JEE Main 2022 — Electromagnetic Induction Question with Solution

The electrical power dissipated due to the current induced in a coil placed in a time-varying magnetic field can be determined by considering Faraday's Law of Induction and the resistance of the coil.

Faraday's Law of Induction

The induced EMF ($\mathcal{E}$) in a coil with $N$ turns experiencing a time-varying magnetic flux (${\Phi }_B$) is given by:

$$\mathcal{E}=-N\frac{d{\Phi }_B}{dt}$$

Resistance of the Coil

The resistance ($R$) of a wire depends on its length ($L$) and cross-sectional area ($A$) as well as the resistivity ($\rho$) of the material:

$$R=\frac{\rho L}{A}$$

For a coil of radius $r$ and with wire of radius $a$, assuming the wire is wound tightly with a length approximated by the circumference of the coil multiplied by the number of turns, we have:

$$L\approx 2\pi rN$$

And the cross-sectional area of the wire is given by:

$$A=\pi a^2$$

Thus, the resistance becomes:

$$R\approx \frac{\rho \cdot 2\pi rN}{\pi a^2}=\frac{2\rho rN}{a^2}$$

Power Dissipated

The electrical power ($P$) dissipated in the coil is related to the induced current ($I$) and the resistance ($R$):

$$P=I^{2}R$$

Using Ohm's Law, the induced current $I$ can be expressed as:

$$I=\frac{\mathcal{E}}{R}$$

Substituting the expressions for $\mathcal{E}$ and $R$:

$$I=\frac{N\frac{d{\Phi }_B}{dt}}{\frac{2\rho rN}{a^2}}=\frac{a^2}{2\rho r}\cdot \frac{d{\Phi }_B}{dt}$$

Hence, the power dissipated:

$$P={\left(\frac{a^2}{2\rho r}\cdot \frac{d{\Phi }_B}{dt}\right)}^2\cdot \frac{2\rho rN}{a^2}$$

$$P=\frac{a^4}{4{\rho }^2{r}^2}{\left(\frac{d{\Phi }_B}{dt}\right)}^2\cdot \frac{2\rho rN}{a^2}$$

$$P=\frac{a^2}{2\rho r}\cdot N{\left(\frac{d{\Phi }_B}{dt}\right)}^2$$

Case when the number of turns is halved and the radius of the wire is doubled

Halving the number of turns:

$$N^{\prime }=\frac{N}{2}$$

Doubling the radius of the wire:

$$a^{\prime }=2a$$

Substituting these into the power formula:

$$P^{\prime }=\frac{(2a{)}^2}{2\rho r}\cdot \frac{N}{2}{\left(\frac{d{\Phi }_B}{dt}\right)}^2$$

$$P^{\prime }=\frac{4a^2}{2\rho r}\cdot \frac{N}{2}{\left(\frac{d{\Phi }_B}{dt}\right)}^2$$

$$P^{\prime }=\frac{4a^2}{2\rho r}\cdot \frac{N}{2}{\left(\frac{d{\Phi }_B}{dt}\right)}^2$$

$$P^{\prime }=2\left(\frac{a^2}{2\rho r}\cdot N{\left(\frac{d{\Phi }_B}{dt}\right)}^2\right)$$

$$P^{\prime }=2P$$

Thus, the electrical power dissipated in the coil would be:

Option D

Doubled