Question
At what angular velocity in rpm will the peak voltage of a generator be 480 V, if its 500-turn, 8.00 cm diameter coil rotates in a 0.250 T field?

$764 \textrm{ rad/s}$

Solution Video

# OpenStax College Physics Solution, Chapter 23, Problem 29 (Problems & Exercises) (1:34) View sample solution

## Calculator Screenshots Video Transcript

This is College Physics Answers with Shaun Dychko. A generator with 500 turns has a peak voltage of 480 volts and the diameter of the coil or the armature inside is 8 centimeters, and the magnetic field inside of it is .25 Tesla. And the question is: "what angular speed does the generator need in order to create this peak voltage of 480 volts? And so we have a formula for peak voltage: it's number of turns times the area of the coil times the magnetic field strength times angular speed. And, we'll replace area with the formula <i>Pi r squared</i>, but we don't know what the radius is, we are given the diameter. So, we'll substitute diameter divided by 2 in place of <i>r</i>. And we end up with <i>Pi d squared over 4</i>. And then we'll substitute that in for area. These coils are circular, we're assuming. It might even say that, actually. Well, it says diameter and so when it says coils in diameter though it implies that it's a circle. So, now let's solve this for <i>omega</i>. And we're going to multiply both sides by <i>4 over N Pi d squared B</i>. And we get <i>omega</i> is 4 times peak voltage, divided by number of turns, times pi, times diameter squared, times magnetic field strength. So that's 4 times 480 volts, divided by 500 turns, times pi, times 8 times 10 to the minus 2 meters diameter squared, times .25 Tesla, which gives an angular speed of 764 radians per second.