Question

A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.052 T, how many turns of wire are needed

Answer 1

The number of turns of wire needed is 573.8 turns

Step-by-step explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

`\omega = 3800 (rev)/(min) * (2\pi)/(1 \ rev) * (1 \min)/(60 \ s ) = 397.99 \ rad/s`

`N = (emf)/(AB\omega ) \\\\N = (190)/(0.016 * 0.052* 397.99) \\\\N = 573.8 \ turns`

Therefore, the number of turns of wire needed is 573.8 turns