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disa [49]
3 years ago
6

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 gen

erator 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
Physics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

The number of turns of wire needed is 573.8 turns

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 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s

N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns

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

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