Force on Moving Charges and Velocity Selector

A current in a wire is really a flow of charged particles. So the force F = BIL on a current-carrying conductor must ultimately be the sum of forces on the individual moving charges inside the wire. In this lesson we make that connection explicit, derive the force on a single moving charge, and explore two classic applications: circular motion in a magnetic field and the velocity selector (an OCR-specific experimental setup that is a favourite for long-answer questions).

This lesson continues OCR H556 Module 6.3.

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Force on a Single Moving Charge

Consider a wire of cross-sectional area A and length L containing n charge carriers per cubic metre, each of charge Q, drifting with velocity v. The current is

I = nAvQ

and from the previous lesson the force on the whole wire (perpendicular field) is

F_wire = BIL = B(nAvQ)L = (nAL)BQv

But nAL is the total number of charge carriers in the wire. So the force per charge is

F = BQv (for velocity perpendicular to field)

If the velocity makes an angle θ with the field, the formula is

F = BQv sin θ

exactly as for wires. The force:

• Is perpendicular to both B and v.
• Is zero if v is parallel to B (sin θ = 0).
• Is maximum when v is perpendicular to B.
• Does no work on the charge, because F · v = 0 at all times.

The last point is crucial: magnetic forces never change the kinetic energy of a charged particle. They can change its direction but not its speed.

Direction — Fleming's Left-Hand Rule Again

For a positive charge moving in direction v, use Fleming's left-hand rule with "current" = direction of motion. For a negative charge (e.g. electron), the force is in the opposite direction, because the equivalent conventional current is opposite to the electron velocity.

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Worked Example — Force on an Electron

An electron travels at 1.5 × 10⁶ m s⁻¹ perpendicular to a magnetic field of 0.20 T. Calculate the force on the electron.

F = BQv
  = 0.20 × 1.6 × 10⁻¹⁹ × 1.5 × 10⁶
  = 4.8 × 10⁻¹⁴ N

Tiny in absolute terms, but enormous per unit mass: a = F/m_e = 4.8 × 10⁻¹⁴ / 9.11 × 10⁻³¹ ≈ 5.3 × 10¹⁶ m s⁻², or about 5 × 10¹⁵ g.

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Circular Motion in a Magnetic Field

Because the magnetic force on a moving charge is always perpendicular to the velocity, it acts exactly like a centripetal force: the particle moves in a circle at constant speed (if the field is uniform and the particle enters perpendicular to it).

Setting magnetic force equal to centripetal force:

BQv = mv² / r

and rearranging,

r = mv / (BQ)

The radius of the circular path is proportional to the momentum mv and inversely proportional to BQ. This is the principle behind:

• Cyclotrons and other particle accelerators.
• Mass spectrometers (heavier particles trace out larger arcs for a given speed).
• Magnetic bottles in fusion research.
• The beautiful curving tracks in bubble chambers and cloud chambers.

The period of the circular motion is

T = 2πr / v = 2π(mv/BQ) / v = 2πm / (BQ)

Notably, T does not depend on v. All particles of the same mass and charge in the same field take the same time to complete a circle. This property is what made the cyclotron possible — the oscillating accelerating voltage can be kept at a constant frequency.

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Worked Example — Radius of an Electron Path

An electron enters a uniform magnetic field of flux density 4.0 mT perpendicular to the field, with speed 6.0 × 10⁶ m s⁻¹. Calculate the radius of the circular path and the period.

Radius:

r = mv / BQ
  = (9.11 × 10⁻³¹ × 6.0 × 10⁶) / (4.0 × 10⁻³ × 1.6 × 10⁻¹⁹)
  = 5.466 × 10⁻²⁴ / 6.4 × 10⁻²²
  = 8.54 × 10⁻³ m
  ≈ 8.5 mm

Period:

T = 2πm / BQ
  = 2π × 9.11 × 10⁻³¹ / (4.0 × 10⁻³ × 1.6 × 10⁻¹⁹)
  = 5.724 × 10⁻³⁰ / 6.4 × 10⁻²²
  = 8.94 × 10⁻⁹ s
  ≈ 8.9 ns

The electron completes a circle of radius 8.5 mm in about 9 nanoseconds — a frequency of about 112 MHz (in the FM radio band).

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The Velocity Selector — OCR-Specific

A velocity selector is a beautifully simple device which uses crossed electric and magnetic fields to filter out particles of a specific speed, irrespective of their charge or mass. OCR H556 specifically requires candidates to understand how it works.