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 consider the factors influencing the induced electromotive force (EMF) and power dissipation in a coil placed in a time-varying magnetic field. According to Faraday's Law of Electromagnetic Induction, the induced EMF (E) in the coil can be expressed as:

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

where:

• N is the number of turns in the coil.
• \Phi is the magnetic flux through the coil.

Now, the resistance (R) of the coil is related to its length and cross-sectional area. The resistance is given by:

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

where:

• \rho is the resistivity of the wire material.
• L is the length of the wire.
• A is the cross-sectional area of the wire.

If the radius of the wire is doubled, the cross-sectional area (A = \pi r^2) will increase by a factor of four, which decreases the resistance by a factor of four since R \propto \frac{1}{A}.

The electrical power (P) dissipated in the coil due to the induced current is given by:

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

Substituting E from Faraday’s Law:

P = \frac{\left(-N \frac{d\Phi}{dt}\right)^2}{R}

If N is halved:

P' = \frac{\left(-\frac{N}{2} \frac{d\Phi}{dt}\right)^2}{\frac{R}{4}}

Simplifying:

P' = \frac{\left(\frac{N}{2}\right)^2 \left(\frac{d\Phi}{dt}\right)^2}{\frac{R}{4}} = \frac{\frac{N^2}{4}\left(\frac{d\Phi}{dt}\right)^2 \times 4}{R} = \frac{N^2\left(\frac{d\Phi}{dt}\right)^2}{R} = 2P

This shows that the power dissipated is doubled when the number of turns is halved and the radius of the wire is doubled.