QJEEM25Ph58f6: JEE Main 2025 Physics - Class 12 JEEM25Ph58f6 Solution: A current of 5 A exists in a square loop of side . Then the…

$B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L}$

For a square loop of side length

$L = \frac{1}{\sqrt{2}} \text{ m},$

the perpendicular distance from the center to any side is

$d = \frac{L}{2} = \frac{1}{2\sqrt{2}} \text{ m}.$

For a finite straight wire, the magnetic field at a point at distance $d$ from the wire is given by the Biot–Savart expression

$B_{\text{side}} = \frac{\mu_0 I}{4\pi d} \left(\sin\theta_1 + \sin\theta_2\right)$

where $\theta_1$ and $\theta_2$ are the angles between the extended wire and the line joining the ends of the wire with the observation point. For one side of the square, by symmetry, the midpoint of the side and the center of the square give

$\tan\theta = \frac{L/2}{L/2} = 1 \quad\Rightarrow\quad \theta = 45^\circ.$

Thus,

$\sin\theta_1 = \sin\theta_2 = \sin 45^\circ = \frac{1}{\sqrt{2}}.$

Then the field due to one side is

$B_{\text{side}} = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} \left(\frac{2}{\sqrt{2}}\right) = \frac{\mu_0 I \sqrt{2}}{2\pi L}.$

Since there are 4 sides with contributions that add vectorially in the same direction at the center, the total magnetic field is

$B = 4 \cdot B_{\text{side}} = \frac{4 \mu_0 I \sqrt{2}}{2\pi L} = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L}.$

Substitute

$L = \frac{1}{\sqrt{2}},$

to obtain

$B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi \left(\frac{1}{\sqrt{2}}\right)} = \frac{2 \sqrt{2}\,\mu_0 I \sqrt{2}}{\pi} = \frac{4\,\mu_0 I}{\pi}.$

Using the provided values

$I = 5 \text{ A} \quad \text{and} \quad \mu_0 = 4\pi \times 10^{-7} \text{ T\,m/A},$

the magnetic field becomes

$B = \frac{4 \times \left(4\pi \times 10^{-7}\right) \times 5}{\pi} = \frac{80\pi \times 10^{-7}}{\pi} = 80 \times 10^{-7} \text{ T} = 8 \times 10^{-6} \text{ T}.$

Thus, the value of $p$ is

$p = 8.$