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JEE Main 2026 (5 Apr, Shift 2) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2026 (5 Apr, Shift 2)

Q21JEE Main 2026 · 5 APR · SHIFT 2Difficult

Magnetic Flux & Faraday's Law

A circular loop of radius 20 cm and resistance 2 Ω2\ \Omega is placed in a time varying magnetic field B⃗=(2t2+2t+3)T\vec{B} = \left(2t^{2}+2t+3\right)T. At t=0t = 0, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at t=3 st = 3\text{ s} is α50 A\frac{\alpha}{50}\text{ A}. The value of α\alpha is ______. (Take π=22/7\pi = 22/7)

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

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Answer: 44

  1. A=πr2=227×(0.2)2=227×0.04A = \pi r^2 = \dfrac{22}{7}\times(0.2)^2 = \dfrac{22}{7}\times0.04 m2^2.
  2. dBdt=4t+2=14\dfrac{dB}{dt} = 4t + 2 = 14 T/s at t=3t = 3 s.
  3. e=AdBdt=227×0.04×14=1.76e = A\dfrac{dB}{dt} = \dfrac{22}{7}\times0.04\times14 = 1.76 V, so I=1.762=0.88I = \dfrac{1.76}{2} = 0.88 A.
  4. 0.88=α50⇒α=440.88 = \dfrac{\alpha}{50} \Rightarrow \alpha = 44.
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