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An electric generator has an 18-cmcm-diameter, 120-turn coil that rotates at 60 HzHz in a uniform magnetic field that is perpendicular to the rotation axis. Part A What magnetic field strength is needed to generate a peak voltage of 330 VV

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Explanation:

OK ok nosepo

[tex]2825.55[/tex]

The magnetic field strength is needed to generate a peak voltage is 0.287 T.

The given parameters;

  • diameter of the generator, d = 18 cm = 0.18
  • number of turn, N = 120 turn
  • frequency of the coil, f = 60 Hz
  • maximum voltage in the coil emf = 330 V;

The maximum voltage in the coil is calculated as follows;

[tex]E_{max} = NAB\omega \\\\[/tex]

where;

  • N is the number of turns
  • A is the area of the coil
  • B is the magnetic field strength
  • ω is angular speed

The magnetic field strength is needed to generate a peak voltage is calculated as;

[tex]B = \frac{E_{max} }{NA \omega } \\\\B = \frac{E_{max} }{N (\frac{\pi d^2}{4} ) \times 2\pi f}\\\\ B = \frac{330}{120 \times (\frac{\pi \times 0.18^2}{4} ) \times 2\pi \times 60} \\\\B = 0.287 \ T[/tex]

Thus, the magnetic field strength is needed to generate a peak voltage is 0.287 T.

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