- #1

David Truong

- 9

- 0

## 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= NBωAsinω

*t*

## The Attempt at a Solution

ω = 2π(60) = 377 rad/s

emf = NBωAsinω

*t*= 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 = NBωAsinω

*t*= 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!