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The divergence should automatically go to zero, as it does:( )11r ∂ r (−r∂ z A θ ) + ∂ zr ∂ r (rA θ )s<strong>in</strong>ce second partial derivitives commute.As a special case, consider r ≪ a. The expression for A θ becomes, for small x= 0A θ = µ Z L/2 Z π04π I 0nŷ dζ adθ ′ cos(θ ′ )√−L/2 −π(x − acosθ ′ ) 2 + a 2 s<strong>in</strong> 2 θ ′ + ζ 2≃ µ Z L/2−z Z π04π I 0nŷ dζ adθ ′ cos(θ ′ )√−L/2−z −π a 2 + ζ 2 − 2axcosθ ′≃ µ 04π I 0nŷ= µ 04 I 0nŷa 2 xZ L/2−z−L/2−zZ L/2−z−L/2−zZ πdζ adθ ′ cos(θ ′ ())√ 1 + axcos(θ′ )−π a 2 + ζ 2 a 2 + ζ 21dζ(a 2 + ζ 2 ) 3/2The <strong>in</strong>tegral is known <strong>in</strong> closed form (look up your favourite table of <strong>in</strong>tegrals. It is <strong>in</strong> all of them):Z βαdz(a 2 + z 2 ) 3/2 = β√β 2 + a − α√ 2 α 2 + a 2Apply<strong>in</strong>g the formula above, we obta<strong>in</strong> A θ and B z for the <strong>in</strong>f<strong>in</strong>ite case:a 2For the f<strong>in</strong>ite case, this becomesA θ = µ 02 I 0nrB z = 1 r ∂ r (rA θ ) = µ 02I 0 nr ∂ (r r2 ) = µ 0 nI 0Z L/2−zB z = µ 0 nI 0 dζ−L/2−z 2(a 2 + ζ 2 ) 3/2⎡L / 2 − z−L / ⎤2 − z= µ 0 nI 0⎣√ (L / )− √ ⎦2 (−L / )2 − z + a 2 2 2 − z + a 2a 2It is worth not<strong>in</strong>g that the fr<strong>in</strong>g<strong>in</strong>g fields of a <strong>solenoid</strong> do not fall off exponentially! Remember the case of the capacitor. There thefr<strong>in</strong>g<strong>in</strong>g fields fell off exponentially. The difference between the two cases is that for the capacitor, we specified voltage rather thanσ(x) on the plates. Here, we have specified the precise currents. When we do that, we do not allow for “self-consistent” currents,and the fr<strong>in</strong>g<strong>in</strong>g fields can penetrate a lot further.3