Ratio of Power Dissipated in Coils in Magnetic Field

To find the value of $\alpha$, we analyze the power dissipated in a coil placed in a time-varying magnetic field.

1. Induced Electromotive Force ($\mathcal{E}$): By Faraday's Law of Electromagnetic Induction, the induced emf $\mathcal{E}$ in a coil with $N$ turns and area $A$ placed in a time-dependent magnetic field $B(t)$ is given by: $\mathcal{E} = -N \frac{d\Phi}{dt} = -N A \frac{dB}{dt}$

Thus, the magnitude of the induced emf is proportional to $N$ and $A$: $\mathcal{E} \propto N A$

2. Resistance of the Coil Wire ($R$): The resistance $R$ of the wire forming the coil is given by: $R = \rho \frac{l}{a_{wire}}$ where:

• $\rho$ is the resistivity of the wire material,
• $l$ is the total length of the wire,
• $a_{wire}$ is the cross-sectional area of the wire.

For a coil with radius $R_{\text{coil}}$, the area is $A = \pi R_{\text{coil}}^2 \implies R_{\text{coil}} = \sqrt{\frac{A}{\pi}}$. The total length of the wire $l$ for $N$ turns is: $l = N (2\pi R_{\text{coil}}) = 2 N \sqrt{\pi A} \propto N \sqrt{A}$

The cross-sectional area of the wire with radius $r$ is: $a_{wire} = \pi r^2 \propto r^2$

Substituting $l$ and $a_{wire}$ into the expression for resistance $R$: $R \propto \frac{N \sqrt{A}}{r^2}$

3. Power Dissipated ($P$): The power dissipated in the coil is: $P = \frac{\mathcal{E}^2}{R}$

Substituting the proportionalities for $\mathcal{E}$ and $R$: $P \propto \frac{(N A)^2}{\left(\frac{N \sqrt{A}}{r^2}\right)} = N A^{3/2} r^2$

4. Ratio of Power Dissipated in the Two Coils: For the first coil, the parameters are $(N, A, r)$, giving a power $P$. For the second coil, the parameters are $(2N, 2A, 3r)$, giving a power $P'$.

The ratio of the power dissipated in the second coil to the first coil is: $\frac{P'}{P} = \left(\frac{N'}{N}\right) \left(\frac{A'}{A}\right)^{3/2} \left(\frac{r'}{r}\right)^2$

Substituting the given values: $\frac{P'}{P} = \left(\frac{2N}{N}\right) \left(\frac{2A}{A}\right)^{3/2} \left(\frac{3r}{r}\right)^2$ $\frac{P'}{P} = 2 \times (2)^{3/2} \times 3^2 = 2 \times 2\sqrt{2} \times 9 = 36\sqrt{2}$

Given that $P' = \sqrt{2} \alpha P$, we have: $\frac{P'}{P} = \sqrt{2} \alpha$

Equating the two expressions: $\sqrt{2} \alpha = 36\sqrt{2} \implies \alpha = 36$