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JEE Main 2026 (21 Jan, Shift 1) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2026 (21 Jan, Shift 1)

Q20JEE Main 2026 · 21 JAN · SHIFT 1Medium

Magnetic Flux & Faraday's Law

A conducting circular loop of area 1.0 m21.0\text{ m}^{2} is placed perpendicular to a magnetic field which varies as B=sin⁡(100t)B = \sin(100t) Tesla. If the resistance of the loop is 100 Ω100\ \Omega, then the average thermal energy dissipated in the loop in one period is ______ J.

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Answer: (c)

  1. e=AdBdt=1×100cos⁡(100t)=100cos⁡(100t)e = A\dfrac{dB}{dt} = 1 \times 100\cos(100t) = 100\cos(100t) V.
  2. P=e2R=104cos⁡2(100t)100=100cos⁡2(100t)P = \dfrac{e^2}{R} = \dfrac{10^4\cos^2(100t)}{100} = 100\cos^2(100t), so Pavg=50P_{avg} = 50 W.
  3. Period T=2π100T = \dfrac{2\pi}{100} s.
  4. Energy in one period =50×2π100=π= 50 \times \dfrac{2\pi}{100} = \pi J. So the answer is (c) π\pi.
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