Q. A square frame of side 1 m carries a current $i$, produces a magnetic field $B$ at its centre. The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is $B'$. The ratio $B/B$’ is
A) $\frac{8}{\pi^{2}}$
B) $\frac{8\sqrt{2}}{\pi^{2}}$
C) $\frac{16}{\pi^{2}}$
D) $\frac{16}{\sqrt{2}\pi^{2}}$
Explanation
(d) $\ OF = CF = \frac{1}{2}$ m
$B\ \ = \frac{\mu_{0}}{4\pi}\ \frac{1}{(1/2)}\ \lbrack\sin{45{^\circ} + \sin{45{^\circ}\rbrack\ \times 4}}$
$= \frac{\mu_{0}}{4\pi}\frac{16i}{\sqrt{2}}$
When the frame is taken as circular of radius $r$, then
$2\pi r = 4\ $
Or $r\ = 2/\pi$
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$B^{''} = \frac{\mu_{0}}{4\pi}\frac{2\pi i}{r} = \frac{\mu_{0}}{4\pi}\ \frac{2\pi i}{2/\pi} = \frac{\mu_{0}}{4\pi}\pi^{2}i$
$\therefore\ \ \ \frac{B}{B^{''}} = \ \frac{16}{\sqrt{2}\ \pi^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.