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JEE Main 2023 (1 Feb, Shift 2) · AC Generator & Transformer

AC Generator & Transformer question from JEE Main 2023 (1 Feb, Shift 2)

Q32JEE Main 2023 · 1 FEB · SHIFT 2Medium

AC Generator & Transformer

A square shaped coil of area 70 cm270\text{ cm}^2 having 600 turns rotates in a magnetic field of 0.4 wbm−20.4\text{ wbm}^{-2}, about an axis which is parallel to one of the sides of the coil and perpendicular to the direction of field. If the coil completes 500 revolutions in a minute, the instantaneous emf when the plane of the coil is inclined at 60∘60^\circ with the field, will be ____________ V. (Take π=227\pi = \frac{22}{7})

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

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Answer: 44

  1. e=NBAωsin⁡θe = NBA\omega\sin\theta, where θ\theta is the angle between the coil's normal and BB.
  2. Plane at 60∘60^\circ to the field, so the normal is at θ=30∘\theta = 30^\circ.
  3. ω=2π×50060=50π3=110021 rad/s\omega = \dfrac{2\pi \times 500}{60} = \dfrac{50\pi}{3} = \dfrac{1100}{21}\text{ rad/s}
  4. NBA=600×0.4×70×10−4=1.68NBA = 600 \times 0.4 \times 70 \times 10^{-4} = 1.68
  5. e=1.68×110021×12=44 Ve = 1.68 \times \dfrac{1100}{21} \times \dfrac{1}{2} = 44\text{ V}
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