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EMF and Peak Voltage of a Generator

  • #1

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!
 

Answers and Replies

  • #2
rude man
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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?
 
  • #3
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!
 

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