309-2008-Solutions8

309-2008-Solutions8 - ECE 309 — Electromagnetic Fields...

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Unformatted text preview: ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 8 Solutions 1. Magnetized cylinder. Take the cylinder (and direction of magnetization) to be along the z-axis, Thus, the bound current densities are ~ J b = ~ ∇ × ~ M = 0 , ~ K b = ~ M × ˆ n = M (ˆ z × ˆ r ) = M ˆ φ Because of the symmetry of the problem, we can find ~ H directly from Ampere’s Law. Because there is no free current in this problem, I ~ H · d ~ ‘ = 0 ⇒ ~ H = 0 , everywhere Using the general relation between ~ H , ~ B , and ~ M , we can find ~ B : ~ B = μ ( ~ H + ~ M ) ~ B = μ ~ M, inside cylinder ~ B = 0 , outside cylinder 2. Long Magnetized cylinder. The cylinder (radius R , oriented along the z axis) has magnetization, ~ M = kr 2 ˆ φ Thus, the bound current densities are ~ K b = ~ M × ˆ n = kR 2 ˆ φ × ˆ r =- kR 2 ˆ z , ~ J b = ~ ∇ × ~ M = ˆ z 1 r d dr ( rM φ ) = ˆ z 1 r d dr ( kr 3 ) = 3 kr ˆ z We can now apply Ampere’s Law to find ~ B , I C ~ B · d ~ ‘ = μ I enc Choosing C to be a circular loop axial with the cylinder (from symmetry considerations), we find B φ ( r < R ) · 2 πr = μ Z r Z 2 π 3 kr · r dr dφ = 3 kμ · 2 π Z r r 2 dr = 2 πkμ r 3 Thus, ~ B ( r ) = μ kr 2 ˆ φ, r < R Outside the cylinder, we have B φ ( r > R ) · 2 πr = 2 πkμ R 3- kμ R 2 Z 2 π Rdφ = 2 πkμ R 3- 2 πkμ R 3 = 0 ~ B ( r ) = 0 , r > R 3. Small cavities inside a magnetized medium....
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309-2008-Solutions8 - ECE 309 — Electromagnetic Fields...

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