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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 square filamentary differential current loop, \(d L\) on a side, is centered at the origin in the \(z=0\) plane in free space. The current \(I\) flows generally in the \(\mathbf{a}_{\phi}\) direction. ( \(a\) ) Assuming that \(r>>d L\), and following a method similar to that in Section \(4.7\), show that $$d \mathbf{A}=\frac{\mu_{0} I(d L)^{2} \sin \theta}{4 \pi r^{2}} \mathbf{a}_{\phi}$$ (b) Show that $$d \mathbf{H}=\frac{I(d L)^{2}}{4 \pi r^{3}}\left(2 \cos \theta \mathbf{a}_{r}+\sin \theta \mathbf{a}_{\theta}\right)$$ The square loop is one form of a magnetic dipole.

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.

Infinitely long filamentary conductors are located in the \(y=0\) plane at \(x=n\) meters where \(n=0, \pm 1, \pm 2, \ldots\) Each carries \(1 \mathrm{~A}\) in the \(\mathbf{a}_{z}\) direction. (a) Find \(\mathbf{H}\) on the \(y\) axis. As a help, $$\sum_{n=1}^{\infty} \frac{y}{y^{2}+n^{2}}=\frac{\pi}{2}-\frac{1}{2 y}+\frac{\pi}{e^{2 \pi y}-1}$$ (b) Compare your result of part \((a)\) to that obtained if the filaments are replaced by a current sheet in the \(y=0\) plane that carries surface current density \(\mathbf{K}=1 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\).

Given the field \(\mathbf{H}=20 \rho^{2} \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m}:(a)\) Determine the current density \(\mathbf{J}\). (b) Integrate \(\mathbf{J}\) over the circular surface \(\rho \leq 1,0<\phi<2 \pi, z=0\), to determine the total current passing through that surface in the \(\mathbf{a}_{z}\) direction. (c) Find the total current once more, this time by a line integral around the circular path \(\rho=1,0<\phi<2 \pi, z=0 .\)

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.

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