GATE 2025 IN Q.30: solution and answer

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Axes. Take ${\hat{a}}_x=\hat{x}$ pointing right along the rails, ${\hat{a}}_y$ pointing up in the plane of the page, and ${\hat{a}}_z$ pointing out of the page, so that ${\hat{a}}_x\times {\hat{a}}_y={\hat{a}}_z$. The field is into the page, so $\mathbf{B}=-0.5{\hat{a}}_z\text{T}$.

2. The bar starts at rest. At the instant the switch closes the bar has not moved yet, so $v=0$ and the motional (back) emf $Blv$ is zero. The bar and rails have zero resistance, so $R_{int}$ is the only thing limiting the current.

3. Ohm's law: $I=\frac{V}{R_{int}}=\frac{60}{0.5}=120\text{A}$.

4. Which way the current flows: out of the + terminal at the top, through $R_{int}$ and the switch, right along the upper rail, down through the bar, then back along the lower rail to the − terminal. The length vector of the bar, taken along the current, is therefore $\mathbf{l}=-1{\hat{a}}_y\text{m}$.

5. Force on a straight current-carrying conductor: $\mathbf{F}=I\mathbf{l}\times \mathbf{B}$.

6. $$\mathbf{F}=(120)\left(-1{\hat{a}}_y\right)\times \left(-0.5{\hat{a}}_z\right)=60\left({\hat{a}}_y\times {\hat{a}}_z\right)=60{\hat{a}}_x\text{N}$$

7. Result: the magnitude is $F=BIl=0.5\times 120\times 1=60\text{N}$, and the force points along $+\hat{x}$, away from the source. As a check, the push enlarges the current loop, which is how a rail gun works. The answer is option (A).