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JEE Main 2026 (6 Apr, Shift 2) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2026 (6 Apr, Shift 2)

Q14JEE Main 2026 · 6 APR · SHIFT 2Medium

Magnetic Flux & Faraday's Law

A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(300t)B = 0.4\sin(300t) Tesla. The normal to the plane of loop makes an angle of 60∘60^\circ with the field. The maximum induced emf produced in the loop is ______ mV.

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Answer: (d)

  1. ϕ=BAcos⁡60∘=0.4sin⁡(300t)×A×12\phi = BA\cos 60^\circ = 0.4\sin(300t) \times A \times \frac{1}{2}, with A=(0.02)2=4×10−4A = (0.02)^2 = 4 \times 10^{-4} m2^2.
  2. e=dϕdt=0.4×300×4×10−4×12cos⁡(300t)e = \frac{d\phi}{dt} = 0.4 \times 300 \times 4 \times 10^{-4} \times \frac{1}{2}\cos(300t)
  3. emax=120×2×10−4=0.024e_{max} = 120 \times 2 \times 10^{-4} = 0.024 V
  4. So the answer is (d) 24 mV.
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