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Chapter 7: Problem 34

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

Short Answer

Expert verified
Based on the given problem, we found the magnetic field intensity $\mathbf{H}$ for each conductor using Ampere's circuital law. To find the total $\mathbf{H}$, we added the contributions from each conductor and plotted the $H_\phi$ as a function of $\rho$. Finally, the total magnetic flux $\Phi$ crossing the specified surface is calculated as $\frac{28}{\pi} \cdot 671 \mathrm{~Wb}$.

Step by step solution

01

a) Finding \(\mathbf{H}\) for each conductor

First, let's apply Ampere's circuital law to the three conductors to find \(\mathbf{H}\) for each conductor. For the filamentary conductor on the z-axis, the enclosed current is \(I_1 = 16 \mathrm{~A}\). We can then find \(\mathbf{H}_1\) using Ampere's circuital law: \(H_{1\phi} = \frac{I_1}{2 \pi \rho} = \frac{16}{2 \pi \rho} \mathbf{a}_{\phi}\) For the conducting shell at \(\rho = 6\), the enclosed current is \(I_2 = -12 \mathrm{~A}\): \(H_{2\phi} = - \frac{I_2}{2 \pi \rho} = \frac{12}{2 \pi \rho} \mathbf{a}_{\phi}\) For the conducting shell at \(\rho = 10\), the enclosed current is \(I_3 = - 4 \mathrm{~A}\): \(H_{3\phi} = - \frac{I_3}{2 \pi \rho} = \frac{4}{2 \pi \rho} \mathbf{a}_{\phi}\)
02

b) Finding the total \(\mathbf{H}\) and plotting \(H_\phi\) versus \(\rho\)

Now, let's sum the contributions from each conductor and find the total \(\mathbf{H}\): \(\mathbf{H} = \mathbf{H}_1 + \mathbf{H}_2 + \mathbf{H}_3 = \left(\frac{16}{2 \pi \rho} + \frac{12}{2 \pi \rho} + \frac{4}{2 \pi \rho}\right) \mathbf{a}_{\phi}\) We can now plot \(H_\phi\) versus \(\rho\): 1. For \(\rho < 6\), the magnetic field is due to \(I_1\) and \(I_2\): \(H_\phi = \frac{16 + 12}{2 \pi \rho}\) 2. For \(6 \leq \rho < 10\), the magnetic field is due to \(I_1\): \(H_\phi = \frac{16}{2 \pi \rho}\) 3. For \(\rho \geq 10\), the magnetic field is due to \(I_1\) and \(I_3\): \(H_\phi = \frac{16 + 4}{2 \pi \rho}\)
03

c) Finding the total flux \(\Phi\) crossing the specified surface

To find the total flux \(\Phi\) crossing the surface \(1 < \rho < 7, 0 < z < 1\), at a fixed \(\phi\), we can calculate the integral of \(\mathbf{H}\) over the surface area: \(\Phi = \int_S \mathbf{H} \cdot d\mathbf{S} = \int_1^7 \int_0^1 \frac{16 + 12}{2 \pi \rho} \rho d\rho dz\) We can simplify the integral and perform the calculations: \(\Phi = \int_1^7 \int_0^1 \frac{28}{2 \pi} d\rho dz = \frac{28}{2 \pi} \int_1^7 \int_0^1 dz d\rho = \frac{28}{2 \pi} \int_1^7 (z = 1) d\rho = \frac{28}{2 \pi} (7 - 1) = \frac{28}{\pi} \cdot 671 \mathrm{~Wb}\) Thus, the total flux \(\Phi\) crossing the specified surface is \(\frac{28}{\pi} \cdot 671 \mathrm{~Wb}\).

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Most popular questions from this chapter

Given \(\mathbf{H}=\left(3 r^{2} / \sin \theta\right) \mathbf{a}_{\theta}+54 r \cos \theta \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m}\) in free space: \((a)\) Find the total current in the \(\mathbf{a}_{\theta}\) direction through the conical surface \(\theta=20^{\circ}, 0 \leq \phi \leq 2 \pi\), \(0 \leq r \leq 5\), by whatever side of Stokes' theorem you like the best. \((b)\) Check the result by using the other side of Stokes' theorem.

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 \(\rho\).

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 \(\mathbf{J}\).

The free space region defined by \(1

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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