JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 13)
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 :
_3rd_September_Morning_Slot_en_13_1.png)
$${{\pi abB\omega } \over R}$$
$${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over R}$$
$${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}}$$
Zero
Explanation
$$\phi $$ = BAcos$$\theta $$
$$ \Rightarrow $$ $$\phi $$ = BAcos($$\omega t$$)
$${{d\phi } \over {dt}} = - BA\omega \sin \left( {\omega t} \right)$$
We know, $$\varepsilon $$ = $$ - {{d\phi } \over {dt}}$$ = $$BA\omega \sin \left( {\omega t} \right)$$
Power (P) = $${{{\varepsilon ^2}} \over R} = {{{B^2}{A^2}{\omega ^2}{{\sin }^2}\left( {\omega t} \right)} \over R}$$
$$ \therefore $$ Pav = $${{{B^2}{A^2}{\omega ^2}} \over 2R}$$
= $${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}}$$ [As A = $$\pi $$ab]
$$ \Rightarrow $$ $$\phi $$ = BAcos($$\omega t$$)
$${{d\phi } \over {dt}} = - BA\omega \sin \left( {\omega t} \right)$$
We know, $$\varepsilon $$ = $$ - {{d\phi } \over {dt}}$$ = $$BA\omega \sin \left( {\omega t} \right)$$
Power (P) = $${{{\varepsilon ^2}} \over R} = {{{B^2}{A^2}{\omega ^2}{{\sin }^2}\left( {\omega t} \right)} \over R}$$
$$ \therefore $$ Pav = $${{{B^2}{A^2}{\omega ^2}} \over 2R}$$
= $${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}}$$ [As A = $$\pi $$ab]
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