Question

The figure shows a square frame of wire placed coplanarly with a long straight wire. The wire carries a current given by $i=i_0\sin \omega t$. Find the expression for mutual inductance between the straight wire and the square loop.

• A. $\frac{{\mu }_{0}a}{2\pi }\ln [1+\frac{a}{b}]$
• B. $\frac{{\mu }_{0}a}{2\pi }\ln [1+\frac{b}{a}]$
• C. $\frac{{\mu }_{0}b}{2\pi }\ln [1+\frac{a}{b}]$
• D. $\frac{{\mu }_{0}b}{2\pi }\ln [1+\frac{b}{a}]$

Answer 1

The correct option is A $\frac{{\mu }_{0}a}{2\pi }\ln [1+\frac{a}{b}]$


Considering an element of thickness $dx$ on square loop at a distance $x$ from the wire.

Flux through the element,

$d\varphi =B_x\left(dA\right)=\frac{{\mu }_{0}i\times adx}{2\pi x}$

Now, total flux,

$\varphi =\int d\varphi ={\int }_b^{a+b}\frac{{\mu }_{0}ia}{2\pi }\frac{dx}{x}$

$=\frac{{\mu }_{0}ia}{2\pi }\ln [1+\frac{a}{b}]$

Emf induced,

$|E|=\left|\frac{d\varphi }{dt}\right|=\frac{d}{dt}\left[\frac{{\mu }_{0}ia}{2\pi }\ln [1+\frac{a}{b}]\right]$

As, $i=i_0\sin \omega t$

$|E|=\frac{{\mu }_{0}a}{2\pi }\ln [1+\frac{a}{b}]\frac{d}{dt}\left(i_0\sin \omega t\right)$

$|E|=\frac{{\mu }_{0}a{i}_0\omega \cos \omega t}{2\pi }\ln [1+\frac{a}{b}]$

Also, $|E|=\frac{Mdi}{dt}$

$\frac{{\mu }_{0}a{i}_0\omega \cos \omega t}{2\pi }\ln [1+\frac{a}{b}]=M{i}_0\omega \cos \omega t$

$\therefore M=\frac{{\mu }_{0}a}{2\pi }\ln [1+\frac{a}{b}]$

Hence, $\left(\text{A}\right)$ is the correct answer.