Question

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

Answer 1

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.