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Faraday's Law - very tricky question

  1. Jun 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A straight infinitly long wire with a current of I upwards is stationary. A rectangular loop with length a and height b is moved to the right with constant speed of v across the wire in the plane of the paper. At t=0 the right edge of the loop is exactly at the same location of the wire.


    2. Relevant equations
    What is the flux through the loop as a function of the time?
    What is the induced emf as a function of the time?


    3. The attempt at a solution
    Well I have 2 main problems here:
    1. At any moment up untill the time that all the loop has passed to the right of the wire I'm facing the fact that some part of the loop (the part that is to the right of the wire) is under the influence of a magnetic field in the direction which is into the paper, and another part (the part that's to the left of the wire) is under the influence of a magnetic field in the direction which is out of the paper - how can I deal with this problem in terms of my integration?
    2. If I want to switch the integral from dA to dr (where r is the distance from the wire) - I have again one part of the loop that is coming towards the wire while another part of the loop is moving away from the wire (At least until the whole loop passes through the wire) - How can I deal with this fact when I try to write the integration limits?

    This is a bonus question and isn't mandatory but it pisses me off!
     

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  2. jcsd
  3. Jun 7, 2013 #2

    tiny-tim

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    hi assaftolko! :smile:

    choose a direction (eg, into the paper), then flux the same way is positive and flux the opposite way is negative

    so just subtract one from the other! :wink:
     
  4. Jun 7, 2013 #3
    I know but I don't get how to really do the calculus here... because the two fluxes are present in the same time... I'm quite lost
     
  5. Jun 7, 2013 #4

    tiny-tim

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    ok, you calculate flux L(t) through the left-hand part at time t, and R(t) through the right-hand part

    then your total flux at time t is F(t) = L(t) - R(t), and that's what you integrate :smile:
     
  6. Jun 7, 2013 #5
    Ok I tried something tell me if so far so good: I said that dA is in the out of the page direction:

    For the flux in the left area:

    ∅L(t) = ∫[itex]\vec{B}[/itex]*[itex]\vec{dA}[/itex] = ∫μ0I/2∏r*dA*cos0 = μ0I/2∏∫[itex]\frac{1}{r}[/itex]dA

    But dA=b*dr and r itself is a function of time, and at t=0 we get (with respect to the left edge of the loop) that r=a, and the final location of the left edge when we deal with the left side flux is r=0 and so:

    ∅L(t)=μ0Ib/2∏∫[itex]^{0}_{a}[/itex][itex]\frac{1}{r}[/itex]*-dr

    -dr and not dr because r is getting smaller as the loop moves towards the wire.

    How am I so far?
     
  7. Jun 7, 2013 #6

    tiny-tim

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    your integrand looks ok, but your limits are wrong

    at each time t, you have to integrate across the whole loop

    your "r" is the distance from the wire

    if the wire is at position w(t) at time t,

    then you have positive flux on the left from r = 0 to r = …

    and negative flux on the right from r = 0 to r = … ? :wink:

    (btw, once you've found what the limits are, there's a fairly obvious way of simplifying the integral)
     
  8. Jun 7, 2013 #7
    well I guess you meant to write r(t) and not w(t) - I think that if r=0 is the location of the wire then r=a-vt is the location at t=0, and for the right part we go from r=0 to r=vt?
     
  9. Jun 7, 2013 #8

    tiny-tim

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    no, you're confusing r(t), which is a function of t and is the position of the wire is at time t

    with r(x,t), the distance of a general point x (0 ≤ x ≤ a) from the wire

    since your flux formula already uses r for the second one, i've used w(t) for the first one, and r(x,t) = |x - w(t)| for the second one :wink:

    (now, to find the flux at time t, integrate over x, with t constant)
     
  10. Jun 7, 2013 #9
    How is r(t) the position of the wire at time t? the position of the wire is constant... it's the loop that moves. I still don;t get what you're trying to tell me :(
     
  11. Jun 7, 2013 #10

    tiny-tim

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    what difference does it make, which one moves?!!

    you have to measure r (in the formula) from the position of the wire, and it goes up to the two sides
     
  12. Jun 7, 2013 #11
    Ok but aren't the limits I've written in my before last post correct? from r=0 to r=vt to the right and from r=0 to r=a-vt to the left (or maybe r=vt-a to the left because this position is at the negative side of the r axis?)?
     
  13. Jun 7, 2013 #12

    tiny-tim

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    ah, that's not what you originally wrote :redface:

    yes that's correct now :smile:

    ok, now do the integration!
    no, it's very confusing if you let r be negative … it makes it too easy to make mistakes

    it's best to keep r positive, even if that means writing two integrals instead of one
     
  14. Jun 7, 2013 #13
    Ok but what I don't get is this - if some of my integration limits are 0 - then we'll end up at some point with ln(0) which is a big no-no... also - I still don't get why it's not correct to say that for the right part we start from r=a (I'm looking at the left edge of the loop with respect to the wire) at t=0 and r=0 at t=the moment the left edge of the loop passes through the wire, and similarly to say that for the right part we start from r=0 (I'm now looking at the right edge of the loop with respect to the wire) and end up at r=a at t=the moment the left edge of the loop passes through the wire. Why is it critical to represent r as a function of t?
     
  15. Jun 7, 2013 #14

    tiny-tim

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    true

    but the smaller part will cancel out with some of the larger part, so you won't need to bother with 0 :wink:
    sorry, i don't understand this :redface:

    the flux is a function of t

    i'm not even sure what you mean by "r"

    the r in your basic formula is the distance frrom the wire to a general point inside the loop, and that distance changes with t
     
  16. Jun 7, 2013 #15
    Yes it changes with time - but if I integrate over dr, and I look at a fixed point in the loop - like it's left edge, then I can say that at t=0: r=a and because the loop moves to the right, r is getting (for the left edge of the loop) smaller and smaller until the left edge of the loop passes through the wire (and then r=0) and from this point on there isn't any area of the loop that is left to the wire, and so there is no flux from here on to the left of the wire. Why isn;t this representing the total flux through the area to the left of the wire?
     
  17. Jun 7, 2013 #16

    tiny-tim

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    yes, but :confused:
    i don't understand :redface:
     
  18. Jun 7, 2013 #17
    I want to write an expression that represents the flux from the area to the left of the wire (the wire being this rectangular area's right edge) and the total flux from the area to the right of the wire - why aren't I doing just that when I integrate dr from r=a to r=0 (let's just focus on the left flux for now)?
     
  19. Jun 7, 2013 #18
    Ok I think I got it - if the loop wasn't moving I think that my integration limits would be correct...
     
  20. Jun 7, 2013 #19

    tiny-tim

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    but r doesn't go from a to 0 (except at t = 0) :confused:
     
  21. Jun 7, 2013 #20
    Yep exactly - if the loop wasn't moving my limits are correct I think
     
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