.

A Betatron is pictured above (with field out of the page). It works by increasing a non-uniform magnetic field $\vec{B}(r)$ in such a way that electrons of charge $e$ and mass $m$ inside the ``doughnut'' tube are accelerated by the $E$-field produced by induction (via Faraday's law) from the ``average'' time-dependent magnetic field $B_1(t)$ inside $a$, while the magnitude of the magnetic field at the radius $a$, $B_2(t) = \vert\vec{B}(a,t)\vert$, bends those same electrons around in the circle of (constant) radius $a$.

This problem solves, in simple steps, for the ``betatron condition'' which relates $B_1(t)$ to $B_2(t)$ such that both things can simultaneously be true.

a) The electrons go around in circles of radius $a$ and are accelerated by an $\vec{E}$ field produced by Faraday's law. We will define the (magnitude of the) average field $B_1$ by $\phi_m = B_1(\pi a^2) = \int_{(r<a)} \vec{B}(r)\cdot \hat{n} dA$. What is the induced $E$ field (tangent to the circle) in terms of $B_1$ and $a$?

(Problem continued on next page!)

b) The electrons (at their instantaneous speed $v$ tangent to the circle) are bent into the circle of radius $a$ by the field $B_2$. Relate $B_2$ to the magnitude of the momentum $p = mv$, the charge $e$ of the electron, and the radius $a$.

c) The force $\vec{F}$ from the $E$-field acting on the electron with charge $e$ in the direction of its motion is equal to the time rate of change of the magnitude of its momentum $p$ (if Newton did not live in vain). Substitute, cancel stuff, and solve for $\frac{dB_1}{dt}$ in terms of $\frac{dB_2}{dt}$. If you did things right, the units will make sense and the relationship will only involve dimensionless numbers, not $e$ or $m$.

Cool! You've just figured out how to build one of the world's cheapest electron accelerators! Or perhaps not....