Charged Particles

01Charged Particles

Force on a moving charge

This section covers the magnitude and direction of the magnetic force on positive and negative charges moving perpendicular to a magnetic field.

Magnetic force on a charged particle

A charged particle in motion produces a magnetic field. If it travels through another magnetic field, the interaction can produce a force on the particle.

When the particle velocity is perpendicular to the magnetic field, the force magnitude is:

F=BQvF=BQv
Symbol Quantity Unit
FF magnetic force N
BB magnetic flux density T
QQ magnitude of the charge C
vv speed of the particle m s1^{-1}

The perpendicular arrangement gives the maximum magnetic force. If the particle travels parallel to the magnetic field, no magnetic force acts on it.

When a force does act, it is perpendicular to both the velocity and the magnetic field. The sign of the charge determines which of the two possible force directions applies.

Positive and negative charges

Fleming's left-hand rule can be applied by relating the moving charge to conventional current:

  • For a positive particle, conventional current points in the same direction as its motion.
  • For a negative particle, conventional current points opposite to its motion.

Consequently, positive and negative particles moving in the same direction through the same magnetic field are deflected in opposite directions.

£££££££££££££££££££££££££££££££££££Binto page+vF¡vF

Here both particles move to the right through a field directed into the page. The positive particle is deflected upwards, whereas the negative particle is deflected downwards.

Worked example: An electron moves at 5.3×107 ms15.3\times10^7\ \mathrm{m\,s^{-1}} perpendicular to a uniform magnetic field with flux density 0.20 T0.20\ \mathrm T. Find the magnitude of the magnetic force.

The magnitude of the electron charge is:

Q=1.60×1019 CQ=1.60\times10^{-19}\ \mathrm C

Use:

F=BQvF=BQv

F=(0.20)(1.60×1019)(5.3×107)F=(0.20)(1.60\times10^{-19})(5.3\times10^7)

F=1.696×1012 NF=1.696\times10^{-12}\ \mathrm N

Therefore:

F1.7×1012 N\boxed{F\approx1.7\times10^{-12}\ \mathrm N}

Exam Tip: Choose the force equation from the object involved. Use F=BQvF=BQv for an individual moving charge; F=BILF=BIL applies to a current-carrying wire.

Create a free account to continue

Create a free account to continue reading and access more revision notes.