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Purcell 7.2

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A wire lies at height z=h carrying current parallel to the y-axis. A square loop, of side length b, is moving in the direction of +x with a velocity v. Find the magnitude of the emf at the moment when the center of the loop crosses the y-axis.

    2. Relevant equations

    3. The attempt at a solution
    Would the fact that the B-field makes circle around the wires mean that flux through the loop is 0 at this moment? I need to find d(Flux)/dt, but that seems like it would be quite a complicated function! Is there a trick that keeps me from having to do that?

    Otherwise, perhaps I would find the z-component of B-field, but that wouldn't involve any dependence on t... so how do I get the t into the equation so that the derivative of my expression for flux is not 0? (If there is no t in the expression, d(anything)/dt is 0...)
  2. jcsd
  3. Oct 26, 2012 #2


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    Hello, NullSpace0. To be clear, the square loop lies in the xy plane according to Purcell's statement of the problem.
    Yes. But, of course, that doesn't mean the emf is zero at that instant.
    It's not too terrible to find the flux through the loop for arbitrary x (the x-coordinate of the center of the loop). You could then find the emf using the chain rule: ##\frac{d\Phi}{dt} = \frac{d\Phi}{dx}\frac{dx}{dt}##

    However, you can avoid this by using Purcell's results in section 7.3 ("A loop Moving Through a Nonuniform Magnetic Field"). In particular, can you see how Purcell's equation (6) might be applied to your problem?

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    Last edited: Oct 26, 2012
  4. Oct 26, 2012 #3
    Perhaps the second method could be applied by considering the Lorentz force on each side of the loop, and then doing the line integral of force per unit charge around the loop at the time of interest?
  5. Oct 26, 2012 #4
    Schaefera and TSny-- thank you for the advice!

    If I do use that second method, would I only want to consider z-components of B-field, since the other component would then lie in the plane of the loop?
  6. Oct 26, 2012 #5


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  7. Oct 26, 2012 #6
    Aha! Working all of this out, I find that Bz=(2Ib)/(cr^2) but the sign is opposite for the sides that are pointing along the wire... I don't actually care about B for the sides of the loop perpendicular to the wire because for them, F and ds will end up being perpendicular and so they come out to 0 in the line integral of force.

    Thus, I get F=(2Ivbq)/(c^2*r^2), and then integrate (1/q)F.ds, I get: emf=(4Ivb^2)/(c^2*(h^2+b^2)). Does this look correct?

    A friend I'm checking answers with somehow got that the b^2 on the bottom is divided by 4 while there is only a factor of 2 multiplying the numerator... not sure where that would come from!

    And if I do this by finding the flux, then taking that integral's time derivative (using the chain rule) and then dividing by c as Faraday's law tells me, I get MY solution to the emf above... but with only a factor of 2 in the numerator, not the 4 I have up there.

    So I have 3 very similar but slightly different solutions...
    Last edited: Oct 26, 2012
  8. Oct 26, 2012 #7


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    I think your friend might be right. You have a triangle with r as hypotenuse and h as one side. What is the length of the other side?
  9. Oct 27, 2012 #8
    I was about to say!

    That should fix it all.
  10. Oct 27, 2012 #9
    Yay, thanks for the help! I needed to include b/2 as the numerator when taking the z-component of the B-field, which eliminates the 4 in the numerator of the final solution and adds the b^2/4 in the bottom.

    Final answer: induced emf= (2Ivb^2)/(c^2(h^2+(b/2)^2))

    So now I should also decide which way the induced current will flow. I think that, if we look down on the square from above, it will flow counter-clockwise... but this based on the fact that as the square moves beneath the wire, flux starts by increasing in the upward direction (+z) and just at the instant it is beneath the wire, net flux is 0 meaning that it is just about to start increasing in the down direction... induced B should point in the +z which requires a counter-clockwise I(ind) but the RHR. Is this reasoning sound?
    Last edited: Oct 27, 2012
  11. Oct 27, 2012 #10


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    I believe that's correct (assuming the current is in the +y direction). Good! :smile:
  12. Oct 27, 2012 #11
    Thank you both so much!
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