Derivation: force on magnetic moment in magnetic field

[robinegberts]

I am trying to derive the equation $\vec{\Gamma} = \vec{m} \times \vec{B}$, where $\vec{m} = I \vec{A}$ is the magnetic moment, and $\vec{A}$ is normal to surface $A$, from the Lorentz force law $\mathrm{d}\vec{F} = I \vec{dl} \times \vec{B}$. For an arbitrary closed current loop $C$ enclosing the area $A$, I have tried

$$ \vec{\Gamma} = \oint_C{\mathrm{d}\vec{\Gamma}} = \oint_C{\vec{r} \times \mathrm{d}\vec{F}} = \oint_C{\vec{r} \times I (\vec{dl} \times \vec{B})} = I \oint_C{\vec{r} \times (\vec{dl} \times \vec{B})}.$$

Now I know that $\vec{A} = \frac{1}{2} \oint{\vec{r} \times \mathrm{d}\vec{l}}$, but I don't know how to complete the proof. I have also tried to consider a square loop giving rise to the magnetic moment $\vec{m}$ to derive the equation, but this looks even more complicated.

 

[Charles Link]

It looks like it should be a simple result from vector identities, but I think this one might be non-trivial. The vector triple product is non-associative, so $B$ doesn't simply factor out. An old E&M textbook that I have, Intermediate Electromagnetic Theory by Schwarz, does their torque derivation by assuming that the magnetic field points in the x-direction, and then writes out the torque integrals for each torque component in terms of x, y, and z. $\\$ @vanhees71 Might you have a simple trick for this one?

 

[robinegberts]

It looks like it should be a simple result from vector identities, but I think this one might be non-trivial. The vector triple product is non-associative, so $B$ doesn't simply factor out. An old E&M textbook that I have, Intermediate Electromagnetic Theory by Schwarz, does their torque derivation by assuming that the magnetic field points in the x-direction, and then writes out the torque integrals for each torque component in terms of x, y, and z. $\\$ @vanhees71 Might you have a simple trick for this one?

Right, I agree completely. I forgot about my undergraduate electrodynamics book, we used Griffiths hehe. In Griffiths, the derivation is done for a rectangular loop which is sloped with respect to the magnetic field. However, problem 2 from the chapter 'Magnetic Fields in Matter' covers this question exactly I see now, and gives the complete derivation based solely on vector calculus. I will write it out here for those interested.

 

[robinegberts]

So here it goes. First, some required vector identities:

The Jabobi identity:
$$\vec{r} \times (d\vec{l} \times \vec{B}) + d\vec{l} \times (\vec{B} \times \vec{r}) + \vec{B} \times (\vec{r} \times d\vec{l}) = \vec{0},$$

and the differential:

$$d[\vec{r} \times (\vec{r} \times \vec{B})] = d\vec{r} \times (\vec{r} \times \vec{B}) + \vec{r} \times (d\vec{r} \times B),$$
since $\vec{B}$ is constant. Since $d\vec{r} = d\vec{l}$, this may be written as

$$d[\vec{r} \times (\vec{r} \times \vec{B})] = d\vec{l} \times (\vec{r} \times \vec{B}) + \vec{r} \times (d\vec{l} \times B).$$

Combining the Jacobian identity with the differential we have

\begin{align}
\vec{r} \times (d\vec{l} \times \vec{B}) &= d[\vec{r} \times (\vec{r} \times \vec{B})] - d\vec{l} \times (\vec{r} \times \vec{B}) \\ &= d[\vec{r} \times (\vec{r} \times \vec{B})] + d\vec{l} \times (\vec{B} \times \vec{r}) \\ &= d[\vec{r} \times (\vec{r} \times \vec{B})] - \vec{r} \times (d\vec{l} \times \vec{B}) - \vec{B} \times (\vec{r} \times d\vec{l}),
\end{align}
and therefore

$$2\vec{r} \times (d\vec{l} \times \vec{B)} = d[\vec{r} \times (\vec{r} \times \vec{B})] - \vec{B} \times (\vec{r} \times d\vec{l}).$$

This gives

\begin{align}
\vec{\Gamma} &= \oint{d\vec{\Gamma}} \\ &= \oint{\vec{r} \times d \vec{F}} \\ &= I \oint{\vec{r} \times (d\vec{l} \times \vec{B})} \\ &= \frac{I}{2} \oint{d[\vec{r} \times (\vec{r} \times \vec{B})]} - \frac{I}{2} \oint{\vec{B} \times (\vec{r} \times d\vec{l})} \\ &= \vec{0} -I \vec{B} \times \frac{1}{2} \oint{\vec{r} \times d\vec{l}} \\ &= -I \vec{B} \times \vec{A} \\ &= I \vec{A} \times \vec{B} \\ &= \vec{m} \times \vec{B}.
\end{align}