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Unformatted text preview: Chapter 28 1. (a) Eq. 283 leads to v F eB B sin . . . sin . . . 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 K mv 1 2 1 2 167 10 4 00 10 134 10 2 27 5 2 16 . . . kg m s J. c hc h This is (1.34 10 16 J) / (1.60 10 19 J/eV) = 835 eV. 2. (a) We use Eq. 283: F B = q vB sin = (+ 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 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) 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 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 6.2 10 N k. B F 4. The magnetic force on the proton is 1135 CHAPTER 28 F = q v B 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 3 m/s, and (b) v y = 7.0 3 m/s. 5. Using Eq. 282 and Eq. 330, we obtain 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 q v v B F B F q v v x y x z x z x y 3 3 d i d i . Substituting V x = 2.0 m/s, v y = 4.0 m/s and q = 1.6 10 19 C, we obtain B x = 2.0 T. 6. Letting F q E v B + d i , we get vB sin = E . We note that (for given values of the fields) this gives a minimum value for speed whenever the sin factor is at its maximum value (which is 1, corresponding to = 90). So v min = E / B = (1.50 10 3 V / m) / (0.400 T) = 3.75 10 3 m / s. 7. Straight line motion will result from zero net force acting on the system; we ignore gravity. Thus, F q E v B + d i . Note that v B so v B vB . Thus, obtaining the speed from the formula for kinetic energy, we obtain ( ) ( ) ( ) ( ) 3 4 3 19 31 100V 20 10 m 2.67 10 T....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics, Energy, Kinetic Energy

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