Question

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

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

Answer 1

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

The magnetic field at a distance x from the wire is $B=\frac{{\mu }_{0}I}{2\pi x}$
The flux due to the magnetic field in the strip of thickness $dx$ and length b is $B.b.dx$
Hence the total magnetic flux in the loop is
$\int d\varphi ={\int }_d^{a+d}\frac{{\mu }_{0}I}{2\pi x}.b.dx$
$\varphi =\frac{{\mu }_{0}Ib}{2\pi }ln\left(\frac{d+a}{a}\right)$
$E=-\frac{d\varphi }{dt}=-\frac{{\mu }_0{I}_0{e}^{-\frac{t}{\tau }}b}{-2\pi \tau }ln\left(\frac{d+a}{a}\right)$
$E=\frac{{\mu }_{0}Ib}{2\pi \tau }ln\left(\frac{d+a}{a}\right)$