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 :

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: