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) = [tex]\frac{2\pi}{60} *3600\ rpm = 377.04 \ rad/s[/tex]
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
