# Thide's treatment of magnetostatics

by muppet
Tags: magnetostatics, thide, treatment
 Sci Advisor PF Gold P: 2,080 Thide's treatment of magnetostatics 1) Thide assumes that each loop is a rigid object, so his force is a single value that acts upon the rigid body. You could ignore the (x) part of F(x), except that he uses it in the next equation below. 2) Hmm, I wonder if my draft version of his book downloaded years ago is different than the current version. I see that Eq. (1.15) is div B = 0. I can't access your link because my firewall at work blocks foreign sites, so I'll assume you are referring to Eq. (1.14) which is $$\mathbf{F}(\mathbf{x}) = J \oint_C d\mathbf{l}\times\mathbf{B}(\mathbf{x}) .$$Thide states earlier that he is considering small loops, so B from loop x' is assumed to be constant over all of loop x and vice versa.
 P: 597 Ah I see. The equations I'm talking about: $$\mathbf{F}^{ms}(\mathbf{x})=\frac{\mu_0 I I'}{4 \pi} \oint_{C} d\mathbf{l} \times \oint_{C'} \mathbf{l'} \times \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3}$$(1.11) $$d \mathbf{B}^{stat}(\mathbf{x}) \equiv \frac{\mathbf{F}^{ms}(\mathbf{x})}{I} =\frac{\mu_0}{4 \pi} d\mathbf{i}'(\mathbf{x'}) \times \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3}$$ (1.15) where $d\mathbf{i}(\mathbf{x}')=I d\mathbf{l}' (\mathbf{x'})$ I'm afraid I still don't understand your answer to 1). Suppose "small" means "small compared to the characteristic scale over which the B-field varies" (which Thide couldn't say at this point). That doesn't make a difference, because the tangent vector to the loop still turns through a complete revolution (!), so no matter how small it is, the corresponding force still pushes different elements in different directions, right?
 Sci Advisor PF Gold P: 2,080 I see your point that Thide is being imprecise (or downright sloppy). To continue with Thide, we have to make an effort to see what he's doing, sloppy though it is, and try to correct it. He's really working backwards from [a global integrated value of F between two rigid loops] to [the infinitesimal contributions dF to that integrated value]. dF, in turn, depends on the dI in loop 1 and on an integral over the currents in loop 2. (Now that I can see the equations, it's clear that we don't need to assume that B is constant across loop 1--it's given at each point by B(x).) Now the confusing part--B(x) is local to a segment of the loop, but F(x) is global to the entire loop. So he really should have said F, without the x dependence. $$\mathbf{F}^{ms}=\frac{\mu_0 I I'}{4 \pi} \oint_{C} d\mathbf{l} \times \oint_{C'} d\mathbf{l'} \times \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3}$$BTW, I don't know what the superscript ms refers to... Then the next equation probably should be in terms of a differential dF(x) $$d \mathbf{B}^{stat}(\mathbf{x}) \equiv \frac{d\mathbf{F}^{ms}(\mathbf{x})}{I} =\frac{\mu_0}{4 \pi} d\mathbf{i}'(\mathbf{x'}) \times \frac{\mathbf{x}-\mathbf{x}'}{|\mathbf{x}-\mathbf{x}'|^3}$$ The expression is consistent this way because now dF(x) is position-dependent around loop 1. It still properly integrates to the force F on the whole loop. You might consider changing books--Griffiths is used widely and is very popular, and there are many other excellent texts that I and others here can recommend.