Generator Physics Example 1

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Example 1

easy
A simple AC generator has a coil of 5050 turns, area 0.04 m20.04 \text{ m}^2, rotating in a 0.5 T0.5 \text{ T} magnetic field. What is the peak EMF if the coil rotates at 60 Hz60 \text{ Hz}?

Solution

  1. 1
    Peak EMF for a generator: E0=NABω\mathcal{E}_0 = NAB\omega, where ω=2πf\omega = 2\pi f.
  2. 2
    Angular frequency: ω=2π×60=377 rad/s\omega = 2\pi \times 60 = 377 \text{ rad/s}.
  3. 3
    E0=50×0.04×0.5×377=377 V\mathcal{E}_0 = 50 \times 0.04 \times 0.5 \times 377 = 377 \text{ V}

Answer

E0=377 V\mathcal{E}_0 = 377 \text{ V}
An AC generator converts mechanical energy to electrical energy by rotating a coil in a magnetic field. The peak EMF increases with more turns, larger area, stronger field, and faster rotation.

About Generator

A device that converts mechanical (kinetic) energy into electrical energy by rotating a coil of wire within a magnetic field, exploiting electromagnetic induction.

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