Question

The current in the outer coil is varying with time as $I=2t^2$. The heat developed in the inner coil between $t=0$ to $t$ seconds is $\frac{k{\mu }_0^2{\pi }^2{a}^4{t}^3}{3b^{2}R}$, where the resistance of the inner coil is $R$ and $b>>a$. Find $k$.

Answer 1

Flux through the inner coil $\varphi =\frac{{\mu }_{0}I}{2b}\left(\pi a^2\right)$

emf induced in the inner coil

$e=\left|\frac{d\varphi }{dt}\right|=\frac{{\mu }_0\pi }{2b}{a}^2\left(\frac{dI}{dt}\right)$

$\Rightarrow e=\frac{{\mu }_0\pi a^2}{2b}\times 4t$

Current in the inner coil $=\frac{e}{R}$

Heat developed $\left(\Delta H\right)$

$\Delta H={\int }_0^t\frac{e^2}{R}dt$

$\Rightarrow \Delta H=\frac{{\mu }_0^2{\pi }^2{a}^4\times 16}{4b^2.R}\frac{t^3}{3}=\frac{4{\mu }_0^2{\pi }^2{a}^4}{b^2.R}\frac{t^3}{3}$

On comparision, $k=4$

Ans $4$