Generator Physics Example 3

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Example 3

medium
A generator coil has 100100 turns and area 0.05 m20.05 \text{ m}^2 in a 0.4 T0.4 \text{ T} field. It needs to produce a peak EMF of 500 V500 \text{ V}. At what frequency must it rotate?

Solution

  1. 1
    From E0=NABω\mathcal{E}_0 = NAB\omega: ω=E0NAB=500100×0.05×0.4=5002=250 rad/s\omega = \frac{\mathcal{E}_0}{NAB} = \frac{500}{100 \times 0.05 \times 0.4} = \frac{500}{2} = 250 \text{ rad/s}.
  2. 2
    Frequency: f=ω2π=2506.28≈39.8 Hzf = \frac{\omega}{2\pi} = \frac{250}{6.28} \approx 39.8 \text{ Hz}

Answer

f≈39.8 Hzf \approx 39.8 \text{ Hz}
The rotation frequency of a generator determines the frequency of the AC output and, together with coil properties and field strength, determines the peak voltage produced.

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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