Question 60 of 153

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$0.5\mu V$

Show answer and solution▾

Answer: (B) $1\mu V$

$d\phi =\dfrac{{\mu }_{0}i}{2\pi y}ady$

$\phi =\dfrac{{\mu }_{0}ia}{2\pi }\int _{x}^{x+a}\dfrac{dy}{y}=\dfrac{{\mu }_{0}ia}{2\pi }\left[ln\left(x+a\right)-lnx\right]$

$\text{E.m.f}=-\dfrac{d\phi }{dt}=-\dfrac{{\mu }_{0}ia}{2\pi }\left[\dfrac{1}{x+a}\dfrac{dx}{dt}-\dfrac{1}{x}\dfrac{dx}{dt}\right]$

$=\dfrac{{\mu }_{0}ia}{2\pi }\dfrac{a.v}{x\left(x+a\right)}=\dfrac{{\mu }_{0}}{2\pi }.\dfrac{i{a}^{2}v}{x\left(x+a\right)}=2\times {10}^{-7}\times \dfrac{1\times \left({0.1}^{2}\times 10\right)}{0.1\times 0.2}=1\mu V$

Alternative method:

Consider two sides which are perpendicular to the velocity as case of motional e.m.f.


$E={E}_{1}-{E}_{2}=\left({B}_{1}–{B}_{2}\right)Vl$

$=\dfrac{{\mu }_{0}I}{2\pi }\left(\dfrac{1}{x}-\dfrac{1}{x+a}\right)Vl$

$=2\times {10}^{-7}\times 1\left(\dfrac{1}{0.1}-\dfrac{1}{0.2}\right)\times 10\times 0.1$

$={10}^{-6}V=1\mu V$