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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

Planar 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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