AcademyMagnetic Fields and Forces
Academy
Charged Particles in Magnetic Fields
Level 1 - Physics topic page in Magnetic Fields and Forces.
Principle
A uniform magnetic field bends perpendicular charged-particle motion into circular motion.
Notation
\(m\)
particle mass
\(\mathrm{kg}\)
\(q\)
particle charge
\(\mathrm{C}\)
\(v_{\perp}\)
speed perpendicular to the field
\(\mathrm{m\,s^{-1}}\)
\(B\)
magnetic field magnitude
\(\mathrm{T}\)
\(r\)
circular-path radius
\(\mathrm{m}\)
\(T\)
period of circular motion
\(\mathrm{s}\)
Method
For velocity perpendicular to a uniform field, the magnetic force is always perpendicular to velocity. It supplies the radial force.
Magnetic force
\[F_B=|q|v_{\perp}B\]
Radial force
\[F_r=m\frac{v_{\perp}^2}{r}\]
Set forces equal
\[|q|v_{\perp}B=m\frac{v_{\perp}^2}{r}\]
Radius
\[r=\frac{mv_{\perp}}{|q|B}\]
The angular speed and period do not depend on speed for nonrelativistic circular motion in a uniform magnetic field.
Angular speed
\[\omega=\frac{v_{\perp}}{r}=\frac{|q|B}{m}\]
Period
\[T=\frac{2\pi}{\omega}=\frac{2\pi m}{|q|B}\]
Rules
Orbit radius
\[r=\frac{mv_{\perp}}{|q|B}\]
Angular speed
\[\omega=\frac{|q|B}{m}\]
Cyclotron period
\[T=\frac{2\pi m}{|q|B}\]
Magnetic work
\[W_B=0\]
Examples
Question
An electron with speed
\[2.0\times10^6\,\mathrm{m\,s^{-1}}\]
moves perpendicular to \[0.010\,\mathrm{T}\]
Find \(r\).Answer
\[r=\frac{mv}{eB}=\frac{(9.11\times10^{-31})(2.0\times10^6)}{(1.60\times10^{-19})(0.010)}=1.14\times10^{-3}\,\mathrm{m}\]
Checks
- Use only the speed component perpendicular to \(\vec B\).
- The magnetic field changes direction of velocity, not speed.
- Larger momentum gives a larger radius; larger charge or field gives a smaller radius.