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JEE Main 2024 (1 Feb, Shift 2) · Rotating Conductors & AC Generator

Rotating Conductors & AC Generator question from JEE Main 2024 (1 Feb, Shift 2)

Q17JEE Main 2024 · 1 FEB · SHIFT 2Medium

Rotating Conductors & AC Generator

A coil of 200 turns and area 0.20 m20.20\text{ m}^2 is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T0.01\text{ T} perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2πβ\frac{2\pi}{\beta} volt. The value of β\beta is _______.

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

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Answer: 5

  1. Peak emf: e0=NBAωe_0 = NBA\omega with ω=2π×0.5=π\omega = 2\pi \times 0.5 = \pi rad/s.
  2. e0=200×0.01×0.20×π=0.4π=2π5 Ve_0 = 200 \times 0.01 \times 0.20 \times \pi = 0.4\pi = \frac{2\pi}{5}\text{ V}
  3. Comparing with 2πβ\frac{2\pi}{\beta}: β=5\beta = 5
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