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CH-28 - Chapter 28 1(a Eq 28-3 leads to v FB eB sin c 6.50...

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Chapter 28 1. (a) Eq. 28-3 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. 28-3: 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 work-energy 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 ˆ ˆ ˆ ˆ 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  4. The magnetic force on the proton is 1135

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CHAPTER 28 F = q v B where q = + e . Using Eq. 3-30 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 0 3 m/s, and (b) v y = 7.0 0 3 m/s. 5. Using Eq. 28-2 and Eq. 3-30, 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 0 , 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 0 . 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. 2 / 2 1.0 10 V 1.60 10 C / 9.11 10 kg e E E B v K m In unit-vector notation, 4 ˆ (2.67 10 T)k B  . 8. We apply F q E v B m a e +  d i to solve for E : 112
E m a q B v e + + +   + 911 10 2 00 10 160 10 400 12 0 150 114 6 00 4 80 31 12 2 19 . .

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CH-28 - Chapter 28 1(a Eq 28-3 leads to v FB eB sin c 6.50...

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