Question

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

Short answer

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

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.

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 \(\mathbf{H}\).

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: $$ \oint_C \mathbf{H} \cdot d\boldsymbol{\ell} = H_x\int_{-w/2}^{w/2}dy + H_x\int_{0}^{3}dz = H_x(w+3) = K \cdot w $$

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 \(-2<y<2\)): $$ H_x = \frac{8 \cdot 4}{4 + 3} = \frac{32}{7}\,\text{A/m} $$

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: $$ \mathbf{H} = H_x \mathbf{a}_{x} = \frac{32}{7}\,\text{A/m} \mathbf{a}_{x} $$