Question 72: Electromagnetic Induction

Question

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.)

Accepted Answer

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

Answer

Halved

Answer

Quadrupled

Answer

The same

Answer

Doubled

Electromagnetic Induction

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.)

Choose the correct answer: