Chapter 7: Problem 3
Two semi-infinite filaments on the \(z\) axis lie in the regions \(-\infty
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Chapter 7: Problem 3
Two semi-infinite filaments on the \(z\) axis lie in the regions \(-\infty
These are the key concepts you need to understand to accurately answer the question.
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A filamentary conductor on the \(z\) axis carries a current of \(16 \mathrm{~A}\)
in the \(\mathbf{a}_{z}\) direction, a conducting shell at \(\rho=6\) carries a
total current of \(12 \mathrm{~A}\) in the \(-\mathbf{a}_{z}\) direction, and
another shell at \(\rho=10\) carries a total current of \(4 \mathrm{~A}\) in the
\(-\mathbf{a}_{z}\) direction. \((a)\) Find \(\mathbf{H}\) for \(0<\rho<12
.\left(\right.\) b) Plot \(H_{\phi}\) versus \(\rho\).
(c) Find the total flux \(\Phi\) crossing the surface \(1<\rho<7,0
Use an expansion in rectangular coordinates to show that the curl of the gradient of any scalar field \(G\) is identically equal to zero.
A square filamentary differential current loop, \(d L\) on a side, is centered at the origin in the \(z=0\) plane in free space. The current \(I\) flows generally in the \(\mathbf{a}_{\phi}\) direction. ( \(a\) ) Assuming that \(r>>d L\), and following a method similar to that in Section \(4.7\), show that $$d \mathbf{A}=\frac{\mu_{0} I(d L)^{2} \sin \theta}{4 \pi r^{2}} \mathbf{a}_{\phi}$$ (b) Show that $$d \mathbf{H}=\frac{I(d L)^{2}}{4 \pi r^{3}}\left(2 \cos \theta \mathbf{a}_{r}+\sin \theta \mathbf{a}_{\theta}\right)$$ The square loop is one form of a magnetic dipole.
Assume that there is a region with cylindrical symmetry in which the conductivity is given by \(\sigma=1.5 e^{-150 \rho} \mathrm{kS} / \mathrm{m}\). An electric field of \(30 \mathbf{a}_{z} \mathrm{~V} / \mathrm{m}\) is present. ( \(a\) ) Find \(\mathbf{J}\). \((b)\) Find the total current crossing the surface \(\rho<\rho_{0}\), \(z=0\), all \(\phi\). ( \(c\) ) Make use of Ampère's circuital law to find \(\mathbf{H}\).
A wire of \(3 \mathrm{~mm}\) radius is made up of an inner material \((0<\rho<2 \mathrm{~mm})\) for which \(\sigma=10^{7} \mathrm{~S} / \mathrm{m}\), and an outer material ( \(2 \mathrm{~mm}<\rho<3 \mathrm{~mm}\) ) for which \(\sigma=4 \times 10^{7} \mathrm{~S} / \mathrm{m}\). If the wire carries a total current of \(100 \mathrm{~mA}\) dc, determine \(\mathbf{H}\) everywhere as a function of \(\rho\).
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