Chapter 7: Problem 9
A current sheet \(\mathbf{K}=8 \mathbf{a}_{x} \mathrm{~A} / \mathrm{m}\) flows
in the region \(-2
Chapter 7: Problem 9
A current sheet \(\mathbf{K}=8 \mathbf{a}_{x} \mathrm{~A} / \mathrm{m}\) flows
in the region \(-2
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Get started for freePlanar current sheets of \(\mathbf{K}=30 \mathbf{a}_{z} \mathrm{~A} /
\mathrm{m}\) and \(-30 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) are located in
free space at \(x=0.2\) and \(x=-0.2\), respectively. For the region \(-0.2
(An inversion of Problem 7.20.) A solid, nonmagnetic conductor of circular cross section has a radius of \(2 \mathrm{~mm}\). The conductor is inhomogeneous, with \(\sigma=10^{6}\left(1+10^{6} \mathrm{\rho}^{2}\right) \mathrm{S} / \mathrm{m}\). If the conductor is \(1 \mathrm{~m}\) in length and has a voltage of \(1 \mathrm{mV}\) between its ends, find: \((a) \mathbf{H}\) inside; \((b)\) the total magnetic flux inside the conductor.
An infinite filament on the \(z\) axis carries \(20 \pi \mathrm{mA}\) in the \(\mathbf{a}_{z}\) direction. Three \(\mathbf{a}_{z}\) -directed uniform cylindrical current sheets are also present: \(400 \mathrm{~mA} / \mathrm{m}\) at \(\rho=1 \mathrm{~cm},-250 \mathrm{~mA} / \mathrm{m}\) at \(\rho=2 \mathrm{~cm}\), and \(-300 \mathrm{~mA} / \mathrm{m}\) at \(\rho=3 \mathrm{~cm}\). Calculate \(H_{\phi}\) at \(\rho=0.5,1.5,2.5\), and \(3.5 \mathrm{~cm}\).
In spherical coordinates, the surface of a solid conducting cone is described by \(\theta=\pi / 4\) and a conducting plane by \(\theta=\pi / 2 .\) Each carries a total current I. The current flows as a surface current radially inward on the plane to the vertex of the cone, and then flows radially outward throughout the cross section of the conical conductor. \((a)\) Express the surface current density as a function of \(r ;(b)\) express the volume current density inside the cone as a function of \(r ;(c)\) determine \(\mathbf{H}\) as a function of \(r\) and \(\theta\) in the region between the cone and the plane; \((d)\) determine \(\mathbf{H}\) as a function of \(r\) and \(\theta\) inside the cone.
Show that the line integral of the vector potential A about any closed path is equal to the magnetic flux enclosed by the path, or \(\oint \mathbf{A} \cdot d \mathbf{L}=\int \mathbf{B} \cdot d \mathbf{S}\).
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