JEE Physics 2025 Question with Solution

Given:

• Current $i = 5 \, \text{A}$
• Side of square loop $= \frac{1}{\sqrt{2}} \, \text{m}$

Find: The value of $p$ in $B = p \times 10^{-6} \, \text{T}$.

For one side of the square, the magnetic field at the centre is written as

$$B = \frac{\mu_0 i}{4\pi d}(\sin \theta_1 + \sin \theta_2)$$

At the centre of the square, for each side,

$$\theta_1 = \theta_2 = 45^\circ$$

So,

$$B = \frac{\mu_0 i}{4\pi d}(\sin 45^\circ + \sin 45^\circ)$$

Using the working shown in the solution,

$$B = \frac{10^{-7} \times 5 \times 2}{\frac{1}{2\sqrt{2}}} = 2 \times 10^{-6} \, \text{T}$$

This is the field due to one side.

Since there are four sides of the square loop,

$$B_{\text{net}} = 4B = 4 \times 2 \times 10^{-6} = 8 \times 10^{-6} \, \text{T}$$

Therefore, comparing with $B = p \times 10^{-6} \, \text{T}$, we get $p = 8$.

Given: The loop is square and all four sides are symmetrically placed about the centre.

Find: Net magnetic field at the centre.

The solution uses the magnetic field formula for a finite straight wire segment. Because the centre is equidistant from all four sides, each side contributes the same magnetic field magnitude, and all contributions are in the same direction perpendicular to the plane of the loop.

For one side,

$$B_{\text{one side}} = 2 \times 10^{-6} \, \text{T}$$

Hence the total field is obtained by multiplying by $4$:

$$B_{\text{net}} = 4 \times B_{\text{one side}}$$ $$B_{\text{net}} = 4 \times 2 \times 10^{-6}$$ $$B_{\text{net}} = 8 \times 10^{-6} \, \text{T}$$

So the required numerical value is $8$.

Note: the solution uses an intermediate distance value in the substitution step; however, its final computed result is clearly stated as $8 \times 10^{-6} \, \text{T}$, so the answer is taken as $8$.