Q. A rectangular loop of length l and breadth b is placed at distance of $x$ from infinitely long wire carrying current i such that the direction of current is parallel to breadth. If the loop moves away from the current wire in a direction perpendicular to it with a velocity $v,$ the magnitude of the emf in the loop is $(\mu =$permeability of free space)
A) $\frac{\mu_{0}iv}{2\pi\ x}\left( \frac{1 + b}{b} \right)$
B) $\frac{\mu_{0}i^{2}v}{4\ \pi^{2}x}\log\left( \frac{b}{l} \right)$
C) $\frac{\mu_{0}ilbv}{2\pi\ x(l + x)}\ $
D)
Explanation
(d) We can show the situation as
[image:6_image71.emf]
Since, loop is moving away from the wire, so the direction of current in the loop will be as shown in the figure.
Net magnetic field on the loop due to wire
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = \frac{\mu_{0}i}{2\pi}\left( \frac{1}{x} - \frac{1}{l + x} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{\mu_{0}il}{2\pi x(l + x)}\ $
So, the magnitude of the emf in the loop
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = vBb = \frac{\mu_{0}il\ v\ b}{2\ \pi\ x\ (1 + x)}\ \ \ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.