Solution The plane of a square loop of wire with edge length a = 0.200 m is perpendicular to the Earth's magnetic field at a point where B = 15.0 µ T , as shown in Figure P31.57. The total resistance of the loop and the wires connecting it to the galvanometer is 0.500 Ω. If the loop is suddenly collapsed by horizontal forces as shown, what total charge passes through the galvanometer? G: For the situation described, the maximum current is probably less than 1 mA. So if the loop is closed in 0.1 s, then the total charge would be Q = I∆t = (1 mA )(0.1 s) = 100 µ C O: We do not know how quickly the loop is collapsed, but we can find the total charge by integrating the change in magnetic flux due to the change in area of the loop ( a2 → 0). A: Q = ∫ Idt = ∫ ε dt R = A2 =0 B  Q = − A R A 1 =a = 2 1 1 B 1  d ΦB  − d ΦB = − d(BA) = −  dt = −  dt  R R R R ∫ ∫ ∫ ∫ A2 =0 A1 =a 2 dA Ba2 (15.0 × 10 −6 T)(0.200 m)2 = = 1.20 × 10 −6 C R 0.500 Ω L : The total charge is less than the maximum charge we predicted, so the answer seems reasonable. It is interesting that this charge can be calculated without knowing either the current or the time to collapse the loop. Note: We ignored the internal resistance of the galvanometer. D’Arsonval galvanometers typically have an internal resistance of 50 to 100 Ω, significantly more than the resistance of the wires given in the problem. A proper solution that includes RG would reduce the total charge by about 2 orders of magnitude ( Q ~ 0.01 µ C). © 2000 by Harcourt, Inc. All rights reserved. 238 Chapter 31 Solutions *31.58 (a) I= dq ε = dt R E = −N where dΦ B dt so and the charge through the circuit will be (b) 31.59 Q= N  π   BAN = BA cos 0 − BA cos  2   R R  so RQ B = NA = (200 Ω)(5.00 × 10– 4 C) (100)(40.0 × 10– 4 m 2 ) ε ∫ dq = Q = N R Φ2 ∫ dΦB Φ1 N (Φ2 – Φ1) R = 0.250 T (a) ε = B lv = 0.360 V (b) FB = IlB = 0.108 N (c) Since the magnetic flux B · A is in effect decreasing, the induced current flow through R is from b to a. Point b is at higher potential. (d) I= R = 0.900 A N o . Magnetic flux will increase through a loop to the left of ab. Here counterclockwise current will flow to produce upward magnetic field. The in R is still from b to a. 31.60 ε = Blv at a distance r from wire ε 31.61 (a) µ I =  0  lv  2π r  At time t , the flux through the loop is At t = 0, Φ B = π ar 2 (b) ε = − dΦB = − π r 2 d( a + bt) = (c) I= (d)  π br 2  π 2b 2 r 4 2 P = ε I = − π br = − R  R  dt dt ε R = − − π br 2 π br 2 R ( ) ( ) Φ B = BA cos θ = ( a + bt) π r 2 cos 0˚ = π ( a + bt)r 2 Chapter 31 Solutions 239 ε = − d (NBA) = −1 dB  π a2 = π a2 K 31.62 dt (a) (b) dt Q = C ε = C π a2K B into the paper is decreasing; therefore, current will attempt to counteract this. Positive charge will go to upper plate . (c) The changing magnetic field through the enclosed area induces an electric field , surrounding the B-field, and this pushes on charges in the wire. The flux through the coil is Φ B = B ⋅ A = BA cos θ = BA cos ω t . The induced emf is 31.63 d cos ω t ) ε = − N dΦB = − NBA ( = NBAω sin ω t. dt dt (a) ε max = NBAω = 60.0(1.00 T)(0.100 × 0.200 m 2 )(30.0 (b) dΦ B ε = , dt N (c) thus dΦ B dt = max ε max = 36.0 V = 0.600 V = N At t = 0.0500 s, ω t = 1.50 rad and (d) The torque on the coil at any time is When 31.64 (a) We use ε = ε max , sin ω t = 1.00 and ε =– N rad s) = 36.0 V 60.0 0.600 Wb/s ε = ε max sin(1.50 rad) = (36.0 V) sin(1.50 rad) = 35.9 V ε ε τ = µ × B = NIA × B = ( NAB) I sin ω t =  max    sin ω t  ω  R τ= 2 ε max = ωR (36.0 V )2 = (30.0 rad s)(10.0 Ω) 4.32 N · m ∆Φ B , with N = 1. ∆t Taking a = 5.00 × 10- 3 m to be the radius of the washer, and h = 0.500 m,  µ0I 1 µ I  a2µ 0 I  1 − µ 0 ahI = − ∆Φ B = B2 A − B1A = A(B2 − B1 ) = π a2  − 0 =   h + a a 2 2(h + a)  2 π (h + a) 2 π a  ∆t = The time for the washer to drop a distance h (from rest) is: (b) Therefore, ε= µ 0 ahI µ 0 ahI = 2(h + a)∆t 2(h + a) g µ 0 aI = 2h 2(h + a) and ε= ( 4π × 10 −7 T ⋅ m/A)( 5.00 × 10 −3 m)(10.0 A) 2(0.500 m + 0.00500 m) 2h g gh 2 (9.80 m/s 2 )(0.500 m) = 97.4 nV 2 Since the magnetic flux going through the washer (into the plane of the paper) is decreasing in time, a current will form in the washer so as to oppose that decrease. Therefore, the current will flow in a clockwise direction . © 2000 by Harcourt, Inc. All rights reserved. 240 Chapter 31 Solutions 31.65 ε = −N dΦB = −N d (BA cos θ ) dt dt  ∆t 31.66  Find an expression for the flux through a rectangular area "swept out" by the bar in time t. The magnetic field at a distance x from wire is B= µ0I and ΦB = ∫ BdA. 2π x ΦB = µ 0 Ivt 2π Then, 31.67 −4 m2   = –10.2 µV 1.80 s  ε = −NBcos θ  ∆A  = − 200(50.0 × 10− 6 T)(cos 62.0˚ ) 39.0 × 10 r+l ∫ dx x = µ 0 Iv  l dΦ B = ln 1 +  2π dt r r ε Therefore, where vt is the distance the bar has moved in time t. The magnetic field at a distance x from a long wire is B = µ0I . 2π x Find an expression for the flux through the loop. dΦ B = µ0I (ldx) 2π x Therefore, 31.68 ΦB = µ 0 Il 2π r+w ∫ r dx µ 0 Il  w = ln 1 +  2π x r ε = − dΦB = µ 0 Ilv dt w 2 π r (r + w) and I= ε R = µ 0 Ilv w 2 π Rr (r + w) As the wire falls through the magnetic field, a motional emf ε = Blv is induced in it. Thus, a counterclockwise induced current of I = ε R = Blv R flows in the circuit. The falling wire is carrying a current toward the left through the magnetic field. Therefore, it experiences an upward magnetic force given by FB = IlB = B2l2 v R. The wire will have attained terminal speed when the magnitude of this magnetic force equals the weight of the wire. Thus, 31.69 so mgR B2l2 vt = mg , or the terminal speed is vt = R B2l2 ΦB = (6.00t 3 – 18.0t 2) T · m2 Maximum and ε occurs when dε dt ε=– dΦB = –18.0t 2 + 36.0t dt = – 36.0t + 36.0 = 0, which gives t = 1.00 s. Therefore, the maximum current (at t = 1.00 s) is I= ε R = (–18.0 + 36.0)V = 6.00 A 3.00 Ω Chapter 31 Solutions 241 ΣF = Mg – T = Ma For the suspended mass, M: 31.70 ΣF = T – I lB = ma, where I = For the sliding bar, m: Mg − v ∫0 B2l2 v = (m + M)a R t dv = ∫ dt 0 (α − β v) ε = –N R = Blv R or a= Mg dv B2l2 v = − dt m + M R(M + m) where α= Mg M+m v= Therefore, the velocity varies with time as *31.71 (a) ε β= and B2l2 R(M + m) 2 2 MgR  α (1 − e −β t ) = 1 − e −B l t/R(M+m)  2 2   β Bl  dΦ B dB d = –NA = –NA (µ nI) dt dt dt 0 where A = area of coil, N = number of turns in coil, and n = number of turns per unit length in solenoid. Therefore, (b) d 4 sin(120π t)] = Nµ 0An(480π ) cos (120π t) dt [ ε = Nµ 0 An ε = 40(4π × 10–7 ) π (0.0500 m)2 (2.00 × 103)(480π) cos(120π t) = (1.19 V) cos(120π t) I= ∆V R [ ] and P = ∆VI = (1.19 V)2 cos2(120π t) (8.00 Ω) 1 1 1 From cos2 θ = 2 + 2 cos 2θ, the average value of cos2θ is 2 , so 31.72 The induced emf is ε = Blv where ( B= P= ( 1 (1.19 V ) = 88.5 mW 2 (8.00 Ω) 2 ) µ0I , v = vi + gt = 9.80 m s 2 t , and 2π y ) y = yi − 21 gt 2 = 0.800 m − 4.90 m s 2 t 2 . ε= (4π × 10 −7 [ ) T ⋅ m A ( 200 A ) ( ) ] 2 π 0.800 m − 4.90 m s 2 t 2 At t = 0.300 s , ε= ( ) (0.300 m) 9.80 m s 2 t = (1.18 × 10 )(0.300) (1.18 × 10 )t −4 [0.800 − 4.90t ] −4 [ 0.800 − 4.90(0.300) 2 ] V = 98.3 µV © 2000 by Harcourt, Inc. All rights reserved. 2 V 242 Chapter 31 Solutions 31.73 The magnetic field produced by the current in the straight wire is perpendicular to the plane of the coil at all points within the coil. The magnitude of the field is B = µ 0 I 2 π r . Thus, the flux linkage is NΦ B = µ 0 NIL h+w dr µ 0 NI max L  h + w  sin(ω t + φ ) ln =  h  2π ∫ h 2π r Finally, the induced emf is ε = −( ε= ) ε = − µ 0 NImax Lω ln 1 + w  cos(ω t + φ ) 2π 4 π × 10 −7 (100)(50.0)(0.200 m)(200 π s −1 ) 2π h 5.00 cm   cos(ω t + φ ) ln 1 +  5.00 cm  − (87.1 mV ) cos(200 π t + φ ) The term sin(ω t + φ )in the expression for the current in the straight wire does not change appreciably when ω t changes by 0.100 rad or less. Thus, the current does not change appreciably during a time interval t< 0.100 = 1.60 × 10 − 4 s . (200 π s −1 ) ct = (3.00 × 108 m / s)(1.60 × 10 −4 s) = 4.80 × 10 4 m equal to the We define a critical length, distance to which field changes could be propagated during an interval of 1.60 × 10 −4 s. This length is so much larger than any dimension of the coilor its distance from the wire that, although we consider the straight wire to be infinitely long, we can also safely ignore the field propagation effects in the vicinity of the coil. Moreover, the phase angle can be considered to be constant along the wire in the vicinity of the coil. If the frequency ω were much larger, say, 200 π × 10 5 s −1 , the corresponding critical length would be only 48.0 cm. In this situation propagation effects would be important and the above expression for ε would require modification. As a "rule of thumb" we can consider field propagation effects for circuits of laboratory size to be negligible for frequencies, f = ω 2 π , that are less than about 106 Hz. 31.74 ΦB = BA cos θ dΦB = –ω BA sinθ ; dt I ∝ – sin θ τ ∝ IB sin θ ∝ – sin2 θ 31.75 The area of the tent that is effective in intercepting magnetic field lines is the area perpendicular to the direction of the magnetic field. This is the same as the base of the tent. In the initial configuration, this is A1 = L(2L cos θ ) = 2(1.50 m)2 cos 60.0˚ = 2.25 m 2 After the tent is flattened, A2 = L(2L) = 2L2 = 2(1.50 m)2 = 4.50 m 2 The average induced emf is: ε = − ∆ΦB = − B( ∆A) = − ( ∆t ∆t 0.300 T )( 4.50 − 2.25) m 2 = – 6.75 V 0.100 s Chapter 32 Solutions ∆I  1.50 A − 0.200 A  = 1.95 × 10–2 V = 19.5 mV = (3.00 × 10– 3 H)   ∆t 0.200 s *32.1 ε 32.2 Treating the telephone cord as a solenoid, we have: = L µ0N 2A (4 π × 10 − 7 T ⋅ m / A)(70.0)2 ( π )(6.50 × 10 − 3 m)2 = = 1.36 µH l 0.600 m L=  0.500 A   ∆I  = (2.00 H)   = 100 V  ∆t   0.0100 s  32.3 ε 32.4 L = µ 0n2 Al so 32.5 L= 32.6 ε 32.7 =+L L = 7.80 × 103 turns/m µ 0 Al n= N ΦB LI → ΦB = = 240 nT · m2 I N =L dI dt Thus, ε where L = (through each turn) µ0N 2A l ( ) ( ) 4 π × 10 −7 T ⋅ m A ( 300) π × 10 − 4 m 2  µ N 2 A  dI = 0 = (10.0 A s) = 2.37 mV  0.150 m  l  dt dI ε back = –ε = L dt =L 2 d (I sin ω t) = Lω Imax cos ω t = (10.0 × 10-3)(120π )(5.00) cos ω t dt max ε back = (6.00π ) cos (120π t) = (18.8 V) cos (377t) © 2000 by Harcourt, Inc. All rights reserved. Chapter 32 Solutions 243 From *32.8 ε From L = L= 32.9  ∆I  , we have  ∆t  dI NΦ B I , we have → Thus, ε 32.11 ε  ∆Ι   ∆t  ΦB = = 24.0 × 10– 3 V = 2.40 × 10– 3 H 10.0 A/s LI (2.40 × 10– 3 H)(4.00 A) = = 19.2 µT · m2 N 500 –6 dI − ε –175 × 10 V = = = – 0.421 A/s dt 4.16 × 10– 4 H L The induced emf is 32.10 L= µ0(420)2(3.00 × 10– 4) µ0 N 2 A = = 4.16 × 10– 4 H 0.160 l ε = –L dt 32.12 =L =L dI ε = − = dt L ε = − L dI , dt − where the self-inductance of a solenoid is given by L = εl µ0N 2A d dI = (90.0 × 10-3) dt (t 2 – 6t) V dt (a) At t = 1.00 s, ε= 360 mV (b) At t = 4.00 s, ε= 180 mV (c) ε = (90.0 × 10-3)(2t – 6) = 0 when (a) 450  B = µ 0nI = µ 0  (0.0400 mA) = 188 µT  0.120 (b) ΦB = BA = 3.33 × 10-8 T · m2 (c) L= t = 3.00 s NΦ B = 0.375 mH I © 2000 by Harcourt, Inc. All rights reserved. µ0N 2A . l 244 Chapter 32 Solutions (d) B and Φ B are proportional to current; L is independent of current Chapter 32 Solutions 245 32.13 µ0(120)2 π ( 5.00 × 10–3)2 µ0 N 2 A = = 15.8 µH l 0.0900 (a) L= (b) Φ ′B = µm N 2 A µm ΦB → L = = 800(1.58 × 10– 5 H) = 12.6 mH l µ0 NBA NA µ 0NI µ 0N 2A NΦ B = ≈ · = I I I 2π R 2π R 32.14 L= 32.15 ε = ε 0e − k t = − L dI dt dI = − ε 0 e − k t dt L If we require I → 0 as t → ∞, the solution is Q = ∫ I dt = ∫ ∞ 0 32.16 I= ε 0 e − k t dt = − ε 0 ε 0 e − k t = dq kL Q = k 2L kL I= dt ε0 k 2L ε (1 − e −Rt/L ) R 0.900 ε R = ε R [1 − e −R(3.00 s)/2.50 H ]  R(3.00 s)  = 0.100 exp −  2.50 H  R= 2.50 H ln 10.0 = 1.92 Ω 3.00 s © 2000 by Harcourt, Inc. All rights reserved. 246 Chapter 32 Solutions L I τ = R = 0.200 s: I = 1 – e–t/τ max 32.17 (a) 0.500 = 1 – e–t/0.200 → t = τ ln 2.00 = 0.139 s (b) 0.900 = 1 – e–t/0.200 → t = τ ln 10.0 = 0.461 s Figure for Goal Solution Goal Solution A 12.0-V battery is about to be connected to a series circuit containing a 10.0-Ω resistor and a 2.00-H inductor. How long will it take the current to reach (a) 50.0% and (b) 90.0% of its final value? G: The time constant for this circuit is τ = L R = 0.2 s, which means that in 0.2 s, the current will reach 1/e = 63% of its final value, as shown in the graph to the right. We can see from this graph that the time to reach 50% of I max should be slightly less than the time constant, perhaps about 0.15 s, and the time to reach 0.9I max should be about 2.5τ = 0.5 s. O: The exact times can be found from the equation that describes the rising current in the above graph and gives the current as a function of time for a known emf, resistance, and time constant. We set time t = 0 to be the moment the circuit is first connected. A: At time t , I (t ) = where, after a long time, I max = ε (1 − e −∞ ) = ε At I (t ) = 0.500I max , (0.500) ε = ε (1 − e Isolating the constants on the right, ln e −t/2.00 s = ln(0.500) and solving for t , − (b) Similarly, to reach 90% of I max , Thus, L: ( ε (1 − e −t/τ ) R R R R −t/0.200 s R ) so 0.500 = 1 − e −t/0.200 s or t = 0.139 s ) t = −0.693 0.200 s 0.900 = 1 − e −t/τ and t = − τ ln(1 − 0.900) t = −(0.200 s) ln(0.100) = 0.461 s The calculated times agree reasonably well with our predictions. We must be careful to avoid confusing the equation for the rising current with the similar equation for the falling current. Checking our answers against predictions is a safe way to prevent such mistakes. Chapter 32 Solutions 247 I = I 0 e −t/τ : Taking τ = L R, 32.18 IR + L dI  1 = I 0 e −t/τ −  τ dt ( ) dI  1 =0 = 0 will be true if I 0 R e −t/τ + L I 0 e −t/τ −  τ dt Because τ = L R, we have agreement with 0 = 0 *32.19 (a) τ = L R = 2.00 × 10– 3 s = 2.00 ms ( ) (b) I = I max 1 − e −t/τ = (c) Imax = ε R = ( )  6.00 V  1 − e −0.250/2.00 = 0.176 A  4.00 Ω  6.00 V = 1.50 A 4.00 Ω (d) 0.800 = 1 – e–t/2.00 ms → t = – (2.00 ms) ln(0.200) = 3.22 ms I= *32.20 ε R (1 – e–t/τ ) = 120 (1 – e–1.80/7.00) = 3.02 A 9.00 ∆VR = IR = (3.02)(9.00) = 27.2 V ∆VL = ε – ∆VR = 120 – 27.2 = 92.8 V 32.21 (a) ∆V R = IR = (8.00 Ω)(2.00 A) = 16.0 V and ∆VL = ε − ∆V R = 36.0 V − 16.0 V = 20.0 V Therefore, (b) ∆V R 16.0 V = = 0.800 ∆VL 20.0 V ∆V R = IR = (4.50 A)(8.00 Ω) = 36.0 V ∆VL = ε − ∆V R = 0 © 2000 by Harcourt, Inc. All rights reserved. Figure for Goal Solution 248 Chapter 32 Solutions Goal Solution For the RL circuit shown in Figure P32.19, let L = 3.00 H, R = 8.00 Ω, and ε = 36.0 V. (a) Calculate the ratio of the potential difference across the resistor to that across the inductor when I = 2.00 A. (b) Calculate the voltage across the inductor when I = 4.50 A. G: The voltage across the resistor is proportional to the current, ∆ VR = IR , while the voltage across the inductor is proportional to the rate of change in the current, ε L = −L dI dt . When the switch is first closed, the voltage across the inductor will be large as it opposes the sudden change in current. As the current approaches its steady state value, the voltage across the resistor increases and the inductor’s emf decreases. The maximum current will be ε /R = 4.50 A, so when I = 2.00 A, the resistor and inductor will share similar voltages at this mid-range current, but when I = 4.50 A, the entire circuit voltage will be across the resistor, and the voltage across the inductor will be zero. O: We can use the definition of resistance to calculate the voltage across the resistor for each current. We will find the voltage across the inductor by using Kirchhoff's loop rule. A: When I = 2.00 A, the voltage across the resistor is (a) ∆V R = IR = ( 2.00 A )(8.00 Ω) = 16.0 V Kirchhoff's loop rule tells us that the sum of the changes in potential around the loop must be zero: ε − ∆VR − ε L = 36.0 V − 16.0 V − ε L = 0 ε L = 20.0 V so and ∆V R εL = 16.0 V = 0.800 20.0 V Similarly, for I = 4.50 A , ∆V R = IR = ( 4.50 A )(8.00 Ω) = 36.0 V (b) ε − ∆VR − ε L = 36.0 V − 36.0 V − ε L = 0 εL = 0 so L : We see that when I = 2.00 A, ∆V R < ε L , but they are similar in magnitude as expected. Also as predicted, the voltage across the inductor goes to zero when the current reaches its maximum value. A worthwhile exercise would be to consider the ratio of these voltages for several different times after the switch is reopened. After a long time, 12.0 V = (0.200 A)R *32.22 Thus, R = 60.0 Ω. Now, τ = L gives R L = τ R = (5.00 × 10– 4 s)(60.0 V/A) = 30.0 mH ( ) I = I max 1 − e −t/τ : 32.23 L 15.0 H τ= R = = 0.500 s : 30.0 Ω ( ) dI  1 = −I max e −t/τ −  τ dt dI R = I e–t/τ dt L max and Imax = ε R dI R ε = 100 V = 6.67 A/s = I e0 = 15.0 H dt L max L (a) t = 0: (b) t = 1.50 s: dI ε e–t/τ = (6.67 A/s)e– 1.50/(0.500) = (6.67 A/s)e–3.00 = 0.332 A/s = dt L Chapter 32 Solutions 249 ( I = I max 1 − e −t/τ 32.24 ) 0.980 = 1 − e − 3.00 ×10 0.0200 = e − 3.00 ×10 τ=− −3 −3 /τ /τ 3.00 × 10 − 3 = 7.67 × 10 − 4 s ln(0.0200) τ = L R, so L = τ R = (7.67 × 10 − 4 )(10.0) = 7.67 mH Name the currents as shown. 32.25 By Kirchhoff’s laws: I1 = I 2 + I 3 (1) +10.0 V − 4.00 I1 − 4.00 I 2 = 0 (2) +10.0 V − 4.00 I1 − 8.00 I 3 − (1.00) dI 3 = 0 (3) dt From (1) and (2), +10.0 − 4.00 I1 − 4.00 I1 + 4.00 I 3 = 0 Then (3) becomes 10.0 V − 4.00(0.500 I 3 + 1.25 A ) − 8.00 I 3 − (1.00) (1.00 H)(dI 3 I1 = 0.500 I 3 + 1.25 A and dI 3 =0 dt dt ) + (10.0 Ω) I 3 = 5.00 V We solve the differential equation using Equations 32.6 and 32.7: I 3 (t ) = [ 5.00 V − 10.0 Ω ) t 1− e ( 10.0 Ω 1.00 H ] = (0.500 A)[1 − e ] −10t/s I1 = 1.25 + 0.500 I 3 = 1.50 A − (0.250 A )e −10t/s 32.26 L , we get R = R L = C 3.00 H = 1.00 × 10 3 Ω = 1.00 kΩ 3.00 × 10 − 6 F (a) Using τ = RC = (b) τ = RC = 1.00 × 10 3 Ω 3.00 × 10 − 6 F = 3.00 × 10 −3 s = 3.00 ms ( )( ) © 2000 by Harcourt, Inc. All rights reserved. 250 Chapter 32 Solutions For t ≤ 0, the current in the inductor is zero. At t = 0, it starts to grow from zero toward 10.0 A with time constant τ = L R = (10.0 mH) (100 Ω) = 1.00 × 10 − 4 s . 32.27 −t/τ ( For 0 ≤ t ≤ 200 µ s, I = I max  1 − e   = 10.00 A 1 − e −10 000t/s ( )  At t = 200 µ s, I = (10.00 A ) 1 − e −2.00 = 8.65 A ( ) Thereafter, it decays exponentially as I = I 0 e − t ′ τ , I = (8.65 A )e 32.28 ε −10 000( t−200 µ s ) s 12.0 V (a) I= (b) Initial current is 1.00 A, : R = 12.0 Ω ) so for t ≥ 200 µ s, ( ) = (8.65 A )e −10 000t s + 2.00 = 8.65 e 2.00 A e −10 000t s = (63.9 A)e −10 000t s = 1.00 A ∆V12 = (1.00 A)(12.00 Ω) = 12.0 V ∆V 1200 = (1.00 A)(1200 Ω) = 1.20 kV ∆V L = 1.21 kV (c) I = Imax e–Rt/L: dI R = – Imax e–Rt/L dt L and –L Solving 12.0 V = (1212 V)e–1212t/2.00 so 9.90 × 10– 3 = e– 606t t = 7.62 ms Thus, τ= 32.29 (a) dI = ∆VL = Imax Re–Rt/L dt L 0.140 = = 28.6 ms; R 4.90 ( I = I max 1 − e −t/τ e −t/τ = 0.820 ) so I max = ε R = 6.00 V = 1.22 A 4.90 Ω ( 0.220 = 1.22 1 − e −t/τ ) t = − τ ln(0.820) = 5.66 ms (b) 10.0   − I = I max  1 − e 0.0286  = (1.22 A) 1 − e −350 = 1.22 A   (c) I = I max e −t/τ ( and ) 0.160 = 1.22e −t/τ so t = –τ ln(0.131) = 58.1 ms Chapter 32 Solutions 251 32.30 (a) (b) For a series connection, both inductors carry equal currents at every instant, so dI/dt is the same for both. The voltage across the pair is L eq dI dI dI = L1 + L 2 dt dt dt so L eq dI dI dI = L 1 1 = L 2 2 = ∆VL dt dt dt where I = I1 + I 2 Thus, (c) L eq ∆VL ∆VL ∆VL = + L eq L1 L2 L eq = L 1 + L 2 and dI dI1 dI 2 = + dt dt dt 1 1 1 = + L eq L 1 L 2 and dI dI dI + R eq I = L 1 + IR 1 + L 2 + IR 2 dt dt dt Now I and dI/dt are separate quantities under our control, so functional equality requires both L eq = L 1 + L 2 (d) ∆V = L eq and R eq = R 1 + R 2 dI dI dI + R eq I = L 1 1 + R 1I1 = L 2 2 + R 2 I 2 dt dt dt where I = I1 + I 2 and dI dI1 dI 2 = + dt dt dt We may choose to keep the currents constant in time. Then, 1 1 1 = + R eq R 1 R 2 We may choose to make the current swing through 0. 1 1 1 = + L eq L 1 L 2 Then, This equivalent coil with resistance will be equivalent to the pair of real inductors for all other currents as well. L= 32.31 32.32 (a) = 42.3 mH 1 1 U = 2 LI 2 = 2 (0.423 H)(1.75 A) 2 = 0.0648 J so The magnetic energy density is given by u= (b) 200(3.70 × 10– 4) N ΦB = I 1.75 B2 (4.50 T)2 = = 8.06 × 106 J/m3 2µ 0 2(1.26 × 10– 6 T · m/A) The magnetic energy stored in the field equals u times the volume of the solenoid (the volume in which B is non-zero). [ ] U = uV = (8.06 × 106 J/m3) (0.260 m)π (0.0310 m)2 = 6.32 kJ © 2000 by Harcourt, Inc. All rights reserved. 252 Chapter 32 Solutions L = µ0 32.33 N 2A (68.0)2 π (0.600 × 10 −2 )2 = µ0 = 8.21 µ H 0.0800 l U = 21 LI 2 = 21 (8.21 × 10 −6 H)(0.770 A)2 = 2.44 µJ 32.34 (a) 2 2 (0.800)(500)2 1 1 Lε   = = = 27.8 J U = 2 LI 2 = 2 L ε  2R  8(30.0)2 8R 2 (b) I= ε 1 − e −(R/L)t  R [ ] R t = ln 2 L ε so 2R u = ε0 *32.36 (a) U = 2 LI 2 = 2 (4.00 H)(0.500 A) 2 = 0.500 J dU = LI = (4.00 H)(1.00 A) = 4.00 J/s = 4.00 W dt (c) P = (∆V)I = (22.0 V)(0.500 A) = 11.0 W (a) ] L 0.800 ln 2 = ln 2 = 18.5 ms R 30.0 1 (b) 32.37 [ B2 u = 2µ = 995 µ J/m3 0 32.35 1 1 ε 1 − e −(R/L)t → e −(R/L)t =  R 2 t= so E2 3 2 = 44.2 nJ/m = From Equation 32.7, I= ε (1 − e − R t The maximum current, after a long time t , is I= ε At that time, the inductor is fully energized and P = I(∆V) = (2.00 A)(10.0 V) = 20.0 W (b) P lost = I 2 R = (2.00 A)2 (5.00 Ω) = 20.0 W (c) P inductor = I(∆Vdrop ) = 0 (d) U= LI 2 (10.0 H)(2.00 A)2 = = 20.0 J 2 2 R R L ) = 2.00 A. Chapter 32 Solutions 253 32.38 We have u = e0 Therefore e0 B = E e0µ 0 = 32.39 E2 2 B2 E2 = 2 2µ 0 B2 2µ 0 and u= so B2 = e0µ 0E 2 6.80 × 10 5 V / m = 2.27 × 10– 3 T 3.00 × 108 m / s The total magnetic energy is the volume integral of the energy density, u = Because B changes with position, u is not constant. For B = B0 ( R / r ) , 2 B2 2µ 0  B 2  R 4 u= 0   2µ 0   r  Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thickness dr. Such a shell has a volume 4π r 2 dr, so the energy stored in it is  2 πB0 2 R 4  dr dU = u 4 π r 2 dr =  µ 0  r 2  ( ) We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the sphere. This gives 2 U= 32.40 2π B 0 R 3 µ0 = 2π (5.00 × 10–5 T)2(6.00 × 106 m)3 = 2.70 × 1018 J (1.26 × 10– 6 T · m/A) I1(t) = I max e − α t sin ω t with I max = 5.00 A, α = 0.0250 s −1 , and ω = 377 rad s . dI1 = I max e − α t ( −α sin ω t + ω cos ω t ) dt [ dI1 = 1.85 × 10 3 A s dt Thus, ] dI1 = ( 5.00 A s)e − 0.0200 − (0.0250) sin(0.800( 377 )) + 377 cos(0.800( 377 )) dt At t = 0.800 s , ε 2 = −M dI1 : dt M= − ε2 dI1 dt = + 3.20 V = 1.73 mH 1.85 × 10 3 A s © 2000 by Harcourt, Inc. All rights reserved. 254 Chapter 32 Solutions ε 2 = −M dI1 = −(1.00 × 10− 4 H)(1.00 × 10 4 A / s) cos(1000t) 32.41 dt (ε 2 )max = M= 32.42 32.43 1.00 V ε2 dI1 dt = 96.0 mV = 80.0 mH 1.20 A / s (a) M= N BΦ BA 700(90.0 × 10 − 6 ) = = 18.0 mH 3.50 IA (b) LA = Φ A 400(300 × 10 − 6 ) = = 34.3 mH 3.50 IA (c) ε B = −M dI A = −(18.0 mH)(0.500 A / s) = dt M= 32.44 [ ( 70.0   π ( 5.00 × 10 ) 0.0500 m   B at center of (larger) loop: B1 = (a) M= (b) M= ] N 2 Φ12 N 2 (B1A1 ) N 2 (µ 0n1I1 )A 1 = = = N 2 µ 0n1A 1 I1 I1 I1 M = (1.00) 4 π × 10 −7 T ⋅ m A 32.45 – 9.00 mV µ 0 I1 2R µ0 π r 2 Φ 2 B1A 2 (µ 0 I1 / 2R)( π r 2 ) = = = 2R I1 I1 I1 µ 0 π (0.0200)2 2(0.200) = 3.95 nH −3 ) 2 m  = 138 nH  Chapter 32 Solutions 255 *32.46 Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at distance x from the wire is B = µ 0 I 2 π x, and this field passes perpendicularly through the plane of the loop. The flux through the loop is Φ B = ∫ B ⋅ dA = ∫ B dA = ∫ B (ldx ) = µ 0 I l 1.70 mm dx µ Il  1.70  = 0 ln  0.400  2 π ∫0.400 mm x 2π The mutual inductance between the wire and the loop is then M= 1(4 π × 10 −7 T ⋅ m A)(2.70 × 10 −3 m) N 2 Φ12 N 2 µ 0 I l  1.70  N 2 µ 0 l = = ln 1.45) = ( (1.45)  0.400  2π I1 2π I 2π M = 7.81 × 10 −10 H = 781 pH 32.47 With I = I1 + I 2 , the voltage across the pair is: ∆V = − L 1 dI dI dI1 dI dI − M 2 = − L 2 2 − M 1 = − L eq dt dt dt dt dt dI1 ∆V M dI 2 = + L 1 L 1 dt dt So, − and − L2 (−L1L2 + M 2 ) 32.48 [1] dI 2 ∆V M dI1 = + L2 L 2 dt dt − leads to (− L1L 2 + M 2 ) dI1 = ∆V (L 2 − M) dt Adding [1] to [2], (− L 1L 2 + M 2 ) dI = ∆V (L 1 + L 2 − 2M) dt So, L eq = − I max = (b) dI 2 = ∆V(L1 − M) dt By substitution, At different times, (a) dI 2 M( ∆V ) M 2 dI 2 + + = ∆V dt L 1 dt L1 [2] L 1L 2 − M 2 ∆V = L 1 + L 2 − 2M dI / dt (UC )max = (U L )max so 1.00 × 10 −6 F C ( ∆V )max = ( 40.0 V) = 0.400 A L 10.0 × 10 −3 H © 2000 by Harcourt, Inc. All rights reserved. [ ] 1 C ∆V )2 2 ( max ( = 21 LI 2 ) max 256 Chapter 32 Solutions 32.49 32.50 [ ] 1 C ∆V )2 2 ( max ( = 21 LI 2 ) ( ∆VC )max = so max L I max = C 20.0 × 10 −3 H (0.100 A) = 20.0 V 0.500 × 10 − 6 F When the switch has been closed for a long time, battery, resistor, and coil carry constant current I max = ε / R . When the switch is opened, current in battery and resistor drops to zero, but the coil carries this same current for a moment as oscillations begin in the LC loop. We interpret the problem to mean that the voltage amplitude of 2 2 . these oscillations is ∆V, in 21 C ( ∆V ) = 21 LI max Then, L = 32.51 C = ( ) 2 2 0.500 × 10 − 6 F (150 V ) ( 250 Ω) C ( ∆V ) C ( ∆V ) R 2 = = = 0.281 H 2 ε2 I max ( 50.0 V )2 2 2 1 1 = = 608 pF 2 (2 π f ) L (2 π ⋅ 6.30 × 106 )2 (1.05 × 106 ) Goal Solution A fixed inductance L = 1.05 µ H is used in series with a variable capacitor in the tuning section of a radio. What capacitance tunes the circuit to the signal from a station broadcasting at 6.30 MHz? G: It is difficult to predict a value for the capacitance without doing the calculations, but we might expect a typical value in the µF or pF range. O: We want the resonance frequency of the circuit to match the broadcasting frequency, and for a simple RLC circuit, the resonance frequency only depends on the magnitudes of the inductance and capacitance. A : The resonance frequency is f 0 = Thus, C= 1 2 π LC 1 1 = = 608 pF 2 (2 π f 0 ) L (2 π )(6.30 × 106 Hz) 2 (1.05 × 10 −6 H) [ ] L : This is indeed a typical capacitance, so our calculation appears reasonable. However, you probably would not hear any familiar music on this broadcast frequency. The frequency range for FM radio broadcasting is 88.0 – 108.0 MHz, and AM radio is 535 – 1605 kHz. The 6.30 MHz frequency falls in the Maritime Mobile SSB Radiotelephone range, so you might hear a ship captain instead of Top 40 tunes! This and other information about the radio frequency spectrum can be found on the National Telecommunications and Information Administration (NTIA) website, which at the time of this printing was at http://www.ntia.doc.gov/osmhome/allochrt.html Chapter 32 Solutions 257 f= 1 : 2 π LC (a) f= 1 1 = = 135 Hz 2 π LC 2 π (0.0820 H)(17.0 × 10 − 6 F) (b) Q = Qmax cos ω t = (180 µ C) cos(847 × 0.00100) = 119 µC (c) I= (a) f= (b) Q = C ε = (1.00 × 10 − 6 F)(12.0 V) = 12.0 µC (c) 1C 2 32.52 32.53 32.54 L= 1 1 = = 0.220 H (2 π f )2 C (2 π ⋅ 120)2 (8.00 × 10 − 6 ) dQ = −ω Qmax sin ω t = −(847)(180) sin(0.847) = – 114 mA dt 1 2 π LC ε2 = = 1 2 π (0.100 H)(1.00 × 10 − 6 F) = 503 Hz 1 LI 2 2 max I max = ε C = 12 V L 1.00 × 10 − 6 F = 37.9 mA 0.100 H (d) At all times U = 21 C ε 2 = 21 (1.00 × 10 − 6 F)(12.0 V)2 = 72.0 µ J ω= 32.55 1 = LC 1 (3.30 H)(840 × 10 Q = Qmax cos ω t, I = −12 F dQ = −ω Qmax sin ω t dt ( [105 × 10 ]cos[(1.899 × 10 −6 (a) Q2 UC = = 2C (b) 2 U L = 21 LI 2 = 21 Lω 2Qmax sin 2 (ω t ) = UL (c) ) = 1.899 × 10 4 rad s (105 × 10 = 4 ( 2 840 × 10 −12 −6 ) 2 [( )( rad s 2.00 × 10 −3 s ) 2 = 6.03 J 2 Qmax sin 2 (ω t ) 2C )( C sin 2 1.899 × 10 4 rad s 2.00 × 10 −3 s ( )] ) 2 840 × 10 −12 F ) )] = 0.529 J U total = UC + U L = 6.56 J © 2000 by Harcourt, Inc. All rights reserved. 258 Chapter 32 Solutions 32.56 (a) ωd = 2 1  R − = LC  2L  fd = Therefore, 32.57 (2.20 × 10 )(1.80 × 10 ) −6 −3  7.60 −  2 2.20 × 10 −3  ( 2 )   = 1.58 × 10 4 rad / s   ωd = 2.51 kHz 2π (b) Rc = 4L = 69.9 Ω C (a) ω0 = 1 1 = = 4.47 krad/s LC (0.500)(0.100 × 10 −6 ) (b) ωd = 1  R − = 4.36 krad/s LC  2L  (c) ∆ω = 2.53% lower ω0 2 Choose to call positive current clockwise in Figure 32.19. It drains charge from the capacitor according to I = – dQ/dt. A clockwise trip around the circuit then gives 32.58 32.59 1 (a) + Q dI − IR − L = 0 C dt + Q dQ d dQ + R+L = 0, identical with Equation 32.29. C dt dt dt Q = Qmax e 0.500 = e t=− (b) − − cos ω dt Rt 2L − Rt 2L so I max ∝ e and Rt = − ln(0.500) 2L so Q = 0.500 Qmax = 0.707Qmax 2L  2L  ln(0.500) = 0.693  R R 2 U 0 ∝ Qmax t=− Rt 2L and U = 0.500U 0 2L  2L  ln(0.707) = 0.347  R R (half as long) Chapter 32 Solutions 259 With Q = Qmax at t = 0, the charge on the capacitor at any time is Q = Qmax cos ω t where ω = 1 LC . The energy stored in the capacitor at time t is then 32.60 U= 2 Q 2 Qmax = cos 2 ω t = U 0 cos 2 ω t. 2C 2C 1 4 cos ω t = When U = U 0 , 1 2 and t π = LC 3 Therefore, π2 t2 = 9 LC or L= The inductance is then: 32.61 (a) (b) d 20.0t ) ε L = − L dI = −(1.00 mH) ( = dt dt 9t 2 π 2C – 20.0 mV Q = ∫ I dt = ∫ ( 20.0t )dt = 10.0t 2 t t 0 0 ∆VC = (c) 1 3 ω t = π rad ( ) −Q −10.0t 2 = = − 10.0 MV s 2 t 2 C 1.00 × 10 −6 F (−10.0t ) ≥ 1 (1.00 × 10 )(20.0t) , 2 2(1.00 × 10 ) 2 2 Q2 1 When ≥ LI 2 , 2 2C or ( −3 −6 ) then 100t 4 ≥ 400 × 10 − 9 t 2 . The earliest time this is true is at 32.62 (a) ε L = − L dI = − L d (b) I= dt dQ , dt dt When t t Q = ∫ I dt = ∫ Kt dt = 21 Kt 2 so 0 ∆VC = Thus 1C 2 ( ∆VC )2 = t = 4.00 × 10 − 9 s = 63.2 µs (Kt) = –LK and (c) 2 1 LI 2 , 2 1C 2 0 −Q Kt2 = − 2C C  K2 t 4  1 2 2  4C 2  = 2 L K t   t = 2 LC © 2000 by Harcourt, Inc. All rights reserved. ( ) 260 Chapter 32 Solutions 2 1  Q 1 Q2 1 = + LI 2 2C  2  2 C 2 32.63 so The flux through each turn of the coil is I= 3Q 2 4CL ΦB = LI = N Q 2N 3L C where N is the number of turns. Equation 30.16: B = 32.64 µ 0 NI 2π r b (a) b µ 0 NI µ NIh dr µ 0 NIh  b  h dr = 0 ln =  a 2π r 2 π ∫a r 2π a Φ B = ∫ B dA = ∫ L= µ 0N 2h  b  NΦ B = ln  a 2π I (b) L= µ 0 (500)2 (0.0100)  12.0  = 91.2 µH ln  10.0  2π (c) Lappx = *32.65 (a) µ 0 N 2  A  µ 0 (500)2  2.00 × 10 − 4 m 2  =   = 90.9 µH 2π  R  2π 0.110   Nµ 0 IR 2 Nµ 0 I = 2 2 3/2 2R 2(R + 0 ) At the center, B= So the coil creates flux through itself Φ B ≈ BA cos θ = When the current it carries changes, ε L = −N d ΦB ≈ − N π Nµ 0 R dI = dt L≈ so Nµ 0 I π π R 2 cos 0° = Nµ 0 IR 2R 2 2 dt −L dI dt π 2 N µ 0R 2 π T · m L ≈ 2 12  4π × 10–7 0.14 m = 2.8 × 10–7 H ~ 100 nH A   (b) 2π r ≈ 3(0.3 m), so r ≈ 0.14 m; (c) 2.8 × 10–7 V · s/A L ≈ = 1.0 × 10– 9 s ~ 1 ns 270 V/A R Chapter 32 Solutions 261 32.66 (a) If unrolled, the wire forms the diagonal of a 0.100 m (10.0 cm) rectangle as shown. The length of this rectangle is L′ = 9.80 m L′ (9.80 m)2 − (0.100 m)2 The mean circumference of each turn is C = 2π r ′ , where r ′ = radius of each turn. The number of turns is then: (9.80 m)2 − (0.100 m)2 N= L′ = C (b) R= −8 ρ l 1.70 × 10 Ω ⋅ m (10.0 m ) = = 0.522 Ω 2 A π 0.322 × 10 −3 m (c) L= 2 µ N 2 A 800µ 0  L′  = π ( r ′ )2 l′ l′  C  2π ( (  24.0 + 0.644  × 10 − 3 m   2 ) ( 24.0 + 0.644 mm is the mean 2 = 127 ) ) 2 2 800 4 π × 10 −7  (9.80 m )2 − (0.100 m )2   24.0 + 0.644   −3 × 10 m L= π      π ( 24.0 + 0.644) × 10 −3 m   2 0.100 m    L = 7.68 × 10 −2 H = 76.8 mH 32.67 From Ampere’s law, the magnetic field at distance r ≤ R is found as:  I  B( 2 π r ) = µ 0 J π r 2 = µ 0  π r 2 , or 2  πR  ( ) ( ) B= µ 0 Ir 2π R2 The magnetic energy per unit length within the wire is then µ0 I 2 R B2 U =∫ 2 π r dr ) = ( 0 2µ l 4π R4 0 0.100 m R ∫0 r 3 dr = µ0 I 2  R4    = 4π R4  4  µ0 I 2 16 π This is independent of the radius of the wire. © 2000 by Harcourt, Inc. All rights reserved. 262 Chapter 32 Solutions The primary circuit (containing the battery and solenoid) is an RL circuit with R = 14.0 Ω , and 32.68 ( ) ( ) 4 π × 10 −7 (12 500) 1.00 × 10 − 4 µ N 2A L= 0 = = 0.280 H 0.0700 l (a) The time for the current to reach 63.2% of the maximum value is the time constant of the circuit: τ= (b) (c) 2 L 0.280 H = = 0.0200 s = 20.0 ms R 14.0 Ω  I f − 0  ∆I  = L   ∆t   ∆t  The solenoid's average back emf is εL where I f = 0.632 I max = 0.632 Thus, ε L = (0.280 H) 2.71 A  = 0.0200 s =L  ∆V   60.0 V  = 0.632 = 2.71 A  R   14.0 Ω  37.9 V The average rate of change of flux through each turn of the overwrapped concentric coil is the same as that through a turn on the solenoid: ( ) ( ) 4 π × 10 −7 T ⋅ m A (12 500 0.0700 m )( 2.71 A ) 1.00 × 10 − 4 m 2 ∆Φ B µ 0n( ∆I )A = = = 3.04 mV 0.0200 s ∆t ∆t (d) The magnitude of the average induced emf in the coil is the average induced current is I= 32.69 εL R = ( εL = N ( ∆Φ B ∆t ) and magnitude of ) N  ∆Φ B  820 3.04 × 10 −3 V = 0.104 A = 104 mA = R  ∆t  24.0 Ω Left-hand loop: E − (I + I 2 )R 1 − I 2 R2 = 0 Outside loop: E − (I + I 2 )R 1 − L Eliminating I 2 gives E ′ − IR′ − L dI =0 dt dI =0 dt E′ (1 − e − R ′t L ) R′ This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7: I (t ) = But R′ = R1R2 / ( R1 + R2 ) and E ′ = R2E / ( R1 + R2 ), so E R 2 / (R 1 + R2 ) E E′ = = R′ R 1R2 / (R 1 + R2 ) R 1 Thus I(t) = E (1 − e − R ′ t L ) R1 Chapter 32 Solutions 263 When switch is closed, steady current I 0 = 1.20 A. W h e n the switch is opened after being closed a long time, the current in the right loop is 32.70 I = I0e 32.71 (a) − R2 t L = I0 I Rt I  = ln 0  I L so e Rt L Therefore, L= R2 t (1.00 Ω)(0.150 s) = 0.0956 H = 95.6 mH = ln( I 0 I ) ln(1.20 A 0.250 A ) and While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives: [ ]( ) + ε 0 − ( 2.00 + 6.00) × 10 3 Ω 9.00 × 10 −3 A = 0 + ε 0 = 72 .0 V with end b at the higher potential (b) (c) After the switch is opened, the current around the outer loop decays as I = I max e − R t L with I max = 9.00 mA , R = 8.00 kΩ , and Thus, when the current has reached a value I = 2.00 mA , the elapsed time is: t=  L   I max   0.400 H   9.00  = ln ln = 7.52 × 10 −5 s = 75.2 µs  R   I   8.00 × 10 3 Ω   2.00  © 2000 by Harcourt, Inc. All rights reserved. L = 0.400 H 264 Chapter 32 Solutions 32.72 (a) The instant after the switch is closed, the situation is as shown i n the circuit diagram of Figure (a). The requested quantities are: ∆VC = 0, - + IR = ε0/R I L = 0, IC = ε 0 R , I R = ε 0 R ∆VL = ε 0 , ∆VL = ε0 IL = 0 Q=0 ∆VC = 0 ∆V R = ε 0 ∆VR = ε0 + - IC = ε0/R ε0 Figure (a) (b) IL = 0 After the switch has been closed a long time, the steady-state conditions shown in Figure (b) will exist. The currents and voltages are: I L = 0, ∆VL = 0, IC = 0, ∆VL = 0 - IR = 0 Q = Cε0 IR = 0 ∆VC = ε 0 , + ∆VR = 0 ∆VC = ε0 + - ∆V R = 0 ε0 Figure (b) 32.73 When the switch is closed, as shown in Figure (a), the current in the inductor is I : 12.0 – 7.50I – 10.0 = 0 → I = 0.267 A When the switch is opened, the initial current in the inductor remains at 0.267 A. IR = ∆V: (0.267 A)R ≤ 80.0 V (a) (b) R ≤ 300 Ω Goal Solution To prevent damage from arcing in an electric motor, a discharge resistor is sometimes placed in parallel with the armature. If the motor is suddenly unplugged while running, this resistor limits the voltage that appears across the armature coils. Consider a 12.0-V dc motor with an armature that has a resistance of 7.50 Ω and an inductance of 450 mH. Assume that the back emf in the armature coils is 10.0 V when the motor is running at normal speed. (The equivalent circuit for the armature is shown in Figure P32.73.) Calculate the maximum resistance R that limits the voltage across the armature to 80.0 V when the motor is unplugged. Chapter 32 Solutions 265 G: We should expect R to be significantly greater than the resistance of the armature coil, for otherwise a large portion of the source current would be diverted through R and much of the total power would be wasted on heating this discharge resistor. O: When the motor is unplugged, the 10-V back emf will still exist for a short while because the motor’s inertia will tend to keep it spinning. Now the circuit is reduced to a simple series loop with an emf, inductor, and two resistors. The current that was flowing through the armature coil must now flow through the discharge resistor, which will create a voltage across R that we wish to limit to 80 V. As time passes, the current will be reduced by the opposing back emf, and as the motor slows down, the back emf will be reduced to zero, and the current will stop. A : The steady-state coil current when the switch is closed is found from applying Kirchhoff's loop rule to the outer loop: + 12.0 V − I (7.50 Ω) − 10.0 V = 0 2.00 V = 0.267 A 7.50 Ω so I= We then require that ∆V R = 80.0 V = (0.267 A )R so R= ∆V R 80.0 V = = 300 Ω 0.267 A I L : As we expected, this discharge resistance is considerably greater than the coil’s resistance. Note that while the motor is running, the discharge resistor turns P = (12 V)2 300 Ω = 0.48 W of power into heat (or wastes 0.48 W). The source delivers power at the rate of about P = IV = [0.267 A + (12 V / 300 Ω)](12 V ) = 3.68 W, so the discharge resistor wastes about 13% of the total power. For a sense of perspective, this 4-W motor could lift a 40-N weight at a rate of 0.1 m/s. 32.74 ( ) ( ) (a) L1 = 2 4 π × 10 −7 T ⋅ m A (1000) 1.00 × 10 − 4 m 2 µ 0 N12 A = = 2.51 × 10 − 4 H = 251 µH l1 0.500 m (b) M= N 2 Φ 2 N 2 Φ1 N 2BA N 2 µ 0 ( N1 l1 ) I1 A µ 0 N1N 2 A = = = = I1 I1 I1 I1 l1 [ (4π × 10 M= (c) −7 dt Q1 = ( T ⋅ m A (1000)(100) 1.00 × 10 − 4 m 2 0.500 m ε 1 = − M dI 2 , Q1 = − ) ] or I1R1 = − M ( dI 2 dt and I1 = ) = 2.51 × 10 − 5 H = 25.1 µH dQ1 M dI 2 =− R1 dt dt ) M I 2i M M tf M dI 2 = − I 2 f − I 2i = − (0 − I 2i ) = R1 R1 ∫0 R1 R1 (2.51 × 10 −5 ) H (1.00 A ) 1000 Ω = 2.51 × 10 −8 C = 25.1 nC © 2000 by Harcourt, Inc. All rights reserved. 266 Chapter 32 Solutions 32.75 2U is non-zero. I2 (a) It has a magnetic field, and it stores energy, so L = (b) Every field line goes through the rectangle between the conductors. (c) Φ = LI so Thus For an RL circuit, 32.76 L= Φ 1 w−a = I I ∫y=a L= w−a  µ0I µ 0 I  2 µ 0 Ix 2µ 0 x 1 w−a x dy + = dy = ln y  2 π y 2 π w −y  I ∫ 2 π y 2π I ∫a ( )  a L= µ 0x  w − a  ln  a  π I(t) = I max e − R t = 10 − 9 L B da R t L : R − t R I(t) = 1 − 10 − 9 = e L ≅ 1 − t L I max Rmax = so (3.14 × 10 − 8 )(10 − 9 ) = 3.97 × 10 − 25 Ω (2.50 yr)(3.16 × 107 s / yr) (If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm 2 , its resistance would be at least 10– 6 Ω). 32.77 1 1 (a) UB = 2 LI 2 = 2 (50.0 H)(50.0 × 10 3 A) (b) Two adjacent turns are parallel wires carrying current in the same direction. Since the loops have such large radius, a one-meter section can be regarded as straight. 2 = 6.25 × 1010 J µ 0I 2π r Then one wire creates a field of B= This causes a force on the next wire of F = IlB sin θ giving F = Il Solving for the force, µ lI 2 µ0I sin 90° = 0 2π r 2π r F = (4π × 10–7 N/A2) (1.00 m)(50.0 × 10 3 A) 2 = 2000 N (2π)(0.250 m) Chapter 32 Solutions 267 P = I ( ∆V ) 32.78 I= P 1.00 × 10 9 W = = 5.00 × 10 3 A ∆V 200 × 10 3 V B( 2 π r ) = µ 0 I enclosed From Ampere’s law, (a) (b) At r = a = 0.0200 m, I enclosed = 5.00 × 10 3 A (4π × 10 B= )( T ⋅ m A 5.00 × 10 3 A 2 π (0.0200 m ) At r = b = 0.0500 m, B= (c) −7 (4π × 10 −7 )( 2 π (0.0500 m ) U = ∫ u dV = ∫ r=a (4π × 10 U= −7 ) = 0.0500 T = b dr ∫a µ 0 I enclosed 2π r 50.0 mT and ) = 0.0200 T = [B(r )]2 (2π rldr ) = µ 0 I 2l B= and I enclosed = I = 5.00 × 10 3 A T ⋅ m A 5.00 × 10 3 A r=b or = 20.0 mT µ 0 I 2l  b  ln  a 4π 2µ 0 4π )( ) (1000 × 10 m) ln  5.00 cm  = 2.29 × 10 T ⋅ m A 5.00 × 10 3 A 4π 2 r 3  2.00 cm    6 J = 2.29 MJ (d) The magnetic field created by the inner conductor exerts a force of repulsion on the current i n the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small rectangular section of the outer cylinder of length l and width w . It carries a current of  w (5.00 × 10 A) 2π (0.0500 m )  3 (5.00 × 10 A)w l(20.0 × 10 T) sin 90.0˚ 2 π (0.0500 m ) 3 and experiences an outward force F = IlBsin θ = The pressure on it is P= ( −3 )( ) 5.00 × 10 3 A 20.0 × 10 −3 T F F = = = 318 Pa A wl 2 π (0.0500 m ) © 2000 by Harcourt, Inc. All rights reserved. 268 Chapter 32 Solutions ( ) *32.79 (a) 4 π × 10 − 7 T ⋅ m A (1400)( 2.00 A ) µ 0 NI B= = = 2.93 × 10 − 3 T (upward) l 1.20 m (b) 2.93 × 10 − 3 T N B2 J  1 N ⋅ m u= = = 3.42 3   = 3.42 2 = 3.42 Pa   2µ 0 2 4 π × 10 −7 T ⋅ m A 1 J m m (c) To produce a downward magnetic field, the surface of the super conductor ( ( ) 2 ) must carry a clockwise current. (d) The vertical component of the field of the solenoid exerts an inward force on the superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field diverges and has a radially outward horizontal component. This component exerts upward force on the clockwise superconductor current. The total force on the core is upward . You can think of it as a force of repulsion between the solenoid with its north end pointing up, and the core, with its north end pointing down. (e) ( ) 2 F = PA = ( 3.42 Pa) π 1.10 × 10 −2 m  = 1.30 × 10 − 3 N   Note that we have not proven that energy density is pressure. In fact, it is not in some cases; see problem 12 in Chapter 21.