Chapter28

# Chapter28 - 1. (a) The force on the electron is ^ FB = qv B...

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1. (a) The force on the electron is () ( ) ( ) 19 6 6 14 ˆˆ ˆ ˆ ij 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. Bx y x y x y y x F q v B qv v B Bj qvB vB = + × + = ⎡⎤ −× × × ⎣⎦ GG G G Thus, the magnitude of G F B is 6.2 × 10 14 N, and G 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, G F B has the same magnitude but points in the negative z direction, namely, 14 ˆ 6.2 10 N k. B F =− × G

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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 G G vF 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 . .. s i n . φ 650 10 160 10 2 60 10 230 400 10 17 19 3 5 N CT ms c hc h (b) The kinetic energy of the proton is () 2 22 7 5 1 6 11 1.67 10 kg 4.00 10 m s 1.34 10 J Km v × × = × , which is equivalent to K = (1.34 × 10 – 16 J) / (1.60 × 10 – 19 J/eV) = 835 eV.

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± k direction (since ±± ± ik 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 ˆˆ ˆ k k ( 0.061 T)k z mg BB qv ⎛⎞ == = ⎜⎟ ⎝⎠ G 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, G B points in the
5. Using Eq. 28-2 and Eq. 3-30, we obtain G Fq v B v B q v B v B xy yx x x =− = di bg ±± kk 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 () 3. 3 z x z x F qv vB F B qv 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 F B × == = −− ×

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6. The magnetic force on the proton is Fq vB = × G G G 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.
() ( ) ( ) 3 4 31 9 3 1 100 V /(20 10 m) 2.67 10 T. 2/ 2 1.0 10 V 1.60 10 C / 9.11 10 kg e EE B v Km −− × == = = × ×× × In unit-vector notation, 4 ˆ (2.67 10 T)k B =− × G . 7. Straight line motion will result from zero net force acting on the system; we ignore gravity. Thus, G G G G Fq EvB =+ × = di 0 . Note that G G v B so G G vBv B ×= . Thus, obtaining the speed from the formula for kinetic energy, we obtain

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8. Letting G G G G Fq EvB =+ × = di 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 factor is at its maximum value (which is 1, corresponding to = 90°). So 3 3 min 1.50 10 V/m 3.75 10 m/s 0.400 T E v B × == = × .
9. We apply G G G G G Fq EvB m a e =+ × = di to solve for G E : G G G G E ma q Bv e × = ×× −× + =− + 911 10 2 00 10

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## This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter28 - 1. (a) The force on the electron is ^ FB = qv B...

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