Question

The armature of a demonstrator
generator consists of a flat square coil of
side 2 cm and 400 turns. The coil rotates
in a magnetic field of 0.6 1. Calculate the
angular speed so that a maximum emf
of 4.8 V is generated​

Answer 1

Given :

Length of the side of a square coil = 2cm

Number of turns of the coil = 400 turns

Magnitude of the magnetic field = 0.6T

Emf generated = 4.8 V

To Find :

The angular speed of the coil = ?

Solution :

• The maximum emf of the coil is

           $E_{max} =NBA\omega$

• By substituting the values in the above equation

           $\omega = \frac{E_{max} }{NBA}$

           $\omega = \frac{4.8}{400\times 0.6\times 0.02^{2} }$

           $\omega = \frac{4.8}{0.096}$

           $\omega=48.48 rev/sec$

The angular speed of the coil is 48.48 rev/s.