Question

A circular coil of area 0.01 m\(^2\) and 40 turns is rotated about its vertical diameter with an angular speed of 50 rad/s in a uniform horizontal magnetic field 0.05 T. If the average power loss due to Joule heating is 25 mW, find the closed loop resistance of the coil.

• A. 50 \(\Omega\)
• B. 12.5 \(\Omega\)
• C. 75 \(\Omega\)
• D. 20 \(\Omega\)

Hint:

For a rotating coil: \(\varepsilon_{\text{rms}} = N A B \omega / \sqrt{2}\), average power \(P = \varepsilon_{\text{rms}}^2 / R\) to find resistance.

Answer 1

The Correct Option is D

Step 1: Find the peak emf first, it comes out as a clean number.
\[ \varepsilon_0 = NAB\omega = (40)(0.01)(0.05)(50) = 1\ \text{V} \]
Step 2: Convert to rms.
\[ \varepsilon_{\text{rms}} = \frac{\varepsilon_0}{\sqrt2}, \qquad \varepsilon_{\text{rms}}^2 = \frac{1}{2}\ \text{V}^2 \]
Step 3: Use the average power formula to isolate R.
\[ P = \frac{\varepsilon_{\text{rms}}^2}{R} \implies R = \frac{\varepsilon_{\text{rms}}^2}{P} = \frac{0.5}{0.025} = 20\ \Omega \]
\[ \boxed{20\ \Omega} \]