Current Electricity

JEE Main level, single correct, medium

252 of 380

Question 252 of 380

An elliptical loop having resistance RR, of semi major axis aa , and semi minor axis bb is placed in a magnetic field as shown in the figure. If the loop is rotated about the xx-axis with angular frequency ω\omega, the average power loss in the loop due to Joule heating is :

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Answer: (A) π2a2b2B2ω22R\dfrac{{\pi }^{2}{a}^{2}{b}^{2}{B}^{2}{\omega }^{2}}{2R}

The expression for induced EMF in the rotating loop in uniform magnetic field is given by
ϵ=NABωcos⁡ωt\epsilon =NAB\omega \cos \omega t
The average power loss
⟨P⟩=<ϵ2R>\left\langle P\right\rangle =\lt \dfrac{{\epsilon }^{2}}{R}\gt =<A2B2ω2cos⁡2ωtR>=\lt \dfrac{{A}^{2}{B}^{2}{\omega }^{2}{\cos }^{2}\omega t}{R}\gt
<cos⁡2θ>=12\lt {\cos }^{2}\theta \gt =\dfrac{1}{2}
<P>=A2B2ω2R(12)\lt P\gt =\dfrac{{A}^{2}{B}^{2}{\omega }^{2}}{R}\left(\dfrac{1}{2}\right)
⇒<P>=π2a2b2B22R(ω2)\Rightarrow \lt P\gt =\dfrac{{\pi }^{2}{a}^{2}{b}^{2}{B}^{2}}{2R}\left({\omega }^{2}\right)

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