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309-2008-Solutions7

309-2008-Solutions7 - ECE 309 Electromagnetic Fields...

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ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 7 Solutions 1. B -field at center of regular N -sided polygon. For an N -sided regular polygon, each side subtends an angle of 2 θ = 2 π/N : The B -field at the center will be the superposition of the B -fields for each side of the polygon. Using the expression (derived in class) for the field a distance b cos θ from the center of a wire of length = 2 b sin θ , we have ~ B one side = μ 0 I 4 πb cos θ · 2 sin θ ˆ n where ˆ n is the direction of the field (into or out of the page). Thus for the N -sided polygon, ~ B ( P ) = N μ 0 I 4 πb cos θ 2 sin θ ˆ n or, ~ B ( P ) = ˆ n μ 0 NI 2 πb tan π N as N → ∞ , π/N becomes small and the small-angle approximation pertains: tan π N π N Thus, ~ B ( P ) = ˆ n μ 0 I 2 b , for N → ∞

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which is the same as the field at the center of a circular loop. 2. Current flowing down a cylindrical wire. This problem exhibits cylindrical symmetry, so we can apply Ampere’s Law to find ~ B , I ~ B · d ~ = μ 0 I enc = μ 0 Z S ~ J · d~a where ~ J = kr ˆ z . Using a coaxial circle (with radius r ) as the Amperian loop, we note that B is azimuthally- directed and constant along such a loop. Thus 2 πrB φ = μ 0 k Z 2 π 0 Z r 0 r 2 dr = 2 πμ 0 k 3 r 3 r < R 2 πμ 0 k 3 R 3 r > R Thus, the B -field is, ~ B = ˆ φ μ 0 k 3 r 2 , for r R ~ B = ˆ φ μ 0 k 3 R 3 r , for r R 3. The Hall Effect. (a) The force on a moving charge is given by the Lorentz force Law, ~ F = q~v × ~ B The current consists of positive charges moving to the right with average velocity ~v , so ~v × ~ B points in the - y direction. Thus,
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