Question

The magnetic field at the point of intersection of diagonals of a square wire loop of side $L$ carrying a current $I$ is

• A. $\frac{\mu_{o}I}{\pi L}$
• B. $\frac{2\mu_{o}I}{\pi L}$
• C. $\frac{\sqrt{2}\mu_{o}I}{\pi L}$
• D. $\frac{2\sqrt{2}\mu_{o}I}{\pi L}$

Answer 1

Answer: (D)

$\frac{2\sqrt{2}\mu_{o}I}{\pi L}$

Answer 2

We can have the figure as follows

The net magnetic field at the point of intersection of the diagonal would be given by
$B_{\text {net }}=\left(B_{L / 2} \sin 45^{\circ}+B_{L / 2} \sin 45^{\circ}\right)$
$+\left(B_{L / 2} \sin 45^{\circ}+B_{L / 2} \sin 45^{\circ}\right)$
$+\left(B_{L / 2} \sin 45^{\circ}+B_{L / 2} \sin 45^{\circ}\right)$
$+\left(B_{L / 2} \sin 45^{\circ}+B_{L / 2} \sin 45^{\circ}\right)$
$=4\left(B_{L / 2} \sin 45^{\circ}+B_{L / 2} \sin 45^{\circ}\right)$
$=8 B_{L / 2} \sin 45^{\circ}=8 \frac{\mu_{0}}{4 \pi} \frac{I}{\left(\frac{L}{2}\right)} \frac{1}{\sqrt{2}}$
$=\frac{\mu_{0}}{\pi} \frac{2 \sqrt{2} I}{L}$
Aliter
$B_{ net }=4 \times \frac{\mu_{0}}{4 \pi} \cdot \frac{I}{\frac{L}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)$
$=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{2 i}{a} \times \frac{2}{\sqrt{2}}$
$=\frac{2 \sqrt{2} \mu_{0} I}{\pi L}$