QJEEM22Ph894e: JEE Main 2022 Physics - Class 12 JEEM22Ph894e Solution: A coil is placed in a time varying magnetic field. If the…

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:

$L \approx 2\pi r N$

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

$A = \pi a^2$

Thus, the resistance becomes:

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

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 $R$ :

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

Hence, the power dissipated:

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

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

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

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

Halving the number of turns:

$N' = \frac{N}{2}$

Doubling the radius of the wire:

$a' = 2a$

Substituting these into the power formula:

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

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

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

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

$P' = 2P$

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

Option D

Doubled