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.
1 answer:
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) =
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
You might be interested in
Answer:
(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.
Explanation:
Given that,
a ball is spun around in circular motion such that it completes 50 rotations in 25 s.
(1). Let T be the period of its rotation. It can be calculated as follows :
(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,
Hence, this is the required solution.