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Magnetic field of revolving disk

  1. Jun 7, 2009 #1
    find the expression for the magnetic field at a distance z on the axis which passes through the centre of a disk, with a radius of R, and a charge density σ, the disk is rotating at an anular velocity of ω.

    http://lh4.ggpht.com/_H4Iz7SmBrbk/SivHVn0m_fI/AAAAAAAABC0/9sgJmbkoJK8/s720/Untitled.jpg [Broken]

    could someone tell me if this is all done correctly?

    the equations i am going to use are
    1)F=ILxB
    2)f=qvxB
    3)biot savar

    1st stage i want to express the charge q,
    q=σπR2

    now i use
    f=qvxB=σπR2vB ( the angle is 90 degrees constant)
    f=σπR3

    F is also ILB so i know that IL=σπR3ω

    now using biot savar

    dB=(μI*dlxr)/4πr3)

    now for I*dl i want to substitute the expression i found earlier, with R being my variable, not that sure about this...

    dB==(μσπR3ωdR)/4π((R2+z2)0.5)3)

    dB=(μσR3ωdR)/4((R2+z2)1.5)

    B=(μσω/4)*∫R3dR/(z2+R2)1.5

    after integration i get

    B=(μσω/4)*(1+z2/(z2+R2))*(z2+R2)0.5 with my limits being R from 0 to R

    and eventually i get
    B=(μσω/4)*[(1+z2/(z2+R2))*(z2+R2)0.5-2z]

    i have never solved anything of this sort and hope everything i have done is okay, thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 7, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    You're pretty close, however your mistake is to assume that the force is constant for every r which it is not. [itex]F=qvB[/itex], but v and q are different at different radii. Therefore the total force would be [itex]\int_0^{2R} vBdq =\int_0^{2R} \omega r B 2 \pi r \sigma dr=\int_0^{2R} I B dr[/tex]. We're integrating over the same range and variable so the integrands need to be the same therefore.

    [tex]
    2 \pi \sigma \omega r^2 B dr=I B dr \Rightarrow I dr= 2 \pi \sigma \omega r^2 dr
    [/tex]

    Then using Biot-Savart yields:
    [tex]
    \frac{\mu_0}{4 \pi} \int_0^{2R} \frac{2 \pi \sigma \omega r^2}{r'^2} \cos \theta dr
    [/tex]

    r' is the distance from a point on the disk to the z-axis, [itex]r'^2=r^2+z^2[/itex]. And [itex]\theta[/itex] is the angle between r and r', so [itex]\cos \theta[/itex] picks out the x component of the magnetic field.

    An easier way would be using the surface current density, [itex]K=\sigma v[/itex]. Biot-Savart then becomes:
    [tex]
    \vec{B}(\vec{r})=\frac{\mu_0}{4 \pi} \int \frac{\vec{K} \times \hat{r}}{r^2}da'
    [/tex]
     
    Last edited: Jun 7, 2009
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