Question

The peak voltage of the generator is 25 V. The side length of the square coil loop is 5.8 cm. The coil rotates at a speed of 340 RPM in a magnetic field with a magnetic flux density of 0.45 T.

a) How many turns must be in the winding?

b) What is the value of the induced voltage at time 0.5 s when the voltage at time zero is zero?

Answer 1

The induced voltage is

$$E=-NBA\omega,\\\space\\ N=\frac{E}{BA\omega}=\frac{25}{0.45·0.058^2·(2\pi·340/60)}=464.$$

The value of the induced voltage at time 0.5 s when the voltage at time zero is zero is

$$E=-NBA\omega\sin(\omega t)=\\\space\\ =464·0.45·0.058^2·\frac{2\pi·340}{60}\sin\bigg(\frac{2\pi·340}{60}·0.5\bigg)=\\=7.65\text{ V}.$$