# Force on infinitesimal loop as stated in Griffiths' Electrodynamics

I've found little to no success in finding a proof of result given by Griffiths at the start of the chapter on Magnetic fields in matter, namely the one for the force acting on an infinitesimal loop with dipole moment $$\mathbf{m}$$ in an external field $$\mathbf{B}$$: $$\mathbf{F} = \nabla(\mathbf{m} \cdot\mathbf{B}). \tag{1}$$ Is there a proof for this result? It does not seem to be an easily provable one, and I believe the proof does not rely much on intuition. Therefore, Griffiths' intuitive (and sometimes not rigorous) approach led him to omit the proof in the end. Anyways, If any of you managed to obtain it either with a more pragmatic or a more formal one, then I'd be grateful if you could post it. Again, whatever explanation is well-accepted.

• You know it doesn't really help to post a question with a nearly identical title to your last question, you should try to be at least a bit more descriptive in the title to differentiate them from one another. Aug 14, 2023 at 17:45

## 1 Answer

The formula is rigorously true for an ideal magnetic dipole and an arbitrary magnetic field (conversely, also for an arbitrary loop and linear magnetic field). If the loop is of finite size and the magnetic field has a varying derivative, you can decompose the big loop into an integral of ideal magnetic dipoles and integrate the force on every one of those dipoles (remarkably, it does not depend on the specific decomposition). This is why Griffiths' proofs (problems 6.4 and 6.22) involving Taylor expansions are valid.

A rigorous way would be to prove it using distributions and the general equations of magnetostatics. The current density of the ideal dipole is: $$j = \nabla\delta\times m$$ since the magnetisation is $$M = m\delta$$ and $$j=\nabla\times M$$.

The magnetic force is given by: $$F = \int j\times Bd^3x$$ Substituting the expression of current: \begin{align} F &= \int (\nabla\delta\times m)\times Bd^3x \\ &= \int [(\nabla \delta\cdot B)m-(m\cdot B)\nabla\delta]d^3x \\ &= \int [\nabla(m\cdot B)-(\nabla\cdot B)m]\delta d^3x \\ &= \nabla(m\cdot B) \end{align} where I expanded the double cross product, integrated by parts, used Maxwell's equation $$\nabla\cdot B = 0$$ and used the definition of the Dirac delta.

Btw, this expression suggests an alternative approach using the principle of virtual work, which is perhaps more intuitive. This approach first starts by calculating the energy of the loop in the external field: $$E = -m\cdot B$$ This can be obtained in a similar approach using the expression of energy: $$E = \int j\cdot Ad^3x$$ in the case of a current loop. Alternatively, the expression of the energy can also be directly obtained using Faraday's law using that $$m\cdot B$$ is $$I\Phi$$ with $$I$$ the current and $$\Phi$$ the magnetic flux.

Hope this helps.