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Flux through a circle not centered at origin.

  1. Dec 13, 2009 #1
    I feel really dumb for this, but I keep getting strange answers so I may be forgetting something.

    1. The problem statement, all variables and given/known data
    Find the magnetic flux through a circle in the xz plane of radius "a" due to a wire. . The center of the circle is (b,0,0).

    2. Relevant equations
    Well I found the magnetic field to be proportional to

    That field has only a component in the phi direction.

    Where rho and phi are cylindrical coordinates.

    3. The attempt at a solution
    I can't figure out the limits of integration for anything unless it's rectangular coordinates, but those become very messy and I obtain an integral which is clearly wrong. I KNOW the answer should reduce to something "nice."

    I know I'm missing something stupid here.
  2. jcsd
  3. Dec 13, 2009 #2


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    Where is the wire? I guess it's the z-axis?
  4. Dec 13, 2009 #3
    Yes, it is. It's an infinite wire carrying current I in the z-hat direction. Basically using the result of a previous problem, I know what the B field is. I just chose not to include the rest of the stuff for that field in this question.
  5. Dec 13, 2009 #4


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    According to wikipedia, [tex]\Phi _B=\iint _S \vec B \cdot d \vec S[/tex].
    I'm not able to solve the problem so I'd love someone to help us.
    If you have the expression of B in all the circle, I think it's easy to solve the integral.
  6. Dec 13, 2009 #5
    I obtained B and I know what the definition of the magnetic flux is. I also know the vector element of area for cylindrical coordinates results in an integral over just [tex]\frac{1}{\rho}[/tex]

    IE, I have

    \Phi _B= k \iint _S \frac{1}{\rho} d\rho dz
    where k is a constant (having trouble texing the constants).
    The issue is I am unsure of what to make the limits of integration.
  7. Dec 13, 2009 #6


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    Are you sure you have a [tex]d\rho dz[/tex] as a differential area? You said you were working cylindrical coordinates, I'm confused.
  8. Dec 13, 2009 #7
  9. Dec 13, 2009 #8


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  10. Dec 13, 2009 #9
    What? Phi is the angle that goes counterclockwise from the x axis. I would NOT be integrating over dphi because phi is constant = 0. The circle lies in the xz plane.

    Draw the circle out on a piece of paper to see what I mean. Clearly rho does not go from 0 to a. Rho goes from (b-a) to (b+a) when z =0, and it goes from b to b when z = a. The limits of integration over the inner integral are clearly not constant.
  11. Dec 13, 2009 #10
    Well, you could divide the circle into infinitesimal horizontal slices and calculate what is the area dA of a slice as a function of x (through trigonometry). On the circle, B is proportional to 1/x so the flux through a slice would be just B(x)(dA/dx)dx which you could integrate along the x axis. However, this gives a complicated integral which I'm not sure how to solve.
  12. Dec 13, 2009 #11
    I tried a similar approach and obtained an integral which Mathematica integrated to have imaginary parts != 0. Im curious if I could shift my origin (make a substitution) and then convert to spherical coordinates (probably the other order). The spherical coordinates integral (after the substitution) for r would be from 0 to a, and for theta it would be 0 to pi. After the origin shift, my B field will still be in phi-hat direction only due to the symmetry.
  13. Dec 13, 2009 #12


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    My error, I thought it was in the x-y plane, sorry about that.
  14. Dec 13, 2009 #13
    It does that because it doesn't know the values of a and b, they might be such that the thing inside the square root is negative. If we assume that b>a>0, however, it's clear that the value of the integral is real. You have to use assumptions, though i don't know how to do that in Mathematica. Anyway, I got (assuming B(x) = B0/x)

    [tex]\Phi = \int_{b-a}^{b+a} \frac{2 B_0 \sqrt{a^2 -(b-x)^2}}{x}dx = 2 B_0 a \int_{\frac{b}{a}-1}^{\frac{b}{a}+1} \frac{ \sqrt{1 -(\frac{b}{a}-x)^2}}{x}dx[/tex]

    which according to Maple is

    [tex]2 B_0 a \,{\frac {\pi \, \left( 1+\frac{b}{a}\,\sqrt {-1+{\frac{b^2}{a^2}}}-{\frac{b^2}{a^2}} \right) }{
    \sqrt {-1+{\frac{b^2}{a^2}}}}}

    which can be simplified to [tex]2 \pi B_0 (b - \sqrt{b^2-a^2}) [/tex]

    Which isn't that bad in the end (assuming I didn't make a mistake). I'm not sure how to calculate the integral by hand though.
    Last edited: Dec 13, 2009
  15. Dec 13, 2009 #14
    YES! Thank you. That's exactly what the result should simplify to (it's very similar to that of a toroid, after all a toroid is roughly just N of those). Thank you. You were right, I didn't make the assumptions in Mathematica, but I'm still quite a novice and don't know how to do that yet.

    Thank you :)
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