Question

A long straight wire is parallel to one edge as shown in figure. If the current in the long wire varies in time as $i=I_0{e}^{-t/\tau }$, what will be the induced emf in the loop ?

• A. $\frac{{\mu }_{0}bi}{\pi \tau }ln\left(\frac{d+a}{d}\right)$
• B. $\frac{{\mu }_{0}bi}{2\pi \tau }ln\left(\frac{d+a}{d}\right)$
• C. $\frac{2{\mu }_{0}bi}{\pi \tau }ln\left(\frac{d+a}{d}\right)$
• D. $\frac{{\mu }_{0}bi}{\pi \tau }ln\left(\frac{d}{d+a}\right)$

Answer 1

The correct option is B $\frac{{\mu }_{0}bi}{2\pi \tau }ln\left(\frac{d+a}{d}\right)$


$\int d\varphi ={\int }_d^{a+d}\frac{{\mu }_{0}i}{2\pi x}.b.dx$
$\varphi =\frac{{\mu }_{0}ib}{2\pi }ln\left(\frac{d+a}{a}\right)$
$\epsilon =-\frac{d\varphi }{dt}=\frac{{\mu }_{0}bi}{2\pi \tau }ln\left(\frac{d+a}{d}\right)$