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MHT CET 2026 (15 May, Shift 2) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from MHT CET 2026 (15 May, Shift 2)

Q18MHT CET 2026 · 15 MAY · SHIFT 2Medium

Magnetic Flux & Faraday's Law

A rectangular wire loop of resistance 1.5 Ω1.5\ \Omega placed in a region of uniform magnetic field of magnitude 0.30.3 T directed normal to the loop. If the area of the loop decreases at a rate of 0.050.05 m2/s{}^{2}/\text{s} in this magnetic field, the heat produced in it is x×10−6x \times 10^{-6} W. The value of xx is

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Answer: (d)

  1. e=BdAdt=0.3×0.05=0.015e = B\frac{dA}{dt} = 0.3 \times 0.05 = 0.015 V
  2. Heat produced per second P=e2R=(0.015)21.5=1.5×10−4P = \frac{e^2}{R} = \frac{(0.015)^2}{1.5} = 1.5 \times 10^{-4} W
  3. P=150×10−6P = 150 \times 10^{-6} W
  4. So the answer is (d) 150.
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