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  • An elliptical loop having resistance R of semi-major axis a, and semi-minor axis b is placed in a Magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency

An elliptical loop having resistance R of semi-major axis a, and semi-minor axis b is placed in a Magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency \omega, the average power loss in the loop due to Joule heating  is  
Option: 1 \frac{\pi ^{2}a^{2}b^{2}B^{2}\omega ^{2}}{2R}
 
Option: 2 Zero
Option: 3 \frac{\pi abB\omega }{R}  
Option: 4 \frac{\pi ^{2}a^{2}b^{2}B^{2}\omega ^{2}}{R}

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\\ \phi=B.A=BAcos \theta=\pi abBcos \theta\\ \\ \varepsilon =-\frac{d\Phi}{dt}=-\frac{d(\pi abBcos \theta)}{dt}=\pi aBsin \theta \frac{d\theta}{dt}=\pi aB \omega bsin \theta\\ \\ i_{max}=\frac{\varepsilon_{max} }{R}=\frac{\pi abB \omega}{R}\\ \\ \because P_{avg}=\frac{i_{max}^2R}{2}\\ \therefore \text{avg power loss in the loop due to joule heating is}=\frac{\pi^2 a^2b^2B^2 \omega^2}{2R}

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