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QJEEM20Ph004f: JEE Main 2020 Physics - Class 12 JEEM20Ph004f Solution: An elliptical loop having resistance R, of semi major axis a,…

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 ω\omega , the average power loss in the loop due to Joule heating is : JEE Main 2020 (Online) 3rd September Morning Slot Physics - Electromagnetic Induction Question 84 English

Multiple Choice Options

  • A πabBωR{{\pi abB\omega } \over R}
  • B π2a2b2B2ω2R{{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over R}
  • C π2a2b2B2ω22R{{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}}
  • D Zero

Step-by-Step Verified Solution

ϕ\phi = BAcos θ\theta

⇒\Rightarrow ϕ\phi = BAcos( ωt\omega t )

dϕdt=−BAωsin⁡(ωt){{d\phi } \over {dt}} = - BA\omega \sin \left( {\omega t} \right)

We know, ε\varepsilon = −dϕdt- {{d\phi } \over {dt}} = BAωsin⁡(ωt)BA\omega \sin \left( {\omega t} \right)

Power (P) = ε2R=B2A2ω2sin⁡2(ωt)R{{{\varepsilon ^2}} \over R} = {{{B^2}{A^2}{\omega ^2}{{\sin }^2}\left( {\omega t} \right)} \over R}

∴\therefore Pav = B2A2ω22R{{{B^2}{A^2}{\omega ^2}} \over 2R}

= π2a2b2B2ω22R{{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}} [As A = π\pi ab]

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