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., ${\oint }_C\mathbf{H}\cdot d\mathit{\ell }=I_{enc}$. 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., $\mathbf{H}=H_x{\mathbf{a}}_{\mathbf{x}}$, 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: $${\oint }_C\mathbf{H}\cdot d\mathit{\ell }=H_x{\int }_{-w/2}^{w/2}dy+H_x{\int }_0^{3}dz=H_x(w+3)=K\cdot w$$

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