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Homework Help: Electromagnetism help

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Two infinite long straight conductors and a rectangular circuit lies in the same plane. The straight conductors is parallel and got the distance 2a from each other. There runs a constant current of I1 and I2 in both of them, and the current is pointed downwards. The rectangular circuit is placed in between the two conductors and got the resistance R. The circuit got the sidelengths a and b, where the side b is parallel with the straight conductors. The sidelength a is constant. The sidelength b grows as a function of time t in the interval 0 < t < [tex]\tau[/tex] which is:

    b(t) = a((1+3(t/[tex]\tau[/tex])^2)-(2(t/[tex]\tau[/tex])^3)),
    where tau is a timeconstant.
    For t < 0, b = a
    For t > [tex]\tau[/tex], b = 2a

    The magneticfields produced by the current in the straight conductors gives a magnetic flux through the rectangular circuit of:

    [tex]\Phi[/tex]B = K * b(t),
    where K is a constant.

    Determine the constant K. The answer must be expressesd in terms of [tex]\mu[/tex]0 and the currents I1 and I2

    2. Relevant equations


    3. The attempt at a solution
    I know the answer is: K = [tex]\mu[/tex](I2-I1)(ln(3)/(2*pi)), but I have no idea how to get there.
    I've tried with Faradays Law, but I couldn't get it there either.
    So, I just need a clue what to do actually :)

  2. jcsd
  3. Aug 13, 2009 #2


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    Since you are asked to find the magnetic flux, I would think that the definition of of magnetic flux would be a relevant equation.:wink:

    [tex]\Phi_B=\int_{\mathcal{S}}\textbf{B}\cdot d\textbf{a}[/tex]

    (Where [itex]d\textbf{a}[/itex] is an infinitesimal area element of the surface [itex]\mathcal{S}[/itex] , directed normal to the surface)

    Look familiar?
  4. Aug 13, 2009 #3
    Yes, but I just figured that since I didn't have a magnetic field to put in that equation, I couldn't use it ?

    Hmm, of course I have B = u0*I/2*pi*r

    So B(pi*r^2) = K * b(t)

    And then integrate what ?
    Last edited: Aug 13, 2009
  5. Aug 13, 2009 #4


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    Aren't there two wires? What is 'r' in this case?

    I recommend drawing the problem out in the x-y plane....let the infinitely long wires run parallel to the y-axis at x=-a and x=+a, then at any point (x,y), the distance from the wire on the left will be |x+a| and the distance from the wire on the right will be |x-a|....what does that make the total magnetic field of the two wires? Which direction does it point?

    Where is this equation coming from?

    Once you find the magnetic field, integrate over the surface you wish to find the flux through....In this case, you want the flux through the surface that is bounded by the rectangular circuit...so x goes from __?__ to __?__ and y goes from __?_ to __?__ and [itex]d\textbf{a}=[/itex]__?__
    Last edited: Aug 13, 2009
  6. Aug 13, 2009 #5
    This is the drawing I came up with (See picture).

    So the magnetic field between the wires is:

    Bnet = u0*I1/2*pi*r+u0*I2/2*pi*r or...?

    I guessed I just add them together, or am I wrong ?

    http://img512.imageshack.us/img512/3199/elecq.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  7. Aug 13, 2009 #6


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    Your attachment is pending approval, and probably will be for some time...it is quicker if you just upload your picture to imageshack.us and then post a link to it...
  8. Aug 13, 2009 #7
  9. Aug 13, 2009 #8


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    Okay, now put in a coordinate system... I suggest having the y-axis run right down the middle of your circuit, and the x-axis go right along the bottom edge of the circuit... with that coordinate system what will the magnetic field due to each wire be, and what are their directions (right-hand rule) at any point (x,y) in between the two wires (i.e. on the interval [itex]-a<x<a[/itex])?
  10. Aug 13, 2009 #9
    The left wire will have a B-field of:
    Bleft = u0*I1/2*pi*(x-a)

    And the right:
    Bright = u0*I2/2*pi*(x+a)

    And since the current is the same direction for each, it's in the +x direction ?
    Last edited: Aug 13, 2009
  11. Aug 13, 2009 #10


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    Actually, you have that backwards...the distance from the left wire is |x+a| and the distance form the right wire is |x-a| (to check this, just plug in x=-a and x=a...at x=a the distance from the right wire should be zero and the distance from the left should be 2a, and vice versa at x=-a)....the absolute value signs are there because distance is a positive quantity....to get rid of them just notice that between the two wires, -a<x<a... so |x+a|=x+a>0 and |x-a|=a-x>0....

    Careful, the fields of each wire circles around the wire (point the thumb of your right hand in the direction of current, and your fingers will curl in the direction of the field) so to the left of each wire the field will point into the page (-z-direction), and to the right of each wire the field will point out of the page (+z-direction) between the two wires , you are to the right of the left wire and to the left of the right wire...
  12. Aug 13, 2009 #11
    Thank you so much :)

    I think I got it now.
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