Question

A wire in the shape of a square frame carries a current $I$ and produces a magnetic field $B_s$, at its centre. Now the wire is bent in the shape of a circle and carries the same current. If $B_c$ is the magnetic field produced at the centre of the circular coil, then $B_s/B_c$ is

• A. $8\pi$
• B. $8{\pi }^2/\sqrt{2}$
• C. $8\sqrt{2}/{\pi }^2$
• D. $8\pi \sqrt{2}$

Answer 1

The correct option is C $8\sqrt{2}/{\pi }^2$

$B$ due to finite wire =$\frac{{\mu }_0}{4\pi }.\frac{i}{d}(\sin {\theta }_1+\sin {\theta }_2)$

For a square $,d=\frac{L}{2}$

$⟹B_s=\frac{2\sqrt{2}{\mu }_{0}i}{\pi L}$

On bending square to circle, perimeter is constant.

$⟹B_c=\frac{{\mu }_{0}i}{2r}$ where $r=\frac{4L}{2\pi }$

$⟹B_c=\frac{{\mu }_{0}i\pi }{4L}$

$\therefore \frac{B_s}{B_c}=\frac{2\sqrt{2}{\mu }_{0}i}{\pi L}\times \frac{4L}{{\mu }_0\pi i}=\frac{8\sqrt{2}}{{\pi }^2}$