Question

A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns was to be quadrupled and the wire radius halved, the electrical power dissipated would be:
A) Halved
B) The same
C) Doubled
D) Quadrupled

Answer 1

The emf induced in a coil when placed in a time-varying magnetic field depends on the number of turns in the coil and the area of the coil. We will use this relation to determine the electric power in the coil.
Formula used: In this solution, we will use the following formula:
Power dissipated in a circuit $P = \dfrac{{{E^2}}}{R}$ where $E$ is the emf and $R$ is the resistance
Emf induced due to changing flux: $E = - \dfrac{{d\phi }}{{dt}} = - \dfrac{{d(B.A)}}{{dt}}$ where $B$ is the magnetic field and $A$ is the area.

Complete step by step answer:
The power dissipated in a circuit is calculated as
$P = \dfrac{{{E^2}}}{R}$
Now the emf induced in the coil when placed in a time-varying magnetic field is calculated as
$E = nA\dfrac{{dB}}{{dt}}$ where $n$ is the number of turns, $A$ is the area of the coil, $B$ is the magnetic field.
The resistance of the wire will depend on the length of the wire and the area according to the relation of resistance and resistivity as $R = \rho \dfrac{l}{A}$.
Now the area of the wire will be $A = \pi {r^2}$. So we can write $R \propto \dfrac{l}{{{r^2}}}$.
The varying terms in the above discussion will be the number of turns in the coil and the area of the wire forming the coil.
Hence the power will be proportional to
$P = \dfrac{{{E^2}}}{R} \propto \dfrac{{{n^2}}}{{l/{r^2}}}$
Since the number of turns is quadrupled and the radius of the wire is halved, we can take the ratio of the new and the old power as
$\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{n_2^2}}{{n_1^2}} \times \dfrac{{r_2^2}}{{r_1^2}} \times \dfrac{{{l_1}}}{{{l_2}}}$
Now since the number of turns is quadrupled, we would need more wire so the new length of the wire will also be ${l_2} = 4{l_1}$ and hence the new power will be
$\dfrac{{{P_2}}}{{{P_1}}} = {\left( {\dfrac{4}{1}} \right)^2} \times {\left( {\dfrac{1}{2}} \right)^2} \times \left( {\dfrac{1}{4}} \right)$
Which gives us
${P_2} = {P_1}$

Hence the power will remain the same so option (B) is the correct choice.

Note: In the question, we’ve been given that the radius of the wire itself is halved and not the area of the coil so it will affect the resistance of the wire. This is because there is no external resistance in the wire but we are considering the resistance of the wire itself. Also, when the number of turns is quadrupled, we must not forget to account for the increase in net resistance due to more length of wire being required.