# What is the direction of an accelerated particle due to a variation of magnetic field?

Let an ion of charge $q$ and mass $m$ traverse a circular orbit of radius $R$ under the influence of a uniform magnetic field $\mathbf{B}=B(r)\,\hat{\boldsymbol{z}}$, where $r$ is the distance from the symmetry axis and $z$ the 'height' or 'vertical' axis (basically a betatron). I'm a bit confused, I've managed to show that when the magnetic field is increased in magnitude, the induced electric field is $$E=-\frac{R}{2}\frac{d\langle{B}\rangle}{dt}\hspace{0.5in}\text{where}\hspace{0.5in}\langle{B}\rangle=\frac{1}{\pi{R}^2}\int_0^R{B}(r)\,2\pi{r}\,dr$$ i.e. $\langle{B}\rangle$ is the spatial average of $B(r)$. And hence, that the electric field introduces a force $\mathbf{F}=q\mathbf{E}$ on the ion that accelerates it as $$\mathbf{a}=-\frac{qR}{2m}\frac{d\langle{B}\rangle}{dt}\,\hat{\boldsymbol{\theta}}$$ where $\theta$ is the azimuthal coordinate. The thing I'm having trouble with is the minus sign, because I know that in a betatron an ion accelerates (positively, not as 'desaccelerate') when the magnetic field is increased. I thought I'd find out what the direction of the velocity of the ion was when the field did not change, and found from Lorentz force that in general for any (positive or negative) charge $q$, $$\mathbf{F}=-qvB\,\hat{\mathbf{r}}\;\Longrightarrow\;\mathbf{v}=-\mathrm{sgn}(q)v\,\hat{\boldsymbol{\theta}}$$ Now I don't know if this last expression is true, I haven't found it anywhere and the sign function seems rare; honestly I just managed, after various attempts, to make the direction of both the velocity (at unchanging $\mathbf{B}$) and the acceleration (at changing $\mathbf{B}$) fit so that the ion would increase or decrease its speed upon an increase or a decrease, respectively, of magnetic field. I'd like to know whether this is true and/or if there's a simpler or more straightforward way of showing this. Thank you in advance.

• @userØØ7 You should use $\sgn(q)$ :-) – Martin Ueding Jun 3 '13 at 17:58

Faraday's law in integral form is: $$\oint\vec{E}\cdot d\vec{l}=-\frac{\partial}{\partial t}\left(\int_{S}\vec{B}\cdot d\vec{a}\right)$$
Referring to the diagram and keeping to the right-hand convention for closed-path line integrals, it seems your signs are correct. The acceleration is in the $\hat{\theta}$ direction and has the same sign as the velocity, just like you have written. 