Chapter-28 - 1109 Chapter 28 1. (a) The force on the...

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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. 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 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 3 m/s, and (b) v y = 7.0 ×1 3 m/s....
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Chapter-28 - 1109 Chapter 28 1. (a) The force on the...

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