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Unformatted text preview: Exercises 1. (a) r = mv/eB = 6.26 m; (b) T = 2πr/v = 1.31 μs 2. v = (2k/m) 1/2 = 1.874 × 10 7 m/s ; (a) r = mv/eB = 2.13 cm; (b) a = v 2 /r = .165 × 10 16 m/s 2 : (c) T = 2πm/eB = 7.16 ns 3. (a) p = qrB = 1.6 × 10 –20 kg.m/s; (b) K = p 2 /2m = 4.8 × 10 5 eV 4. v 1 = 10 6 i , F 1 =+0.05k N v 2 = (–i + j)10 6 /(2) 1/2 ; F 2 = –0.035k N v 3 = (i – k) 10 6 /(2) 1/2 : F 3 = 0.035(i + k ) N 5. Since the forces are along ±z, we infer B z =0. Since the forces have the same magnitude and opposite directions the field must be directed midway between v 1 and v 2 . Thus B is directed at 30 ° to the +y axis. 6. (a) F = qvBsin45 ° and RHR lead to 0.0106 j N; (b) Know B y = 0. Then sin θ = F/qvB = 4/15, so θ = 15.5 ° Direction of B is (45 – 15.5) = 29.5 ° to the +z axis in the xz plane. (Note that the charge is negative.) 7. F = qv x B = –0.16i – 0.32 j –0.64k N 8. F = qv x B : (3i + j + 2k ) = (–2) (–i + 3 j ) x (B y j +B z k ); So, 3i + j +2k = 2B y k – 2B z j – 6B z i , which gives B = j – 0.5k T 9. (–2i + 6 j ) × 10 –13 = (–3) (v x i + v y j ) x (–1.2k ) Find v = (–3.13i – 1.04 j ) × 10 6 m/s 10. (a) B z = 0.25 T, B y = 0, B x is unknown; (b) B x = 0, thus B = 0.25k T 11. With v in the xy plane, F is along z, so B z =0. CHAPTER 29 If v is along the z axis, F is along x , so B x = 0. –1.228 × 10 –13 k = (1.6 × 10 –13 )(2i + 3 j ) × (B y j ). B = –0.4 j T 12. F = ILB = (10 3 )(1)(5 × 10 –5 ) = 0.05 N, upward 13. F 1 = IdB 1 j , F 2 = –IdB 1 i , F 3 = IdB 1 (i – j ) 14. F 1 = 0, F 2 = IdB 2 k N; F 3 = –IdB 2 k N 15. F 1 =IdB 3 k ; F 2 =0 ; F 3 = –IdB 3 k 16. With I along –z, the magnetic force is along +x. Equate force components along incline : mgsin θ = ILBcos θ Thus, I = mgtan θ /LB = 5.9 A....
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.
 Spring '11
 EE

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