Question 91: Electromagnetic Induction

Question

An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in
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 :
![](https://imagex.cdn.examgoal.net/jgy2xukf16nx50/0cc7d670-bf47-4605-98a2-8f65672b2a85/eba27440-f5cb-11ea-8293-3b6cc591631c/file-jgy2xukf16nx51-472w.png)

Accepted Answer

$\phi $ = BAcos$	heta $
$ \Rightarrow $ $\phi $ = BAcos($\omega t$)
${{d\phi } \over {dt}} =  - BA\omega \sin \left( {\omega t} ight)$
We know, $\varepsilon $ = $ - {{d\phi } \over {dt}}$ = $BA\omega \sin \left( {\omega t} ight)$
Power (P) = ${{{\varepsilon ^2}} \over R} = {{{B^2}{A^2}{\omega ^2}{{\sin }^2}\left( {\omega t} ight)} \over R}$
$ 	herefore $ P
av
= ${{{B^2}{A^2}{\omega ^2}} \over 2R}$
= ${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}}$ [As A = $\pi $ab]

Answer

${{\pi abB\omega } \over R}$

Answer

${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over R}$

Answer

${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}}$

Answer

Zero

Electromagnetic Induction

An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in 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 :

Choose the correct answer: