GATE 2023 EE Q.61: solution and answer

Official answer: 2.49 to 2.51Source: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Field of an infinite current sheet: apply Ampere's circuital law to a rectangular loop that straddles the sheet, with sides of length $L$ parallel to ${\hat{a}}_y$ above and below the sheet. The enclosed current is $KL$, and by symmetry the field has the same magnitude on both sides, so $2HL=KL$. In vector form $$\mathbf{H}=\frac{1}{2}\mathbf{K}\times {\hat{a}}_n$$ where ${\hat{a}}_n$ is the unit normal pointing from the sheet toward the field point.

2. The magnitude $|\mathbf{H}|=K/2$ does not depend on the distance from the sheet, so only the side of the sheet matters. The point $(1,1,1)$ has $z=1>0$, so ${\hat{a}}_n=+{\hat{a}}_z$.

3. Cross product: ${\hat{a}}_x\times {\hat{a}}_z=-{\hat{a}}_y$, so $$\mathbf{H}=\frac{1}{2}(5{\hat{a}}_x)\times {\hat{a}}_z=-2.5{\hat{a}}_y\text{A/m}$$

4. Magnitude: $|\mathbf{H}|=\frac{K}{2}=\frac{5}{2}=2.50\text{A/m}$. The $x$ and $y$ coordinates of the point make no difference.