Field in a finite solenoid.

So going back to Eq. 1, we obtain∇ ·⃗B = 1 r ∂ r (rB r ) + 1 r ∂ θB θ + ∂ z B z = 1 r ∂ r (rB r ) = 0Hence B r = B 0/r, which is not allowed if the domain includes r = 0. i.e., Br ≡ 0. So,This magnetic field is obtained from⃗B(⃗r) = µ Z04π⃗B(⃗r) = B θ ˆθ + B z ẑ⃗j(⃗r ′ ) × ∇ 1R 12dV ′where ⃗j = I 0 n ˆθ ′ δ(r ′ − a). To make progress we go to the Vector Potential⃗A(⃗r) = µ Z0 ⃗ j(⃗r ′ )dV ′4π R 12Since the problem is symmetric in θ and z, we assume θ = 0 and z = 0. ⃗j(⃗r ′ ) is along ˆθ ′ , i.e., along − ˆxsinθ ′ + ŷcosθ ′ . Then,⃗A(r) = µ Z ∞ Z π0dz ′ adθ ′ I 0 n(−sin(θ ′ ) ˆx + cos(θ ′ )ŷ)√4π −∞ −π(x − acosθ ′ ) 2 + a 2 sin 2 θ ′ + z ′2= µ Z ∞ Z π0dz ′ adθ ′ I 0 ncos(θ ′ )ŷ√4π −∞ −π(x − acosθ ′ ) 2 + a 2 sin 2 θ ′ + z ′2= µ Z ∞ Z π04π I 0nŷ dz ′ adθ ′ cos(θ ′ )√−∞ −π(x − acosθ ′ ) 2 + a 2 sin 2 θ ′ + z ′2For θ = 0, ŷ is the same as ˆθ, since ˆr is along ˆx. Thus, this shows us that ⃗A(r) is along ˆθ which is what we wanted to get out of thisexercise.We are now almost done. If ⃗A is along θ and depends only on r, its curl∣ ∣∣∣∣∣1ˆr rˆθ ẑr det ∂ r ∂ θ ∂ z0 rA θ 0 ∣ = 1 r ∂ r (rA θ )ẑ⃗B is purely along ẑ and depends only on r. We could use the above formulae to calculate it, but there is now an easier way. We useStoke’s theorem on a loop in the r − z plane, with unit length along z, and its two sides at r = 0 and at r.I⃗B · ⃗dl = B z (0) − B z (r)= µ 0 I enclThus, we immediately conclude (assuming B z → 0 as r → ∞){µ0 nIB z (r) = 0 r < a0 r > aFinite SolenoidWhat happens when the solenoid is not infinitely long?∂ z is no longer zero, and so, B r can now be present:∇ ·⃗B = 1 r ∂ r (rB r ) + ∂ z B z = 0 (2)The arguments we used for ⃗A do not change; the only difference is that the limits in z are now finite. So we have⃗A(⃗r) = A θ (r,z)ˆθTaking the curl yields⃗B(⃗r) = −ˆr∂ z A θ + ẑ 1 r ∂ r (rA θ )2

The divergence should automatically go to zero, as it does:( )11r ∂ r (−r∂ z A θ ) + ∂ zr ∂ r (rA θ )since 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 sin 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 integral is known in closed form (look up your favourite table of integrals. It is in all of them):Z βαdz(a 2 + z 2 ) 3/2 = β√β 2 + a − α√ 2 α 2 + a 2Applying the formula above, we obtain A θ and B z for the infinite case:a 2For the finite 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 noting that the fringing fields of a solenoid do not fall off exponentially! Remember the case of the capacitor. There thefringing 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 fringing fields can penetrate a lot further.3