JEE Physics 2025 Question with Solution

Given: The resistance of path ABC is $r$ and that of path ADC is $2r$. The total current entering at A is $I$.

Find: The magnitude of the resultant magnetic field at the center of the square loop.

Since the two paths are in parallel, the current divides inversely in proportion to their resistances. Therefore,

$$i_1 = \frac{2I}{3}, \qquad i_2 = \frac{I}{3}$$

where $i_1$ flows through ABC and $i_2$ flows through ADC.

The magnetic fields at the center due to these two current paths are in opposite directions, so the net field is obtained by subtraction. Using the expression given in the solution working,

$$B_{\text{centre}} = \frac{2\mu_0 \sqrt{2}}{4\pi \left(\frac{a}{2}\right)} \left[\frac{2I}{3} - \frac{I}{3}\right]$$

Simplifying,

$$B_{\text{centre}} = \frac{\mu_0 \sqrt{2}}{\pi a} \cdot \frac{I}{3} = \frac{\sqrt{2}\mu_0 I}{3\pi a}$$

Therefore, the correct option is A.

Note: The printed Option A appears as $\frac{\sqrt{2\mu_0 I}}{3 \pi a}$, but the solution working clearly gives $\frac{\sqrt{2}\mu_0 I}{3\pi a}$. The answer has been resolved using the solution.

Given: A current $I$ reaches point A and splits into two branches of a square loop. The branch ABC has resistance $r$ and branch ADC has resistance $2r$.

Find: Resultant magnetic field at the center.

From current division in parallel branches,

$$\frac{i_1}{i_2} = \frac{2r}{r} = 2$$

and

$$i_1 + i_2 = I$$

Hence,

$$i_1 = \frac{2I}{3}, \qquad i_2 = \frac{I}{3}$$

Each branch produces magnetic field at the center, but because the currents circulate around the center in opposite senses, their fields oppose each other. Therefore, the effective current contribution is

$$i_1 - i_2 = \frac{2I}{3} - \frac{I}{3} = \frac{I}{3}$$

Using the field expression quoted in the extracted solution,

$$B_{\text{centre}} = \frac{2\mu_0 \sqrt{2}}{4\pi \left(\frac{a}{2}\right)} \left(\frac{I}{3}\right)$$

So,

$$B_{\text{centre}} = \frac{\sqrt{2}\mu_0 I}{3\pi a}$$

Thus, the magnitude of the resultant magnetic field at the center is $\frac{\sqrt{2}\mu_0 I}{3\pi a}$, corresponding to Option A by the solution.