Question

The magnetic field at the centre of coil of $n$ turns, 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

The correct option is A $\frac{\sqrt{2}{\mu }_{0}ni}{\pi l}$

The magnetic field at the center of the coil of $n$ turns bent in the form of a square of side $2l$ as shown in the figure.

The biot-savart law,

$dB=\frac{{\mu }_0}{4\pi }\frac{idlsin\theta }{r^2}$

From the above figure.

$B=\frac{{\mu }_0}{4\pi }\frac{2il}{l^2}(sin{45}^0+sin{45}^0)$

$B=\frac{\sqrt{2}{\mu }_{0}i}{4\pi l}$ (inward direction)

The magnetic field at $O$ due to $AB$, $BC$, $CD$ and $DA$ is same and direction is also same, so the magnetic field at the center of the square is $4$ times of $B$.

$B^{\prime }=4\frac{\sqrt{2}{\mu }_{0}i}{4\pi l}=\frac{\sqrt{2}{\mu }_{0}i}{\pi l}$

For $n$ turns ,magnetic field is n times

The correct option is A.