HW11 - Physics 3122 Fall 2010 Homework Set 11 Solutions 5.1 R s 2 a 2 = R 2 F = qvB = mv 2 R R 2 2 Rs s2 a 2 = R 2 R =(s2 a 2 2s mv = qBR = qB s2 a

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Physics 3122, Fall 2010: Homework Set 11 Solutions. 5.1 ( R s ) 2 + a 2 = R 2 R 2 2 Rs + s 2 + a 2 = R 2 R = ( s 2 + a 2 ) /2 s F = qvB = mv 2 / R mv = qBR = qB ( s 2 + a 2 ) /2 s 5.4 B = ( kz )/ x = 0 OK Bottom F 1 = I ˆ y dy × k ( a /2) ˆ x a / 2 a / 2 = 1 2 Ika 2 ˆ z Right F 2 = I ˆ z dz × kz ˆ x a / 2 a / 2 = 1 4 Ika 2 ˆ y Top F 3 = I ( ˆ y ) dy × k ( a /2) ˆ x a / 2 a / 2 = 1 2 Ika 2 ˆ z Left F 4 = I ( ˆ z ) dz × kz ˆ x a / 2 a / 2 = 1 4 Ika 2 ˆ y F = F 1 + F 2 + F 3 + F 4 = Ika 2 ˆ z 5.8( a ) B = μ 0 I ˆ z 4 π y ( x a ) x a ) 2 + y 2 ( x b ) x b ) 2 + y 2 Ρ Σ ΢ ΢ Τ Φ Υ Υ x = 0 a = R b = R y = R B 1 = 0 I ˆ z 4 R R 2 + R 2 ( R ) R 2 + R 2 Ρ Σ ΢ Τ Φ Υ = 0 I ˆ z 4 R B = 4 B 1 = 2 0 I ˆ z R or B = 0 I (sin θ 2 sin 1 ) ˆ n /4 s s = R 1 = /4 2 = B 1 = 0 I 1/ ( ) [ ] ˆ n R = 0 I ˆ n R B = 4 B 1 = 0 I ˆ n / R 5.11 d B = 0 dIa 2 /2[ a 2 + ( z ʹ z ) 2 ] 3 / 2 dI = nId ʹ z α < ʹ z < β B = 0 nI 2 a 2 d ʹ z [ a 2 + ( z ʹ z ) 2 ]
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This note was uploaded on 01/22/2012 for the course PHYS 3122 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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