6.3.2(c)Particles and Medical PhysicsCore

Velocity Selectors

Crossed electric and magnetic fields can select a particle speed. Balance the forces for an undeflected beam, then distinguish that speed selection from a later mass measurement.

40 min 6.3 Magnetic Fields

What you'll be able to do

Choose field directions for opposing forces
Derive and use v=E/B
Explain selection limits and magnetic analysis

Balance two transverse forces

Arrange velocity, electric field and magnetic field mutually perpendicular, with electric and magnetic forces in opposite directions for the intended beam. The electric-force magnitude is |q|E and the perpendicular magnetic-force magnitude is |q|vB. At one speed they cancel and the particle passes straight through.

Reversing charge reverses both forces. The condition for no deflection depends on their magnitude ratio, so the selected speed is independent of charge magnitude and mass. The fields themselves must still be oriented correctly: two forces in the same direction cannot cancel.

|q|E=|q|vB\quad\Rightarrow\quad v=\frac EB

Explain rejection, not just the equation

At a higher speed the magnetic-force magnitude exceeds the electric force; at a lower speed it is smaller. Those particles deflect in opposite senses and may be blocked by the exit slit. A finite-width slit and finite field length allow a range of nearby speeds rather than an infinitely precise single speed.

Use the particle charge and a stated magnetic-field direction to draw the force arrows. Do not assign “up” to one force without specifying the geometry. A speed selector does not speed every incoming particle up to E/B: it transmits particles already close to that speed.

Combine selection with a magnetic analyser

After the selector, a separate magnetic field can bend transmitted particles into circles. At known speed, radius r=mv/(|q|B) measures mass-to-charge ratio. Ions with the same speed and charge magnitude but different masses have different radii.

If electric and magnetic fields overlap, use both forces in that region. Applying the magnetic-only radius formula inside an undeflected selector is inconsistent because the net force there is zero. State where each equation applies in a multi-stage instrument.

Work through the reasoning

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.

Example 1A selector has E=

4.0\times10^4

V m⁻¹ and B=0.20 T. Find selected speed.

v=E/B

v=(4.0\times10^4)/0.20=2.0\times10^5

m s⁻¹.

Answer

2.0\times10^5

m s⁻¹.

Example 2Singly charged ions at that speed enter a separate 0.10 T field and follow radius 0.050 m. Estimate their mass.

m=|q|Br/v

m=(1.60\times10^{-19})(0.10)(0.050)/(2.0\times10^5)=4.0\times10^{-27}

kg.

Answer

4.0\times10^{-27}

kg for the stated charge state.

Equation recap

|q|E=|q|vB\quad\Rightarrow\quad v=\frac EB

Equation recap

Common mistakes to avoid

Adding force magnitudes when they point in opposite directions.

Draw vectors and take their signed resultant.

Calling E/B a mass measurement by itself.

It selects speed; another stage can determine mass-to-charge ratio.

Key takeaways

Opposing field forces select v=E/B.
The ideal selected speed is independent of mass and charge magnitude.
Magnetic analysis must be treated as a separate field region.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 23 lessons