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Estimate induced emf in the coil

  • Thread starter Cade
  • Start date
  • #1
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[Solved] Estimate induced emf in the coil

Homework Statement



Consider a single-loop coil whose magnetic flux is given by this image:
xjz8c.png

Estimate the induced emf in the coil at times near t = 0.3s, t = 0.4s and t = 0.5s

This is for algebra-based physics.

Homework Equations



emf = -N*(change in flux)/change in time

The Attempt at a Solution



At t = 0.3s
emf = -1(4- -4)/(0.4 - 0.2) = -8/0.2 = -40V

At t = 0.4s
Rate of change = 0

At t = 0.5s
emf = -1(-4 - 4)/(0.6 - 0.4) = 8/0.2 = 40V

My first and third answers are wrong. What've I missed? The text book has -0.06kV, 0 and 0.06kV but those answers are also incorrect.

Edit: Solved it. This is the graph of
4 Cos[(2 Pi/(0.4))*x]
Differentiating this gives y = -20pi*Sin[5pi*x]
Evaluating it at t = 0.3 s gives 63V. I do not know why differentiation is used for an algebra-based physics problem when my TA explicitly told me to not use calculus, but there it is.
 
Last edited:

Answers and Replies

  • #2
993
13
make a better estimate because the gradient at 0.3s is rather different from the gradient of the line joining the max at 0.4s and the min at 0.2s.
 
  • #3
92
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How would I do that?

Using t = 0.25s and t = 0.35s as my points for t = 0.3s,
-(3 - -3)/(0.35 - 0.25) = -6/0.1 = -60V, which is also wrong.

Edit: The correct answer is -63V, 0V, 63V. I have no idea how to calculate this. The textbook has -60V, 0V, 60V, which is apparently incorrect.

Edit: Solved it. This is the graph of
4 Cos[(2 Pi/(0.4))*x]
Differentiating this gives y = -20pi*Sin[5pi*x]
Evaluating it at t = 0.3 s gives 63V. I do not know why differentiation is used for an algebra-based physics problem when my TA explicitly told me to not use calculus, but there it is.
 
Last edited:
  • #4
993
13
So your second estimate, 60V, was correct!
I do not think you can use differentiation if an estimate was asked for.
 
  • #5
92
0
Yes, however the homework system won't accept it for some reason. I tried 60V, as the textbook said that, then I tried a wider window and got 40V, then I gave up and fitted it to a cosine curve.
 
  • #6
993
13
You can get a better estimate than 60V if a tangent is drawn on the curve at the required point and its gradient is calculated. The longer this tangent is drawn the better is the estimate. It is still an estimate since the axes are not subdivided further.
 
  • #7
92
0
Thanks, I will keep that in mind for next time.
 

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