Q. A square loop of side a placed in the same plane as a long straight wire carrying a current $i.\ $The centre of the loop is at a distance r from the wire, where $r > > a$, figure. The loop is moved away from the wire with a constant velocity $v$. The induced emf in the loop is

A) $\frac{\mu_{0}i\ a\ v}{2\ \pi\ r}$
B) $\frac{\mu_{0}i\ a^{3}\ v}{2\ \pi\ r^{3}}$
C) $\frac{\mu_{0}i\ \ v}{2\ \pi\ }$
D) $\frac{\mu_{0}i\ a^{2}\ v}{2\ \pi\ r^{2}}$
Explanation

(d) Magnetic field intensity at a distance r from the straight wire carrying current is
$\ \ \ \ B = \frac{\mu_{0}i}{2\pi r}\ $
As area of loop, $A = a^{2}$
And magnetic flux, $\phi\ = BA$
$\therefore\ \phi = \frac{\mu_{0}ia^{2}}{2\pi r}\ $
The induced emf in the loop is
$e = \ \left| \frac{d\phi}{dt} \right| = \ \left| \frac{d}{dt}\ \frac{\mu_{0}\ i\ a^{2}v}{2\ \pi\ r} \right|$
$e = \ \frac{\mu_{0}ia^{2}}{2\ \pi\ r^{2}}\frac{dr}{dt} = \frac{\mu_{0}i\ a^{2}v}{2\ \pi\ r^{2}}$
Where $v\ \ = \frac{dr}{dt}\ \ $is velocity .

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.