A four-pole alternator has a uniform magnetic flux density of 0,5 T. A square coil with 200 turns and sides of 18cm is rotated at 20 r/s inside the field. The frequency is equal to 40 Hz. Determine the time period of the emf.
We can start by finding the magnetic flux through the coil. The magnetic flux is given by the product of the magnetic flux density and the area of the coil. The area of the square coil is (side)^2 = (0.18 m)^2 = 0.0324 m^2.
The magnetic flux is therefore:
Flux = Magnetic flux density * Area
= 0.5 T * 0.0324 m^2
= 0.0162 Wb
Next, we can find the rate of change of the magnetic flux, which is equal to the emf induced in the coil. The rate of change of magnetic flux is given by the product of the number of turns in the coil, the rate of rotation, and the magnetic flux.
Emf = Number of turns * Rate of change of magnetic flux
= 200 * (20 r/s) * 0.0162 Wb
= 64.8 V
Finally, we can calculate the time period of the emf. The time period is equal to the reciprocal of the frequency.
Time period = 1 / Frequency
= 1 / 40 Hz
= 0.025 s
Therefore, the time period of the emf is 0.025 seconds.