ch28 - Chapter 28 1. (a) Equation 28-3 leads to v eB sin FB...

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1097 Chapter 28 1. (a) Equation 28-3 leads to v F B eB sin 6.50 10 17 N 1.60 10 19 C 2.60 10 3 T sin23.0 4.00 10 5 m s . (b) The kinetic energy of the proton is     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. 2. 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 right-hand rule, B points in the k direction (since ? i   ? k   ? 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. 28-3 (with = 90°). Therefore, with 2 1.2 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     . 3. (a) The force on the electron is                 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 Thus, the magnitude of F B is 6.2 10 14 N, and 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, F B has the same magnitude but points in the negative z direction, namely,   14 ˆ 6.2 10 N k. B F   (-0.074 T) ˆ k
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CHAPTER 28 1098 4. (a) We use Eq. 28-3: F B = |q| vB sin = (+ 3.2 10 –19 C) (620 m/s) (0.045 T) (sin 52°) = 7.0 10 –18 N. (b) The acceleration is a = F B / m = (7.0 10 – 18 N) / (6.6 10 – 27 kg) = 1.06 10 9 m/s 2 . (c) Since it is perpendicular to v F B , does not do any work on the particle. Thus from the work-energy theorem both the kinetic energy and the speed of the particle remain unchanged. 5. Using Eq. 28-2 and Eq. 3-30, we obtain r F q v x B y v y B x ? k q v x 3 B x v y B x ? k 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     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 where q = + e . Using Eq. 3-30 this becomes (4 10 17 )i ^ + (2 10 17 )j ^ = e [(0.025 v y + 40)i ^ + (20 – 0.025 v x )j ^ – (0.02 v x + 0.01 v y )k ^ ] with SI units understood. Equating corresponding components, we find (a) v x = 4.2  0 3 m/s, and (b) v y = 8.4  0 3 m/s.
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1099 7. We apply r F q r E r v r B m e r a to solve for E : r E m e r a q r B r v 9.11 10 31 kg 2.00 10 12 m s 2 ?
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ch28 - Chapter 28 1. (a) Equation 28-3 leads to v eB sin FB...

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