Magnetic Field Deflection¶
How to estimate the deflection of an electron beam due to a magnetic field¶
This problem came up recently at work, while integrating an Excillum E1 MetalJet. The question was: to what extent should effort be made to shield a beam of electrons from an external magnetic field?
As it turns out, this sort of problem is relatively easy to solve using only a couple elementary physical principles:
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charged particles in a magnetic field move in a circle
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charged particles accelerate when an external voltage is applied
We’ll first briefly review these two principles, and then finish by calculating the magnetic field needed to produce a significant deflection in beam of electrons.
Deflection in a magnetic field¶
As an electron passes through a magnetic field, it gets tugged in the direction perpendicular to both its motion and the magnetic field according to the Biot-Savard law, which is the magnetic equivalent of Coulomb’s law. This force produces exactly the centripetal force needed to produce circular (or sometimes helical) motion. Since this motion is periodic, it will have an angular frequency related to it, known as the “cyclotron frequency”.
The cyclotron angular frequency $\omega$ is given by
$$ \omega = \frac vr = \frac{qB}m $$
where $v$ is the speed, $r$ is the radius of curvature, $q$ is the electric charge, $m$ is the particle mass, and $B$ is the magnetic field. This angular frequency is related to the amount of time $T$ for a charged particle to traverse the entire circle according to $T = 2\pi / \omega$.
Additionally, a deflection by some amount $x$ over a distance $D$ corresponds to a radius of curvature
$$ r \simeq \frac{D^2}{2\,x} \,. $$
To understand why, see Figure 1 below.
With a little trigonometry, we can show that the radius of curvature $r$ is related to the deflection amount $x$ as follows:
$$ r = \frac{t}{\sin\theta} = \frac {2\,t^2}x = \frac{D^2 + x^2}{2x} \simeq \frac{D^2}{2x} \quad (D \gg x)\,. $$
So, the magnetic field $B$ needed to deflect an electron (with charge $e$ and mass $m_e$) by an amount $x$ over a distance $D$ is
$$ B = \frac{2\,m_e v x}{eD^2} \,. $$
This is helpful, but to go any further we need to establish an estimate for the velocity of the electrons, $v$.
Acceleration in an electric potential¶
To determine the electron's speed, note that an electron passing through a potential $V$ will experience an increase in kinetic energy
$$ \frac12 m_e v^2 = eV \quad \implies \quad v = \sqrt{\frac{2eV}{m_e}} \,. $$
This relationship allows us to trade knowledge of the accelerating voltage for knowledge of the velocity, which simplifies things a bit since the applied voltage is much more readily known than the velocity.
Putting it all together¶
We’re now ready to substitute these expressions into one another. Taking nominal values of $V$, $x$, $D$ to be 100 kV, 10 $\mu$m, 100 mm, respectively, we find
$$ B = 2.2 \,{\rm Gauss}\times \left(\frac{V}{100\,{\rm kV}}\right)^{1/2}\left(\frac x{10\,\mu{\rm m}}\right)\left(\frac{100 \,{\rm mm}}D\right)^2 \,. $$
This answer shows that, to avoid a deflection of 100 kV electrons by more than 10 microns over a distance of 100mm, we need to keep magnetic fields below about 2 Gauss. Since the typical refrigerator magnet produces a magnetic field with strength 50-100 Gauss, this is an important concern when designing hardware that operates near this type of X-ray source: since the liquid metal target is only about 10 microns wide, it’s possible to miss it entirely if the ambient magnetic fields are too strong.