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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2026 · 21 January · Morning Shift · Physics · Electromagnetic Induction

A conducting circular loop of area 1.0 ~m^ 2 is placed...

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

Options

  1. Aπ2\pi^{2}
  2. Bπ\pi
  3. Cπ2\frac{\pi}{2}
  4. D2π2 \pi

Correct answer

B. π\pi

Step-by-step solution

Flux: ϕ=BA=sin⁡(100t)\phi = BA = \sin(100t) Wb
Induced EMF: ε=−dϕdt=−100cos⁡(100t)\varepsilon = -\frac{d\phi}{dt} = -100\cos(100t) V
Induced current: I=εR=−cos⁡(100t)I = \frac{\varepsilon}{R} = -\cos(100t) A
Power dissipated: P=I2R=100cos⁡2(100t)P = I^2R = 100\cos^2(100t) W
Average power: Pavg=12×100=50P_{avg} = \frac{1}{2} \times 100 = 50 W
Time period: T=2π100T = \frac{2\pi}{100} s
Energy in one period: E=Pavg×T=50×2π100=πE = P_{avg} \times T = 50 \times \frac{2\pi}{100} = \pi J
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