Question

An electric generator is intended to output 10 kJ of energy during each cycle. A 50-mT magnetic field will penetrate the circular coil, which is 0.5 m in radius and consists of 4000 turns of wire. The net average electrical resistance of the entire system is 10 ohms. Find the angular speed at which the coil (or magnetic field) must oscillate. Recall that Watt's Law gives the instantaneous power of any electrical element, and one cycle of time is two pi over the angular frequency.

Answer 1

Due to the presence of magnetic field an emf will be induced in the loop. From Faraday's law, we know that,

= - N d/dt = - N BdA/dt = B d( pi r^2 cos(t) )/dt ;

= N B pi r^2 sin(t)

We know that the maximum value of sin(t) = 1

=  N B pi r^2

= / N B pi r^2

We have, E = 10 kJ per cycle =>

Power is energy per unit time, so

P = 10000 / 1s = 10000 Watts

We know that, P = I^2 R = V^2/R

V^2 = P x R = 10000 x 10 => V = 316.23 Volts

V = = 316.23 V

=  /N B pi r^2

N = 4000 ; B = 50 mT = 0.05 T ; r = 0.5 m

= 316.23 / 4000 x 0.05 x 3.14 x 0.5 x 0.5 = 2.014 rad/s

Hence, = 2.014 rad/s.