Question

A simple generator is used to generate a peak output voltage of 23.0 V . The square armature consists of windings that are 5.1 cm on a side and rotates in a field of 0.500 T at a rate of 55.0 rev/s .

How many loops of wire should be wound on the square armature?

Answer 1

Answer: 51

Step-by-step explanation:

Given

Output is 23 V

The square armature side is
`a=5.1\ cm`

Magnetic field
`B=0.5\T`

Rate of revolution
`n=55\ rev/s`

Angular speed

`\omega =2\pi n\\\omega=2\pi * 55=110\pi\ rad/s`

Peak voltage is given by

`E_(peak)=NB\omega A\quad [\text{N=Number of windings; A=area of cross-section}]\\\\N=(E_(peak))/(B\omega A)\\\\N=(23)/(0.5* 110\pi* (0.051)^2)\approx 51`

So, there are approximate 51 loops