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I'd like to ask for some clarifications on the formulas for the force on a loop with magnetic dipole moment immersed in a non-uniform magnetic field, given by Griffiths in Introduction to Electrodynamics (4th ed, page 268 for reference).

First, he analyzes the peculiar case of a simple circular loop of radius $R$ with current $I$ suspended above a finite, small solenoid in what he calls the fringing region. In this configuration, the magnetic field lines exiting the upper surface (basically a cylinder with a ring above it, both aligned on the same vertical axis, which is not specified) go through the circular loop. He then says that $\mathbf{B}$ has a radial component and so the loop experiences a net downward force equal to: $$ F = 2\pi R I B \cos(\theta) \tag{1}.$$ Now is there a way to obtain this formula rigorously?

I tried doing it but didn't go much far. Thanks to the figure the author adds, I assumed that for every $d\mathbf{l}'$, the magnetic field exiting the loop at that particular point (which follows a curved trajectory) can be approximated, wlog, to a straight line forming a certain angle with the horizontal axis, which we can call $\alpha$. Now since the loop is perfectly horizontal, we can assume this angle to be the same for every $d\mathbf{l}'$. So basically we have that $$d\mathbf{F} = I \,d\mathbf{l}' \times \mathbf{B},\quad \mathbf{B} = B_z\hat{\mathbf{z}} + B_r \mathbf{\hat{r}},\quad d\mathbf{l}' = \hat{\mathbf{\phi}}\,dl'.$$ The cross product yields $$d\mathbf{F} = \hat{\mathbf{r}}IB{_z}\,dl' -\mathbf{\hat{z}}IB_r \,dl'.$$ Now for what concerns the radial component of the force, we can find a symmetric infinitesimal segment $d\mathbf{l}_{{\rm opposite}}'$, so that the radial component of the force can be ignored since it vanishes upon integration. The downward component is equal for every $d\mathbf{l}'$ so at the end one just integrates along the circular path and obtains $(1)$. I couldn't translate these reasonings into anything concrete, unfortunately, so I'd like someone to help me obtain Griffith's formula.

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"Net force" means the total force on the entire loop. From a simple physics perspective the force created by the even magnetic field acting on the current in the wire on one side is exactly balanced by the force in the opposite direction created by the current in the wire on the other side.

enter image description here

(Apologies for getting the force in the wrong direction here. ⊕ indicates current flowing towards the viewer.)

When the magnetic field has a radial component (a component which is not perpendicular to the magnet) then you get a net force which is upward.

enter image description here

The equation is nothing but the standard formula for the force on a current-carrying wire in a magnetic field.

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  • $\begingroup$ Everything you said is ok, that's exactly what Griffiths does by inserting the figure, namely an intuitive explanation of why the net force acting on the loop tries to "pull" it towards the solenoid. Unfortunately, this answer actually does not help me much, since I'm trying to find a way to formalize the problem and get the result by applying the well-known formula which involves the contour integral around the loop. I can't, therefore, approve your answer. I appreciate your efforts nevertheless $\endgroup$ Aug 14, 2023 at 7:57
  • $\begingroup$ It's an absolutely trivial calculation from Coulomb's law though. $\endgroup$ Aug 14, 2023 at 8:06
  • $\begingroup$ How is it possible? Coloumbs law does not involves contour integrals,could you formalize it a bit more $\endgroup$ Aug 14, 2023 at 8:47
  • $\begingroup$ Sorry not Coulomb's law, I got the name confused. It is the usual force of a magnetic field on a wire carrying a current. $\endgroup$ Aug 14, 2023 at 8:56
  • $\begingroup$ Yes I thought so. I think my approach was right I just didn't finish the computation: we have infinitesimal downwards force component, equal for every infinitesimal segment of the loop, which upon integration, leaves us with something of the sort $$|\mathbf{F}|=IB_r \oint dl=2I \pi RB\cos\theta, B_r=B\cos\theta$$ $\endgroup$ Aug 14, 2023 at 9:08

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