Question

A plane loop, shaped as two squares of sides $a=1m$ and $b=0.4m$ is introduced into a uniform magnetic field perpendicular to the plane of loop. The magnetic field varies as $B=10^{-3}$ sin

$\left(100t\right)T$. The amplitude of the current (in A) induced in the loop if its resistance per unit length is $r=5m\Omega m^{-1}$ is

Answer 1

Answer: 3

$\varphi$ (flux linked) $=a^{2}B$ cos $0^{\circ }-b^{2}B$ cos $180^{\circ }$
$E=-\frac{d\varphi }{dt}=-\left(a^2-b^2\right)\frac{dB}{dt}=\left(a^2-b^2\right)B_0\omega$ cos $\omega t$
where $B=B_0$ sin $\omega t,B_0=10^{-3}T,\omega =100$
$\therefore I_{max}=\left(a^2-b^2\right)\frac{B_0\omega }{R}$
and $R=\left(4a+4b\right)r=4\left(a+b\right)r$
$\therefore I_{max}=\frac{\left(a-b\right)B_0\omega }{4r}=\frac{\left(1-0.4\right)\times 10^{-3}\times 100}{4\times 5\times 10^{-3}}=3A$