When a coil is placed in a time dependent magnetic field the power...

The induced emf in the coil is given by Faraday's law:

$E = N A \dfrac{dB}{dt}$

The resistance of the coil is $R = \rho \dfrac{l}{a}$, where $l$ is the total length of the wire and $a$ is its cross-sectional area.

The length of the wire is $l = N (2\pi R_c)$, where $R_c$ is the radius of the coil. Since the area of the coil is $A = \pi R_c^2$, we have $R_c = \sqrt{\dfrac{A}{\pi}}$.

Thus, $l = 2N\sqrt{\pi A}$.

The cross-sectional area of the wire is $a = \pi r^2$.

Therefore, the resistance is:

$R = \rho \dfrac{2N\sqrt{\pi A}}{\pi r^2} \propto \dfrac{N\sqrt{A}}{r^2}$

The power dissipated in the coil is:

$P = \dfrac{E^2}{R} \propto \dfrac{(N A)^2}{\dfrac{N\sqrt{A}}{r^2}} = N A^{3/2} r^2$

For the second coil, the parameters are $N' = 2N$, $A' = 2A$, and $r' = 3r$. The new power dissipated is:

$P' = P \left(\dfrac{2N}{N}\right) \left(\dfrac{2A}{A}\right)^{3/2} \left(\dfrac{3r}{r}\right)^2$

$P' = P (2) (2\sqrt{2}) (9) = 36\sqrt{2} P$

Given that $P' = \sqrt{2}\alpha P$, we can compare the expressions to find:

$\sqrt{2}\alpha = 36\sqrt{2} \Rightarrow \alpha = 36$

Answer: $36$