# Physics exercises_solution: Chapter 28

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## Physics exercises_solution: Chapter 28

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 28

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## Nội dung Text: Physics exercises_solution: Chapter 28

1.  28.1: For a charge with velocity v  (8.00  10 6 m s) ˆ, the magnetic field produced at j  μ qv  rˆ a position r away from the particle is B  0 . So for the cases below: 4 r 2  ˆ ˆ a) r  (  0.500 m)i  v  r   k , r02  1 ˆ ˆ 4 μ0 qv ˆ μ (6.0  106 C)(8.0  106 m s) ˆ ˆ ˆ B 2 k 0 2 k   (1.92  105 T)k   B0 k . 4π r0 4π (0.50 m)  b) r  (0.500 m) ˆ  v  r  0  B  0. j ˆ ˆ  ˆ ˆ 1 c) r  (0.500 m)k  v  r   i , r02  . ˆ ˆ 4  qv ˆ ˆ  B   0 2 i  B0 i . 4 r0  1 d) r   (0.500 m) ˆ  (0.500 m)k  v  r   i , r 2   2r0 j ˆ ˆ ˆ ˆ 2  qv i ˆ B i ˆ Bi ˆ B 0 2  0  0 4 r 2 2 2 2 2 0  qv q v  28.2: Btotal  B  B   2  2  4 d d  μ0 (8.0  106 C)(4.5  106 m s) (3.0  106 C)(9.0  106 m s)  B      4 (0.120 m) 2 (0.120 m) 2   B  4.38  10  4 T, into the page.
2.    μ0 qv  r 28.3: B  4π r 3    ˆ   a) v  vi , r  ri ; v  r  , B  0 0  ˆ    b) v  vi , r  rˆ; v  r  vrk , r  0.500 m j ˆ 7 6  μ  q v 1  10 N  s C )(4.80  10 C)(6.80  10 m s) 2 2 5 B 0  2  2  1.31  10  6 T  4π  r (0.500 m)  ˆ q is negative, so B   (1.31  0  6 T)k  ˆ    c) v  vi , r  (0.500 m)(i  ˆ); v  r  (0.500 m)vk , r  0.7071 m ˆ j ˆ μ    1  107 N  s 2 C 2 )(4.80  106 C)(0.500 m)(6.80  105 m s )  B   0  q v  r r3   (0.7071 m)3  4π   B  4.62  10 7 T; B   (4.62  10 7 T)k ˆ  ˆ  ˆ   ˆ d) v  vi , r  rk ; v  r   vrj , r  0.500 m 7 6  μ  q v 1  10 N  s C )(4.80  10 C)(6.80  10 m s ) 2 2 5 B 0  2   4π  r (0.500 m)2  B  1.31  10  6 T; B  (1.31  10  6 T) ˆ j 28.4: a) Following Example 28.1 we can find the magnetic force between the charges: μ qqvv (8.00  106 C)(3.00  106 C)(4.50  106 m s )(9.00  106 m FB  0  (10 7 T  m A ) 4π r 2 (0.240 m) 2  1.69  10 3 N (the force on the upper charge points up and the force on the lower charge points down). The Coulomb force between the charges is 12 2 F  k qr1 q2  (8.99  10 9 N  m 2 C 2 ) (8.00)(3.00)  102 C  3.75 N (the force on the upper 2 (0.240 m) charge points up and the force on the lower charge points down). The ratio of the Coulomb force to the magnetic force is 1.693.75 N3 N  2.22  10 3   10  c2 v1 v 2 . b) The magnetic forces are reversed when the direction of only one velocity is reversed but the magnitude of the force is unchanged.
3. 28.5: The magnetic field is into the page at the origin, and the magnitude is μ  qv qv  B  B  B  0  2  2  4π  r r  0 (4.0  10 6 C)(2.0  105 m s) (1.5  10 6 C)(8.0  105 m s )  B      4π (0.300 m) 2 (0.400 m) 2  6  B  1.64  10 T, into the page. μ0 qv μ qv 28.6: a) q   q; Bq  2 into the page; Bq   0 2 out of the page. 4πd 4πd v μ0 qv (i) v   B  into the page. 2 4π (2d 2 ) (ii) v  v  B  0.  qv (iii) v  2v  B  0 2 out of the page. 4d  μ q 2vv b) F  qv   B q  0 and is attractive. 4π (2d ) 2 μ0 q 2vv q2 F c) FB  2 , FC  2  B  μ0 ε0vv  μ0 ε0 (3.00  105 m s) 2 4π (2d ) 4πε0 (2d ) FC  1.00  10 6. 28.7: a) r  cos θi  sin θˆ  cos(150)i  sin (150) ˆ   (0.866)i  (0.500) ˆ. ˆ ˆ j ˆ j ˆ j  b) dl  r  (  dl i )  (  (0.866)i  (0.500) ˆ)   dl (0.500)k   (5.00  10 3 m)k ˆ ˆ ˆ j ˆ ˆ  μ I dl  r ˆ μ I dl (0.500 m) ˆ μ (125 A)(0.010 m)(0.500 m) ˆ c) d B  0 2  0 2 k 0 k 4π r 4π r 4π (1.20 m) 2  d B   (4.3  10 8 T)k. ˆ
4. 28.8: The magnetic field at the given points is: μ I dl sin θ μ0 (200 A) (0.000100 m) dBa  0 2  2  2.00  10 6 T. 4π r 4π (0.100 m) μ I dl sin θ μ0 (200 A) (0.000100 m) sin 45 dBb  0 2  2  0.705  10 6 T. 4π r 4π 2(0.100 m) μ I dl sin θ μ0 (200 A) (0.000100 m) dBc  0   2.00  10 6 T. 4π r2 4π (0.100 m) 2 μ I dl sin θ μ0 I dl sin (0) dBd  0   0. 4π r2 4π r2 μ I dl sin θ dBe  0 4π r2  (200 A) (0.00100 m) 2  dBe  0 4 3(0.100 m) 2 3  dBe  0.545  10 6 T.
5. 28.9: The wire carries current in the z-direction. The magnetic field of a small piece of  μ0 I dl  r ˆ wire d B  2 at different locations is therefore: 4π r  a) r  (2.00 m)i  lˆ  r  ˆ ˆ ˆ j μ0 I dl sin θ ˆ μ0 (4.00 A) (5  104 m) sin 90  dB  2 j  5.00  1011 Tˆ. j 4π r 4π (2.00 m)2  b) r  (2.00 m) ˆ  lˆ  r   i . j ˆ ˆ  μ0 I dl sin θ ˆ  μ0 (4.00 A) (5  104 m) sin (90) ˆ  dB  i i 4π r2 4π (2.00 m) 2 ˆ   5.00  1011 Ti .  1 ˆ ˆ ˆ c) r  (2.00 m) i  (2.00 m) ˆ  lˆ  r  j ˆ ( j  i) 2 μ0 I dl sin θ 1 ˆ ˆ μ0 (4.00 A) (5.0  104 m) 1 ˆ ˆ d B ( j  i)  ( j  i) 4π r2 2 4π (2.00 m)2  (2.00 m)2 2  1.77  10 11 T( ˆ  i ) j ˆ  ˆ ˆ ˆ d) r  (2.00 m)k  l  r  0 d μI  1 1  μ0 I  8  4 μ0 I ˆ 28.10: a) At x  :B 0   d 2  3d 2   2π  3d   3πd , in the j direction.  2 2π     d b) The position x   is symmetrical with that of part (a), so the magnetic 2 4 0 I field there is B  , in the ˆ direction. j 3d 28.11: a) At the point exactly midway between the wires, the two magnetic fields are in opposite directions and cancel. b) At a distance a above the top wire, the magnetic fields are in the same  I ˆ  I ˆ  I ˆ  0 I ˆ 2 0 I ˆ direction and add up: B  0 k  0 k  0 k  k k. 2r1 2r2 2a 2 (3a ) 3a c) At the same distance as part (b), but below the lower wire, yields the same 2μ I ˆ magnitude magnetic field but in the opposite direction: B   0 k . 3πa
6. 28.12: The total magnetic field is the vector sum of the constant magnetic field and the wire’s magnetic field. So: a) At (0, 0, 1 m):  Iˆ ˆ  (8.00 A) i   (1.0  10 7 T)i . ˆ ˆ B  B 0  0 i  (1.50  10 6 T)i  0 2r 2 (1.00 m) b) At (1 m, 0, 0): μI ˆ ˆ μ (8.00 A) k ˆ B  B 0  0 k  (1.50  10 6 T)i  0 2πr 2π (1.00 m) ˆ ˆ  B  (1.50  10 6 T) i  (1.6  10 6 T) k  2.19  10 6 T, at θ  46.8 from x to z. 0 I ˆ ˆ μ (8.00 A) i ˆ c) At (0, 0, – 0.25 m): B  B 0  i  (1.50  10 6 T)i  0 2πr 2π (0.25 m) ˆ  (7.9  10 6 T) i . a μI a xdy μ Ix y μ0 I 2a 28.13: B  0 4π a ( x 2  y 2 )3 2  0 4π x ( x  y 2 )1 2 2 2  4π x( x  a 2 )1 2 2 . a μ0 I 2πrB0 2π (0.040 m) (5.50  104 T) 28.14: a) B0  I   110 A. 2πr μ0 μ0 μI B b) B  0 , so B(r  0.080 m)  0  2.75  10 4 T, 2πr 2 B0 B(r  0.160 m)   1.375  10 4 T. 4 μ0 I μ (800 A) 28.15: a) B   0  2.90  10 5 T, to the east. 2πr 2π (5.50 m) b) Since the magnitude of the earth’s magnetic filed is 5.00  10 5 T, to the north, the total magnetic field is now 30 o east of north with a magnitude of 5.78  10 5 T. This could be a problem!
7. 28.16: a) B = 0 since the fields are in opposite directions. μI μ I  I 1 1 b) B  Ba  Bb  0  0  0    2πra 2πrb 2π  ra rb    (4π  107 Tm A) (4.0 A)  1 1     0.3 m  0.2 m   2    6.67  10  6 T  6.67 T c) Note that B a  ra and B b  rb B  Ba cos θ  Bb cos θ  2 Ba cos θ 5 tan θ   θ  14.04 : ra  (0.20 m)2  (0.05 m)2 20 I B  2 0 cos θ 2ra (4  10 7 Tm A) (4.0 A) 2 cos 14.04 2 (0.20 m)  (0.05 m) 2 2  7.53  10 6 T  7.53 T, to the left.
8. 28.17: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires. Consider a point a distance x from the wire carrying I 2 = 75.0 A. Btot will be zero where B1  B2 . μ0 I1 μI  0 2 2π (0.400 m  x) 2πx I 2 (0.400 m  x)  I1 x; I1  25.0 A, I 2  75.0 A x = 0.300 m; Btot  0 along a line 0.300 m from the wire carrying 75.0 A amd 0.100 m from the wire carrying current 25.0 A. b) Let the wire with I 1  25.0 A be 0.400 m above the wire with I 2 = 75.0 A. The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires. But to have B1  B2 must be closer to wire #1 since I 1 < I 2 , so can have Btot  0 only at points above both wires. Consider a point a distance x from the wire carrying I 1  25.0 A. Btot will be zero where B1  B2 . μ0 I1 μ0 I 2  2πx 2π (0.400 m  x) I 2 x  I1 (0.400 m  x); x  0.200 m Btot  0 along a line 0.200 m from the wire carrying 25.0 A and 0.600 m from the wire carrying current I 2  75.0 A.
9. 28.18: (a) and (b) B = 0 since the magnetic fields due to currents at opposite corners of the square cancel. (c) B  Ba cos 45  Bb cos 45  Bc cos 45  Bd cos 45 μI  4 Ba cos 45  4  0  cos 45  2πr  r  (10 cm)2  (10 cm)2  10 2 cm  0.10 2 m (4π  10 7 Tm A) (100 A) B4 cos 45 2 (0.10 2 m)  4.0  10 4 T, to the left.
10. 28.19:    B 1 , B 2 , B 3 ⊙  I B  0 ; r  0.200 m for each wire 2r B1  1.00  10 5 T, B2  0.80  10 5 T , B3  2.00  10 5 T Let ⊙ be the positive z-direction. I 1  10.0 A, I 2  8.0 A, I 3  20.0 A B1z   1.00  105 T, B2 z   0.80  105 T, B3z   2.00  105 T B1z  B2 z  B3 z  B4 z  0 B4 z   ( B1z  B2 z  B3 z )   2.0  10 6 T To give B 4 in the  direction the current in wire 4 must be toward the bottom of the page. μ0 I rB4 (0.200 m) (2.0  106 T) B4  so I 4    2.0 A 2πr ( μ0 2π ) (2  10 7 T  m A) F 0 I 2  1 1  0 I 2 28.20: On the top wire:     , upward. L 2  d 2d  4d On the middle wire, the magnetic fields cancel so the force is zero. F 0 I 2  1 1  0 I 2 On the bottom wire:     , downward. L 2  d 2d  4d 28.21: We need the magnetic and gravitational forces to cancel: 0 I 2 L  I2  λLg  h 0 2h 2λg
11.  0 I 1 I 2 L  0 (5.00 A) (2.00 A) (1.20 m) 28.22: a) F    6.00  10 6 N, and the force is 2r 2 (0.400 m) repulsive since the currents are in opposite directions. b) Doubling the currents makes the force increase by a factor of four to F  2.40  10 5 N. F  0 I1 I 2 F 2r 2 (0.0250 m) 28.23:   I2   (4.0  10 5 N m)  8.33 A. L 2r L  0 I1  0 (0.60 A) b) The two wires repel so the currents are in opposite directions. 28.24: There is no magnetic field at the center of the loop from the straight sections. The magnetic field from the semicircle is just half that of a complete loop: 1 1  I   I B  Bloop   0   0 , 2 2  2R  4R into the page. 28.25: As in Exercise 28.24, there is no contribution from the straight wires, and now we have two oppositely oriented contributions from the two semicircles: 1  B  ( B1  B2 )   0  I 1  I 2 , 2  2R  into the page. Note that if the two currents are equal, the magnetic field goes to zero at the center of the loop. 28.26: a) The field still points along the positive x-axis, and thus points into the loop from this location. b) If the current is reversed, the magnetic field is reversed. At point P the field would then point into the loop. c) Point the thumb of your right hand in the direction of the magnetic moment, under the given circumstances, the current would appear to flow in the direction that your fingers curl (i.e., clockwise).
12. 2aBx 2(0.024 m) (0.0580 T) 28.27: a) Bx  μ0 NI 2a , so I    2.77 A μ0 N (4π  10 7 T  m A) (800) b) At the center, Bc   0 NI 2a . At a distance x from the center,  0 NIa 2   NI   a3   a3  Bx   0  2  ( x  a 2 )3 2   Bc   2  ( x  a 2 )3 2   2( x  a ) 2 2 32  2a      a3 1 Bx  1 Bc means 2  (x  a ) 2 2 32 2 ( x 2  a 2 )3  4a 6 , with a  0.024 m, so x  0.0184 m  NI  0 (600) (0.500 A) 28.28: a) From Eq. (29-17), Bcenter  0   9.42  10 3 T. 2a 2(0.020 m) b) From Eq. (29-16),  0 NIa 2  (600) (0.500 A) (0.020 m) 2 B( x)   B (0.08 m)  0  1.34  10  4 T. 2( x  a ) 2 2 32 2((0.080 m)  (0.020 m) ) 2 2 32 μ0 NIa 2 2 B( x) ( x 2  a 2 )3 2 28.29: B( x)  N 2( x 2  a 2 )3 2 μ0 Ia 2   2(6.39  104 T) (0.06 m) 2  (0.06 m) 2  32  69 μ0 (2.50 A) (0.06 m) 2  28.30:  B  dl  μ I0 encl  3.83  10 4 T  m  I encl  305 A.   b)  3.83  10 4 since dl points opposite to B everywhere. 28.31: We will travel around the loops in the counterclockwise direction.  a) I encl  0   B  dl  0.  b) I encl   I1  4.0 A   B  dl   μ0 (4.0 A)  5.03  10 6 T  m.  c) I encl   I1  I 2  4.0 A  6.0 A  2.0 A   B  dl  μ0 (2.0 A)  2.51  10 6 T  m.  d) I encl   I1  I 2  I 3  4.0 A   B  dl   μ0 (4.0 A)  5.03  106 T  m. Using Ampere’s Law in each case, the sign of the line integral was determined by using the right-hand rule. This determines the sign of the integral for a counterclockwise path.
13. 28.32: Consider a coaxial cable where the currents run in OPPOSITE directions.  μI a) For a  r  b, I encl  I   B  dl  μ0 I  B 2πr  μ0 I  B  0 . 2πr b) For r  c, the enclosed current is zero, so the magnetic field is also zero. 28.33: Consider a coaxial cable where the currents run in the SAME direction.  μI a) For a  r  b, I encl  I1   B  dl  μ0 I1  B 2r  μ0 I1  B  0 1 . 2πr  b) For r  c, I encl  I1  I 2   B  dl  μ0 ( I1  I 2 )  B 2r  μ0 ( I1  I 2 ) μ0 ( I 1  I 2 ) B . 2πr 28.34: Using the formula for the magnetic field of a solenoid: μ NI μ0 (600) (8.00 A) B  μ0 nI  0   0.0402 T. L (0.150 m) μ0 NI BL (0.0270 T) (0.400 m) 28.35: a) B  N   716 turns L μ0 I 0 (12.0 A) N 716 turns n   1790 turns m. L 0.400 m b) The length of wire required is 2 rN  2 ( 0 .0140 m ) (116 )  63 m . N 28.36: B  0 I L BL I 0 N (0.150 T) (1.40 m)  (4  10 7 Tm A)(4000)  41.8 A
14. μ0 I Br 28.37: a) B  , so I   3.72  106 A 2πr (  0 2π ) μ NI 2aBx b) Bx  0 , so I   2.49  105 A 2a μ0 N c) B  μ0 nI   0 ( N L) I , so I  BL μ0 N  237 A 28.38: Outside a toroidal solenoid there is no magnetic field and inside it the magnetic μ NI field is given by B  0 . 2πr a) r = 0.12 m, which is outside the toroid, so B = 0.  NI  0 (250) (8.50 A) b) r = 0.16 m  B  0   2.66  10 3 T. 2r 2 (0.160 m) c) r = 0.20 m, which is outside the toroid, so B = 0  0 NI  0 (600) (0.650 A) 28.39: B    1.11  10 3 T. 2r 2 (0.070 m)  NI K m  0 NI  0 (80) (400) (0.25 A) 28.40: a) B     0.0267 T. 2r 2r 2 (0.060 m) b) The fraction due to atomic currents is B  80 B  80 (0.0267 T)  0.0263 T. 79 79 K m μ0 NI 2πr B 2π (0.0290 m) (0.350 T) 28.41: a) If K m  1400  B  I    2πr K m μ0 N μ0 (1400)(500) 0.0725 A. 1400 b) If K m  5200  I  I part(a)  0.0195 A. 5200 K m μ0 NI 2πrB 2π (0.2500 m) (1.940 T) 28.42: a) B   Km    2021. 2πr μ0 NI μ0 (500) ( 2.400 A) b) X m  K m  1  2020.
15. 28.43: a) The magnetic field from the solenoid alone is: (i) B0  μ0 nI  μ0 (6000 m 1 ) (0.15 A)  B0  1.13  10 3 T. K 1 5199 (ii) But M  m B0  (1.13  10 3 T)  M  4.68  10 6 A m. 0 0 (iii) B  K m B0  (5200)(1.13  10 3 T)  5.88 T. b)  J   N  m  C  m    2 28.44:        Am . 2 T   N  s C  m   s  28.45: The material does obey Curie’s Law because we have a straight line for temperature against one over the magnetic susceptibility. The Curie constant from the graph is 1 1 C   1.55  10 5 K  A T  m.  0 (slope)  0 (5.13) 28.46: The magnetic field of charge q  at the location of charge q is into the page.    ˆ  μ0 qv  r  (qv)i    0 qv sin θ  ( k )   μ0 qq sin θ  ˆ ˆ ˆ  F  qv  B  (qv)i 2  ˆ  j 4π r  4π r2   4π r 2  where  is the angle between v  and r . ˆ  μ0 (8.00  10 6 C)(5.00  10 6 C)(9.00  10 4 m s)(6.50  104 m s)  0.4   ˆ F   4   j   (0.500 m) 2  0.5    F  (7.49  10 8 N ) ˆ. j
16. 19  μ I  μ (1.60  10 C)(6.00  10 m s)(2.50 A) 4 28.47: F  qvB  qv 0   0  2πr  2π (0.045 m) 19  1.07  10 N. Let the current run left to right, the electron moves in the opposite direction, below the wire, then the magnetic field at the electron is into the page, and the electron feels a force upward, toward the wire, by the right-hand rule (remember the electron is negative). F qvB sin  ev   0 I  28.48: (a) a      m m m  2πr  (1.6  1017 C)(250,000 m s)(4π  107 Tm A)(25 A) a (9.11  10 31 kg)(2π )(0.020 m)  1.1  1013 m s 2 , away from the wire. b) The electric force must balance the magnetic force. eE = eVB i E  vB  v 0 2r (250, 000 m/s)(4π  107 Tm/A)(25 A)  2π (0.020 m)  62.5 N/C, away from the wire. (c) mg  (9.11  10 31 kg)(9.8 m/s 2 )  10 29 N Fel  eE  (1.6  10 19 C)(62.5 N/C)  10 17 N Fel  1012 Fgrav , so we can neglect gravity.
17. 28.49: Let the wire connected to the 25.0  resistor be #2 and the wire connected to the 10.0  resistor be #1. Both I 1 and I 2 are directed toward the right in the figure, so at the location of the proton I 2 is  and I1  ⊙ μ0 I1 I B1  and B2  0 2 , with r  0.0250 m. 2πr 2πr B1  8.00  10 T, B2  3.20  10 5 T and B  B1  B2  4.80  10  5 T 5 and in the direction ⊙. Force is to the right. F  qvB  (1.602  10 19 C)(650  10 3 m/s)(4.80  10 5 T)  5.00  10 18 N 28.50: The fields add   0 IR 2  B  B1  B2  2 B1  2  2 3/ 2   2( R  x )  2 7 (4π  10 Tm/A)(1.50 A)(0.20 m)2   5.75  10 6 T [(0.20 m)  (0.125 m) ] 2 2 3/ 2 F  qv B sin   (1.6  10 19 C)(2400 m/s)(5.75  10 6 T) sin 90  2.21  10 21 N, perpendicular to the line ab and to the velocity. 28.51: a) Along the dashed line, B1 and B 2 are in opposite directions. If the line has slope  1.00 then r1  r2 and B1  B2 , so Btot  0.
18. ˆ i ˆ j ˆ k μ0 q v0  r μ0 q ˆ 28.52: a) B  v0 x v0 y v0 z 4π r 2 4π r 2 1 0 0  μ0 q 4π r 2  j  v0 z ˆ  v0 y k  (6.00  10 6 T) ˆ ˆ j μ q  |q|  0 2 v0 y  0  v0 y  0 and  0 2 v0 z  6.00  10 6 T 4π r 4π r 6 4π (6.00  10 T)(0.25 m)2  v0 z    521 m/s.  0 (7.20  10 3 C) 2 2 And v0 x   v0  v0 y  v0 z   (800 m/s) 2  (521 m/s) 2  607 m/s. 2 ˆ i ˆ j ˆ k b) B (0, 0.250 m, 0)  μ0 q v 0  r  0 q 4π r 2 ˆ  4π r 2 v0 x v0 y v0 z   μ0 q 4π r 2  ˆ ˆ v0 x k  v0 z i  0 1 0 μ0 | q | μ (7.20  103 C)  B (0, 0.250 m, 0)  2 v0  0 2 800 m/s  9.2  10 6 T. 4π r 4π (0.250 m) 28.53: Choose a cube of edge length L , with one face on the y-z plane. Then: B0 x ˆ B L B L3 0   B dA   B d A   i  d A  0   dA  0  B0  0, xL xL a a xL a so the only possible field is a zero field.
19. 28.54: a)   I ˆ μ I  b) B 2    0 22 i  2πr  ˆ B 3   0 3  (sin i  cosˆ) j  2 r   3 And so  μ   I I ˆ I  B   0    2  3 sin θ  i  3 cos θ ˆ  j  2π   r2 r3    r3   μ   I2 I3 ˆ I3 ˆ  B   0    (0.030 m)  (0.050 m) (0.6) i  (0.050 m) (0.8) j    2π       μ   2π   ˆ   B   0  (12 I 3  33.3I 2 ) i  (16 I 3 ) ˆj μ 2π  ˆ  0 (12)(4.00 A)  (33.3)(2.00 A)) i  (16)(4.00 A) ˆ j  ˆ  3.72  10 6 Ti  1.28  105 Tˆ. j    c) F  I l  B  I lB ˆ  I lB i 1 j 1 x ˆ 1 y  1.00 A 0.010 m [(3.72  106 T) ˆ  (1.28  10 5 T) i ] j ˆ  3.72  108 Tˆ  1.28  10 7 Ti ; F  1.33  10 7 N, 16.2 counterclockwise j ˆ from +x-axis.
20. 28.55: a) If the magnetic field at point P is zero, then from Figure (28.46) the current I 2 must be out of the page, in order to cancel the field from I 1 . Also: μI μI r (0.500 m) B1  B2  0 1  0 2  I 2  I1 2  6.00 A   2.00 A. 2πr1 2πr2 r1 (1.50 m) b) Given the currents, the field at Q points to the right and has magnitude μ  I I  μ  6.00 A 2.00 A  BQ  0  1  2   0   r r  2π  0.500 m  1.50 m   2.13  10 T.  6 2π  1 2    c) The magnitude of the field at S is given by the sum of the squares of the two fields because they are at right angles. So: 2 2 2 μ  I1   I 2  μ  6.00 A   2.00 A  2 BS  B  B2 1 2  0      0 r  r      6  0.60 m    0.80 m   2.1  10 T. 2π  1  2 2π     28.56: a) b) At a position on the x-axis: μI μ0 I a Bnet  2 0 sin  2πr π x2  a2 x  a2 2 μ0 Ia  Bnet  ,  π x2  a2 in the positive x-direction , as shown at left. c) d) The magnetic field is a maximum at the origin, x = 0. μ Ia e) When x  a, B  0 2 . πx