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Emf induced by swinging rod in magnetic field (I got 2 answers?)

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    auccxv.png


    2. Relevant equations



    3. The attempt at a solution

    Their answer assumes that the B-field is perpendicular to the velocity at all times, which is clearly not the case.

    1g0uxj.png

    And in the next part of the question, won't the answer be zero? since now velocity is parallel to B-field? (So rod is simply moving along the field)
     
  2. jcsd
  3. Apr 17, 2013 #2

    rude man

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    θyes, "in the next part of the question", since B and v are in the same direction the answer is zero volts.

    I can understand what you tried to do better with your first go, even though the answer was wrong & was right in the second go.

    So: r = R sin θ i ... never mind the j part, it's in line with B ...
    For small angles, how can you simplify (approximate) this eq?

    Then v = ?

    Then do your emf = v x B * L

    Finally, what is the equation for θ(t) for a simple pendulum?
     
  4. Apr 17, 2013 #3
    approx r = Rθ i

    then v = R dθ/dt

    ε = BR*(dθ/dt)

    θ = θ0 cos(ω0t)
    (dθ/dt) = θ0ω0 sin(ω0t)

    okay it all works out now..thanks alot!

    In my first part, I assumed (dθ/dt) = ω0 = 2∏/T which is wrong as (dθ/dt) is not constant.
     
  5. Apr 17, 2013 #4

    rude man

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    Good work! Wish every OP was as sharp & interested as you.
     
  6. Apr 18, 2013 #5
    However, I just realized that the last part indicated that an emf was induced (8mV)...How is that possible? Are they assuming the oscillations are not perpendicular to the B-field now?
     
  7. Apr 18, 2013 #6

    rude man

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    There is also a vertical component to v , right? I was looking at your expression for r and I don't think I agree with your j component. r is the vector going from the pendulum at rest to its position at an angle θ. So if θ = 0 then the j component of r is zero, not R as you have it.

    This is the crux of the issue since once you get the j component of v right the rest is just repeating what you did before but with B = B i instead of B j . Let's work on getting v right.
     
  8. Apr 18, 2013 #7
    the origin is taken at the top of the swing, so at θ = 0, x = 0, y = -R as indicated.
     
  9. Apr 18, 2013 #8

    rude man

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    OK, but we need to get the right expression for v in the y direction. I don't quite dig you expressions for r and then v = dr/dt. If you go with yours, v in the y direction approximates to
    vy = R θ dθ/dt whereas I make it 2R θ dθ/dt. Factor of 2 off.

    My way: origin at pendulum mass when θ = 0. Then for small θ, s = arc length = Rθ, x = s cosθ and y = s sinθ. So y = Rθ sinθ ~ R2θ and therefore vy = d/dt(Rθ2) = R d/dt(θ2) = 2Rθ dθ/dt.

    We need to put this to bed, or you could proceed with the problem using your or my vy.

    Anyway, after that it's just more of the same as before, using the expression for θ(t) for a simple pendulum.
     
  10. Apr 18, 2013 #9

    haruspex

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    At angle θ, the vertical displacement is R(1-cos(θ)) ≈ Rθ2/2.
     
  11. Apr 18, 2013 #10

    rude man

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    Haruspex looks like he has it right, so we go with your expression for v_y = R*theta d(theta)/dt.
     
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