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 $ω$ , the average power loss in the loop due to Joule heating is:

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$\frac{π a b B ω}{R}$

$Z e r o$

$\frac{π^{2} a^{2} b^{2} B^{2} ω^{2}}{R}$

$\frac{π^{2} a^{2} b^{2} B^{2} ω^{2}}{2 R}$

answer is B.

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Detailed Solution

$\begin{array}{l} M a g n t i c f l u x = ϕ = BA \cos ω t = B (π ab) \cos ω t \\ I n d u c e d e m f = V = |\frac{d ϕ}{dt}| = B π ab ω \sin ω t \\ Peak emf = V_{0} = B π ab ω \\ R m s e m f = V_{r m s} = \frac{V_{0}}{\sqrt{2}} = \frac{B π ab ω}{\sqrt{2}} \\ A v e r a g e P o w e r = P = \frac{{V_{rms}}^{2}}{R} = {(\frac{B π ab ω}{\sqrt{2}})}^{2} \frac{1}{R} \end{array}$