Question

A rectangular loop of length $l$ and breadth $b$ is placed at a distance of $x$ from an 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 e.m.f. in the loop is (${\mu }_0$ permeability of free space)

• A. $\frac{{\mu }_{0}iv}{2\pi x}\left(\frac{l+b}{b}\right)$
• B. $\frac{{\mu }_0{i}^{2}v}{4{\pi }^{2}x}log\left(\frac{b}{l}\right)$
• C. $\frac{{\mu }_{0}iblv}{2\pi x(l+x)}$
• D. $\frac{{\mu }_{0}iblv}{2\pi }log\frac{x+l}{x}$

Answer 1

The correct option is C $\frac{{\mu }_{0}iblv}{2\pi x(l+x)}$

Consider a strip of width $dy$ at $y$ distance from the wire

for infinite long wire
$B=\frac{{\mu }_o}{4\pi }\frac{2I}{y}$
area of strips$=bdy$
$d\varphi =2I\frac{{\mu }_o}{4\pi }\frac{b}{y}dy$
$\varphi ={\int }_x^{x+l}2I\frac{{\mu }_o}{4\pi }\frac{b}{y}dy$
$\varphi =\frac{2I{\mu }_{o}b}{4\pi }ln\frac{x+l}{x}$
now Induced emf $=\frac{d\varphi }{dt}=\frac{I{\mu }_{o}b}{2\pi }\left(\frac{x}{x+l}\right)(-\frac{l}{x^2})\frac{dx}{dt}$

Magnitude of emf $=\frac{I{\mu }_{o}bl}{2\pi }\left(\frac{v}{x(x+l)}\right)$