HW12 - Physics 3122, Fall 2010: Homework Set 12 Solutions....

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Unformatted text preview: Physics 3122, Fall 2010: Homework Set 12 Solutions. 5.13 ∫ B⋅ d = µ0Ienc (a) s < a ˆ B = B( s)ϕ =0 B =0 Ienc ∫ Jda = ∫ (b) J = ks s<a Ienc = ∫ s>a € I= a s>a Ienc = I B2πs = µ0 I ks2πsds = 2πka 3 / 3 ˆ B = µ0 Iϕ / 2πs k = 3I / 2πa 3 same as (a) 0 s ks+ 2πs+ ds+ = 2πks3 / 3 = Is3 / a 3 0 ˆ B = µ0 Is2ϕ / 2πa 3 B2πs = µ0 Is3 / a 3 5.22 (a) µ0 Id ˆˆ ˆ ˆ A( r ) = ∫ ℜ r = ss + zz r + = z+ z d = dz+ z ℜ = [s2 + (z − z+ )2 ]1/ 2 4π µ Iz z 2 ˆ ˆ dz+ µ Iz Ⱥ Ⱥ z2 A( r ) = 0 ∫ 2 = − 0 Ⱥln ( z − z+ ) + s2 + ( z − z+ ) 2 Ⱥ Ⱥ z1 4 π z1 [ s + ( z − z+ ) 2 ]1 / 2 4 π Ⱥ { } µ Iz Ⱥ ( z − z1 ) + s2 + ( z − z1 ) 2 Ⱥ ˆ Ⱥ A( r ) = 0 ln Ⱥ 4 π Ⱥ ( z − z2 ) + s2 + ( z − z2 ) 2 Ⱥ Ⱥ Ⱥ € 5.25 ˆ B = B( s)ϕ ˆ A = A( s) z In B2πs = µ0 ( I / πR 2 )(πs 2 ) B = µ0 Is / 2πR 2 Out B2πs = µ0 I B = µ0 I / 2πs ˆ ∇⋅ A = ∂A( s) /∂z = 0 B = ∇ × A = [ −∂A( s) /∂s]ϕ In ∂A( s) /∂s = − µ0 Is / 2πR 2 Out ∂A( s) /∂s = − µ0 I / 2πs € € A( s) = − µ0 Is2 / 4 πR 2 + C A( R) = 0 A( s) = µ0 I ( R 2 − s2 ) / 4 πR 2 A( s) = −µ0 I ln( s) / 2π + C A( R) = 0 A( s) = − µ0 I ln( s / R) / 2π 5.34 ˆ ˆ ˆˆ ˆ ˆˆ ˆ m = IAn = IπR 2 z B( r ) ≈ µ0 [ 3( m⋅ r ) r − m] / 4 πr 3 = µ0 IR 2 [ 3( z ⋅ r ) r − z ] / 4 r 3 ˆ ˆ ˆ r = zz B( r ) ≈ µ0 IR 2 z / 2 z 3 c.f. B( r ) = µ0 R 2 z / 2( R 2 + z 2 ) 3 / 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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