JEE PYQ: Magnetic Effects Of Current Question 13
Question 13 - 2020 (03 Sep 2020 Shift 1)
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:
(1) $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$
(2) zero
(3) $\frac{\pi a b B \omega}{R}$
(4) $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{R}$
Show Answer
Answer: (1) $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$
Solution
Average power $\langle P \rangle = \frac{(\pi ab)^2 B^2 \omega^2}{2R} = \frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$.
BETA
