JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Magnetic Effects of Current

For a finite straight wire carrying current $i$, the magnetic field at a point a perpendicular distance $d$ from the wire is
$B=\frac{{\mu }_{0}i}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

where ${\theta }_1,{\theta }_2$ are the angles subtended by the wire's ends at that point.

For a square loop of side $a=\frac{1}{\sqrt{2}}$ m, the perpendicular distance from the center to each side is $d=\frac{a}{2}=\frac{1}{2\sqrt{2}}$ m, and each end of a side subtends $45^{\circ }$ at the center, so ${\theta }_1={\theta }_2=45^{\circ }$.

Field due to one side:
$B_1=\frac{{\mu }_{0}i}{4\pi d}\times 2\sin 45^{\circ }=\frac{(4\pi \times 10^{-7})(5)}{4\pi \left(\frac{1}{2\sqrt{2}}\right)}\times \frac{2}{\sqrt{2}}=10^{-7}\times 5\times 4=2\times 10^{-6}$ T

All four sides contribute equally and in the same direction at the center:
$B=4B_1=8\times 10^{-6}$ T

The value of p is 8.