A Lorentz force with both an electric and a magnetic part can cancel itself. Arrange $\vec{E}$ and $\vec{B}$ at right angles, and for one particular speed the two pushes are equal and opposite. That is the heart of two classic instruments: the velocity selector and the mass spectrometer, which goes on to bend the selected ions into circles.
Key idea
Velocity selector: with $\vec{E}\perp\vec{B}$, only charges with $$v = \frac{E}{B}$$ pass straight through, whatever their charge or mass.
Mass spectrometer: those charges then enter a field $B'$ alone and circle with radius $r = \dfrac{mv}{qB'}$, so $\dfrac{m}{q} = \dfrac{rB'}{v}$.
Mass spectrometer: those charges then enter a field $B'$ alone and circle with radius $r = \dfrac{mv}{qB'}$, so $\dfrac{m}{q} = \dfrac{rB'}{v}$.
Why it works
Straight through means no net force (Newton's first law): $qE = qvB$. The charge cancels from both sides, and the mass never appears, because nothing is accelerating. A faster particle feels a bigger magnetic force and curves one way; a slower one is pulled the other way by the electric force. Only $v = E/B$ survives the slit at the end. After the selector, every ion has the same speed, so the radius of its circle depends only on $m/q$.
Worked examples
Worked example: Choosing a speed
A selector has $E = 2.0 \times 10^{4}\ \text{V/m}$ and $B = 0.25\ \text{T}$. Which speed passes?
- $v = E/B = 2.0 \times 10^{4}/0.25 = 8.00 \times 10^{4}\ \text{m/s}$.
- Electrons, protons and heavy ions all pass at this same speed.
Answer: $v = 8.00 \times 10^{4}\ \text{m/s}$
Worked example: Separating isotopes
Singly charged neon-20 and neon-22 ions ($1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$) leave that selector and enter a $0.50\ \text{T}$ field, completing half circles. How far apart do they land?
- $r_{20} = \dfrac{(20)(1.66 \times 10^{-27})(8.00 \times 10^{4})}{(1.60 \times 10^{-19})(0.50)} = 3.32\ \text{cm}$ and likewise $r_{22} = 3.65\ \text{cm}$.
- Each lands one diameter from the entrance, so the separation is $2(r_{22} - r_{20}) = 6.64\ \text{mm}$.
Answer: about $6.64\ \text{mm}$
Common mistake
"The selected speed depends on the particle's charge or mass." Write the balance: $qE = qvB$. The $q$ divides out, and $m$ is not there at all. A selector set for protons passes electrons of the same speed too (both forces reverse together for negative charge). Charge and mass matter only afterward, in the spectrometer's curve, $r = mv/(qB)$.
In 1897 J. J. Thomson used crossed electric and magnetic fields to measure the charge-to-mass ratio of the particles in cathode rays, the first measurement of a property of the electron.