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Chapter 7: Problem 9

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 current sheet, \(\mathbf{K}=20 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\), is located at \(\rho=2\), and a second sheet, \(\mathbf{K}=-10 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\), is located at \(\rho=4 .(a)\) Let \(V_{m}=0\) at \(P(\rho=3, \phi=0,\), \(z=5\) ) and place a barrier at \(\phi=\pi\). Find \(V_{m}(\rho, \phi, z)\) for \(-\pi<\phi<\pi\) \((b)\) Let \(\mathbf{A}=0\) at \(P\) and find \(\mathbf{A}(\rho, \phi, z)\) for \(2<\rho<4\)

A long, straight, nonmagnetic conductor of \(0.2 \mathrm{~mm}\) radius carries a uniformly distributed current of 2 A dc. \((a)\) Find \(J\) within the conductor. (b) Use Ampère's circuital law to find \(\mathbf{H}\) and \(\mathbf{B}\) within the conductor. (c) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) within the conductor. \((d)\) Find \(\mathbf{H}\) and \(\mathbf{B}\) outside the conductor. \((e)\) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) outside the conductor.

( \(a\) ) 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.

A cylindrical wire of radius \(a\) is oriented with the \(z\) axis down its center line. The wire carries a nonuniform current down its length of density \(\mathbf{J}=b \rho \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}^{2}\) where \(b\) is a constant. ( \(a\) ) What total current flows in the wire? \((b)\) Find \(\mathbf{H}_{i n}(0<\rhoa)\), as a function of \(\rho ;(d)\) verify your results of parts \((b)\) and \((c)\) by using \(\nabla \times \mathbf{H}=\mathbf{J}\).

Two semi-infinite filaments on the \(z\) axis lie in the regions \(-\infty
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