JEE Main 2025 Jan 24 Shift 1, Physics Q24: Magnetic Field due to Electric Current

For a finite straight current-carrying segment, the magnetic field at perpendicular distance $d$ from its midpoint is

$B_{\text{side}}=\frac{{\mu }_{0}i}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

For a square of side $L$ and current $i$, the centre lies at perpendicular distance $d=L/2$ from each side, and each side subtends ${\theta }_1={\theta }_2=45^{\circ }$ from the centre, so $\sin {\theta }_1+\sin {\theta }_2=\sqrt{2}$.

$B_{\text{side}}=\frac{{\mu }_{0}i}{4\pi (L/2)}\cdot \sqrt{2}=\frac{{\mu }_{0}i\sqrt{2}}{2\pi L}$

All four sides produce fields in the same direction at the centre, so the total field is

$B=4B_{\text{side}}=\frac{2\sqrt{2}{\mu }_{0}i}{\pi L}$

Substitute ${\mu }_0/(4\pi )=10^{-7}$ T·m/A, $i=5$ A, $L=1/\sqrt{2}$ m:

$B_{\text{side}}=10^{-7}\cdot \frac{5\cdot 2}{1/\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=10^{-7}\cdot \frac{5\cdot 2}{1/\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=10^{-7}\cdot 10=2\times 10^{-6}\text{T}$

(Plug in numerically: with $d=L/2=1/(2\sqrt{2})$, $B_{\text{side}}=\frac{10^{-7}\cdot 5}{1/(2\sqrt{2})}\cdot \sqrt{2}=10^{-7}\cdot 5\cdot 2\sqrt{2}\cdot \sqrt{2}=10^{-7}\cdot 20=2\times 10^{-6}$ T.)

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

So $p=8$.

$\therefore$ Answer: $8$.