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PHY 5346 Homework Set 10 Solutions – Kimel 2. 5.2 a) The system is described by First consider a point at the axis of the solenoid at point z0. Using the results of problem 5.1, dφm = μ0 4π NIdzΩ From the figure, Ω = ∫ r̂ ⋅ dA⃗ r2 = ∫ dAcosθ r2 = 2πz ∫ 0 R ρdρ ρ2 + z2 3/2 = 2π − z R2 + z2 + 1 φm = μ0 2 NI ∫ z0 ∞ z − 1 R2 + z2 + 1z dz = μ0 2 NI −z0 + R2 + z0 2 Br = − μ0 2 NI ∂ ∂z0 −z0 + R2 + z0 2 = μ0 2 NI −z0 + R2 + z0 2 R2 + z0 2 In the limit z0 → 0 Br = μ0 2 NI By symmetry, thej loops to the left of z0 give the same contribution, so B = B l + Br = μ0NI H = NI By symmetry, B⃗ is directed along the z axis, so δφm = −δρ⃗ ⋅ B⃗ = 0 if δρ⃗ is directed ⊥ to the z axis. Thus for a given z, φm is independent of ρ, and consequently H = NI everywhere within the solenoid. If you are on the outside of the solenoid at position z0, by symmetry the magnetic field must be in the z direction. Thus using the above argument, φm must not depend on ρ. Set us take ρ far away from the axis of the solenoid, so that we can replace the loops by elementary dipoles m⃗ directed along the z axis. Thus for any point z0 we will have a contributions φm m⃗ ⋅ r⃗1 r1 3 + m⃗ ⋅ r⃗2 r2 3 where m⃗ ⋅ r⃗1 = −m⃗ ⋅ r⃗2 and r1 = r2. Thus H = 0
