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Unformatted text preview: 1109 Chapter 28 1. (a) The force on the electron is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 19 6 6 14 ? ? i j i k = 1.6 10 C 2.0 10 m s 0.15 T 3.0 10 m s 0.030 T ˆ 6.2 10 N k. B x y x y x y y x F qv B q v v B B j q v B v B = × = + × + =  × × × = × r r r r Thus, the magnitude of r F B is 6.2 × 10 14 N, and r F B points in the positive z direction. (b) This amounts to repeating the above computation with a change in the sign in the charge. Thus, r F B has the same magnitude but points in the negative z direction, namely, ( 29 14 ˆ 6.2 10 N k. B F =  × r 2. (a) We use Eq. 283: F B = q vB sin φ = (+ 3.2 × 10 –19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 × 10 –18 N. (b) a = F B / m = (6.2 × 10 – 18 N) / (6.6 × 10 – 27 kg) = 9.5 × 10 8 m/s 2 . (c) Since it is perpendicular to r r v F B , does not do any work on the particle. Thus from the workenergy theorem both the kinetic energy and the speed of the particle remain unchanged. 3. (a) Eq. 283 leads to v F eB B = = × × × ° = × sin . . . sin . . . φ 650 10 160 10 2 60 10 230 4 00 10 17 19 3 5 N C T m s c hc h (b) The kinetic energy of the proton is ( 29 ( 29 2 2 27 5 16 1 1 1.67 10 kg 4.00 10 m s 1.34 10 J 2 2 K mv = = × × = × , which is equivalent to K = (1.34 × 10 – 16 J) / (1.60 × 10 – 19 J/eV) = 835 eV. 4. The force associated with the magnetic field must point in the $ j direction in order to cancel the force of gravity in the  $ j direction. By the righthand rule, r B points in the CHAPTER 28 1110  $ k direction (since $ $ $ i k j ×  = e j ). Note that the charge is positive; also note that we need to assume B y = 0. The magnitude  B z  is given by Eq. 283 (with φ = 90°). Therefore, with 2 1.0 10 kg m = × , 4 2.0 10 m/s v = × and 5 8.0 10 C q = × , we find ? ? k k ( 0.061 T)k z mg B B qv = =  =  r 5. Using Eq. 282 and Eq. 330, we obtain r F q v B v B q v B v B x y y x x x y x = = d i b g d i $ $ k k 3 where we use the fact that B y = 3 B x . Since the force (at the instant considered) is F z $ k where F z = 6.4 × 10 –19 N, then we are led to the condition ( 29 ( 29 3 . 3 z x y x z x x y F q v v B F B q v v = ⇒ = Substituting v x = 2.0 m/s, v y = 4.0 m/s and q = –1.6 × 10 –19 C, we obtain 19 19 6.4 10 N 2.0 T. (3 ) ( 1.6 10 C)[3(2.0 m/s) 4.0 m] z x x y F B q v v × = = =  × 6. The magnetic force on the proton is F qv B = × r r r where q = + e . Using Eq. 330 this becomes (4 × 10 17 )i ^ + (2 × 10 17 )j ^ = e [(0.03 v y + 40)i ^ + (20 – 0.03 v x )j ^ – (0.02 v x + 0.01 v y )k ^ ] with SI units understood. Equating corresponding components, we find (a) v x =  3.5 ×1 3 m/s, and (b) v y = 7.0 ×1 3 m/s....
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 Summer '10
 Prof.Yang
 Physics, Force

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