Question
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 coils number of turns, area of the coil and radius of the coil wire are 2 N , 2A and 3 r 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 $\_\_\_\_$ .
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Solution
Sol. [Key Concept: $P = \varepsilon^{2}/R$ with $\varepsilon \propto \ NA$ and $R \propto \ N/r^{2}$, giving $P \propto \ {NA}^{3/2}r^{2}$.]
Power dissipated : $P = \frac{\varepsilon^{2}}{R}$
Induced emf $\varepsilon = - NA\frac{dB}{dt}$
Resistance of coil : $R = \rho\frac{\mathcal{l}}{\pi r^{2}}$
Now : $\mathcal{l} = N\left( 2\pi R_{\text{coil}\text{~}} \right)$
Also : $A = \pi R_{\text{coil}\text{~}}^{2} \Rightarrow R_{\text{coil}\text{~}} = \sqrt{\frac{A}{\pi}}$
Thus $;R = \frac{\rho N(2\pi)}{\pi r^{2}}\sqrt{\frac{\text{ }A}{\pi}}$
$$\begin{array}{r} P = \frac{N^{2}{\text{ }A}^{2}\left( \frac{\text{ }dB}{dt} \right)^{2}}{\frac{2\rho\text{ }N}{r^{2}}\sqrt{\frac{\text{ }A}{\pi}}} \Rightarrow P \propto {NA}^{3/2}r^{2}\#(1) \end{array}$$
$$\begin{array}{r} P^{'} \propto (2\text{ }N)(2\text{ }A)^{3/2}(3r)^{2}\#(2) \end{array}$$
$$\frac{(2)}{(1)}:\frac{P^{'}}{P} = 36\sqrt{2} \Rightarrow P^{'} = (\sqrt{2})(36)P $$Thus : $\alpha = 36$
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