The Lorentz Force — Charges in Magnetic Fields
The Lorentz Force
When a charged particle moves through a magnetic field, it experiences a force perpendicular to both its velocity and the field. This is the magnetic part of the Lorentz force, and it is responsible for some of the most important phenomena in physics and technology: the operation of electric motors, the bending of particle beams in accelerators, the trapping of charged particles in the Van Allen belts, and the aurora borealis.
The Lorentz force law states that the force on a charge q moving with velocity v in a magnetic field B is F equals q times the cross product of v and B. The magnitude is F equals q v B sine theta, where theta is the angle between the velocity and field vectors. Several key properties follow immediately. The force is zero when the charge is stationary, because v equals zero. The force is maximum when the velocity is perpendicular to the field, because sine of 90 degrees equals one. And the force is always perpendicular to the velocity, which means it changes the direction of the particle without changing its speed — magnetic fields do no work.
Because the force is always perpendicular to the velocity, a charged particle moving perpendicular to a uniform magnetic field follows a circular path. The radius of this circle — called the cyclotron radius or Larmor radius — is r equals mv over qB. This relationship is the basis for mass spectrometers, cyclotrons, and the magnetic confinement of plasma in fusion reactors. The cyclotron frequency f equals qB over 2 pi m is independent of speed, a remarkable fact exploited in the design of particle accelerators.
In this lesson you will learn to apply the Lorentz force law, calculate cyclotron radii and frequencies, and understand the motion of charges in magnetic fields. The mathematics uses vectors, cross products, and circular motion — concepts that connect algebra, trigonometry, and physics in elegant ways.
The force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$:
$$\vec{F} = q\vec{v} \times \vec{B}$$
Magnitude: $F = qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.
Direction: given by the right-hand rule (curl fingers from $\vec{v}$ toward $\vec{B}$; thumb points along $\vec{F}$ for positive charges).
A charge moving perpendicular to a uniform field follows a circle of radius:
$$r = \frac{mv}{qB}$$
The cyclotron frequency (independent of speed!):
$$f = \frac{qB}{2\pi m}$$
A proton ($m = 1.67 \times 10^{-27}$ kg, $q = 1.6 \times 10^{-19}$ C) moves at $v = 3 \times 10^6$ m/s perpendicular to a $B = 0.5$ T field. Find the cyclotron radius and frequency.
Step 1: Apply the cyclotron radius formula.
$$r = \frac{mv}{qB} = \frac{(1.67 \times 10^{-27})(3 \times 10^6)}{(1.6 \times 10^{-19})(0.5)}$$
Step 2: Compute numerator and denominator.
Numerator: $5.01 \times 10^{-21}$ kg·m/s
Denominator: $8.0 \times 10^{-20}$ C·T
Step 3: Divide.
$$r = \frac{5.01 \times 10^{-21}}{8.0 \times 10^{-20}} = 0.0626 \text{ m} \approx 6.3 \text{ cm}$$
Step 4: Find the cyclotron frequency.
$$f = \frac{qB}{2\pi m} = \frac{(1.6 \times 10^{-19})(0.5)}{2\pi (1.67 \times 10^{-27})} = \frac{8.0 \times 10^{-20}}{1.049 \times 10^{-26}} \approx 7.63 \times 10^6 \text{ Hz} = 7.63 \text{ MHz}$$
An electron ($m = 9.11 \times 10^{-31}$ kg) enters a 0.2 T field at $v = 5 \times 10^7$ m/s at an angle of $30°$ to the field. What is the force on the electron?
Step 1: Use $F = qvB\sin\theta$.
$$F = (1.6 \times 10^{-19})(5 \times 10^7)(0.2)\sin(30°)$$
Step 2: $\sin(30°) = 0.5$.
$$F = (1.6 \times 10^{-19})(5 \times 10^7)(0.2)(0.5) = 8.0 \times 10^{-13} \text{ N}$$
Tiny by human standards, but for an electron this produces an acceleration of $a = F/m \approx 8.8 \times 10^{17}$ m/s², almost a trillion g's!
In a mass spectrometer, singly charged ions ($q = 1.6 \times 10^{-19}$ C) are accelerated to $v = 1 \times 10^5$ m/s and enter a 0.8 T field. Ion A has a radius of 12 cm and Ion B has a radius of 14 cm. Find the mass of each ion.
Step 1: Solve the cyclotron radius for mass: $m = \frac{qBr}{v}$
Step 2: Ion A: $m_A = \frac{(1.6 \times 10^{-19})(0.8)(0.12)}{1 \times 10^5} = 1.536 \times 10^{-25}$ kg
Step 3: Ion B: $m_B = \frac{(1.6 \times 10^{-19})(0.8)(0.14)}{1 \times 10^5} = 1.792 \times 10^{-25}$ kg
Converting to atomic mass units ($1$ u $= 1.66 \times 10^{-27}$ kg): $m_A \approx 92.5$ u, $m_B \approx 107.9$ u.
Practice Problems
Show Answer Key
1. $F = (1.6\times10^{-19})(2\times10^6)(1.0) = 3.2 \times 10^{-13}$ N
2. $r = (9.11\times10^{-31})(4\times10^7)/[(1.6\times10^{-19})(0.3)] = 7.6 \times 10^{-4}$ m $\approx 0.76$ mm
3. $\sin\theta = 0.5 \implies \theta = 30°$ or $150°$
4. $r = (1.67\times10^{-27})(10^7)/[(1.6\times10^{-19})(50\times10^{-6})] \approx 2088$ m $\approx 2.1$ km
5. The force is always perpendicular to velocity, so $W = \vec{F}\cdot d\vec{s} = 0$. It changes direction, not speed.
6. $r = (6.64\times10^{-27})(10^6)/[(3.2\times10^{-19})(2)] = 0.01038$ m $\approx 1.04$ cm
7. $r_1/r_2 = m_1/m_2$ (since $r \propto m$)
8. $f = (1.6\times10^{-19})(1)/[2\pi(9.11\times10^{-31})] \approx 2.80 \times 10^{10}$ Hz $= 28.0$ GHz