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 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.
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For a rotating coil: \(\varepsilon_{\text{rms}} = N A B \omega / \sqrt{2}\), average power \(P = \varepsilon_{\text{rms}}^2 / R\) to find resistance.
Updated On: Jul 18, 2026
- 50 \(\Omega\)
- 12.5 \(\Omega\)
- 75 \(\Omega\)
- 20 \(\Omega\)
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The Correct Option is D
Solution and Explanation
Step 1: Recall induced emf for rotating coil.
\[ \varepsilon_{\text{rms}} = N A B \omega / \sqrt{2} \]
Step 2: Substitute known values.
\[ N = 40, \, A = 0.01 \, \text{m}^2, \, B = 0.05 \, \text{T}, \, \omega = 50 \, \text{rad/s} \]
\[ \varepsilon_{\text{rms}} = \frac{40 \cdot 0.01 \cdot 0.05 \cdot 50}{\sqrt{2}} \approx 0.707 \, \text{V} \]
Step 3: Recall power in resistor.
\[ P = \frac{\varepsilon_{\text{rms}}^2}{R} \implies R = \frac{\varepsilon_{\text{rms}}^2}{P} \]
Step 4: Substitute power.
\[ P = 25 \, \text{mW} = 0.025 \, \text{W} \]
\[ R = \frac{0.707^2}{0.025} \approx 20 \, \Omega \]
Step 5: Verify reasoning.
RMS emf formula correct for rotation about diameter; power formula consistent.
Step 6: Final conclusion.
Hence, the closed loop resistance of the coil is:
\[ \boxed{20 \, \Omega} \]
\[ \varepsilon_{\text{rms}} = N A B \omega / \sqrt{2} \]
Step 2: Substitute known values.
\[ N = 40, \, A = 0.01 \, \text{m}^2, \, B = 0.05 \, \text{T}, \, \omega = 50 \, \text{rad/s} \]
\[ \varepsilon_{\text{rms}} = \frac{40 \cdot 0.01 \cdot 0.05 \cdot 50}{\sqrt{2}} \approx 0.707 \, \text{V} \]
Step 3: Recall power in resistor.
\[ P = \frac{\varepsilon_{\text{rms}}^2}{R} \implies R = \frac{\varepsilon_{\text{rms}}^2}{P} \]
Step 4: Substitute power.
\[ P = 25 \, \text{mW} = 0.025 \, \text{W} \]
\[ R = \frac{0.707^2}{0.025} \approx 20 \, \Omega \]
Step 5: Verify reasoning.
RMS emf formula correct for rotation about diameter; power formula consistent.
Step 6: Final conclusion.
Hence, the closed loop resistance of the coil is:
\[ \boxed{20 \, \Omega} \]
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