Question

A long straight wire is parallel to one edge as in fig. If the current in the long wire is varies in time as $I=I_0{e}^{-1/x}$, 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 $\frac{{\mu }_{0}bI}{2\pi \tau }ln\left(\frac{d+a}{d}\right)$

As shown in figure a strip of length $dx$ is in the loop at a distance $x$ from wire.Magnetic field across $dx$ thickness is $db=\frac{{\mu }_{0}I}{2\pi x}$ $\hat{K}$ outward direction.Now flux passing through $dx$ thickness is

$\varphi =\int Bdsd\varphi =dBbdx{\int }_0^{\varphi }d\varphi ={\int }_d^{(d+a)}\frac{{\mu }_{0}Ib}{2\pi x}dx$

Now $\varphi$ is the flux passing through the loop.

$\varphi =\frac{{\mu }_{0}lb}{2\pi }(lnx{)}^{d+a}\varphi =\frac{{\mu }_{0}bI}{2\pi }ln\frac{(d+a)}{\left(d\right)}\varphi =\frac{{\mu }_{0}b}{2\pi }ln\frac{(d+a)}{d}.I_0{e}^{\frac{-t}{\tau }}$

putting value of I

Induced emf $e=\frac{-d\varphi }{dt}$

$e=-\frac{{\mu }_{0}b}{2\pi }ln\left(\frac{d+a}{d}\right)I_0{e}^{\frac{-1}{\tau }}\times \left(\frac{-t}{\tau }\right)e=\frac{{\mu }_{0}bI}{2\pi \tau }ln\frac{(d+a)}{d}$