Question

A square loop of side 20 cm is placed in a magnetic field $B=(5+2t^2)$; such that, the magnetic field is perpendicular to the plane of the loop. If resistance of loop is $2\Omega$, heat generated in 5 sec in joules is (write until two digits after the decimal point)

Answer 1

$\epsilon =\frac{d\varphi }{dt}=\frac{d}{dt}BA=\frac{d}{dt}(5+2t^2)a^2$
$\epsilon =4a^{2}t=i=\frac{\in }{R}=\frac{4a^{2}t}{2}=2a^{2}t$
Heat generated $={\int }_0^5{i}^{2}Rdt$
= ${\int }_0^{5}4a^4{t}^{2}Rdt=4a^{4}R{\left(\frac{t^3}{3}\right)}_0^5$
Heat $=\frac{4\times {0.2}^4\times 2\times 5^3}{3}=0.533J$