D.3 Motion in electromagnetic fields

Read the trajectory

A particle initially moving perpendicular to a uniform electric field has constant velocity in the direction perpendicular to the field and constant acceleration along the field, so its path is parabolic. A particle initially at rest accelerates along a field line.

Solve the motion

Find the force and acceleration direction first, then use constant-acceleration equations. The time in the field comes from motion along the entry direction; the transverse displacement comes from the electric acceleration.

Explain and measure the circular path

For velocity perpendicular to a uniform magnetic field, the force is always perpendicular to the velocity and supplies the centripetal force. The direction changes continuously while speed and kinetic energy remain constant.

Worked example — charge-to-mass ratio

A particle beam with $v=2.5\times10^7\,\mathrm{m\,s^{-1}}$ follows a circle of radius $r=7.8\,\mathrm{cm}=0.078\,\mathrm{m}$ in $B=1.8\,\mathrm{mT}=1.8\times10^{-3}\,\mathrm{T}$. Then $|q|/m=v/(Br)=1.8\times10^{11}\,\mathrm{C\,kg^{-1}}$. The path direction is needed separately to determine the sign of the charge.

Use the full motion picture

A velocity component parallel to the field is unchanged, while the perpendicular component produces circular motion. Together they can form a helical path. Magnetic force does no work because it is perpendicular to instantaneous velocity, so kinetic energy stays constant.

Find the undeflected speed

For the geometry in which the two forces oppose, a particle travels straight when their magnitudes are equal. Charge magnitude and mass do not determine this selected speed.

Worked example — crossed-field selector

For $v=5.9\times10^6\,\mathrm{m\,s^{-1}}$ and $B=42\,\mathrm{mT}=0.042\,\mathrm{T}$, the balancing field is $E=vB=(5.9\times10^6)(0.042)=2.5\times10^5\,\mathrm{V\,m^{-1}}$. Across plates $0.10\,\mathrm{m}$ apart, $V=Ed=2.5\times10^4\,\mathrm{V}$.

Check the geometry

The cancellation condition depends on velocity being perpendicular to both fields and on the electric and magnetic force directions being opposite. If the particle is deflected, compare the vector forces rather than applying E/B blindly.

Worked example — oblique proton motion

For $|q|=1.60\times10^{-19}\,\mathrm{C}$, $v=3.4\times10^5\,\mathrm{m\,s^{-1}}$, $B=5.3\times10^{-3}\,\mathrm{T}$ and $\theta=32^\circ$, $F=|q|vB\sin\theta=1.5\times10^{-16}\,\mathrm{N}$. Only the velocity component perpendicular to the field contributes.

Find the direction

The magnetic force is perpendicular to both velocity and field. Use the right-hand rule for a positive charge; reverse the result for a negative charge. This force bends the path but does no work.

Connect to circular motion

For perpendicular motion, set $F=|q|vB$ equal to $mv^2/r$ to obtain $r=mv/(|q|B)$. Increasing $|q|$ or $B$ reduces the radius; increasing $m$ or $v$ increases it.

Worked example — measuring a field

A perpendicular wire carries $I=1.64\,\mathrm{A}$ through $L=8.13\,\mathrm{cm}=0.0813\,\mathrm{m}$ and experiences $F=4.12\times10^{-4}\,\mathrm{N}$. Thus $B=F/(IL)=(4.12\times10^{-4})/(1.64\times0.0813)=3.09\times10^{-3}\,\mathrm{T}$.

Read the angle limits

The force is zero when the wire is parallel to the field and maximum when it is perpendicular. Use conventional current direction for the force rule, not the electron drift direction.

Connect the models

The conductor formula is the combined magnetic force on moving charge carriers: $I=q/t$ and $L=vt$ lead from $F=qvB\sin\theta$ to $F=BIL\sin\theta$.

Worked example — opposite currents

For $I_1=3.7\,\mathrm{A}$, $I_2=1.6\,\mathrm{A}$ and $r=12\,\mathrm{cm}=0.12\,\mathrm{m}$, $F/L=\mu_0I_1I_2/(2\pi r)=9.9\times10^{-6}\,\mathrm{N\,m^{-1}}$. Opposite current directions mean the wires repel; each force has the same magnitude and opposite direction.

Read attraction and repulsion

Parallel currents in the same direction attract; currents in opposite directions repel. Each wire experiences an equal-magnitude force in the opposite direction. The result follows because each wire lies in the magnetic field produced by the other.

Check the scaling

The force per unit length increases with either current and decreases inversely with separation. Keep $F/L$ units as $\mathrm{N\,m^{-1}}$, equivalently $\mathrm{kg\,s^{-2}}$.