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A simple generator is used to generate a peak output voltage of 25.0 V . The square armature consists of windings that are 7.0 cm on a side and rotates in a field of 0.490 T at a rate of 60.0 rev/s . Part A How many loops of wire should be wound on the square armature

User ASanch
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1 Answer

1 vote

Answer:

The 28 loops wound on the square armature

Step-by-step explanation:

Peak output voltage
\epsilon _(peak) = 25 V

Area of square armature
A = (7 * 10^(-2) )^(2) = 49 * 10^(-4)

Magnetic field
B = 0.490 T

Angular frequency
\omega = 2\pi f = 2 \pi * 60 = 120\pi

According to the law of electromagnetic induction,


\epsilon _(peak) = NBA \omega

Where
N = number of loops of wire.


N = (25)/(49 * 10^(-4) * 0.49 * 120\pi )


N = 27.6 ≅ 28

Thus, 28 loops of wire should be wound on the square armature.

User Maarten Docter
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8.5k points
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