1. Not at all. The charge exerts weak, attractive polarization forces on both ends of the compass needle, but in this configuration the forces will balance and have no net effect.

2. d. Point your right thumb in the direction of the current and curl your fingers around the wire.

3. b. Point your right thumb out of the page, in the direction of $\vec{v}$. Your fingers are pointing down as they curl around the left side.

4. b. The right-hand rule gives a downward $\vec{B}$ for a clockwise current. The north pole is on the side from which the field emerges.

5. c. For a field pointing into the figure, $\vec{v}\times \vec{B}$ is to the right. But the electron is negative, so the force is in the direction of $-(\vec{v}\times \vec{B})$.

6. b. Repulsion indicates that the south pole of the loop is on the right, facing the bar magnet; the north pole is on the left. Then the right-hand rule gives the current direction.

7. a or c. Any magnetic field to the right, whether leaving a north pole or entering a south pole, will align the magnetic domains as shown.

1. $B_2=40\text{ mT, }{B}_3=0\text{ T,} B_4=40\text{ mT}$

3.

1. 0 T

2. $1.6\times {10}^{-15}\widehat{k}\text{T}$

3. $-4.0\times {10}^{-16}\widehat{k}\text{T}$

5. $(-2.8\times {10}^{-16}\text{T})\widehat{k}$

7. 2.5 A, 250 A, 5000 A to 50,000 A, 500,000 A

9. $0.040\mu \text{A}$

11.

1. 20 A

2. $1.6\times {10}^{-3}\text{m}$

13. $\vec{B}{}_{\text{a}}=2.0\times {10}^{-4}î\text{T,}\vec{B}{}_{\text{b}}=4.0\times {10}^{-4}î\text{T,}\vec{B}{}_{\text{c}}=2.0\times {10}^{-4}î\text{T}$

15. 11 A

17.

1. $2.5\text{A}{\text{m}}^2$

2. $150\mu \text{T}$

19. 0

21. 23.0 A, into the page

23. $1.26\times {10}^{-6}\text{T}\text{m}$

25. 2.4 kA

27.

1. $8.0\times {10}^{-13}\text{N in}-ĵ$

2. $5.7\times {10}^{-13}\text{N in}ĵ$

29.

1. 1.4442 MHz

2. 1.6450 MHz

3. 1.6457 MHz

31.

1. 86 mT

2. $1.62\times {10}^{-14}\text{J}$

33. $3.0\text{Ω}$

35. $\vec{F}{}_{\text{on}1}=(2.5\times {10}^{-4}\text{N, up})\text{,}\vec{F}{}_{\text{on}2}=0\text{N,}\vec{F}{}_{\text{on}3}=(2.5\times {10}^{-4}\text{N, down})$

37. 240 A

39.

1. $1.26\times {10}^{-11}\text{N}\text{m}$

2. $0\text{°}\text{or}180\text{°}$

41.

1. $x=0.50\text{cm}$

2. $x=8.0\text{cm}$

43. $4.1\times {10}^{-4}\text{T,}\text{into the page}$

45. ${\displaystyle \frac{{\mu }_{0}I\theta }{4\pi R}}$

47. $5.0\times {10}^{-3}\text{T}$

49. #18, 4.1 A

51. $1.1\mu \text{A}$

53. ${\displaystyle \frac{{\mu }_{0}I}{4R}}$

55.

• a. $I_{\text{through}}=\frac{I{r}^2}{R_1^2}\text{; }{I}_{\text{through}}=I\text{,}{I}_{\text{through}}=0\text{ A}$

• b.

  

  Full Alternative Text

57. $0\text{; }\frac{{\mu }_{0}I}{2\pi r}\left(\frac{r^2-R_1^2}{R_2^2-R_1^2}\right)\text{; }\frac{{\mu }_{0}I}{2\pi r}$

59. 2.0 mT, into the page

61. $2.4\times {10}^{10}\text{m}\text{/}{\text{s}}^2\text{,}\text{up}$

63.

1. $4.6\times {10}^{-13}\text{J}$

2. 2850 rev

65. 0.82 mm, 3.0 mm

67. 0.69 T

69. 2.10 T

71. 2.0 A

73.

75.

1. $B=\frac{\mu g\text{tan }\theta }{I}$

2. $\vec{B}=(11\text{mT, down})$

77.

• a.

  

  Full Alternative Text

• b. $\frac{1}{2}IL{B}_0ĵ$

• c. $\frac{1}{3}I{L}^2{B}_0$

79. 0.16 T

81.

1. ${\displaystyle \frac{{\mu }_{0}IL}{4\pi d\sqrt{(L\text{/}2{)}^2+d^2}}}$

2. ${\displaystyle \frac{\sqrt{2}{\mu }_{0}I}{\pi R}}$

83.

1. Horizontal and to the left above the sheet; horizontal and to the right below the sheet

2. $\frac{1}{2}{\mu }_0{j}_s$