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JEE Main 2021 (25 Jul, Shift 1) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2021 (25 Jul, Shift 1)

Q66JEE Main 2021 · 25 JUL · SHIFT 1Medium

Magnetic Flux & Faraday's Law

A circular conducting coil of radius 1 m is being heated by the change of magnetic field B⃗\vec{B}passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2 μΩ\mu\Omega. The magnetic field is slowly switched off such that its magnitude changes in time as

B=4π×10−3 T(1−t100)B = \frac{4}{\pi} \times 10^{-3}\,T\left(1-\frac{t}{100}\right)

The energy dissipated by the coil before the magnetic field is switched off completely is E = ___________ mJ.

Numerical answer type. Work it out, then open the answer.

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Answer: 80

  1. A=π(1)2=πA = \pi(1)^2 = \pi m2^2, ∣dBdt∣=4π×10−3×1100=4π×10−5\left|\dfrac{dB}{dt}\right| = \dfrac{4}{\pi}\times 10^{-3}\times\dfrac{1}{100} = \dfrac{4}{\pi}\times10^{-5} T/s.
  2. e=AdBdt=4×10−5e = A\dfrac{dB}{dt} = 4\times10^{-5} V (constant).
  3. P=e2R=16×10−102×10−6=8×10−4P = \dfrac{e^2}{R} = \dfrac{16\times10^{-10}}{2\times10^{-6}} = 8\times10^{-4} W.
  4. Field reaches zero at t=100t = 100 s, so E=8×10−4×100=0.08E = 8\times10^{-4}\times100 = 0.08 J =80= 80 mJ.
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