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Homework Help: Magnetic field of two wires.

  1. May 7, 2006 #1
    Hello guys. I just had a question about the below problem. I am not sure if this is right but more likely it is wrong. I have been trying this problem for an extensive period of time now. If anyone could help out I would greatly appreciate it.

    Following the dashed lines I showed the work that I did to solve this.

    Question: Figure below is an end view of two long, parallel wires perpendicular to the xy-plane. Each carries a current I, but in opposite directions. (X means into the mage and the dot is out of the page).

    Derive the expression for the magnitude of B at any point on the x-axis in terms of the x-coordinate of the point. What is the direction of B?
    Click on this link to view the picture I copied.

    38648?db=v4net.gif


    --------

    This is what I tried~~~

    By right hand rule the vector would be upward and both of the magnetic fields from the two wires will add.

    The field due to a long, straight, current-carrying conductor is B=mu_0I/2(pi)r where r is the distance from the wire to the point P.

    The distance there is r=sqrt(a^2+x^2) from both conductors. By using superposition and adding vectors we can find the field B.

    Thus Btotal=B1+B2

    B1=Mu_0*I/(2*pi*sqrt(a^2+x^2))
    B2=Mu_0*I/(2*pi*sqrt(a^2+x^2))

    B1+B2= 2*Mu_0*I/(2*pi*sqrt(a^2+x^2))
    B1+B2= Mu_0*I/(pi*sqrt(a^2+x^2))

    Is this not correct?
     
    Last edited: May 7, 2006
  2. jcsd
  3. May 7, 2006 #2

    nrqed

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    Your magnitudes are ok but the directions you give are wrong. The B fields produced by each wire at point P is *not* upward. You have to use a certain version of the right hand rule (with now the thumb pointing in the direction of the current and your hand curling around the wire, showing the direction of the B field ). For the top wire for example, the B field at P will be somewhere in the first quadrant (draw a circle centered around the wire and at point P the B field must be tangent to the circle, pointing in a direction in the first quadrant (if a was equal to x, the B field would be pointing straight northeast). Likewise, the B field produced by the lower wire at point P is somewhere in the fourth quadrant (would be straight southeast if a was equal to x).

    It's hard to explain without a blackboard!
     
  4. May 7, 2006 #3
    Thanks! The field would then point along the +x axis using the right hand rule.

    I am trying to use the answer that I got for magnitude and my system does not accept Mu_0*I/(pi*sqrt(a^2+x^2)) for the magnitude of the field at point P.
    Is it possible I made an error in the calculation? Thanks!!
     
    Last edited: May 7, 2006
  5. May 7, 2006 #4

    nrqed

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    Right...and because the currents are identical and the distances are the same, you should see that the *y* component of the two B fields cancel out exactly in this problem, leaving only an *x* component (the final result will therefore be that the total B field at P has only an x component which is equal to twice the x component of the B field of one wire)
     
  6. May 7, 2006 #5
    Thanks, also
    Still unable to solve the part with the magnitude though, it seems that it might be wrong, maybe I am missing something important
     
    Last edited: May 7, 2006
  7. May 7, 2006 #6

    nrqed

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    nice drawing.

    But did you calculate the x component of each B field and added them up together? You need to do that!

    It's really hard to explain without a drawing, but let's say you look at the B field produced by the top wire at P. It is pointing in the first quadrant, right? Now, it makes an angle with the x axis. You will have to mulitply your magnitude of B by the cos of that angle.

    Now, if you draw a line from the top wire to point B, you should be able to convince yourself that the angle between that line and a vertical line is the same as the angle between the B field and the x axis. Then, using trigonometry, you can figure out the cos of that angle
     
    Last edited by a moderator: Apr 16, 2017
  8. May 7, 2006 #7
    So B1 will actually B1cos(theta) in which case will be

    B=mu*I/(2*pi*r)*a/(sqrt(x^2+a^2)
    and thus

    B1+B2 = mu*I*a/(pi*(x^2+a^2)

    ? Still not there :(
     
    Last edited: May 7, 2006
  9. May 7, 2006 #8

    nrqed

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    What is the correct answer?? This looks right to me
     
  10. May 7, 2006 #9
    That one seemed to do the trick. Thanks alot for all your help!! :biggrin: :) :) :)
     
  11. May 7, 2006 #10

    nrqed

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    Great! You are welcome.

    (btw, it is always nice to check some limits in expressions like those...as a goes to zero, you can expect from the figure that the B field will go to zero and that agreea with the expression. As x goes to zero, you expect the two B field to be parallel and in the x direction so that the total B field will be twice the magnitude of each wire, and that also checks out)

    Regards


    Patrick
     
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