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.)

• A. Halved
• B. Quadrupled
• C. The same
• D. Doubled

Answer 1

The Correct Option is D

To solve this problem, we need to understand how changes in the coil's dimensions affect the electrical power dissipated due to the current induced. This involves Faraday's law of electromagnetic induction and the resistance formula.

1. **Faraday's Law:** The induced electromotive force (EMF) in the coil due to a change in magnetic flux is given by:

\(E = -N \frac{d\Phi}{dt}\) 

where \(E\) is the EMF, \(N\) is the number of turns, and \(\frac{d\Phi}{dt}\) is the rate of change of magnetic flux.

2. **Resistance of the Coil:** The resistance \((R)\) of the coil is given by:

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

where \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. The cross-sectional area for a wire of radius \(r\) is \(A = \pi r^2\). With the radius doubled, \(A\) becomes \(4 \pi r^2\).

3. **Effect on Resistance:** Given that the radius is doubled, the new resistance becomes:

\(R_{new} = \rho \frac{L}{4 \pi r^2} = \frac{R}{4}\)

4. **Induced Current**: The induced current \(I\) can be expressed as:

\(I = \frac{E}{R}\) or \(I_{new} = \frac{E_{new}}{R_{new}}\) 
As \(N\) is halved, \(E_{new} = \frac{E}{2}\)

5. **Electrical Power Dissipated**: Power \((P)\) dissipated is given by \(P = I^2 R\). For the new configuration:

\(P_{new} = (I_{new})^2 R_{new}\)

Substituting the expressions for \(I\) and \(R\):

\(P_{new} = \left(\frac{\frac{E}{2}}{\frac{R}{4}}\right)^2 \cdot \frac{R}{4} = \left(\frac{2E}{R}\right)^2 \cdot \frac{R}{4} = \frac{4E^2}{R^2} \cdot \frac{R}{4} = \frac{E^2}{R} \cdot 2 = 2P\)

This indicates the power is doubled.

Conclusion: The correct answer is Doubled, confirming that the electrical power dissipated in the coil, under the given changes, is doubled.

Answer 2

The correct answer is (D) : Doubled
As number of turns are halved so length of wire is halved, and radius is doubled, then area will be 4 times the previous one if previous resistance is R then new resistance is R/8 and if previous emf is E then new emf will be E/2 so
\(P_i=\frac{E_2}{R}\)
\(P_f=\frac{(\frac{E}{2})^2}{\frac{R}{8}}\)
\(=\frac{2E^2}{R}=2P_i\)
Because of the changes in answer key, students can challenge this question.