Flux and Induced EMF vs displacement graphs

  • Thread starter Taniaz
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  • #1
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Homework Statement


A small coil of wire is situated in a non-uniform magnetic field. The coil consists of 40 turns of wire and moves with a constant speed in a straight line. The coil has displacement x from a fixed point P. The variation with x of the magnetic flux in the coil is shown in the first graph.
Sketch the variation with x of the e.m.f E induced in the coil for values of x from x=0 to x=6.0 cm.

Homework Equations


E = (delta N*flux) / delta t

The Attempt at a Solution


I know the slope of the flux vs. x graph is the emf so in the region where the flux vs. x graph has a negative yet constant slope, I drew a horizontal line in the negative region for the emf vs x graph. And the region where the flux vs x graph has a positive slope, I drew a horizontal line in the positive region for the emf vs x graph.

I'm confused as to what happens at point 3? Will the emf vs x graph be vertical as I drew it (second graph)?
Thank you.
 

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Answers and Replies

  • #2
413
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you are essentially correct with your analysis. Strictly speaking the induced emf is given by -d(NΦ)/dt so perhaps your lines should be inverted.
Do you realise that the lines shown have different gradients?
At point 3 the change from one to the other is instantaneous....not much more to say about that !!
 
  • #3
364
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Yes that makes sense. Yes they do have different gradients (shown by their steepness, that's why the second one was a bit higher than the first one).
So there should be a vertical line at 3 joining the two together?

Should we take the minus into account? Because sometimes they just write Faraday's law without the minus sign to indicate the magnitude only?
 
  • #4
cnh1995
Homework Helper
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So there should be a vertical line at 3 joining the two together?
Yes.
How are you calculating the magnitude of the emf? What is the speed of the coil?
Should we take the minus into account? Because sometimes they just write Faraday's law without the minus sign to indicate the magnitude only?
Minus sign is not necessary, as it is up to you to choose which direction is plus or minus for induced emf.
 
  • #5
364
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In the first part of the question they told us that "The coil is moved at constant speed between point P and the point where x = 3.0 cm" and then asked us to calculate the flux linkage which I did by multiplying the number of turns of the coil by the change in the flux from x=0 till x=3.0 cm. In the next part of the question they gave us the induced emf and asked for the speed so I did E = delta N(flux) / delta t but this can be written as E = (delta N(flux) / delta x) ( delta x / delta t) and we can solve for delta x / delta t which is the speed.
 
  • #6
cnh1995
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"The coil is moved at constant speed between point P and the point where x = 3.0 cm" and then asked us to calculate the flux linkage which I did by multiplying the number of turns of the coil by the change in the flux from x=0 till x=3.0 cm.
Why did you calculate the change in flux? Is it specifically asked? Because "flux linkage" of the coil can be computed at any single value of x. You can compute it at x=0 and at x=6. Why take the difference of them?
In the next part of the question they gave us the induced emf and asked for the speed so I did E = delta N(flux) / delta t but this can be written as E = (delta N(flux) / delta x) ( delta x / delta t) and we can solve for delta x / delta t which is the speed.
Correct.
 
  • #7
364
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They asked for it from x=0 till x=3

Thank you
 
  • #8
I know the slope of the flux vs. x graph is the emf
Shouldn't the slope of the flux vs. x graph be emf * velocity

V = (dN*flux/dt) = (dN*flux / dx) * (dx / dt)
dx / dt = velocity or speed
=> ( V / velocity ) = ( dN*flux / dx)
 
Last edited:

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