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Crossed-field questions look intimidating: a charge, an electric field, a magnetic field, all at right angles. Underneath is one balance of forces. This post explains that balance, works through a proton beam with numbers, and shows the direction check that catches most errors.
Is the velocity selector in the syllabus?
The CBSE 2026-27 curriculum and the JEE (Main) 2026 syllabus both list "force on a moving charge in uniform magnetic and electric fields". The current NCERT Class 12 Chapter 4, Moving Charges and Magnetism gives the Lorentz force, F = q(E + v × B), but the rationalised text does not have a separate section on the velocity selector. Treat it as an application of the Lorentz force that the syllabus wording covers, and check your year's syllabus.
How does a velocity selector work?
Send a positive charge q in the +x direction with speed v. Put a magnetic field B along +z. The magnetic force is qv × B, and x̂ × ẑ = −ŷ, so it pushes the charge in −y with size qvB. Now add an electric field E along +y. It pushes the charge in +y with size qE.
The charge goes straight if the two forces cancel:
qE = qvB, so v = E/B.
The charge cancels out, and mass never appears, so the selected speed is the same for protons, electrons, alpha particles or ions. Flip the sign of the charge and both forces reverse, so they still cancel.
Worked example 1: selecting protons
A selector has E = 3.0 × 10⁴ V/m and B = 0.15 T, with plates 1.0 cm apart.
Selected speed: v = E/B = 3.0 × 10⁴/0.15 = 2.0 × 10⁵ m/s.
Voltage across the plates: V = Ed = 3.0 × 10⁴ × 0.010 = 300 V.
Now a proton enters at 3.0 × 10⁵ m/s, faster than the selected speed.
Magnetic force: qvB = 1.6 × 10⁻¹⁹ × 3.0 × 10⁵ × 0.15 = 7.2 × 10⁻¹⁵ N.
Electric force: qE = 1.6 × 10⁻¹⁹ × 3.0 × 10⁴ = 4.8 × 10⁻¹⁵ N.
Net: 2.4 × 10⁻¹⁵ N in the magnetic force's direction (−y).
A faster particle bends towards the magnetic force; a slower one bends towards the electric force. Either way, it hits the plates or misses the exit slit.
Worked example 2: what happens after the selector?
The selected protons leave the plates and enter a region with only a magnetic field, also 0.15 T, perpendicular to their motion. They move in a circle of radius r = mv/(qB):
r = (1.67 × 10⁻²⁷ × 2.0 × 10⁵)/(1.6 × 10⁻¹⁹ × 0.15) ≈ 1.39 × 10⁻² m, about 1.4 cm.
A particle with the same charge and about twice the mass, such as a deuteron, comes through the same selector at the same speed and follows a circle about twice as large, roughly 2.8 cm. Because the selector fixed v, the radius now depends only on m/q. That is the idea behind a mass spectrometer. For circular motion in a field on its own, see our guide to cyclotron and charged-particle motion.
Where did the idea come from?
In his 1897 paper Cathode Rays, J. J. Thomson deflected the rays with an electric field and with a magnetic field. He adjusted the magnetic field until its deflection matched the electric deflection, which gave the speed of the rays as a ratio of the two fields, v = F/H in his notation. He also found that m/e was the same whichever gas was in the tube. Our profile of J. J. Thomson and the electron tells that story.
Which mistakes cost marks?
Thinking the selector depends on charge or mass. It does not; only E and B matter.
Getting the cross product wrong. Our post on right-hand-rule mistakes gives one reliable method.
Assuming the speed changes inside. For a particle that goes straight, neither force does work. A deflected particle does change speed, because the electric force then does work.
Mixing units. E in V/m and B in tesla give v in m/s. Convert kV/cm before dividing.
A practice routine
Draw the axes and mark v, B, qv × B and qE for a positive and a negative charge.
Do three selector problems: find v, then the net force on a faster and a slower particle.
Chain two questions: selector speed, then radius in a magnetic field alone.
Explain in one sentence why an electron and a proton at the same speed both pass.
When a crossed-field set-up stalls you, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: redo example 1 for an electron entering at 1.0 × 10⁵ m/s and decide which plate it hits.
