If power dissipated in a coil having total number of turns N, cross-section area Aand radius of coil R when kept in a time varying magnetic field is P. Now if another coil having total number of turns 2N, cross-section area 2A and radius 3R is placed in the same time varying magnetic field, the power dissipated is αP. Find value of α

The correct option is 108.

Let us derive the relation for the power dissipated in a coil placed in a time-varying magnetic field step-by-step.

1. Resistance of the coil ($R_e$):
The electrical resistance of the wire forming the coil depends on its resistivity ($\rho$), its total length ($l$), and its cross-sectional area ($A$).
The total length of the wire for a coil of radius $R$ with $N$ turns is given by:
$l=N\cdot \left(2\pi R\right)$
Thus, the resistance is:
$R_e=\rho \frac{l}{A}=\rho \frac{2\pi RN}{A}$

2. Induced electromotive force (EMF) ($V$):
According to Faraday's Law of Electromagnetic Induction, the induced EMF ($V$) is proportional to the rate of change of magnetic flux ($\Phi$) through the coil:
$V=-\frac{d\Phi }{dt}$
The magnetic flux through a coil with $N$ turns, each of loop area $A_{\text{loop}}=\pi R^2$, in a magnetic field $B$ is:
$\Phi =N\cdot B\cdot \left(\pi R^2\right)$
Therefore, the induced EMF is:
$V=-N\pi R^2\frac{dB}{dt}$

3. Power dissipated ($P$):
The power dissipated in the coil is given by:
$P=\frac{V^2}{R_e}$
Substituting the expressions for $V$ and $R_e$ into this equation:
$P=\frac{{\left(N\pi R^2\frac{dB}{dt}\right)}^2}{\rho \frac{2\pi RN}{A}}$
Simplifying the expression:
$P=\frac{N^2{\pi }^2{R}^4{\left(\frac{dB}{dt}\right)}^2\cdot A}{2\pi \rho RN}$
$P=\left(\frac{\pi {\left(\frac{dB}{dt}\right)}^2}{2\rho }\right)\cdot \left(N{R}^{3}A\right)$
Since the time-varying magnetic field and resistivity remain the same, we have the proportionality:
$P\propto N{R}^{3}A$

4. Calculating the ratio ($\alpha$):
For the first coil:
$P\propto N{R}^{3}A$
For the second coil, the parameters are modified as follows:
Number of turns $N^{\text{'}}=2N$
Radius of coil $R^{\text{'}}=3R$
Cross-sectional area of wire $A^{\text{'}}=2A$
Let $P^{\text{'}}=\alpha P$ be the new power dissipated:
$P^{\text{'}}\propto N^{\text{'}}{\left(R^{\text{'}}\right)}^3{A}^{\text{'}}$
Substitute the values of the new parameters:
$P^{\text{'}}\propto \left(2N\right)\cdot {\left(3R\right)}^3\cdot \left(2A\right)$
$P^{\text{'}}\propto 2\cdot 27\cdot 2\cdot \left(N{R}^{3}A\right)$
$P^{\text{'}}\propto 108\cdot \left(N{R}^{3}A\right)$
Therefore:
$P^{\text{'}}=108P$
Comparing this with $P^{\text{'}}=\alpha P$, we find:
$\alpha =108$