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# Chapter-28 - Chapter 28 1(a The force on the electron is r...

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1109 Chapter 28 1. (a) The force on the electron is ( ( ( ( 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, ( 14 ˆ 6.2 10 N k. B F - = - × r 2. (a) We use Eq. 28-3: 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 work-energy theorem both the kinetic energy and the speed of the particle remain unchanged. 3. (a) Eq. 28-3 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 right-hand rule, r B points in the

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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. 28-3 (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. 28-2 and Eq. 3-30, 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. 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 ×1 0 3 m/s, and (b) v y = 7.0 ×1 0 3 m/s. 7. Straight line motion will result from zero net force acting on the system; we ignore gravity. Thus, r r r r F q E v B = + × = d i 0 . Note that r r v B so r r v B vB × = . Thus, obtaining the speed from the formula for kinetic energy, we obtain
1111 ( 29 ( 29 ( 29 3 4 3 19 31 100 V /(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 - = - × r . 8. Letting r r r r F q E v B = + × = d i 0 , we get sin vB E φ = . We note that (for given values of the fields) this gives a minimum value for speed whenever the sin φ factor is at its maximum value (which is 1, corresponding to φ = 90°). So 3 3 min 1.50 10 V/m 3.75 10 m/s 0.400 T E v B × = = = × .

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