Question

When a coil is placed in a time dependent magnetic field the power dissipated in it is \(P\). The number of turns, area of the coil and radius of the coil wire are \(N, A\) and \(r\) respectively. For a second coil the number of turns, area of the coil and radius of the coil wire are \(2N, 2A\) and \(3r\) respectively. If the first coil is replaced with second coil the power dissipated in it is \(\sqrt{2}\alpha P\). The value of \(\alpha\) is:

• A. \(36\)
• B. \(128\sqrt{2}\)
• C. \(16\)
• D. \(64\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Power dissipated in a coil is \(P = \frac{\mathcal{E}^2}{R}\).
Induced EMF \(\mathcal{E} = N \cdot A \cdot \frac{dB}{dt}\).
Resistance \(R = \rho \frac{l}{s}\), where \(l\) is the total length of the wire and \(s\) is its cross-sectional area.
Step 2: Key Formula or Approach:
1. \(\mathcal{E} \propto N \cdot A\).
2. \(l \propto N \cdot \sqrt{A}\) (as length of one turn is \(2\pi R_{coil} = 2\sqrt{\pi A}\)).
3. \(s \propto r^2\).
4. \(R \propto \frac{N \sqrt{A}}{r^2}\).
5. \(P = \frac{\mathcal{E}^2}{R} \propto \frac{(NA)^2}{N\sqrt{A}/r^2} = N A^{3/2} r^2\).
Step 3: Detailed Explanation:
Let \(P_1 = P\).
For the second coil: \(N' = 2N, A' = 2A, r' = 3r\).
\[ P_2 = (2N) (2A)^{3/2} (3r)^2 \times \text{constant} \]
\[ P_2 = (2) (2\sqrt{2}) (9) \times (N A^{3/2} r^2 \times \text{constant}) \]
\[ P_2 = 36\sqrt{2} P \].
Given \(P_2 = \sqrt{2} \alpha P\), we equate:
\[ \sqrt{2} \alpha = 36\sqrt{2} \implies \alpha = 36 \].
Step 4: Final Answer:
The value of \(\alpha\) is 36.