JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency , the average power loss in the loop due to Joule heating is : 

- A
- B
- C
- DZero
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Correct answer: C
- Magnetic flux through the rotating loop
The area of the elliptical loop is
If the loop rotates about the -axis with angular speed , then the angle between the magnetic field and the area vector of the loop changes as
Hence the magnetic flux through the loop is
So,
- Induced emf
By Faraday's law,
Therefore,
= \pi abB\omega \sin(\omega t).$$ Thus the instantaneous emf magnitude is $$e(t)=\pi abB\omega \sin(\omega t).$$ 3. **Instantaneous current** Since the loop has resistance $R$, $$i(t)=\frac{e(t)}{R}=\frac{\pi abB\omega}{R}\sin(\omega t).$$ 4. **Instantaneous Joule heating power** Power dissipated is $$P(t)=i^2R=\frac{e^2}{R}.$$ So, $$P(t)=\frac{(\pi abB\omega)^2}{R}\sin^2(\omega t).$$ 5. **Average power over one cycle** The average value of $\sin^2(\omega t)$ over a complete cycle is $$\langle \sin^2(\omega t)\rangle = \frac12.$$ Hence, $$P_{\text{avg}}=\frac{(\pi abB\omega)^2}{R}\cdot \frac12 =\frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}.$$ 6. **Option check** - **A:** $\dfrac{\pi abB\omega}{R}$ — incorrect (dimensionally power does not match) - **B:** $\dfrac{\pi^2 a^2 b^2 B^2 \omega^2}{R}$ — misses factor $\frac12$ - **C:** $\dfrac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$ — correct - **D:** Zero — incorrect Therefore, the correct answer is **C**.More from Electromagnetic Induction
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