# EMF and Peak Voltage of a Generator

• David Truong
In summary, the conversation discusses a simple electric generator with a 30-turn coil and a magnetic field of 0.80 T at a frequency of 60 Hz. The peak output voltage is calculated using the equation emf = NBωAsinωt, with a known angular frequency of 377 rad/s. The time (t) is assumed to be 1 second since frequency is measured in Hz. The peak output voltage is found to be 1.75 V. To maintain the same output voltage at a frequency of 50 Hz, the number of turns in the coil must be adjusted. The amplitude of the signal is dependent on the angular frequency (w) and is equal to 1 at peak voltage.

## Homework Statement

A simple electric generator contains a 30-turn coil of area 6.6 x 10^-4 meters squared. The coil spins in a magnetic field of 0.80 T at a frequency of 60 Hz.

a) What is the peak output voltage?
b) How must the number of turns be changed to maintain the same output voltage, but operate at a frequency of 50 Hz.

## Homework Equations

ω = 2πf
emf/peak voltage= NBωAsinωt

## The Attempt at a Solution

ω = 2π(60) = 377 rad/s

emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)t

My problem is that I do not know how/where to obtain the time (t) to finish the calculation. My first assumption is to use t = 1 second, since frequency is in Hz, which is revolutions per second.

emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)(1) = 1.75 V

Is this correct? If so, then I am sure I can do part b) on my own. Thanks for the help!

David Truong said:
emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)t

My problem is that I do not know how/where to obtain the time (t) to finish the calculation. My first assumption is to use t = 1 second, since frequency is in Hz, which is revolutions per second.

emf = NBωAsinωt = 30(0.80T)(377)(6.6x10^-4)sin(377)(1) = 1.75 V

Is this correct? If so, then I am sure I can do part b) on my own. Thanks for the help!
If a signal is Asin(wt), what is its amplitude? Does it depend on w or t?

So I believe I figured it out. At peak voltage, the amplitude sinwt is at its peak too, which is equal to 1. Thanks for the help!