Question

The magnetic field at the centre of coil of n turn, bent in the form of a square of side $2l,$ carrying current $i,$ 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. $\frac{2{\mu }_{0}ni}{\pi l}$

Answer 1

Answer: (B)

Magnetic field due to one side of the square at centre $O$
$B_1=\frac{{\mu }_0}{4\pi }.\frac{2i\sin 45{}^{\circ }}{a/2}$
$B_1=\frac{{\mu }_0}{4\pi }.\frac{2\sqrt{2}i}{a}$
Hence, magnetic field at centre due to all side
$B=4B_1=\frac{{\mu }_0(2\sqrt{2}i)}{\pi a}$
Magnetic field at centre due to n turns
$B=nB=\frac{{\mu }_0(2\sqrt{2}i)}{\pi a}$
$=\frac{{\mu }_{0}2\sqrt{2}ni}{\pi (2l)}=\frac{\sqrt{2}{\mu }_{0}ni}{\pi l}$
$(\because a=2l)$