Classical Electrodynamics: Computing Scalar Potential and Magnetic Field in a Cylinder, Assignments of Physics

Download Classical Electrodynamics: Computing Scalar Potential and Magnetic Field in a Cylinder and more Assignments Physics in PDF only on Docsity! PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 8 7 We can directly compute the contribution of the top face to the scalar potential. Here, we have that ϕ1(z) = 1 4π ∫ n′ ·M(x′) |x− x′| da ′, = 1 4π M02π ∫ a 0 ρ dρ ( ρ2 + (z − L/2)2 )1/2 , = M0 4 ∫ a2+(z−L/2)2 (z−L/2)2 du√ u , = M0 2 (√ a2 + (z − L/2)2 − |z − L/2| ) . From the obvious symmetry of the problem we see that ϕ2 = M0 2 (√ a2 + (z + L/2)2 − |z + L/2| ) . Let us now determine the magnetic field. We know that Hz = −∂zϕM = −∂z (ϕ1 + ϕ2). Notice that inside the cylinder ϕ1 + ϕ2 contains an additional M0z relative to when |z| > L/22. Computing this trivial derivative, we see that Hz = −M02   z + L/2( a2 + (z + L/2)2 )1/2 + z − L/2 ( a2 + (z − L/2)2 )1/2  − { 0 |z| > L/2 M0 |z| ≤ L/2 Now, notice that Bz = µ (Hz + M0) inside the cylinder and Bz = µHz outside the cylinder. Therefore, the additional ambiguity is cancelled and we see that, both inside and outside the cylinder, ∴ Bz = −µM02   z + L/2( a2 + (z + L/2)2 )1/2 + z − L/2 ( a2 + (z − L/2)2 )1/2   . ‘óπ²ρ ’²́δ²ι δ²ιξαι Problem 5.20 Starting from Jackson’s problem (5.12) and the fact the magnetization M inside a volume Ω bounded by a surface ∂Ω is equivalent to a volume current density JM = (∇×M) and a surface current density (M× n), we are to show that in the absence of macroscopic conduction currents, the total magnetic force on the body can be written F = − ∫ Ω (∇ ·M)Be d3x + ∫ ∂Ω (M · n)Be da, where Be is the applied magnetic induction (not including the body in question) . We will make use of a large number of trivial identities listed in the inside cover of Jackson’s text and elsewhere. Our derivation begins with the expression F = − ∫ Ω JM ×Be d3x, and using the form of J given in the problem, we can write F = ∫ Ω (∇×M)×Be d3x + ∫ ∂Ω (M× n)×Be da. In the first integrand, we have the expression (∇×M)×Be = −Be × (∇×M). Now, we know that −Be × (∇×M) = (M · ∇)Be + (Be · ∇)M + M× (∇×Be)−∇(M ·Be). Notice that the curl of Be vanishes. Along similar lines, we see that (M× n)×Be = −Be × (M× n) = −(B · n)M + (M ·Be)n. 2In the other two cases, above and below the cylinder, ϕ1 + ϕ2 does not depend linearly on z and the constant factor will vanish upon differentiation.