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
$=4 \cdot \frac{\mu I }{4 \pi\left(\frac{ a }{2}\right)}\left(2 \sin 45^{\circ}\right)=\frac{2 \sqrt{2} \mu_{0} I }{\pi a }$