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The magnetic field of a spinning spherical shell of uniform charge

  1. Nov 23, 2013 #1
    In chapter 5, magnetostatics, of Griffiths' Introduction to Electrodynamics (third edition), there's a problem in the back of the chapter that asks you to calculate the force of attraction between the northern and southern hemispheres of a spinning charged spherical shell.

    The problem in its exact words:

    Problem 5.42: Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell.

    Well, we haven't gotten so far in our class that I can actually do this. But I'm reviewing for the test, so I figured there was at least some part of the problem that I could do. I decided to calculate the magnetic force at the southern tip P. Seeing as this isn't the problem, I don't know if my answer is correct. Here's what I did:

    I used the Biot-Savart law to do so. I let the distance from P to the point of current dI be r, the height (which goes from 0 to 2R) be z, and the distance from z to dI be a. Radius = R.

    I used these values to work out a formula for r, namely r^2= z^2 + (2Rz)^2. My current I = σω(2Rz - z^2)^(1/2) (though I'm thinking it should be line charge λ). The diameter is equal to 2∏(2Rz - z^2)^(1/2)
    My Biot savart law has me integrate I/r^2 cos∅ which turns into:

    B = σμω/2 * Integral[(z*diameter)/(r^(3/2))]​

    Resulting in the magnetic field B equal to μσω/3 (which seems unlikely to me.)

    If formatting this or scanning my work to this post would motivate someone to actually verify my answer, then I'll do that... just let me know.
     
  2. jcsd
  3. Nov 24, 2013 #2

    Philip Wood

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    Gold Member

    It's brave of you to have a go at this. I'm afraid I haven't gone right through your attempt, but I've spotted a couple of things near the beginning.
    Neither of these can be right, because the units don't match. Always check units!

    In the first equation the units of the first two terms are m2, but the unit of [itex](2Rz)^2[/itex] is m4. For what it's worth, I got [itex]r^2 = 2Rz[/itex].

    The right hand side of the second equation has the units Am-1. What you're calling I is the current per unit arc length of a line of longitude as the sphere rotates.

    Hope this is of some help.
     
  4. Nov 25, 2013 #3
    Oh my, I'll have to go over it again at some point today. I'll report back if I make any more progress. Thanks!
     
  5. Nov 27, 2013 #4

    Philip Wood

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    I thought the original problem you posted looked hard, and a brief search of the internet using "Griffiths spinning charged hemispheres question" confirmed that it was! Apparently – I don't possess Griffiths – you should first study example 5.11. This uses vector potential, A, though in my limited experience many problems that can most neatly be done using A can be done more laboriously using the ordinary field vector, B.
    Good luck!
     
    Last edited: Nov 27, 2013
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