9478 / 2027
Electromagnetic Forces overview

Lesson 9 of 9

Select a beam speed

  • I can explain velocity selection by mutually perpendicular electric and magnetic fields.Syllabus

    Syllabus 9478, 17(m). Explain velocity selection by mutually perpendicular electric and magnetic fields.

Make a guess. You are not marked.

In a velocity selector, ions of one speed pass straight through. A faster ion enters. What happens to it?

  1. It is deflected in the direction of the magnetic force, which grows with speed.
  2. It is deflected in the direction of the electric force, which grows with speed.
  3. It passes straight through, because the selector works at any speed.
Show the answer

It is deflected in the direction of the magnetic force, which grows with speed.

The electric force qE is the same at any speed, but the magnetic force Bqv grows with speed. For a faster ion the magnetic force wins, so it is deflected that way.

Electric and magnetic forces can oppose one another for particles travelling in a chosen direction. Only the matching incident speed makes their magnitudes equal and leaves the charged beam undeviated.

Let the incident velocity be right, E downward and B into the page. These directions are mutually perpendicular. A positive charge feels electric force down and magnetic force up. An electron has both force directions reversed: electric force up and magnetic force down.

Establish that the forces oppose before equating their magnitudes. For a nonzero charge travelling in the stated direction:

|Q|E = B|Q|v
v = E/B

The charge magnitude cancels, and mass does not enter this ideal force-balance condition. Neutral particles do not satisfy this nonzero-charge selection argument. Reversing the incident direction without changing either field makes the two forces reinforce rather than cancel.

Calculate the selected speed

A field E of a few thousand N/C divided by B of a few 10-4 T suggests a selected speed of order 107 m/s. Use the supplied E = 4000 N/C and B = 0.200 mT = 2.00 × 10-4 T:

v = 4000/(2.00 × 10-4)
= 2.00 × 107 m/s

This B is different from the 1.00 mT field in the magnetic semicircle example. For an electron with |Q| = 1.60 × 10-19 C at the selected speed, both opposing forces have magnitude 6.40 × 10-16 N.

The selected speed passes through both apertures

The selected speed passes through both aperturesAn electron beam enters right through two aligned apertures. The electric field is four thousand newtons per coulomb down and the magnetic field is 0.200 millitesla into the page. At twenty million metres per second, electric force is upward and magnetic force downward, both 6.40 times ten to the minus sixteen newton. Their sixty-four-unit arrows use the same force scale as the next two panels and cancel. The path is straight. Field and velocity arrows have separate schematic scales.E-eElectricMagneticCrosses: B into the pageAligned entrance and exit aperturesv = 2.00 × 10⁷ m/s

The opposing forces are each 6.40 x 10^-16 N, so this electron travels straight at 2.00 x 10^7 m/s. In all three panels, force-arrow length uses 10 drawing units per 10^-16 N; field and velocity arrows use separate scales.

Slower electron: initial force upward

Slower electron: initial force upwardOnly the initial forces are shown for an electron entering right at fifteen million metres per second. E is down and B into the page, unchanged from the selected-speed case. Electric force is 6.40 times ten to the minus sixteen newton upward, drawn sixty-four units long. Magnetic force is 4.80 times ten to the minus sixteen newton downward, drawn 48 units long. The initial resultant is upward with magnitude 1.60 times ten to the minus sixteen newton. There is no drawn trajectory or claim that the subsequent force direction remains fixed.Same E down and B into pageEB in-evElectricMagneticv = 1.50 × 10⁷ m/s

Initial electric force: 6.40 x 10^-16 N up. Initial magnetic force: 4.80 x 10^-16 N down. These arrows predict the initial deflection only; the selector does not set every incident speed to E/B.

Faster electron: initial force downward

Faster electron: initial force downwardOnly the initial forces are shown for an electron entering right at twenty-five million metres per second. E is down and B into the page, unchanged from the selected-speed case. Electric force is 6.40 times ten to the minus sixteen newton upward, drawn sixty-four units long. Magnetic force is 8.00 times ten to the minus sixteen newton downward, drawn 80 units long. The initial resultant is downward with magnitude 1.60 times ten to the minus sixteen newton. There is no drawn trajectory or claim that the subsequent force direction remains fixed.Same E down and B into pageEB in-evElectricMagneticv = 2.50 × 10⁷ m/s

Initial electric force: 6.40 x 10^-16 N up. Initial magnetic force: 8.00 x 10^-16 N down. These arrows predict the initial deflection only; the selector does not set every incident speed to E/B.

The selected electron beam has equal opposing forces and follows the straight aperture line. The slower and faster panels compare initial forces on the same local force scale; they are not calculated full paths through the combined fields.

The selected beam remains straight when gravity, collisions, fringing and interactions between beam particles are negligible. A selector does not automatically accelerate every incident particle to E/B; it identifies the speed for an undeviated path in this geometry.

Decide the initial deflection of other speeds

For the same electron charge and fields, electric force stays at 6.40 × 10-16 N upward. Magnetic force depends on incident speed:

  • At 1.50 × 107 m/s, magnetic force is 4.80 × 10-16 N down. The initial resultant is 1.60 × 10-16 N up.
  • At 2.50 × 107 m/s, magnetic force is 8.00 × 10-16 N down. The initial resultant is 1.60 × 10-16 N down.

Positive charges entering with these same velocities have the opposite initial force directions. These statements describe the initial resultants. Once a particle deflects, its velocity changes and so does its magnetic force; a complete off-speed trajectory is not generally a constant-acceleration parabola.

Check your understanding

Electrons enter right through E = 4000 N/C downward and B = 0.200 mT into the page. The selected speed is 2.00 x 107 m/s. What happens initially to an electron entering at 1.50 x 107 m/s?

  1. It initially deflects downward because the magnetic force is larger.
  2. It initially deflects upward because the electric force is larger.
  3. It travels straight because any perpendicular electric and magnetic fields cancel.
Show the answer

It initially deflects upward because the electric force is larger.

The slower electron has less downward magnetic force, while its upward electric force is unchanged. The initial resultant is 1.60 x 10-16 N upward. The device selects a speed rather than automatically bringing every particle to it.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

In a velocity selector, crossed electric and magnetic fields act on a beam of positive ions. The electric field strength is 3.0 x 104 V m-1 and the magnetic flux density is 0.015 T. Explain why only ions of one speed pass straight through, and calculate that speed. [3]

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