Induced EMF and a circular loop of wire

AI Thread Summary
The discussion revolves around calculating the induced electromotive force (emf) in a 25-turn circular loop of wire flipped 180 degrees in a magnetic field. The first method, using the equation NABwsin(wt), yields zero, while the second method, -(dB/dt)(A)=V, produces a non-zero result. Participants debate the validity of both equations and their derivations, questioning why they do not yield the same answer. It is suggested that the first equation may only calculate the emf at a specific instant rather than over the entire time interval. The conversation emphasizes the importance of understanding the change in magnetic flux and the correct application of the equations.
anonymousphys
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Homework Statement


A 25-turn circular loop of wire has a diameter of 1m. In 0.2 seconds it is flipped 180 degrees at a location where the magnitude of the Earth's magnetic field is 50 micro T. What is the emf generated in the loop?

Homework Equations


-(dB/dt)(A)=V

The Attempt at a Solution


When I use NABwsin(wt)=V, I get zero. When I use -(dB/dt)(A)=V, I get a non-zero number. I believe the 2nd equation works but why doesn't the first work?

Thanks for any help.
 
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anonymousphys said:

Homework Statement


A 25-turn circular loop of wire has a diameter of 1m. In 0.2 seconds it is flipped 180 degrees at a location where the magnitude of the Earth's magnetic field is 50 micro T. What is the emf generated in the loop?


Homework Equations


-(dB/dt)(A)=V


The Attempt at a Solution


When I use NABwsin(wt)=V, I get zero. When I use -(dB/dt)(A)=V, I get a non-zero number. I believe the 2nd equation works but why doesn't the first work?

Thanks for any help.

Where does the first equation come from? If you used d/dt[BA] = V is B changing?
 
Thanks for the reply. NABwsin(wt)=V comes from taking the derivative of NABcos(wt)=magnetic flux. The 2nd equation should be taking into account the change in angle. Shouldn't both equations lead to the same answer?
 
anonymousphys said:
Thanks for the reply. NABwsin(wt)=V comes from taking the derivative of NABcos(wt)=magnetic flux. The 2nd equation should be taking into account the change in angle. Shouldn't both equations lead to the same answer?

I believe they should. How did you get 0 for the first equation? I do not get 0.
 
Show the work on both because when I use the 2nd equation all that happens is I derive the first equation.
 
zachzach said:
I believe they should. How did you get 0 for the first equation? I do not get 0.

w=pi/(.2)
sin(wt) when t=(.2) equals 0.

Hm..I think I'm not using the equation correctly?
 
Doh! I see what you mean :/.
 
Here's my thoughts: By plugging in wt only for the final time, you only are calculating the emf produced at that instant. So maybe you should use:

V = -d/dt(BA) = -[BA(Final) - BA(Initial)]/[t(Final) - t(Initial)]

sorry cannot get fraction in latex to work, I am a newb.
 
Last edited:
<br /> <br /> V = -\frac{d}{dt}[BA] = -[\frac{BA_f - BA_i}{t_f - t_i}]<br /> <br />

I did it in latex :).
 
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