Lorentz Force on a Moving Charge

A charged particle moving through a magnetic field feels a force perpendicular to both its velocity and the field. This is the Lorentz force:

$\vec F = q\,\vec v \times \vec B$

The cross product means the force is always perpendicular to the velocity — so it can change the direction of the particle but not its speed (and therefore not its kinetic energy).

For a uniform magnetic field, the result is circular motion at a frequency that depends only on the field and the particle's charge-to-mass ratio.

Step-by-Step Solution

Given: A proton ($q = +1.6 \times 10^{-19}\;\text{C}$, $m = 1.67 \times 10^{-27}\;\text{kg}$) moving at $v = 10^{6}\;\text{m/s}$ perpendicular to a magnetic field $B = 0.01\;\text{T}$ (100 gauss).

Find: The Lorentz force, the radius of the circular path, and the period.

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Step 1 — Magnitude of the Lorentz force.

For perpendicular $\vec v$ and $\vec B$:

$F = qvB$

$= (1.6 \times 10^{-19})(10^{6})(0.01)$

$= 1.6 \times 10^{-15}\;\text{N}$

A truly tiny force — but acting on a tiny mass.

Step 2 — Compute the centripetal acceleration.

$a = \dfrac{F}{m} = \dfrac{1.6 \times 10^{-15}}{1.67 \times 10^{-27}} \approx 9.58 \times 10^{11}\;\text{m/s}^{2}$

That's about $10^{11}$ g — a number so large it's hard to even visualize. But it makes sense: protons in cyclotrons routinely undergo accelerations of this magnitude.

Step 3 — Radius of the circular orbit.

The Lorentz force is the centripetal force. Setting $qvB = mv^{2}/r$:

$r = \dfrac{mv}{qB} = \dfrac{(1.67 \times 10^{-27})(10^{6})}{(1.6 \times 10^{-19})(0.01)}$

$= \dfrac{1.67 \times 10^{-21}}{1.6 \times 10^{-21}}$

$\approx 1.044\;\text{m}$

So the proton orbits in a circle a bit over 1 meter in radius. That's why early cyclotrons were built quite large.

Step 4 — Cyclotron period.

$T = \dfrac{2\pi m}{qB} = \dfrac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.01)}$

$= \dfrac{1.05 \times 10^{-26}}{1.6 \times 10^{-21}}$

$\approx 6.55 \times 10^{-6}\;\text{s} = 6.55\;\text{μs}$

The proton completes one orbit every 6.55 microseconds, or about 152{,}000 revolutions per second.

Step 5 — Notice $T$ is independent of speed.

The period $T = 2\pi m/(qB)$ depends only on mass, charge, and magnetic field — not on the particle's speed. This is the key insight behind the cyclotron: faster protons orbit in larger circles, but they do it at the same period, so a single oscillating accelerating field can keep boosting them until they reach high energies.

Step 6 — Right-hand rule for direction.

For a positive charge moving in $+x$ with $\vec B$ in $+z$, the force is in $+y$ (point fingers along $v$, curl toward $B$, thumb shows $\vec F$). For a negative charge, flip the direction.

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Answer: A proton at $10^{6}\;\text{m/s}$ in a 0.01 T field experiences a Lorentz force of $1.6 \times 10^{-15}\;\text{N}$, which curves it into a circular orbit of radius 1.044 m with period 6.55 μs (≈ 152 kHz). The cyclotron period is independent of the particle's speed.

Try It

• Watch the charge spiral as the magnetic field deflects it.
• The force vector (green) is always perpendicular to the velocity (cyan).
• Adjust the field strength $B$ — the radius shrinks proportionally.
• Adjust the speed — the radius grows proportionally, but the period stays the same (cyclotron miracle).