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Induced EMF in a solenoid

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data
    The current in a solenoid (you may treat it as a long solenoid of length 2.0 m, turns, and radius 0.5 m) is decreasing at a rate of 1.7 A/s.
    What is the EMF at a point .35m inside the solenoid?

    2. Relevant equations

    1. B=N(mu_0*I)/(2a)
      • where B=magnetic field, N=turns per length, I=current, and a=radius
    2. Epsilon=d(phi)/dt
      • where Epsilon=EMF, phi=magnetic flux, and t=time
    3. Phi=int(B dot dA)
      • where A=cross sectional area and you take the dot product of the two vectors
    4. A=pi*r^2
      • where r=new radius (.35m inside the solenoid)
    3. The attempt at a solution

    • I have tried saying that Epsilon = -N(BA/t) and then substituting (mu_0*I)/2a in for B and pi*r^2 in for A so that Epsilon= -N*((mu_0*I)/2a)*(pi*r^2)*(1/t) and then pulling out I/t as my decreasing current over time.
    • I have also tried when r=a and they are both .5m and when they are both .15m
    • I have also tried finding the flux at some random current (11A) and then finding the flux after one second due to the decreasing current and then plugging those values in to my Epsilon=d(phi)/dt equation
     
  2. jcsd
  3. Nov 3, 2014 #2

    collinsmark

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    Hello physics24sam,

    I suspect that you haven't received a reply as of yet because the problem statement was not typed in correctly, word for word. [Edit: or if the problem statement was typed in correctly, it is ambiguous.]

    (a) It seems that you left out how many turns are in the solenoid.

    (b) What do you mean by ".35m inside the solenoid"? Do you mean that there is a wire loop in the center of the solenoid (coaxial with the solenoid) that has a radius of 0.35 m? Or are you asking about the emf of some sort of coil or something that lies 0.35 m from the edge of the solenoid? It is not clear from the original problem statement.

    (c) From you response it seems to me that you have tried a number of approaches that did not work. Do you know what the correct answer is? (And how do you know that your attempted solutions were not correct?)

    (d) Please show more detail in your attempted solutions. Perhaps we can help point out where something went wrong.
     
    Last edited: Nov 3, 2014
  4. Nov 3, 2014 #3
    Sorry about that...

    a) there are 1.1x10^4 turns in the solenoid

    b) In regards to the .35m, I'm not sure. That is how it is worded in the problem. All it says is .35m inside which I assumed to mean .35m radially inward.

    c) My professor posted an answer key, so I do know the right answer. He did not post a solution though, so I still have to work the problem out. The answer is 4.5x10^-3 V
     
  5. Nov 4, 2014 #4

    andrevdh

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    I would have thought it to came to 71 volt?
     
  6. Nov 4, 2014 #5

    collinsmark

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    I feel sorry to hear that your instructor worded it that way. The emf induced by the current in the solenoid, simply does not exist at a "point" inside the solenoid. You need to define a path in order to compute or measure the induced emf.

    But I've worked the problem backwards to determine what your instructor likely meant:

    Approximate the induced emf on a circular loop of radius 0.35 m, inside the solenoid, which shares the same axis as the solenoid. (You may assume the loop is near the center of the solenoid.)​

    Start by calculating [itex] \frac{d \vec B}{dt} [/itex] inside the solenoid. Then you can find the rate of change of flux through the 0.35 m loop. It should be easy to get the emf using Lenz's law [Edit: or Faraday's law].
     
    Last edited: Nov 4, 2014
  7. Nov 4, 2014 #6

    NascentOxygen

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    My interpretation would be: determine the voltage across all 2m of solenoid, then calculate proportionally how much you'd measure across 0.35m of it.

    How does that gel with the required answer? :cool:
     
  8. Nov 4, 2014 #7

    collinsmark

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    That's an idea for the interpretation of the problem: Treat the solenoid as an inductor. Calculate its self inductance, and then calculate the voltage at a tap located 0.35 m from the end, knowing the rate of change of the current. ([itex] v = L \frac{di}{dt} [/itex]).

    But that approach/interpretation doesn't match the answer. :(
     
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