When a coil is placed in a time dependent magnetic field the power dissipated in it is $P$. The number of turns, area of the coil and radius of the coil wire are $N$, $A$ and $r$ respectively. For a second coil number of turns, area of the coil and radius of the coil wire are $2N$, $2A$ and $3r$ respectively. When the first coil is replaced with second coil the power dissipated in it is $\sqrt2\,\alpha P$. The value of $\alpha$ is ______.
Answer: (A) $36$
Induced emf: $\varepsilon=NA\dfrac{dB}{dt}\propto NA$.
Resistance of the wire: length $\propto N\times$(perimeter of one turn) $\propto N\sqrt A$, cross-section $\propto r^2$, so $R\propto\dfrac{N\sqrt A}{r^2}$.
$$P=\frac{\varepsilon^2}{R}\propto\frac{N^2A^2r^2}{N\sqrt A}=NA^{3/2}r^2$$
$$\frac{P_2}{P_1}=2\times2^{3/2}\times3^2=2\times2\sqrt2\times9=36\sqrt2$$
So $P_2=\sqrt2\times36P$ and $\alpha=36$.
Solution by Sreeraj P, M.Sc Physics