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309-2008-Solutions9

# 309-2008-Solutions9 - ECE 309 Electromagnetic Fields...

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ECE 309 — Electromagnetic Fields University of Virginia Fall 2008 Homework # 9 Solutions 1. From either Gauss’s Law or Laplace’s equation, the field between the inner and outer con- ductors of a cylindrical capacitor is, ~ E = V r ln( b/a ) ˆ r where V is the voltage across the capacitor, a is the inner radius and b is the outer radius. The displacement current density is thus, ~ J d = ~ E ∂t ~ J d = V r ln( b/a ) where the second relation above holds for phasor representations of ~ J d and V . The total current flowing across a cylindrical surface of radius r is thus, I d = V ln( b/a ) Z 2 π 0 Z 0 rdφdz r = 2 π‘ V ln( b/a ) which is obviously independent of r . The capacitance of the coaxial capacitor is given by, C = Q V = λ‘ V = 2 π ‘ ln( b/a ) where λ is the charge per unit length on the coaxial capacitor. Thus, the charging current for the capacitor is given (from circuit theory) by, I c = C dV dt I c = jωCV = 2 π ‘ V ln( b/a ) which is the same as the displacement current found above. 2. Given the electric field, ~ E = ˆ xA ( x + y ) + ˆ yB ( x - y ) cos ωt

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is in a source-free medium, we know that it is divergenceless, ~ ∇ · ~ E = 0 . Thus, ~ ∇ · ~ E = ∂E x ∂x + ∂E y ∂y + ∂E z ∂z = A - B = 0 Thus, the relation between the coefficients is, A = B 3. To find the ~ H -field, let’s first express the electric field in phasor form, ~ E = ˆ yE 0 sin πx a e - z z ( * ) From Faraday’s Law, ~ ∇ × ~ E = - jωμ ~ H, ~ H = j ωμ ~ ∇ × ~ E so that ~ H = j ωμ h - ˆ x ∂E y ∂z + ˆ z ∂E y ∂x i = j ωμ h z E 0 sin πx a e - z z ˆ x + π a E 0 cos πx a e - z z i ˆ z = - β z ωμ E 0 sin πx a e - z z ˆ x + j π ωμa E 0 cos πx a e - z z ˆ z To find the instantaneous field, we multiply by exp( jωt )
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