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Homework Help: Magnetized Toroid

  1. Mar 29, 2006 #1

    I needed some help with a problem from the Griffiths Book on EM. It's problem number 6.10 if anyone has the book. Here is the problem and I have attached a crude drawing using MSPaint.

    An iron rod of length L and square cross section (side a), is given a uniform longitudinal magnetization M, and then bent around into a circle with a narrow gap (width w). Find the magnetic field at the center of the gap, assuming w << a << L. [Hint: treat it as the superposition of a complete torus plus a square loop with reversed current.]

    The field inside a toroid is: [tex]B= \frac{\mu_0 NI}{2\pi s}[/tex]. The hint would lead me to believe that I can take NI -> M. The field at the center of the gap due to a loop of current in the opposite direction would be:[tex]B=-\frac{2\sqrt{2}\mu_0 I}{\pi a} [/tex].

    This means the answer would be:
    [tex]B=\frac{\mu_0 NI}{L} - \frac{2\sqrt{2}\mu_0 I}{\pi a}[/tex].

    However I don't see how to resolve the current(I) in this problem. Is there a way to relate M and I or am I going about this problem in the completely wrong direction.

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    Last edited: Mar 29, 2006
  2. jcsd
  3. Mar 29, 2006 #2


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    Not quite. NI would be the TOTAL current in the toroid. You've probably figured out the bound surface current is [itex]K_b=M[/itex]. So knowing you have this bound surface currentm you can solve for the total current and equate that NI.
  4. Mar 30, 2006 #3
    Ok it seems then that the field should be

    [tex]B = \frac{\mu_0 M}{L} - \frac{2\sqrt{2}\mu_0 Mw}{\pi a}


    B = \mu_0 M \left[ \frac{1}{L} - \frac{2\sqrt{2}w}{\pi a} \right]

    However since L >> a and [tex]2\sqrt{2}w >> 1[/tex] the field would be negative. But that seems proposterous considering that w is just a minute width compared to the whole of the toroid.
    Last edited: Mar 30, 2006
  5. Mar 31, 2006 #4


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    That ain't right. How'd you get the L? The field shouldn't depend on L. (It also doesn't add up unit wise).

    For the whole loop the total current is [itex]K_b(2\pi s)[/itex] so [itex]B=\mu_0M[/itex].
    The rest is all ok.
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