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A current sheet \(\mathbf{K}=8 \mathbf{a}_{x} \mathrm{~A} / \mathrm{m}\) flows in the region \(-2

Short Answer

Expert verified
Answer: The magnetic field intensity \(\mathbf{H}\) at point \(P(0,0,3)\) is \(\frac{32}{7}\,\text{A/m} \mathbf{a}_{x}\).

Step by step solution

01

Understand Ampere's Circuital Law and Biot-Savart's Law

Ampere's Circuital Law states that the circulation of the magnetic field intensity \(\mathbf{H}\) around a closed path is equal to the current enclosed by that path, i.e., \(\oint_{C} \mathbf{H} \cdot d\boldsymbol{\ell} = I_{enc}\). Biot-Savart's Law relates the infinitesimal magnetic field intensity \(d\mathbf{H}\) generated by an infinitesimal current element to the current and the position vectors.
02

Determine a suitable Amperian loop

For a current sheet with thickness located in the region \(-2<y<2\), choose a rectangular Amperian loop with edges parallel to the y-axis and z-axis, and with one side lying in the y-z plane. Let the loop have height \(h=3\) and width \(w\) (to be found). The loop is symmetric with respect to the y-axis, and hence the magnetic field at point \(P\) is along the \(x\)-axis, i.e., \(\mathbf{H} = H_x \mathbf{a_x}\), where \(H_x\) is the magnitude of \(\mathbf{H}\).
03

Apply Ampere's Circuital Law to the loop

Consider the current enclosed by the loop. As the current is uniformly distributed along the y-axis for \(-2<y<2\), we can write the total current enclosed as \(I_{enc} = K \cdot w\). Applying Ampere's Circuital Law, we get: $$ \oint_C \mathbf{H} \cdot d\boldsymbol{\ell} = H_x\int_{-w/2}^{w/2}dy + H_x\int_{0}^{3}dz = H_x(w+3) = K \cdot w $$
04

Solve for H_x

From the above equation, we can solve for the magnitude of the magnetic field intensity \(H_x\): $$ H_x = \frac{K \cdot w}{w + 3} $$ Now, substitute the given values of \(K=8 \,\text{A/m}\), and the width \(w = 4\) (from \(-2<y<2\)): $$ H_x = \frac{8 \cdot 4}{4 + 3} = \frac{32}{7}\,\text{A/m} $$
05

Write the final answer

We found the magnitude of the magnetic field intensity at point \(P\), which is along the \(x\)-axis. Therefore, the magnetic field intensity \(\mathbf{H}\) at \(P(0,0,3)\) is: $$ \mathbf{H} = H_x \mathbf{a}_{x} = \frac{32}{7}\,\text{A/m} \mathbf{a}_{x} $$

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Most popular questions from this chapter

A toroid having a cross section of rectangular shape is defined by the following surfaces: the cylinders \(\rho=2\) and \(\rho=3 \mathrm{~cm}\), and the planes \(z=1\) and \(z=2.5 \mathrm{~cm}\). The toroid carries a surface current density of \(-50 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) on the surface \(\rho=3 \mathrm{~cm}\). Find \(\mathbf{H}\) at the point \(P(\rho, \phi, z):(a) P_{A}(1.5 \mathrm{~cm}, 0\), \(2 \mathrm{~cm}) ;\left(\right.\) b) \(P_{B}(2.1 \mathrm{~cm}, 0,2 \mathrm{~cm}) ;\) (c) \(P_{C}(2.7 \mathrm{~cm}, \pi / 2,2 \mathrm{~cm}) ;\) (d) \(P_{D}(3.5 \mathrm{~cm},\), \(\pi / 2,2 \mathrm{~cm})\)

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.

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.

Show that \(\nabla_{2}\left(1 / R_{12}\right)=-\nabla_{1}\left(1 / R_{12}\right)=\mathbf{R}_{21} / R_{12}^{3}\).

( \(a\) ) Find \(\mathbf{H}\) in rectangular components at \(P(2,3,4)\) if there is a current filament on the \(z\) axis carrying \(8 \mathrm{~mA}\) in the \(\mathbf{a}_{z}\) direction. ( \(b\) ) Repeat if the filament is located at \(x=-1, y=2\). ( \(c\) ) Find \(\mathbf{H}\) if both filaments are present.

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