20.3 Force on a moving charge
- Syllabus
- 9702–2028–2029
- Topic
- 20.3
- Level
- A2
Learning objectives
20.3.1The direction of the force on a charge moving in a magnetic field• determine the direction of the force on a charge moving in a magnetic field20.3.2F = BQv sin θ• recall and use F = BQv sin θ20.3.3The origin of the Hall voltage and derive and use the expression V H = BI /• understand the origin of the Hall voltage and derive and use the expression V H = BI / (ntq), where t = thickness20.3.4The use of a Hall probe to measure magnetic flux density• understand the use of a Hall probe to measure magnetic flux density20.3.5The motion of a charged particle moving in a uniform magnetic field• describe the motion of a charged particle moving in a uniform magnetic field perpendicular to the direction of motion of the particle20.3.6How electric and magnetic fields can be used in velocity selection• explain how electric and magnetic fields can be used in velocity selection
Determine magnetic-force direction on a moving positive or negative charge
Magnetic force on a moving charge is perpendicular to both its velocity v and magnetic field B. It is also perpendicular to the plane containing v and B.
| Step | Direction action |
|---|---|
| 1 | identify B and the particle velocity v |
| 2 | for a positive charge, treat conventional current as pointing along v |
| 3 | use Fleming's left hand: First finger B, seCond finger v/current, thuMb force |
| 4 | for a negative charge, reverse the force found for a positive charge |
On diagrams, × means into the page and • means out of the page. Reversing v, B or charge sign reverses force; reversing any two leaves force direction unchanged.
Because force is perpendicular to velocity, it does no work on an isolated particle: speed and kinetic energy stay constant while direction changes.
Do not put a negative charge's velocity directly into the conventional-current finger and stop there. First find the positive-charge force, then reverse it for q<0.
Calculate magnetic-force magnitude with F=B|Q|v sinθ
forcemagnitudeF=B∣Q∣vsinθ=B∣Q∣vperpendicular
| Symbol | Meaning |
|---|---|
| B | magnetic flux density in T |
| Q | |
| v | particle speed in m s⁻¹ |
| θ | angle between v and B |
Force is maximum, B|Q|v, for perpendicular motion. It is zero for v parallel/antiparallel to B and for a stationary particle.
A proton moving at 2.0×10⁶ m s⁻¹ perpendicular to B=0.30 T experiences F=(0.30)(1.60×10⁻¹⁹)(2.0×10⁶)=9.6×10⁻¹⁴ N.
Use |Q| for force magnitude and determine direction separately from charge sign. θ is between v and B; magnetic force itself is perpendicular to both.
Derive Hall voltage from magnetic deflection and charge separation
Charge carriers drifting through a conductor in a magnetic field are deflected sideways. Opposite charges build on the two side faces, creating a transverse Hall electric field and Hall voltage. Separation stops growing when electric and magnetic forces balance.
Let current I flow along the conductor, B be perpendicular to the broad face, width across the Hall contacts be w, and thickness parallel to B be t. Carrier number density is n, carrier charge magnitude q and drift speed v.
forcebalance:qEH=qvB,soEH=vBHallvoltage:VH=EHw=vBwcurrent:I=nqAv=nq(wt)v,sov=I/(nqwt)ThereforeVH=BI/(ntq).
| Symbol | Meaning |
|---|---|
| n | number density of mobile charge carriers, m⁻³ |
| t | conductor/probe thickness parallel to B |
| q | magnitude of one carrier's charge |
For B=4.0×10⁻⁶ T, I=5.4 A, n=1.5×10¹⁶ m⁻³, t=1.8×10⁻³ m and q=1.60×10⁻¹⁹ C, V_H=BI/(ntq)=5.0 V.
A semiconductor such as silicon has much smaller n than copper, so it produces a larger, easier-to-measure Hall voltage for the same B,I,t and q.
Hall voltage is transverse, not the ordinary voltage drop along current. Its polarity reverses if B, I or carrier sign reverses; the displayed formula gives magnitude when q is a magnitude.
A calibrated Hall probe measures the magnetic field component normal to its face
A Hall probe carries a fixed known current. Its transverse Hall voltage is proportional to the magnetic flux density component perpendicular to the active probe face, so calibration converts voltage to B.
| Probe orientation | Hall reading |
|---|---|
| active plane perpendicular to B | maximum magnitude |
| active plane parallel to B | zero |
| rotate through 180° from maximum | same magnitude, opposite sign |
| Measurement step | Action |
|---|---|
| 1 | zero the probe away from the field or remove offset |
| 2 | keep probe current and calibration range fixed |
| 3 | rotate to maximum magnitude to align the face normal with B |
| 4 | convert Hall voltage using sensitivity; sign gives field direction |
With sensitivity 20 mV T⁻¹, a maximum reading of +6.0 mV gives B=6.0/20=+0.30 T along the calibrated positive normal.
A zero reading may mean the probe is parallel to the field, not that no field exists. A Hall probe infers B from carrier deflection; it does not directly measure magnetic force.
A charge moving perpendicular to uniform B follows a constant-speed circle
When v is perpendicular to a uniform magnetic field, magnetic force has constant magnitude and is always perpendicular to v. It acts as centripetal force, continuously changing direction but not speed, so the path is circular.
B∣Q∣v=mv2/rThereforer=mv/(B∣Q∣).
T=2πr/v=2πm/(B∣Q∣):forafixedparticleandB,periodisindependentofspeedandradius.
| Change with others fixed | Circular path |
|---|---|
| larger momentum mv | larger radius |
| larger B or | Q |
| reverse charge sign or B | opposite curvature |
| double v | radius doubles, period unchanged |
If the field occupies only part of space, the particle follows a circular arc inside and then continues in a straight line tangent to the arc after leaving the field.
Magnetic force does not speed the particle up. The circular result requires v perpendicular to B; a parallel velocity component persists and produces a helical path instead.
Crossed electric and magnetic fields select particles with v=E/B
In a velocity selector, electric and magnetic forces oppose. Particles pass undeflected when qE=qvB, so v=E/B.
This condition assumes perpendicular, uniform fields and the correct orientation; faster or slower particles deflect.
E=2.0×10⁴ N C⁻¹ and B=0.50 T select v=4.0×10⁴ m s⁻¹.
The selector does not select charge sign or mass directly; it selects speed, with curvature after the selector used for further analysis.