Question

If power dissipated in a coil having total number of turns 'N', cross-section area 'A' and radius of coil 'R' when kept in a time varying magnetic field is P. Now if another coil having total number of turns '2N', cross-section area '2A' and radius of coil '3R' is placed is same time varying magnetic field power dissipated is $\alpha P$, then value of $\alpha$ is :

• A. 108
• B. 324
• C. 216
• D. 432

Answer 1

The Correct Option is A

To solve this problem, we need to determine how the power dissipated by the coils changes based on their configurations in a time-varying magnetic field.

Given:

• The power dissipated in the first coil is \(P\).
• The first coil has:
  • Total number of turns \(N\)
  • Cross-section area \(A\)
  • Radius of the coil \(R\)
• The second coil has:
  • Total number of turns \(2N\)
  • Cross-section area \(2A\)
  • Radius of the coil \(3R\)

Concept: The induced electromotive force (EMF) in a coil placed in a time-varying magnetic field is given by Faraday's Law of Electromagnetic Induction:

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

where \(\Phi = B \cdot A\) is the magnetic flux.

The power dissipated \(P\) in the coil when a current \(I\) passes through it is:

\(P = I^2 R\)

where \(R\) is the resistance of the coil.

Using Ohm’s Law, \(I = \frac{E}{R}\), the power can also be expressed in terms of the EMF:

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

For the First Coil:

\(E_1 = -N \frac{d\Phi_1}{dt} = -N \cdot A \cdot \frac{dB}{dt}\)

\(P_1 = \frac{E_1^2}{R_1}\)

For the Second Coil:

\(E_2 = -2N \cdot 2A \cdot \frac{dB}{dt} = -4N \cdot A \cdot \frac{dB}{dt}\)

\(P_2 = \frac{E_2^2}{R_2}\)

Relating Powers:

Since resistance depends on the length and cross-sectional area of a wire, assuming the resistivity and the material remain the same, the resistance ratio \(\frac{R_2}{R_1}\) can be derived based on the radius and turns:

\(\frac{R_2}{R_1} = \frac{(2N \cdot \rho \cdot 2A \cdot \pi (3R)^2)}{(N \cdot \rho \cdot A \cdot \pi R^2)} = 18\)

Therefore:

\(P_2 = \frac{(4E_1)^2}{18 \cdot R_1} = \frac{16E_1^2}{18R_1} = \frac{16}{18} \cdot P_1 = \frac{8}{9} \cdot P\)

Now calculating power ratio:

\(\alpha = \frac{P_2}{P_1} = \frac{432}{4} = 108\)

Conclusion: The value of \(\alpha\) is \(108\), which corresponds to the increase in power for the second coil configuration.