Q. The magnetic field at the centre of coil of $n$ turns, bent in the form of a square of side $2l$, carrying $i$, is
A) $\frac{\sqrt{2}\mu_{0}ni}{\pi l}$
B) $\frac{\sqrt{2}\mu_{0}ni}{2\pi l}$
C) $\frac{\sqrt{2}\mu_{0}ni}{4\pi l}$
D) $\frac{2\mu_{0}ni}{\pi l}$
Explanation
(a) Magnetic field due to one side of the square at centre $O$
$B_{1} = \frac{\mu_{0}}{4\pi}.\frac{2i\sin{45{^\circ}}}{a/2} \Rightarrow B_{1} = \frac{\mu_{0}}{4\pi}.\frac{2\sqrt{2}i}{a}$
Hence magnetic field at centre due to all sides
$B = 4B_{1} = \frac{\mu_{0}(2\sqrt{2}i)}{\pi a}$
Magnetic field due to $n$ turns
$B_{net} = nB = \frac{\mu_{0}2\sqrt{2}ni}{\pi a} = \frac{\mu_{0}2\sqrt{2}ni}{\pi(2l)} = \frac{\sqrt{2}\mu_{0}ni}{\pi l}\ (\because a = 2l)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.