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JEE Main 2020 (3 Sep, Shift 1) · Rotating Conductors & AC Generator

Rotating Conductors & AC Generator question from JEE Main 2020 (3 Sep, Shift 1)

Q27JEE Main 2020 · 3 SEP · SHIFT 1Medium

Rotating Conductors & AC Generator

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:

Figure

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Answer: (c)

  1. Area of the ellipse A=πabA = \pi ab; rotating about the x-axis, ϕ=Bπabcos⁡ωt\phi = B\pi ab\cos\omega t.
  2. e=Bπab ωsin⁡ωte = B\pi ab\,\omega\sin\omega t, so e0=πabBωe_0 = \pi abB\omega.
  3. Average Joule power: P=e022R=π2a2b2B2ω22RP = \frac{e_0^2}{2R} = \frac{\pi^2a^2b^2B^2\omega^2}{2R}
  4. So the answer is (c).
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