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JEE Main 2014 (11 Apr, Shift 1) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2014 (11 Apr, Shift 1)

Q78JEE Main 2014 · 11 APR · SHIFT 1Easy

Magnetic Flux & Faraday's Law

A coil of circular cross-section having 1000 turns and 4 cm2 face area is placed with its axis parallel to a magnetic field which decreases by 10−-2 Wb m−2{}^{-2} in 0.01 s. The e.m.f. induced in the coil is:

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Answer: (a)

  1. e=NAΔBΔte = NA\frac{\Delta B}{\Delta t}
  2. N=1000N = 1000, A=4×10−4A = 4\times 10^{-4} m2^{2}, ΔBΔt=10−20.01=1\frac{\Delta B}{\Delta t} = \frac{10^{-2}}{0.01} = 1 T/s
  3. e=1000×4×10−4×1=0.4e = 1000\times 4\times 10^{-4}\times 1 = 0.4 V =400= 400 mV
  4. So the answer is (a) 400 mV.
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