# Force on an electric dipole moving through a magnetic field

A dipole of moment $\vec{p}$ where p is fixed, moves with velocity $\vec{v}$ though a magnetic field $\vec{B}$. Show that the force on the dipole is $\vec{v}\times(\vec{p}\cdot\vec{\nabla})\vec{B}+\dot{\vec{p}}\times\vec{B}$.

I need some help getting started. I know $\vec{F}=(\vec{p}\cdot\vec{\nabla})\vec{E}~$ but I can't immediately see how this would lead to an expression involving $\vec{B}$. I could try and use $\vec{F}=e(\vec{E}+\vec{v}\times\vec{B})$ but this doesn't really apply to an electric dipole.

## 1 Answer

The Lorentz Force does apply to an electric dipole.

An electric dipole is two charges of charge $\pm q$ separated by a distance $d=p/q.$ So start with the force $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$ if there is no magnetic field you just get $\vec{F}=q\vec{E}$ so you get two forces, one is $(-q)\vec E(\vec r,t)$ and the other is $q\vec E(\vec r+\vec p/q).$

So the total force is $q\left(\vec E(\vec r+\vec p/q)-\vec E(\vec r,t)\right)$ which equals $q d\frac{\left(\vec E(\vec r+\vec p/q)-\vec E(\vec r,t)\right)}{d}$ which is $p\frac{\left(\vec E(\vec r+\vec p/q)-\vec E(\vec r,t)\right)}{d}$ and in the limit as $d$ goes to zero but $q$ goes to infinity in such a way that $p,$ and $\vec p$ are constant you get $p\left(\hat d\cdot \vec \nabla\right)\vec E(\vec r,t)$ and this equals $p\left(\hat p\cdot \vec \nabla\right)\vec E(\vec r,t)$ or $\left(\vec p\cdot \vec \nabla\right)\vec E(\vec r,t).$ So even that formula is just the Lorentz Force.

So now we have to add a magnetic field instead of an electric field and we need to include that the dipole's motion since the force depends on the motion of the parts. If the dipole has two charges one with velocity $\vec v_1$ and the other with velocity $\vec v_2$ we can start by looking at the force on the average velocity $\vec v=(\vec v_1+\vec v_2)/2.$ This is actually the correct total force if the parts of the dipole don't move relative to each other.

So $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$ becomes $\vec{F}=q(\vec{v}\times\vec{B})$ so we get So the total force is $q\vec v\times\left(\vec B(\vec r+\vec p/q)-\vec B(\vec r,t)\right)$ which equals $q d\vec v\times\frac{\left(\vec B(\vec r+\vec p/q)-\vec B(\vec r,t)\right)}{d}$ which is $p\vec v\times\frac{\left(\vec B(\vec r+\vec p/q)-\vec B(\vec r,t)\right)}{d}$ and in the limit as $d$ goes to zero but $q$ goes to infinity in such a way that $p,$ and $\vec p$ are constant you get $p\vec v\times\left(\hat d\cdot \vec \nabla\right)\vec B(\vec r,t)$ and this equals $p\vec v\times\left(\hat p\cdot \vec \nabla\right)\vec B(\vec r,t)$ or $\vec v\times\left(\vec p\cdot \vec \nabla\right)\vec B(\vec r,t).$ So that term is just the magnetic force on the dipole due to the average velocity.

So each particle has a peculiar velocity. But we also get that the magnitude $p$ doesn't change. This means that $\dot{\vec p}$ is orthogonal to $\vec p$ (because $0=\frac{\mathrm d}{\mathrm d t}(\vec p\cdot \vec p)=2\vec p\cdot\dot{\vec p}$).

So what needs to be done? You can relate the peculiar motion $\vec v_1-\vec v$ and $\vec v_2-\vec v$ to $\dot{\vec p}$ and then compute the force due to that velocity and add it to the other force to get the total force.