# Induced EMF and a circular loop of wire

#### anonymousphys

1. 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?

2. Homework Equations
-(dB/dt)(A)=V

3. 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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#### zachzach

1. 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?

2. Homework Equations
-(dB/dt)(A)=V

3. 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?

#### anonymousphys

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?

#### zachzach

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.

#### zachzach

Show the work on both because when I use the 2nd equation all that happens is I derive the first equation.

#### anonymousphys

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?

#### zachzach

Doh! I see what you mean :/.

#### zachzach

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:

#### zachzach

$$V = -\frac{d}{dt}[BA] = -[\frac{BA_f - BA_i}{t_f - t_i}]$$

I did it in latex :).

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