Q 12-06-123JEE MainJEE Main 2020 (3 Sep, Shift 1)Medium
An elliptical loop having resistance $R$, of semi major axis $a$, and semi minor axis $b$ is placed in a uniform magnetic field $B$. The loop lies in the $xy$-plane with its major axis along the $x$-axis, and the field is along the $z$-axis. 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:
Answer: (A) $\dfrac{\pi^{2}a^{2}b^{2}B^{2}\omega^{2}}{2R}$
Area of the ellipse $A = \pi ab$. Flux $\Phi = BA\cos\omega t$, so $\varepsilon = BA\omega\sin\omega t$.
$$\langle P\rangle = \frac{\langle\varepsilon^{2}\rangle}{R} = \frac{(\pi abB\omega)^{2}}{2R}$$
Solution by Sreeraj P, M.Sc Physics