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PHY 5346 Homework Set 9 Solutions – Kimel 3. 5.19 The system is described by The effective volume magnetic charge density is zero, since M⃗ is constant within the cylinder. The effective surface charge density (n̂ ⋅ M⃗ from Eq. (5.99)) is M0, on the top surface and −M0 on the bottom surface. From the bottom surface the potential is (for z > 0 Φb = 1 4π −M0 2π ∫ 0 a ρdρ ρ2 + z2 1/2 = − M0 2 a2 + z2 − z By symmetry, the potential from the top surface is (on the inside) Φ t = M0 2 a2 + L − z2 − L − z The total magnetic potential is Φ = Φb + Φ t = − M0 2 a2 + z2 − z + M0 2 a2 + L − z2 − L − z So, on the inside of the cylinder, Hz = − ∂∂z − M0 2 a2 + z2 − z + M0 2 a2 + L − z2 − L − z Hz = − M0 2 2 − z a2 + z2 − L − z a2 + L − z2 while above the cylinder, Hz = − M0 2 − z a2 + z2 + z − L a2 + L − z2 with a similar expression below the cylinder. B⃗ = μ0 H⃗ + M⃗ Thus inside the cylinder, Bz = μ0 − M0 2 2 − z a2 + z2 − L − z a2 + L − z2 + M0 Bz = μ0M0 2 z a2 + z2 + L − z a2 + L − z2 while above the cylinder, Bz = μ0M0 2 z a2 + z2 − z − L a2 + L − z2 First we plot Bz in units of a for L = 5a gz = 1 2 z 1+z2 + 5−z 1+5−z2 if z < 5 1 2 z 1+z2 − z−5 1+5−z2 if 5 < z gz 0 0.2 0.4 0.6 0.8 1 2 4 6 8 10 z And similarly, Hz in units of a for L = 5a. fz = − 12 2 − z 1+z2 − 5−z 1+5−z2 if z < 5 − 12 − z 1+z2 + z−5 1+5−z2 if 5 < z fz -0.4 -0.2 0 0.2 0.4 2 4 6 8 10 z
