Question

Suppose you wanted to use the Earth's magnetic field to make an AC generator at a location where the magnitude of the field is 0.50 mT. Your coil has 1000.0 turns and a radius of 5.0 cm. At what angular velocity would you have to rotate it in order to generate an emf of amplitude 1.0 V?

Answer 1

Answer:

The angular velocity I would have to rotate it in order to generate an emf of amplitude 1.0 V is 254.65 rad/s

Explanation:

given information:

B = 0.5 mT = 0.0005 T

N = 1000

r = 5 cm = 0.05 m

emf, ε = 1 V

according to Faraday's law

ε = -N dΦ/dt, Φ = B A

  = - N d( B A)/dt

  = - N d( B A cos ωt)/dt

   = - N B A d(cos ωt)/dt

   = N B A ω sin ωt

A = πr², so

ε = N B πr² ω sin ωt

where

ε = emf

N = number of coil turn

B = magnetic field

r = radius

ω = angular velocity

Φ = magnetic flux

emf maximum, sin ωt = 1. So,

ε = N B πr² ω

ω = ε/N B πr²

   = 1/[(1000) (0.0005) π (0.05)²

   = 254.65 rad/s