Answer to Question #237617 in Electricity and Magnetism for Arvin Shopon

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

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