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Magnetic field above a thin charged disc

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin disc of radius R carries a surface charge [tex]\sigma[/tex]. It rotates with angular frequency [tex]\omega[/tex] about the z axis, which is perpendicular to the disc and through its center. What is B along the z axis?

    2. Relevant equations
    General Biot-Savart law:
    B(x) = [tex]\frac{\mu_{0}}{4\pi}\int\frac{J(x') x (x-x')}{|x-x'|}d^{3}x'[/tex]

    K [tex]\equiv \frac{dI}{dl_{perpendicular}}[/tex]
    K = [tex]\sigma[/tex]v

    3. The attempt at a solution
    I'm wondering if the general form Biot-Savart law can be 'generalized' to a 2-D surface current density K instead, and if the form would be the same?
    B(x) = [tex]\frac{\mu_{0}}{4\pi}\int\frac{K(x') x (x-x')}{|x-x'|}d^{2}x'[/tex]
  2. jcsd
  3. Oct 13, 2008 #2


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    I don't follow your equations, you aren't using a standard notation. Regardless, you don't need to use a surface current. You can treat every infinitesimal interval of radius dr as a loop of current. Integrate the equation for field from a current loop over r from 0 to R
  4. Oct 13, 2008 #3


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    Hi russdot, try using \vec{B} instead of B in your LaTeX equations, and put the whole equation in LaTeX to make it easier to read.

    The Biot-Savart Law for a surface current is:

    [tex]\vec{B}(\vec{r})=\frac{\mu _0}{4 \pi} \int \frac{\vec{K}(\vec{r'}) \times \widehat{\vec{r}-\vec{r'}}}{|\vec{r}-\vec{r'}|^2} d^3r'[/tex]

    Such integrals are tedious to evaluate, so if you are allowed to use Ampere's law or calculate the vector potential [itex]\vec{A}[/itex] first, I would do that.
  5. Oct 13, 2008 #4
    Hi gabbagabbahey,
    Thanks for the tips, I'm still getting used to the LaTeX notation.

    Ok, and then each infinitesimal loop will have a current I = [tex]\sigma 2 \pi R dr[/tex].
  6. Oct 13, 2008 #5


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    You're welcome:smile:

    Shouldn't each loop have a current of [itex]2 \pi r \sigma \omega dr[/itex]?:wink:
  7. Oct 13, 2008 #6
    Ah yes, I really should eat some food...
    but since [tex]\vec{I} = \lambda \vec{v}[/tex]
    and [tex]\lambda = \sigma 2 \pi dr[/tex]
    and [tex]\vec{v} = r \vec{\omega}[/tex]
    then [tex]\vec{I} = \sigma 2 \pi \vec{\omega} r dr[/tex], correct?
  8. Oct 13, 2008 #7


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    Yep, looks good to me (my sigma didn't show up right the first time)....what then is the magnetic field of each loop? What do you get for the total magnetic field?
  9. Oct 13, 2008 #8
    For the magnetic field of each loop (along Z axis), I get:
    [tex]\vec{B}(z) = \frac{\mu_{0}}{4 \pi} \int \frac{\sigma 2 \pi \omega x'^{2} (\hat{\phi} \times \hat{r})dx'd\phi}{x'^{2} + z^{2}}[/tex]
    where [tex]\hat{\phi} \times \hat{\r} = \hat{k}cos\psi + \hat{R}sin\psi[/tex]
    ([tex]\psi[/tex] is the angle between[tex]\vec{r} = \vec{x} - \vec{x}'[/tex] and [tex]\hat{R}[/tex] where [tex]\hat{R}[/tex] is the cylindrical radial unit vector)

    the [tex]sin\psi[/tex] term integrates out to zero, because [tex]\hat{R} = \hat{i}cos\phi + \hat{j}sin\phi[/tex] and I used the trig substitution [tex]cos\psi = \frac{x'}{\sqrt{x'^{2} + z^{2}}}[/tex]

    [tex]\vec{B}(z) = \mu_{0} \pi \sigma \omega \int_{0}^{R} \frac{ x'^{3} dx'}{(x'^{2} + z^{2})^{3/2}} \hat{k}[/tex]

    Which gives a total magnetic field:
    [tex]\vec{B}(z) = \mu_{0} \pi \sigma \omega (\frac{2z^{2} + R^2}{\sqrt{z^{2} + R^{2}}} - 2z) \hat{k}[/tex]
  10. Apr 21, 2011 #9
    Uhm... wouldn't the current flowing through an infinitesimal loop be the charge divided by a period T=[tex]\frac{2\pi}{\omega}[/tex] ?
    [tex]i=\sigma\omega rdr[/tex]

    P.s: how did you resolve the integral? Substituction with sinh(t) ? Are there any faster method?
    Last edited: Apr 21, 2011
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