EMF and Peak Voltage of a Generator

AI Thread Summary
The discussion focuses on calculating the peak output voltage of a simple electric generator with a 30-turn coil in a magnetic field. The user correctly calculates the angular frequency (ω) as 377 rad/s and applies the formula for electromotive force (emf) to find the peak voltage, arriving at a value of 1.75 V. There is uncertainty about the time variable (t) in the calculation, but it is clarified that the peak voltage occurs when sin(ωt) equals 1. The user expresses confidence in proceeding with the second part of the problem regarding the number of turns needed for a different frequency. Overall, the conversation highlights the application of physics equations in determining generator output.
David Truong
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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!
 
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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!
 
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