Accelerating an electron in a betatron

Electricity and Magnetism Level 5

In a betatron, electrons move in circular orbits and are accelerated by a time dependent magnetic field, perpendicular to the plane of the orbits. Suppose that the flux through an orbit of radius $r=25 \textrm{cm}$ grows during the acceleration time at a constant rate of $\frac{d\Phi}{dt}=5 \textrm{Wb/s}$ . In the acceleration process an electron in this orbit, acquires a kinetic energy $\Delta{E_{k}}=25 \textrm{MeV}$ . Find the distance covered in meters by the electron during the acceleration time. Keep in mind that that the radius of the electron's orbit does not change.

Details and assumptions

$e= 1.6 \times 10^{-19} C$ $1\textrm{eV}= 1.6 \times 10^{-19} \textrm{J}$

The answer is 7.85E+6.

1 solution

David Mattingly Staff
May 13, 2014

From Faraday's law, we know that a change in flux is always accompanied by an emf. This emf is what accelerates the electrons in a betatron. We have that $\mathcal{E}= 2\pi r E =- \frac{d\Phi}{dt}$ where $r$ is the radius of the orbit and E is the electric field generated by the varying magnetic field. Note that the electric field is tangential to the electron's orbit. The electric force acting on the electron is $F_{e}= e E= \frac{e}{2\pi r} \frac{d\Phi}{dt} .$ The work done by the electric force is equal to the change in kinetic energy $W=F_{e} s=\Delta E_{k} .$ Thus we find that the distance traveled by the particle is equal to $s= \frac{ 2 \pi r \Delta E_{k}}{ e \frac{d\Phi}{dt}}=7.85\times 10^{3} \textrm{k m}=7.85 \times 10^{6} \textrm{m}.$

Accelerating an electron in a betatron

The answer is 7.85E+6.

1 solution

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