Question

Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.

Answer 1

Answer:

The peak voltage is 473.86 V

Explanation:

Given;

number of turns of generator, N = 200  turn

diameter of the coil, d = 0.1 m

radius of the coil, r = 0.05 m

Magnitude of the magnetic field, B = 0.8 T

angular frequency, ω = 3600 rpm

angular frequency, ω (rad/s) = $\frac{2\pi}{60} *3600\ rpm = 377.04 \ rad/s$

The peak voltage is given by;

E = NBAω

Where;

A is the area of the coil = πr² =  π (0.05)² = 7.855 x 10⁻³ m²

E = (200)(0.8)(7.855 x 10⁻³)(377.04)

E = 473.86 V

Therefore, the peak voltage is 473.86 V