A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be : (Assume the coil to be short circuited.)

Electromagnetic Induction PYQ — JEE Main 2022 · 28 Jun · Shift 2 · Q63 with Solution

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Correct answer: D

For a coil of $N$ turns in a time-varying magnetic field,

\mathcal{E} = N\frac{d\Phi}{dt}

where $\Phi$ is magnetic flux through one turn.

So, if the number of turns is halved:

N' = \frac{N}{2}

then the induced emf becomes

\mathcal{E}' = \frac{1}{2}\mathcal{E}

Resistance is

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

where:

Now analyze the changes:

L' = \frac{L}{2}

A' = \pi (2r)^2 = 4\pi r^2 = 4A

Hence new resistance:

R' = \rho \frac{L'}{A'} = \rho \frac{L/2}{4A} = \frac{1}{8}\rho \frac{L}{A} = \frac{R}{8}

Since the coil is short-circuited, current induced is

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

Power dissipated:

P = I^2R = \frac{\mathcal{E}^2}{R}

So the new power is

P' = \frac{(\mathcal{E}')^2}{R'}

Substitute $\mathcal{E}' = \mathcal{E}/2$ and $R' = R/8$ :

P' = \frac{(\mathcal{E}/2)^2}{R/8} = \frac{\mathcal{E}^2/4}{R/8} = \frac{\mathcal{E}^2}{4}\cdot \frac{8}{R} = 2\frac{\mathcal{E}^2}{R} = 2P

The electrical power dissipated becomes

\boxed{\text{Doubled}}

So the correct option is D.