ssm_ch26 - Chapter 26 7 (a) The magnitude of the current...

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Chapter 26 7 (a) The magnitude of the current density is given by J = nqv d ,w h e r e n is the number of particles per unit volume, q is the charge on each particle, and v d is the drift speed of the particles. The particle concentration is n =2 : 0 £ 10 8 cm ¡ 3 =2 : 0 £ 10 14 m ¡ 3 , the charge is q =2 e =2(1 : 60 £ 10 ¡ 19 C) = 3 : 20 £ 10 ¡ 19 C, and the drift speed is 1 : 0 £ 10 5 m = s. Thus J =(2 £ 10 14 m ¡ 3 )(3 : 2 £ 10 ¡ 19 C)(1 : 0 £ 10 5 m = s) = 6 : 4A = m 2 : (b) Since the particles are positively charged, the current density is in the same direction as their motion, to the north. (c) The current cannot be calculated unless the cross-sectional area of the beam is known. Then i = JA can be used. 17 The resistance of the wire is given by R = ½L=A ,whe re ½ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire. The cross-sectional area is A = ¼r 2 = ¼ (0 : 50 £ 10 ¡ 3 m) 2 =7 : 85 £ 10 ¡ 7 m 2 .H e r e r =0 : 50 mm = 0 : 50 £ 10 ¡ 3 m is the radius of the wire. Thus ½ = RA L = (50 £ 10 ¡ 3 ­ )(7 : 85 £ 10 ¡ 7 m 2 ) 2 : 0m =2 : 0 £ 10 ¡ 8 ­ ¢ m : 19 The resistance of the coil is given by R = ½L=A ,whe re L is the length of the wire, ½ is the resistivity of copper, and A is the cross-sectional area of the wire. Since each turn of wire has length 2 ¼r ,where r is the radius of the coil, L =(250)2 ¼r = (250)(2 ¼ )(0 : 12 m) = 188 : 5m. If r w is the radius of the wire, its cross-sectional area is A = ¼r 2 w = ¼ (0 : 65 £ 10 ¡ 3 m) 2 =1 :
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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_ch26 - Chapter 26 7 (a) The magnitude of the current...

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