# emsol09 - Electromagnetic Theory I Solution Set 9 Due 31...

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Electromagnetic Theory I Solution Set 9 Due: 31 Octber 2011 33. Consider the magnetic feld produced by a current I in an infnitely long wire lying on the z axis -∞ < z < . a) Use symmetry arguments and Amp` ere’s law to obtain the B feld everywhere outside the wire. Express the Cartesian components oF B as explicit Functions oF x, y, z . Solution : Ampere’s law with a circular current loop centered on the z axis yields, with r = p x 2 + y 2 , B = μ 0 I 2 πr ˆ ϕ = μ 0 I 2 πr ( - ˆ x sin ϕ + ˆ y cos ϕ ) B x = - μ 0 Iy 2 π ( x 2 + y 2 ) , B y = μ 0 Ix 2 π ( x 2 + y 2 ) , B z = 0 b) By direct integration oF each component oF ∇× A = B , fnd the vector potential A For this B in Coulomb gauge, ∇· A = 0. Solution : We can fnd a vector potential with A x = A y = 0 y A z = B x , A z = - μ 0 I 4 π ln( x 2 + y 2 ) + f ( x, z ) B y = - x A z = μ 0 Ix 2 π ( x 2 + y 2 ) + x f ( x, z ) , x f ( x, z ) = 0 We may set f = constant by fxing a gauge ∇ · A = z A z = 0. We can choose this constant to involve a distance r 0 to make the argument oF the ln dimensionless A z = - μ 0 I 4 π ln x 2 + y 2 r 2 0 c) Since ∇× B = 0 “almost everywhere” we should be able to fnd a scalar potential such that B = -∇ φ “almost everywhere”. By explicitly integrating the components oF this equation, fnd a candidate For φ as an explicit Function oF x, y, z . Solution : - ∂φ ∂z = 0 , - ∂φ ∂x = B x , φ = μ 0 I 2 π tan - 1 x y + f ( y ) - ∂φ ∂y = - f 0 ( y ) + μ 0 Ix 2 π ( x 2 + y 2 ) = B y , f 0 ( y ) = 0 φ = μ 0 I 2 π tan - 1 x y (1) 1

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d) In view of the fundamental theorem of calculus φ ( y ) - φ ( x ) = Z y x d l ·∇ φ = - Z y x d l · B , (2) explain how your result for part c) does not run afoul of Amp`
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emsol09 - Electromagnetic Theory I Solution Set 9 Due 31...

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