Lorentz force and charged-particle motion

The Lorentz force describes the action of electric and magnetic fields on a point charge: $\vec F = q(\vec E + \vec v \times \vec B).$ The magnetic part is $\vec F_B = q\,\vec v \times \vec B$, and its magnitude is $F_B = |q|vB\sin\theta$, where $\theta$ is the angle between $\vec v$ and $\vec B$. For a positive charge the force direction is that of $\vec v \times \vec B$ (right-hand rule); for a negative charge it is opposite. In a uniform magnetic field, the velocity component perpendicular to $\vec B$ produces circular motion of radius $r = \frac{mv_\perp}{|q|B}$, while the parallel component is unchanged, so the general trajectory is a helix. The period $T = \frac{2\pi m}{|q|B}$ is independent of speed in the nonrelativistic limit. The magnetic force does no work, so the speed and kinetic energy stay constant; for superposed fields the total induction $\vec B = \vec B_1 + \vec B_2 + \ldots$ is used in the formula.

A proton ($q=+1.6\times10^{-19}$ C) moves east with speed $v=2.0\times10^5$ m/s in a uniform magnetic field $B=0.50$ T directed north. Find the magnitude and direction of the magnetic force.

An electron ($|q|=1.6\times10^{-19}$ C, $m=9.11\times10^{-31}$ kg) enters perpendicular to a uniform magnetic field $B=2.5$ mT with speed $v=4.0\times10^6$ m/s. Find the radius of the trajectory and the period.

A positive ion with charge $q$ and mass $m$ moves to the right through crossed fields: $\vec E$ is downward and $\vec B$ is into the page. Find the condition on $E$, $B$, and $v$ for the ion to move in a straight line, and calculate $v$ if $E=3.0\times10^4$ V/m and $B=0.20$ T. Which way are the electric and magnetic forces directed?

Use a safe online simulation of a charge moving in a magnetic field (for example, PhET “Charged Particle in a Magnetic Field” or an equivalent). Set a uniform $\vec B$, direct the initial velocity perpendicular to the field, and observe the circular trajectory. Vary the sign of the charge, the magnitude of $B$, and the speed $v$; record the radius and period. Compare the measurements with $r=mv/(|q|B)$ and $T=2\pi m/(|q|B)$. Then set the velocity at an angle to $\vec B$ and describe the helical path. Do not use real high-voltage sources or strong magnets; perform observations only in the simulation.

Show how the relative orientation of $\vec v$ and $\vec B$ determines the direction of the Lorentz force and the shape of a charged particle’s trajectory.

Diagram: vector $\vec v$ points to the right, $\vec B$ points into the page; for a positive charge $\vec F$ points upward by the right-hand rule, and the trajectory is a circle; for a negative charge the force and the sense of circulation are opposite. Beside it, the case where $\vec v$ makes an angle with $\vec B$ is shown: the trajectory is a helix.

A microwave oven heats food with radiation at $2.45\ \text{GHz}$. That radiation is produced by the magnetron, in which electrons ($m = 9.11\times10^{-31}\ \text{kg}$, $q = 1.6\times10^{-19}\ \text{C}$) move perpendicular to a magnetic field of induction $B = 0.088\ \text{T}$. The magnetic part of the Lorentz force bends them into a circle with period $T = \frac{2\pi m}{qB} = \frac{2\pi \cdot 9.11\times10^{-31}}{1.6\times10^{-19}\cdot 0.088} = 4.1\times10^{-10}\ \text{s},$ that is, with frequency $1/T = 2.45\ \text{GHz}$.

What acts here is the purely magnetic Lorentz force $\vec F = q\,\vec v\times\vec B$. It is always perpendicular to the velocity, so it changes only the direction of the velocity, not its magnitude: the electron goes round a circle and its energy neither grows nor falls. That is exactly why the period $T = 2\pi m/(qB)$ does not depend on the electron's speed — the magnetron works because this circulation frequency matches the frequency of its resonator.

Above the polar regions electrons plunge towards the atmosphere with a velocity component across the field lines of $v_\perp = 1.0\times10^{7}\ \text{m/s}$. Near Earth the field induction is $B = 5\times10^{-5}\ \text{T}$, and the perpendicular part of the velocity turns into a circle of radius $r = \frac{m v_\perp}{qB} = \frac{9.11\times10^{-31}\cdot 1.0\times10^{7}}{1.6\times10^{-19}\cdot 5\times10^{-5}} \approx 1.1\ \text{m}.$ The velocity component along $\vec B$ does not change, so the trajectory is a helix that runs down the field line as far as the air at $90\!-\!150\ \text{km}$ altitude.

This is the Lorentz force with no electric field: $\vec F = q\,\vec v\times\vec B$. Because the force is perpendicular to both $\vec v$ and $\vec B$, it does no work, and the speed $1.0\times10^{7}\ \text{m/s}$ is preserved — only the direction of motion turns. That gives the helix: circular motion in the plane across $\vec B$ plus steady motion along $\vec B$, so the aurora traces the same lines as Earth's magnetic field.

A sample of an athlete's urine or blood is ionised and accelerated by a voltage $U = 1000\ \text{V}$. A singly charged fragment with $q = 1.6\times10^{-19}\ \text{C}$ and $m = 3.0\times10^{-26}\ \text{kg}$ (about $18\ \text{u}$) leaves the accelerator with speed $v = \sqrt{2qU/m} \approx 1.0\times10^{5}\ \text{m/s}$ and enters a field $B = 0.5\ \text{T}$ at right angles. Its path is a circle of radius $r = \frac{mv}{qB} = \frac{3.0\times10^{-26}\cdot 1.0\times10^{5}}{1.6\times10^{-19}\cdot 0.5} \approx 3.8\ \text{cm}.$ A heavier ion is deflected less, so the detector registers each mass separately.

The work $qU$ done by the electric field sets the ion's speed, and then the magnetic part of the Lorentz force $\vec F = q\,\vec v\times\vec B$ merely turns that velocity without changing its magnitude. The radius turns out to be proportional to $m/q$, so one and the same instrument sends molecules of different mass to different points of the detector — that is how a banned substance is picked out among the thousands of other molecules in the sample.