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Field induced by solenoid

  1. May 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A solenoid has 500 turns per meter and cross-sectional area 4cm2
    Current through it is increasing at 100A/s and flows in a clockwise direction. There is a wire loop, concentric with the solenoid which sits outside the solenoid. Given that the magnetic field inside the solenoid is B=μnI, find the magnitude and direction of the induced emf in the wire loop.

    2. Relevant equations


    3. The attempt at a solution
    I know that emf is given by Faraday's law of induction as -N dΦ/dt, Φ being magnetic flux. I don't know how to calculate the flux. I know equations for it, Φ=∫B.dl=μI, and I can't really get my head around what dl is! Or where the area has to do with it...
     
  2. jcsd
  3. May 25, 2015 #2

    G01

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    I think you have the wrong equation for magnetic flux. Flux is not defined by the line integral of the magnetic field. Magnetic flux is a measure of the amount of magnetic field that passes through a given surface/area. If the area of the Surface is a and the differential unit vector normal to the surface is [itex]d\vec{a}[/itex], then the flux, by definition, is

    [tex]\Phi= \int \vec{B} \cdot d\vec{a}[/tex]

    A good analogy here is that of a water pipe. The total "flux" of water passing through the pipe will depend on the total integrated flow of water through the crosssectional area of the pipe.


    If the magnetic field vectors through the surface is directed normal to that surface and constant in magnitude, the flux is given by:

    [tex]\Phi= \int \vec{B} \cdot d\vec{A}= B*A [/tex]

    Magnetic flux on hyperphysics: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/fluxmg.html
     
  4. May 26, 2015 #3
    I thought that integral was always supposed to be zero...
     
  5. May 26, 2015 #4

    G01

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    Through a closed surface, yes, the integral is zero. That is not necessarily the case for an open surface, like the area enclosed by the ring in this problem.
     
  6. May 26, 2015 #5
    Oh, ok. I'm going based on a single closed loop of wire round the outside of the solenoid here, so my area would be given by the 4cm2 then. B and dA are parallel, so flux = BA = μnIA. I want change in flux and since only I is changing, will the change in flux over change in time just be 100μnA?
     
  7. May 26, 2015 #6

    G01

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    Yeah, you're on the right track. Remember [itex]\mu[/itex] and [itex]n[/itex] are not units but quantities that have specific values for the problem.
     
  8. May 26, 2015 #7
    μ=IA, far as I can remember. As for the value of n, I'm not really sure! I would say one, because I assume my wire loop encloses just the one, but then why would 500 turn per metre be given?
     
  9. May 26, 2015 #8

    G01

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    You're confusing terminology. Here, [itex]\mu[/itex] is the magnetic permeability of the medium inside the solenoid. It's safe to assume vacuum if a specific material is not given.

    As for n: Remember that the single wire loop encircles the total magnetic field generated by the solenoid, which depends upon n, the turns per unit length of the solenoid.
     
  10. May 26, 2015 #9
    It's μ0 then? Rather than magnetic moment, yes, on reflection of course that makes more sense! Sorry. ε=100μ0nA.
     
  11. May 26, 2015 #10

    G01

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    OK! I think you have a good understanding of this problem now. Let me know if you have any other questions.
     
  12. May 26, 2015 #11
    Thanks for your help! :)
     
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