Question

A current I flows in the anticlockwise direction through a square loop of side a lying in the xoy plane with its center at the origin. The magnetic induction at the center of the square loop is

• A. $\frac{2\sqrt{2}{\mu }_{0}I}{\pi a}{\hat{e}}_x$
• B. $\frac{2\sqrt{2}{\mu }_{0}I}{\pi a}{\hat{e}}_z$
• C. $\frac{2\sqrt{2}{\mu }_{0}I}{\pi a^2}{\hat{e}}_z$
• D. $\frac{2\sqrt{2}{\mu }_{0}I}{\pi a^2}{\hat{e}}_x$

Answer 1

Answer: (B)

Field due to one side of loop at $O$
$=4\cdot \frac{\mu I}{4\pi \left(\frac{a}{2}\right)}\left(2\sin 45^{\circ }\right)$
Field at $O$ due to all four sides is along unit vector ${\hat{e}}_z$
$\therefore$ Total field
$=4\cdot \frac{\mu I}{4\pi \left(\frac{a}{2}\right)}\left(2\sin 45^{\circ }\right)=\frac{2\sqrt{2}{\mu }_{0}I}{\pi a}$