Q. A wire in the form of a square of side ‘$a$’ carries a current $i$. Then the magnetic induction at the centre of the square wire is (Magnetic permeability of free space$= \mu_{o}$)
A) $\frac{\mu_{0}i}{2\pi a}$
B) $\frac{\mu_{0}i\sqrt{2}}{\pi a}$
C) $\frac{2\sqrt{2}\mu_{0}i}{\pi a}$
D) $\frac{\mu_{0}i}{\sqrt{2}\pi a}$
Explanation
(c) $B_{0} = 4 \times \frac{\mu_{0}}{4\pi} \times \frac{i}{(a/2)}\left\lbrack \sin{45{^\circ} +}\sin{45{^\circ}} \right\rbrack$
$= 4 \times \frac{\mu_{0}}{4\pi} \times \frac{2i}{a} \times \frac{2}{\sqrt{2}}$
$= \frac{\mu_{0}i2\sqrt{2}}{\pi a}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.