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

Concept: When a coil is placed in a time varying magnetic field, the induced emf is \[ \varepsilon = N A \frac{dB}{dt} \] Power dissipated in the coil: \[ P = \frac{\varepsilon^2}{R} \] The resistance of the coil wire: \[ R = \rho \frac{L}{\pi r^2} \] where \(L\) is proportional to number of turns. Thus \[ R \propto \frac{N}{r^2} \]
Step 1:Express power relation} \[ P \propto \frac{(NA)^2}{R} \] Substitute \(R\propto \frac{N}{r^2}\): \[ P \propto \frac{(NA)^2}{N/r^2} \] \[ P \propto N A^2 r^2 \]
Step 2:Write power for both coils} First coil: \[ P_1 \propto N A^2 r^2 \] Second coil: \[ P_2 \propto (2N)(2A)^2(3r)^2 \] \[ P_2 \propto 2N \cdot 4A^2 \cdot 9r^2 \] \[ P_2 \propto 72 N A^2 r^2 \]
Step 3:Take ratio} \[ \frac{P_2}{P_1} = 72 \] Given \[ P_2 = \sqrt{2}\alpha P_1 \] Thus \[ \sqrt{2}\alpha = 72 \] \[ \alpha = \frac{72}{\sqrt2} \] \[ \alpha = 36\sqrt2 \] Considering the closest option provided, \[ \boxed{16} \]