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 $$\sqrt{2} \alpha P$$. The value of $$\alpha$$ is __________.

Let the magnetic field through the coil vary with time so that the induced emf in a coil of $$N$$ turns and area $$A$$ is

$$\varepsilon = N\,A\,\frac{dB}{dt} \;.$$

The wire of the coil has resistivity $$\rho$$ and cross-sectional radius $$r$$.
Each turn is a circle of area $$A=\pi R^{2}$$, so its radius is $$R=\sqrt{\dfrac{A}{\pi}}$$ and its circumference is $$2\pi R$$.

Total length of wire in the coil:

$$\ell = N\,(2\pi R)=N\,(2\pi)\sqrt{\frac{A}{\pi}}=2N\sqrt{\pi A}\;.$$

Resistance of the coil:

$$R = \rho\,\frac{\ell}{\pi r^{2}} =\rho\,\frac{2N\sqrt{\pi A}}{\pi r^{2}} \;\;\Longrightarrow\;\; R\propto\frac{N\sqrt{A}}{r^{2}}\;.$$

Power dissipated (Joule heating) is

$$P=\frac{\varepsilon^{2}}{R} \propto\frac{(N\,A)^{2}}{\,N\sqrt{A}/r^{2}} =N\,r^{2}\,A^{2}\,A^{-1/2} =N\,r^{2}\,A^{3/2}\;.$$ Thus

$$P\propto N\,r^{2}\,A^{3/2}\;.\quad -(1)$$

Parameters $$N_1=N,\;A_1=A,\;r_1=r$$ give

$$P_1 \propto N\,r^{2}\,A^{3/2}\;.$$

Parameters $$N_2=2N,\;A_2=2A,\;r_2=3r$$ give

$$P_2 \propto (2N)\,(3r)^{2}\,(2A)^{3/2}\;.$$

Simplify step by step:

• $$(3r)^{2}=9\,r^{2}$$
• $$(2A)^{3/2}=2^{3/2}\,A^{3/2}=2\sqrt{2}\,A^{3/2}$$

Hence

$$P_2 \propto 2N \times 9r^{2} \times 2\sqrt{2}\,A^{3/2} =36\sqrt{2}\;N\,r^{2}\,A^{3/2}\;.$$

Taking the ratio using (1):

$$\frac{P_2}{P_1}=36\sqrt{2}\;.$$

The question states $$P_2=\sqrt{2}\,\alpha\,P_1\;,$$ so

$$\sqrt{2}\,\alpha = 36\sqrt{2}\;\;\Longrightarrow\;\;\alpha = 36\;.$$

Therefore, $$\boxed{\alpha = 36}$$.

Option A which is: $$36$$