# ssm_ch28 - Chapter 28 3(a The magnitude of the magnetic...

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Chapter 28 3 (a) The magnitude of the magnetic force on the proton is given by F B = evB sin Á ,whe re v is the speed of the proton, B is the magnitude of the magnetic field, and Á is the angle between the particle velocity and the field when they are drawn with their tails at the same point. Thus v = F B eB sin Á = 6 : 50 £ 10 ¡ 17 N (1 : 60 £ 10 ¡ 19 C)(2 : 60 £ 10 ¡ 3 T)sin23 : 0 ± =4 : 00 £ 10 5 m = s : (b) The kinetic energy of the proton is K = 1 2 mv 2 = 1 2 (1 : 67 £ 10 ¡ 27 kg)(4 : 00 £ 10 5 m = s) 2 =1 : 34 £ 10 ¡ 16 J : This is (1 : 34 £ 10 ¡ 16 J) = (1 : 60 £ 10 ¡ 19 J = eV) = 835 eV. 17 (a) Since the kinetic energy is given by K = 1 2 mv 2 ,whe re m is the mass of the electron and v is its speed, v = r 2 K m = s 2(1 : 20 £ 10 3 eV)(1 : 60 £ 10 ¡ 19 J = eV) 9 : 11 £ 10 ¡ 31 kg =2 : 05 £ 10 7 m = s : (b) The magnitude of the magnetic force is given by evB and the acceleration of the electron is given by v 2 =r ,whe re r is the radius of the orbit. Newton’s second law is evB = mv 2 =r ,so B = mv er = (9 : 11 £ 10 ¡ 31 kg)(2 : 05 £ 10 7 m = s) (1 : 60 £ 10 ¡ 19 C)(25 : 0 £ 10 ¡ 2 m) =4 : 68 £ 10 ¡ 4 T=468 ¹ T : (c) The frequency f is the number of times the electron goes around per unit time, so f = v 2 ¼r = 2 : 05 £ 10 7 m = s 2 ¼ (25 : 0 £ 10 ¡ 2 m) =1 : 31 £ 10 7 Hz = 13 : 1MHz : (d) The period is the reciprocal of the frequency: T = 1 f = 1 1 : 31 £ 10 7 Hz =7 : 63 £ 10 ¡ 8 s=76 : 3ns : 29 (a) If v is the speed of the positron, then v sin Á is the component of its velocity in the plane that is perpendicular to the magnetic field. Here Á is the angle between the velocity and the field (89 ± ). Newton’s second law yields eBv sin Á = m ( v sin Á ) 2 =r ,where r is the radius of the orbit. Thus r =( mv=eB )sin Á . The period is given by T = 2 ¼r v sin Á = 2 ¼m eB = 2 ¼ (9 : 11 £ 10 ¡ 31 kg) (1 : 60 £ 10 ¡ 19 C)(0 : 100 T) =3 : 58 £ 10 ¡ 10 s : The expression for r was substituted to obtain the second expression for T

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(b) The pitch p is the distance traveled along the line of the magnetic field in a time interval of one period. Thus p = vT cos Á . Use the kinetic energy to find the speed: K = 1 2 mv 2 yields v = r 2 K m = s 2(2 : 0 £ 10 3 eV)(1 : 60 £ 10 ¡ 19 J = eV) 9 : 11 £ 10 ¡ 31 kg =2 : 651 £ 10 7 m = s : Thus p =(2 : 651 £ 10 7 m = s)(3 : 58 £ 10 ¡ 10 s) cos 89 : 0 ± =1 : 66 £ 10 ¡ 4 m : (c) The orbit radius is r = mv sin Á eB = (9 : 11 £ 10 ¡ 31 kg)(2 : 651 £ 10 7 m = s) sin 89 : 0 ± (1 : 60 £ 10 ¡ 19 C)(0 : 100 T) =1 : 51 £ 10 ¡ 3 m : 41 (a) The magnitude of the magnetic force on the wire is given by F B = iLB sin Á ,whe re i is the current in the wire, L
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## This note was uploaded on 01/19/2009 for the course PHYS 171.102 taught by Professor Heckman during the Fall '98 term at Johns Hopkins.

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ssm_ch28 - Chapter 28 3(a The magnitude of the magnetic...

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