Magnetic field at the centre of a spinning disk

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  1. dknt's Avatar
    1 24-05-2012  00:24
    • Overlord in Training
    • Location: UK
    • Posts: 2,234
    Magnetic field at the centre of a spinning disk
    If someone could check what I've done it would be much appreciated.

    I've derived the magnetic field on the axis of a loop of current, with radius a, to be

    B= \dfrac{ \mu _0 I}{2} \cdot \dfrac{a^2}{(a^2 +z^2)^{\frac{3}{2}}}

    I'm then told there is a rotating disk of uniform charge density, \sigma and radius R, rotating with constant angular speed \omega and found that the current due to a ring of the disk at a distance r from the centre is \mathrm{d}I=\omega \sigma r \mathrm{d} r

    I'm now asked to find an expression for the magnetic field at the centre.

    What I've done is;

    At the centre of a loop of current z=0, so B= \dfrac{ \mu _0 I}{2a}

    I've then said that a disk is a sum of loops over the radius, so

    \mathrm{d}B= \dfrac{ \mu _0 \mathrm{d}I}{2r}

    Substituting in for dI I get,

    \mathrm{d}B= \dfrac{ \mu _0 \omega \sigma r \mathrm{d}r}{2r}

    \mathrm{d}B= \dfrac{ \mu _0 \omega \sigma \mathrm{d}r}{2} ,

    so

    B= \displaystyle\int^R_0 \dfrac{ \mu _0 \omega \sigma \mathrm{d}r}{2}

    B= \dfrac{ \mu _0 \omega \sigma R}{2}

    Is that correct? Thanks for any help.
    Last edited by dknt; 24-05-2012 at 01:49.
  2. dknt's Avatar
    2 24-05-2012  23:01
    • Overlord in Training
    • Location: UK
    • Posts: 2,234
    Re: Magnetic field at the centre of a spinning disk
    Bump
  3. jk5430's Avatar
    3 25-05-2012  13:43
    • New Member
    • Posts: 18
    Re: Magnetic field at the centre of a spinning disk
    Looks good to me
    Last edited by jk5430; 25-05-2012 at 13:56.
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