Chapter 7: Problem 3
Two semi-infinite filaments on the \(z\) axis lie in the regions \(-\infty
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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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Get started for freeA disk of radius \(a\) lies in the \(x y\) plane, with the \(z\) axis through its center. Surface charge of uniform density \(\rho_{s}\) lies on the disk, which rotates about the \(z\) axis at angular velocity \(\Omega \mathrm{rad} / \mathrm{s}\). Find \(\mathbf{H}\) at any point on the \(z\) axis.
Let \(\mathbf{A}=(3 y-z) \mathbf{a}_{x}+2 x z \mathbf{a}_{y} \mathrm{~Wb} / \mathrm{m}\) in a certain region of free space. (a) Show that \(\nabla \cdot \mathbf{A}=0 .(b)\) At \(P(2,-1,3)\), find \(\mathbf{A}, \mathbf{B}, \mathbf{H}\), and \(\mathbf{J}\).
Consider a sphere of radius \(r=4\) centered at \((0,0,3)\). Let \(S_{1}\) be that portion of the spherical surface that lies above the \(x y\) plane. Find \(\int_{S_{1}}(\nabla \times \mathbf{H}) \cdot d \mathbf{S}\) if \(\mathbf{H}=3 \rho \mathbf{a}_{\phi}\) in cylindrical coordinates.
A current filament carrying \(I\) in the \(-\mathbf{a}_{z}\) direction lies along the entire positive \(z\) axis. At the origin, it connects to a conducting sheet that forms the \(x y\) plane. (a) Find \(\mathbf{K}\) in the conducting sheet. \((b)\) Use Ampere's circuital law to find \(\mathbf{H}\) everywhere for \(z>0 ;(c)\) find \(\mathbf{H}\) for \(z<0\).
The magnetic field intensity is given in a certain region of space as \(\mathbf{H}=\) \(\left[(x+2 y) / z^{2}\right] \mathbf{a}_{y}+(2 / z) \mathbf{a}_{z} \mathrm{~A} / \mathrm{m} .(a)\) Find \(\nabla \times \mathbf{H} .(b)\) Find \(\mathbf{J} .(c)\) Use \(\mathbf{J}\) to find the total current passing through the surface \(z=4,1 \leq x \leq 2,3 \leq z \leq 5\), in the \(\mathbf{a}_{z}\) direction. ( \(d\) ) Show that the same result is obtained using the other side of Stokes' theorem.
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