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Homework Help: Induced current in a coil

  1. Jan 3, 2010 #1
    1. The problem statement, all variables and given/known data
    As shown in this picture:
    2igneir.png
    a 18 cm long solenoid with 1500 loops (N = 1500) and radius 2 cm is located at the center of wire frame (coil) that is 5(width) by 8(height) cm. There is a current inside of solenoid that is growing from 0 to 2.5 A in 7.5 seconds that rotates counter clock wise. The frame is tilted and there is 70° angle between the frame and the axis of the solenoid.

    Find the induced voltage in the frame.

    2. Relevant equations
    [tex]|\mathcal{E}| = \left|{{d\Phi_B} \over dt}\right|[/tex]
    B = µnI, n = N/h
    Phi = AB

    3. The attempt at a solution
    First I find magnetic Flux = A * µnI:
    Phi = (Pi * 0.02^2) * (µ * (1500/0.18m) * 2.5A) = 3,28997e-5 Wb

    emf = dPhi/dt
    emf = 3,28997e-5 Wb/7.5s = 4.39 V

    The way I understand it, the size of the frame, 5 by 8 cm doesn't affect the induced current, since the only source of the magnetic flux is the solenoid and it is located completely inside of the wire frame. Neither does the angle of the frame have effect in this case, since all the magnetic flux is passed through the fire frame. Is that correct?
     
    Last edited: Jan 3, 2010
  2. jcsd
  3. Jan 3, 2010 #2

    diazona

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    Where did the [itex]4\pi[/itex] come from, and why did you divide by [itex]0.18\mathrm{m}[/itex] (I assume 0.18 meters is what you meant)?

    That sounds right.
    That part isn't right. One of your formulas is incomplete, actually: the formula for flux is
    [tex]\Phi = \iint \vec{B}\cdot\mathrm{d}^2\vec{A}[/tex]
    There is a dot product between the magnetic field and the normal vector to the surface. That dot product contributes a factor of [itex]\cos\theta[/itex], which depends on the angle between the field and the frame.
     
  4. Jan 3, 2010 #3
    I'm sorry, there shouldn't be [itex]4\pi[/itex].
    Division by 0.18m is from: B = µnI, where n = N/h (N is the number of loops and h is length of the solenoid).

    Yes, there is a dot product, but in this case, I think the flux through the wire frame is the same no matter at what angle the solenoid is placed, as long as all the flux generated by the solenoid goes through the wire frame.

    If we calculate the flux at the angle of 70° between the solenoid and frame, then the flux is:
    Phi = B * A * cos(90°-70°).
    The area of the solenoid that cuts through the plane of the wire frame also grows by the factor of 1/cos(90°-70°). So in this case the flux is the same at both 90° and 70°. Am I wrong?
     
  5. Jan 3, 2010 #4

    diazona

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    Oh yeah, I guess I wasn't quite paying attention. That sounds right.

    So why are you asking this question? Did you get the wrong answer?
     
  6. Jan 4, 2010 #5
    Excuse me,it does seem correct to write the flux as
    Resulting BAcos(20°)
    And no further correction would be necessary in my view
     
  7. Jan 4, 2010 #6
    Excuse me,it does seem correct to write the flux as a dot product
    Resulting BAcos(20°)
    And no further corrrection would be necessary in my view
     
  8. Jan 4, 2010 #7
    I wasn't so sure, because the exercise gives the size and the angle of the wire frame. It seems they are there just to confuse though, they are not needed to find the induced current.
     
  9. Jan 4, 2010 #8
    I asked someone about this and they say:
    "this applies only to static systems, i.e. those which do not generate ANY flux themselves"
    meaning emf = dPhi/dt

    And:
    "the angle of wireframe wires will be set differently vs angles of electrons travelling in solenoid, thus lorentz contraction will occur with different ratio"

    So the angle of the wire frame does matter in this case?
     
  10. Jan 4, 2010 #9
    The point is ,again in my view, that the flux through a normal section A of the solenoid is not the flux through the frame.
     
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