Question

A uniform circular loop of radius a and resistance $R$ is pulled at a constant velocity $v$ out of a region of uniform magnetic field whose magnitude is $B$. The plane of loop and the velocity are both perpendicular to $\overrightarrow{B}$. Then the electrical power in the circular loop at the instant when the arc (of circular loop) outside the region of magnetic field subtends an angle $\frac{\pi }{3}$ at centre of the loop is

• A. $\frac{B^2{a}^2{v}^2}{R}$
• B. $\frac{2B^2{a}^2{v}^2}{R}$
• C. $\frac{B^2{a}^2{v}^2}{2R}$
• D. None of these

Answer 1

Answer: (A)

$\frac{B^2{a}^2{v}^2}{R}$

Here the loop is coming out of the magnetic field at constant velocity.
We first find out the rate of change of the magnetic flux to calculate the induced emf.
We have the area enclosed by the magnetic field at any instant as
$A=\pi a^2-\left[\pi a^2\right(\frac{2\theta }{2\pi })-\frac{x}{2}2a\sin \theta ]$
$=\pi a^2-a^2\theta +ax\sin \theta$
$=\pi a^2-a^2{\cos }^{-1}\left(\frac{x}{a}\right)+ax\frac{\sqrt{x^2-a^2}}{a}$

Thus we get the emf as $ϵ=\frac{d\varphi }{dt}=B\frac{dA}{dt}$
We calculate $\frac{dA}{dt}=0+\frac{a^2}{\sqrt{1-\frac{x^2}{a^2}}}\frac{1}{a}\frac{dx}{dt}+[\sqrt{a^2-x^2}-\frac{x}{2}\frac{2x}{\sqrt{x^2-a^2}}]\frac{dx}{dt}$
here $\theta =\pi /3$ and $x=\frac{\sqrt{3}a}{2}$

Solving this we get $\frac{dA}{dt}=av$
Thus we get $P=\frac{{ϵ}^2}{R}=\frac{B^2}{R}(\frac{dA}{dt}{)}^2=\frac{B^2{a}^2{v}^2}{R}$