Question

A plane loop, shaped as two squares of sides $a = 1 m$ and $b = 0.4 \,m$ 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 $r =5 m\Omega m^{-1}$ is

Answer 1

$\phi$ (flux linked) $ =a^{2}B$ cos $0^{\circ} - b^{2}B$ cos $180^{\circ}$
$E = - \frac{d\phi}{dt} = -\left(a^{2} -b^{2}\right) \frac{dB}{dt} = \left(a^{2} -b^{2}\right) B_{0}\omega$ 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 $R =\left(4a +4b\right)r =4 \left(a+ b\right)r$
$\therefore I_{max} =\frac{\left(a-b\right)B_{0} \omega}{4r} = \frac{\left(1-0.4\right)\times10^{-3} \times100}{4 \times5 \times10^{-3}} = 3A$