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