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Magnetic Field of a thin disk

  1. Dec 1, 2003 #1
    I'm not quite getting the following problem:

    A thin disk of dielectric material, having a total charge +Q uniformly distributed over its surface, and having radius a, rotates n times per second about an axis through the center of the disk and perpendicular to the disk.

    Show that the magnetic field produced at the center of the disk is unQ/a.

    I know that B at the center of a wire ring is uI/2r. If I start here then dB = udI/(2dr).

    I = dq/dt. In this case I = nQ/sec, so can the equation become
    db = undQ/(2dr)? I'm sure this isn't right but here goes.

    Q/(pi)a^2 = dq/2(pi)rdr so dq = 2Qrdr/a^2 ?

    If I plug this into my equation the dr's cancel, so I must be doing something wrong?

    Thanks in advance for the help.
  2. jcsd
  3. Dec 1, 2003 #2

    Doc Al

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    Staff: Mentor

    Hint: Find the current for a rotating ring of charge of width dr. (Find the charge per unit area; from that find the charge in each ring.)
  4. Dec 1, 2003 #3


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    Actually, your original equations don't quite make sense:
    dB= μdI/(2dr)has a differential on one side and a derivative on the other. Also, I don't believe I= dq/dt= nQ/sec. Since the disk is rotating at n revolutions per second, it is rotating at 2πn radians per second and a piece at distance r (linear units) from the center is moving at 2πnr linear units per second. A thin ring, distance r from the center and width dr would have area 2πrdr and so charge (2πrdr)(Q/(&pia2)= 2rQdr/a2.
    I= dq/dt= 2πnr(2rQdr/a2)= (4πQnr2/a2)dr.
    That makes dB= μI/(2r)= 2πμQnr/a2 (this is "dB" because it is a small part of the whole magnetic field due to a small part of the entire radius. To find B, integrat from r= 0 to r= a.
  5. Dec 1, 2003 #4
    OK, let's see if this works.

    Doc Al, didn't I already kind of work out your hint.

    Charge per unit area for the whole disk is Q/(pia^2).
    The ring's charge per unit aread would be dQ/2pirdr.

    Therefore dQ = 2Qrdr/(a^2).

    Since every second ndQ charges flow through what could be considered a cross sectional area, can I substitute ndQ for I?

    If I can then dB = nu2Qrdr/(2ra^2) = nuQdr/a^2.

    If I integrate this from 0 to a, I get the correct answer:

    But is the math and reasoning OK or did I force something that I shouldn't have.

    Thanks for your help.
  6. Dec 1, 2003 #5

    Doc Al

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    Staff: Mentor

    I wasn't clear what you were doing the first time.
    Last edited: Dec 1, 2003
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