Answer:
The peak output voltage of this generator is 314.2 V.
Explanation:
Given;
the number of turns of the coil, N = 1000 turns
area of the coil, A = 0.001 m²
angular frequency of the coil, f = 100 cycles/seconds
magnitude of the magnetic field, B = 0.5 T
The peak output voltage of this generator is given by;
E = NBAω
Where;
ω is the angular velocity = 2πf
E = NBA(2πf)
E = 1000 x 0.5 x 0.001(2 x π x 100)
E = 314.2 V
Therefore, the peak output voltage of this generator is 314.2 V.
