D.3.3—Crossed electric and magnetic fields

Separate the two forces

With perpendicular uniform electric and magnetic fields, a charged particle can experience an electric force $F_E=qE$ parallel to the electric field and a magnetic force $F_B=qvB$ perpendicular to both velocity and magnetic field. For the correct geometry, these forces can point in opposite directions.

Find the undeflected speed

For the geometry in which the two forces oppose, a particle travels straight when their magnitudes are equal. Charge magnitude and mass do not determine this selected speed.

Worked example — crossed-field selector

For $v=5.9\times10^6\,\mathrm{m\,s^{-1}}$ and $B=42\,\mathrm{mT}=0.042\,\mathrm{T}$, the balancing field is $E=vB=(5.9\times10^6)(0.042)=2.5\times10^5\,\mathrm{V\,m^{-1}}$. Across plates $0.10\,\mathrm{m}$ apart, $V=Ed=2.5\times10^4\,\mathrm{V}$.

Check the geometry

The cancellation condition depends on velocity being perpendicular to both fields and on the electric and magnetic force directions being opposite. If the particle is deflected, compare the vector forces rather than applying E/B blindly.

Common trap

Do not use the electric-field direction alone to predict the path, and do not insert the particle’s mass into v=E/B. This is a force-balance condition.