Magnetized Toroid Problem 6.10 - Griffiths EM

[buhthestuh]

Hello,

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: B= \frac{\mu_0 NI}{2\pi s}. 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:B=-\frac{2\sqrt{2}\mu_0 I}{\pi a}.

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

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.

 

[Galileo]

The field inside a toroid is: B= \frac{\mu_0 NI}{2\pi s}. The hint would lead me to believe that I can take NI -> M.

Not quite. NI would be the TOTAL current in the toroid. You've probably figured out the bound surface current is K_b=M. So knowing you have this bound surface currentm you can solve for the total current and equate that NI.

 

[buhthestuh]

Ok it seems then that the field should be

B = \frac{\mu_0 M}{L} - \frac{2\sqrt{2}\mu_0 Mw}{\pi a}<br />

or

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

However since L >> a and 2\sqrt{2}w &gt;&gt; 1 the field would be negative. But that seems proposterous considering that w is just a minute width compared to the whole of the toroid.

 

[Galileo]

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 K_b(2\pi s) so B=\mu_0M.
The rest is all ok.