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Magnetic Field due to a Long Straight Wire (Proof)

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data

    vf0ox.png

    Whilst revising the Biot-Savart law, I came across the result:
    magcur3.gif
    for the magnetic field due to a long straight wire. I understand that this can be found by integrating the Biot-Savart equation over the length of the wire, however after attempting the integral myself and looking at the integration in my notes I find myself unable to understand how to come to the result.

    2. Relevant equations

    72c45a37f39a11fd433909be250d7260.png

    3. The attempt at a solution

    Outlined above. Any help will be appreciated, thank you.
     
  2. jcsd
  3. Apr 29, 2012 #2
    Could you clarify what confuses you about the integral? The way to set it up would be as an integral over [itex]z[/itex], say, along the wire, with each wire element [itex]dz[/itex] contributing to the field at P based on the distance to point P and the angle between the radius vector and the z-axis.
     
  4. Apr 29, 2012 #3
    This is the proof from my lecture notes:
    2gspbfm.png
    I understand where the cross product has been replaced with dssinθ up to the integral of sinθ/r2.

    I also understand the first relationship r2 = s2 + R2 as just being Pythagoras, however I do not understand it's necessity, also I do not understand the second relationship involving sinθ either.

    So I suppose you could say I don't understand the relationships, and also why and how they are used.
     
  5. Apr 29, 2012 #4

    jtbell

    User Avatar

    Staff: Mentor

    In order to do the integral, you have to write it in terms of a single variable. Here, the writer chose s as the variable of integration, so he has to eliminate ##\theta## and r in favor or s. R is a constant as far as the integral is concerned.

    This is not the only way to do it. You can also do it by using ##\theta## as the integration variable, in which case you have to eliminate s and r. You should of course end up with the same final result.

    You can also (at least in principle) use r as the integration variable. I've never done it that way in this particular example, but I've done something similar in other situations.
     
  6. Apr 29, 2012 #5
    Oh ok yeah I see, however I still do not understand this relationship then:

    sin(∏-θ) = R/(s2 + R2)0.5
     
  7. Apr 29, 2012 #6
    Theta is the outside angle in that diagram, pi-theta makes it the inside angle so that you can use the simple definition (sin theta = opp / hyp) to remove the theta from the equation.
     
  8. May 1, 2012 #7
    Thank you, I fully understand the derivation now!
     
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