23 HMK #6 Chp 27a

23 HMK #6 Chp 27a - Ch. 27 PHYSICS 2B Spring 2011 Homework...

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Ch. 27 PHYSICS 2B Spring 2011 Homework Solutions – Chapter 27 Author: Evan Grohs - [email protected] Ch. 27 Problems: 55, 56, 58, 59, 61, 65, 70, 71, 74, 78, 82, 85, 86 Page 1/10
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Ch. 27 page 2/10 M AGNETIC F IELD AND M AGNETIC F ORCES 27.55. (a) I DENTIFY : Use Eq.(27.2) to relate , , and . vB F S ET U P : The directions of 11 and vF are shown in Figure 27.53a. q F says that F is perpendicular to and . The information given here means that B can have no z -component. Figure 27.53a The directions of 22 and are shown in Figure 27.53b. F is perpendicular to and , so B can have no x -component. Figure 27.53b Both pieces of information taken together say that B is in the y -direction; ˆ . y B Bj E XECUTE : Use the information given about 2 F to calculate 2 222 ˆˆˆ : , . yy FF v B  F iv kB j  2 2 2 2 2 ˆˆ ˆ ˆ says ( ) and y q F qvB F  Fv B i k j i 21 /( ) /( ). has the maginitude /( ) and is in the -direction. y BF q v F q v B F q v y  (b) 2 sin / 2 / 2 y Fq v B q v B F  E VALUATE : 12 2 . vv v is perpendicular to B whereas only the component of 1 v perpendicular to B contributes to the force, so it is expected that , as we found. 27.56. I DENTIFY : Apply . q F S ET U P : B x 0.650 T, 0 y B and 0. z B E XECUTE : () 0 . xy z z y v Bv B F y q ( v z B x v x B z ) (9.45 10 8 C)(5.85 10 4 m/s)(0.650 T) 3.59 10 3 N. F z q ( v x B y v y B x ) (9.45 10 8 C)( 3.11 10 4 m/s)(0.650 T) 1.91 10 3 E VALUATE : F is perpendicular to both v and . B We can verify that 0.  Since B is along the x -axis, x v does not affect the force components. 27.58. I DENTIFY : The period is 2/ Tr v , the current is / Qt and the magnetic moment is IA S ET U P : The electron has charge e . The area enclosed by the orbit is 2 r . 27
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Ch. 27 page 3/10 E XECUTE : (a) 16 21 . 5 1 0 s Tr v  (b) Charge e passes a point on the orbit once during each period, so 1.1mA IQ te t . (c) 22 4 2 9.3 10 A m IA I r     E VALUATE : Since the electron has negative charge, the direction of the current is opposite to the direction of motion of the electron. 27.59 See Addendum #1 27.61. (a) I DENTIFY and S ET U P : The maximum radius of the orbit determines the maximum speed v of the protons. Use Newton's 2nd law and 2 / c av R for circular motion to relate the variables. The energy of the particle is the kinetic energy 2 1 2 . K mv E XECUTE : m F =a gives 2 (/) qvB mv R 19 7 27 (1.60 10 C)(0.85 T)(0.40 m) 3.257 10 m/s.
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This note was uploaded on 12/18/2011 for the course PHYS 2B 2b taught by Professor Hirsch during the Spring '10 term at UCSD.

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23 HMK #6 Chp 27a - Ch. 27 PHYSICS 2B Spring 2011 Homework...

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