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=4{{B}_{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) $