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Unformatted text preview: Proficiency Exam, Fall 2001 Friday, September 14 Part 1 9:00 a.m. to 1:00 p.m. Useful constants: e = 1 . 60 10 19 C hc = 1240 eV nm c = 3 . 00 10 8 m s m e = 0 . 511 MeV c 2 k = 1 4 = 8 . 99 10 9 Nm 2 C 2 = 8 . 85 10 12 C 2 Nm 2 4 = 10 7 N A 2 g = 9 . 81 m s 2 1. Four infinitely long wires parallel to the zaxis are arranged with ( x, y ) = (0 , d ) , (0 , d ) , ( d, 0) , and ( d, 0) , respectively. (Here d > .) The wires at (0 , d ) and (0 , d ) carry a current of magnitude i in the positive zdirection, and the wires at ( d, 0) and ( d, 0) carry a current of magnitude i in the negative zdirection. (a) What are the magnitude and direction of the magnetic field at any point along the xaxis? For better understanding we try to explicitly calculate almost all part of the problem. First of all consider a point on the xaxis and a point on each one of the wires. We have to determine contribution from each wire to the total magnetic field for a point on the xaxis. 1. First from the wire with coordinate (0 , d, z ). Here we denote the source points with prime and the field points unprimed, see figure. The magnetic field from an element of length dl at point ( x, , 0) is given by BiotSavart law. d ~ B 1 = I 4 ~ dl ~ R R 3 (1) ~ R = ~ r ~ r (2) ~ r = ( x, , 0) , ~ r = (0 , d, z ) (3) ~ R = ( x, d, z ) (4) d ~ l = z dl (5) d ~ l ~ R = ( x d + y x ) (6) R r r d l y z x I Substitute Eqs.25 in Eq.1 and integrate ~ B 1 = I 4 Z  dz x d + y x x 2 + d 2 + z 2 3 2 (7) let x 2 + d 2 = a 2 and tan( ) = z a dz a = (1 + tan 2 ( )) d d = dz a (1 + tan 2 ( )) = dz a (1 + ( z a ) 2 ) note also 1 1 + ( z a ) 2 = 1 1 + tan 2 ( ) = cos 2 ( ) and, z ] [ ] 2 2 [ (8) Rewrite Eq.7 we have ~ B 1 = I 4 a 2 ( x d + y x ) Z  dz a (1 + ( z a ) 2 ) s (1 + z a ) 2 integrand is even function of z f ( z ) = f ( z ) ~ B 1 = I 4 x d + y x x 2 + d 2 2 Z 2 d cos( ) note that cos( ) 0 for...
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