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MaRussiya [10]
3 years ago
8

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 c

m 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
Physics
1 answer:
4vir4ik [10]3 years ago
6 0

Answer:

<h3>The 28 loops wound on the square armature</h3>

Explanation:

Peak output voltage \epsilon _{peak}  = 25 V

Area of square armature A = (7 \times 10^{-2} )^{2}  = 49 \times 10^{-4}

Magnetic field B = 0.490 T

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

According to the law of electromagnetic induction,

     \epsilon _{peak} = NBA \omega

Where N = number of loops of wire.

  N = \frac{25}{49 \times 10^{-4} \times 0.49 \times 120\pi  }

  N = 27.6 ≅ 28

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

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