Question

A square frame of side $10cm$ and a long straight wire carrying current $1A$ are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of $10m{s}^{-1}$ (see figure). The e.m.f induced at the time the left arm of the frame is at $x=10cm$ from the wire is

• A. $0.75\mu V$
• B. $1\mu V$
• C. $2\mu V$
• D. $0.5\mu V$

Answer 1

Answer: (B)

$d\varphi =\frac{{\mu }_{0}i}{2\pi y}ady$
$\varphi =\frac{{\left(\mu \right)}_{0}ia}{2\pi }{\int }_x^{x+a}\frac{dy}{y}=\frac{{\left(\mu \right)}_{0}ia}{2\pi }\left[ln\left(x+a\right)-lnx\right]$
$\text{E.m.f}=-\frac{d\varphi }{dt}=-\frac{{\mu }_{0}ia}{2\pi }\left[\frac{1}{x+a}\frac{dx}{dt}-\frac{1}{x}\frac{dx}{dt}\right]$
$=\frac{{\left(\mu \right)}_{0}ia}{2\pi }\frac{a.v}{x\left(x+a\right)}=\frac{{\left(\mu \right)}_0}{2\pi }.\frac{i{a}^{2}v}{x\left(x+a\right)}=2\times {\left(10\right)}^{-7}\times \frac{1\times \left({\left(0.1\right)}^2\times 10\right)}{0.1\times 0.2}=1\mu V$
Alternative method:
Consider two sides which are perpendicular to the velocity as case of motional e.m.f.

$E=E_1-E_2=\left(B_1-B_2\right)Vl$
$=\frac{{\left(\mu \right)}_{0}I}{2\pi }\left(\frac{1}{x}-\frac{1}{x+a}\right)Vl$
$=2\times {\left(10\right)}^{-7}\times 1\left(\frac{1}{0.1}-\frac{1}{0.2}\right)\times 10\times 0.1$
$=10^{-6}V=1\mu V$