Question

when the armature of an ac generatr rotates at 15.0 rad/s, the amplitude of the induced emf is 27.0 V. What is the amplitude of the induced emf when the armature roates at 10.0 rad/s

Answer 1

To solve this problem we will apply the concepts related to the electric field. This is defined as the product between the angular frequency, the number of turns of the body (solenoid in this case) the magnetic field and the sine of the angular frequency and time. Mathematically this can be described as

$E = \omega NBA |sin \omega t|$

Here,

$\omega$ = Angular frequency

N = Number of turns

B = Magnetic field

The emf has its maximum value when $sin \omega t = \pm 1$

Thus the amplitude of the emf is

$E = \omega NBA$

When number of turns of armature, area and applied magnetic field remains constant, induced emf is proportional to angular speed.

$E \propto \omega$

Further it can be written as follows,

$\frac{E_1}{E_2} \propto \frac{\omega_1}{\omega_2}$

$E_2 = \frac{\omega_2}{\omega_1}E_1$

$E_2 = \frac{10rad/s}{15rad/s}(27.0V)$

$E_2 = 18V$

Therefore the maximum amplitude of induced emf when armature rotates at 10.0rad/s is 18V