Question

A long straight wire is parallel to one edge as in 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. $\frac{{\mu }_{0}bl}{\pi \tau }\ln \left(\frac{d+a}{d}\right)$
• B. $\frac{{\mu }_{0}bl}{2\pi \tau }\ln \left(\frac{d+a}{d}\right)$
• C. $\frac{2{\mu }_{0}bl}{\pi \tau }\ln \left(\frac{d+a}{d}\right)$
• D. $\frac{{\mu }_{0}bl}{\pi \tau }\ln \left(\frac{d}{d+a}\right)$

Answer 1

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


As shown in figure, let us consider a strip of length $b$ and width $dx$ in the loop at a distance $x$ from wire.

Magnetic field across thickness is , $dB=\frac{{\mu }_{0}I}{2\pi x}\hat{k}$ outward direction.

Now flux passing through-thickness is , $d\varphi =\overrightarrow{dB}\cdot \overrightarrow{A}$

Integrating on both sides we get,

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

Now $\varphi$ is the flux passing through the loop , $\varphi =\frac{{\mu }_{0}Ib}{2\pi }[\ln \left(x\right){]}_d^{d+a}$

$\varphi =\frac{{\mu }_{0}Ib}{2\pi }\ln \left(\frac{d+a}{a}\right)$

Substituting the value of $I$,

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

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

$e=\frac{{\mu }_{0}bI}{2\pi \tau }\ln \left(\frac{d+a}{a}\right)$

Hence, option (b) is the correct answer.