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MHT CET 2026 (12 May, Shift 2) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from MHT CET 2026 (12 May, Shift 2)

Q17MHT CET 2026 · 12 MAY · SHIFT 2Medium

Magnetic Flux & Faraday's Law

A closed coil consists of 500500 turns on a rectangular frame of area 4.04.0 cm2^{2} and has a resistance of 50 Ω50\ \Omega. The coil is kept with its plane perpendicular to a uniform magnetic field of 0.20.2 T. The amount of charge flowing through the coil if it is turned through 180∘180^\circ will be 1.6×10−x1.6 \times 10^{-x} C. The Value of xx is

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Answer: (c)

  1. Turning through 180∘180^\circ: Δϕ=2NBA=2×500×0.2×4×10−4=0.08\Delta\phi = 2NBA = 2 \times 500 \times 0.2 \times 4 \times 10^{-4} = 0.08 Wb.
  2. q=ΔϕR=0.0850=1.6×10−3q = \frac{\Delta\phi}{R} = \frac{0.08}{50} = 1.6 \times 10^{-3} C
  3. So x=3x = 3, answer (c).
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