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fa01em_sol

# fa01em_sol - Proficiency Exam Fall 2001 Friday September 14...

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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 π 0 = 8 . 99 × 10 9 Nm 2 C 2 0 = 8 . 85 × 10 - 12 C 2 Nm 2 μ 0 4 π = 10 - 7 N A 2 g = 9 . 81 m s 2 1. Four infinitely long wires parallel to the z -axis are arranged with ( x, y ) = (0 , d ) , (0 , - d ) , ( d, 0) , and ( - d, 0) , respectively. (Here d > 0 .) The wires at (0 , d ) and (0 , - d ) carry a current of magnitude i in the positive z -direction, and the wires at ( d, 0) and ( - d, 0) carry a current of magnitude i in the negative z -direction. (a) What are the magnitude and direction of the magnetic field at any point along the x -axis? For better understanding we try to explicitly calculate almost all part of the problem. First of all consider a point on the x -axis 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 x-axis. 1. First from the wire with coordinate (0 , d, z 0 ). Here we denote the source points with prime and the field points unprimed, see figure. The magnetic field from an element of length dl 0 at point ( x, 0 , 0) is given by Biot-Savart law. d ~ B 1 = μ 0 I 4 π ~ dl 0 × ~ R R 3 (1) ~ R = ~ r - ~ r 0 (2) ~ r = ( x, 0 , 0) , ~ r 0 = (0 , d, z 0 ) (3) ~ R = ( x, - d, z 0 ) (4) d ~ l 0 = ˆ z dl 0 (5) d ~ l 0 × ~ R = x d + ˆ y x ) (6)

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R r r’ d l’ y z x I Substitute Eqs.2-5 in Eq.1 and integrate ~ B 1 = μ 0 I 4 π Z -∞ dz 0 ˆ x d + ˆ y x x 2 + d 2 + z 0 2 3 2 (7) let x 2 + d 2 = a 2 and tan( α ) = z 0 a dz 0 a = (1 + tan 2 ( α )) = dz 0 a (1 + tan 2 ( α )) = dz 0 a (1 + ( z 0 a ) 2 ) note also 1 1 + ( z 0 a ) 2 = 1 1 + tan 2 ( α ) = cos 2 ( α ) and, z 0 ] - ∞ ∞ [ α ] - π 2 π 2 [ (8) Rewrite Eq.7 we have ~ B 1 = μ 0 I 4 π a 2 x d + ˆ y x ) Z -∞ dz 0 a (1 + ( z 0 a ) 2 ) s (1 + z 0 a ) 2 integrand is even function of z 0 f ( z ) = f ( - z ) ~ B 1 = μ 0 I 4 π ˆ x d + ˆ y x x 2 + d 2 2 Z π 2 0 cos( α ) note that cos( α ) 0 for α ] - π 2 π 2 [ (9) ~ B 1 = μ 0 I 2 π ˆ x d + ˆ y x x 2 + d 2 (10) 2. For the wire at (0 , - d, z 0 ) the calculation will be the same except that we
have to change d → - d .

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