Solutions6

Solutions6 - Solutions to Problem Set 6 1. a) The current...

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Solutions to Problem Set 6 1. a) The current and charge as functions of time are I ( t ) = I 0 cos ωt Q ( t ) = Q 0 sin ωt = I 0 ω sin ωt Electric field points in z-direction but depends just on the radial coordinate (and time), E ( r,t ) = E ( r z sin ωt satisfies the Bessel’s differential equation 2 E - 1 c 2 2 E ∂t 2 = 0 1 r ∂r r ∂E ( r ) ∂r + k 2 E ( r ) = 0 The solution is E ( r ) = KJ 0 ( kr ) where K is some constant. We can determine it by noticing that Q 0 = Z σ dS where σ = ± 0 E ( r ) Using also the fact that Q 0 = I 0 ω we get the solution for E ( r ), E ( r ) = I 0 2 πa± 0 c J 0 ( kr ) J 1 ( ka ) In order to determine B ( r ) we use ∇ × E = - B ∂t = ωB 0 ( r z sin ωt which gives B 0 ( r ) = I 0 2 πac 2 ± 0 J 1 ( kr ) J 1 ( ka ) The first two terms in the expansions are E ( r ) = I 0 πa 2 ω± 0 ± 1 - ω 2 r 2 4 c 2 + ω 2 a 2 8 c 2 + O ( ω 4 ) ² B ( r ) = μ 0 I 0 r 2 πa 2 ω ± 1 - ω 2 r 2 8 c 2 + ω 2 a 2 8 c 2 + O ( ω 4 ) ² b) We just use the definition of w e and w m
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Solutions6 - Solutions to Problem Set 6 1. a) The current...

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