Difficulty: Intermediate | Prerequisites: Vectors, cross products, Newton's second law, circular motion from PHY 211.
This section introduces how magnetic fields interact with moving charges. Unlike electric forces, magnetic forces depend on the velocity of the charge and are always perpendicular to both the velocity and the field. This perpendicularity means magnetic forces change direction but never speed, which leads directly to circular motion. Understanding the Lorentz force and the radius of curvature in a uniform B field is essential for everything from mass spectrometers to particle accelerators, and it forms the basis for the force on current-carrying wires covered in the next topic.
A magnetic field exerts a force on a moving charge given by F = qv × B. The force is always perpendicular to the velocity, so it changes direction but not speed. In a uniform magnetic field, a charged particle moves in a circle with radius R = mv/(qB).
Magnetic field (B)
A vector field produced by moving charges (electric currents) that exerts forces on other moving charges. Units: Tesla (T).
Lorentz force
The force on a charged particle moving through a magnetic field: F = qv × B. In simple terms, it is the "sideways push" a magnetic field gives to a moving charge.
Cross product (A × B)
A vector operation that produces a vector perpendicular to both inputs. The magnitude is |A||B|sinθ, where θ is the angle between A and B. Direction is given by the right-hand rule.
Right-hand rule
Point your fingers in the direction of the first vector (v), curl them toward the second vector (B), and your thumb points in the direction of v × B. If the charge is negative, the force is opposite to your thumb.
Cyclotron radius (R)
The radius of the circular path a charged particle follows in a uniform magnetic field: R = mv/(qB). Also called the Larmor radius or radius of gyration.
Momentum (p)
p = mv. Appears in the cyclotron radius formula as R = p/(qB), which is useful when momentum is given directly.
Magnetic fields are created by electric currents (moving charges).
Magnetic fields exert forces on electric currents (charges in motion).
A stationary charge in a magnetic field experiences zero magnetic force.
A × B is proportional to the component of B that is perpendicular to A.
Magnitude: |A × B| = |A||B|sinθ
Direction: determined by the right-hand rule.
A × B = −(B × A): order matters.
If A is parallel to B (θ = 0° or 180°), the cross product is zero.
F = qv × B
The force is perpendicular to both v and B.
For a positive charge, use the right-hand rule directly.
For a negative charge, the force is opposite to the right-hand rule result.
Because F is always perpendicular to v, the magnetic force does no work. It can change the direction of motion but never the speed (kinetic energy stays constant).
When a charged particle enters a uniform B field with velocity perpendicular to B:
The force is always perpendicular to the velocity, so the particle moves in a circle.
Speed does not change (only direction changes).
Setting up the equation:
The magnetic force provides the centripetal force: qvB = mv²/R
Solving for the radius: R = mv/(qB)
Using momentum: R = p/(qB)
Larger mass or higher speed → bigger circle.
Stronger field or larger charge → smaller (tighter) circle.
A particle of charge q and mass m is accelerated from rest through an electric field E over a distance d, then enters a region of uniform magnetic field B.
Find the velocity after acceleration:
Work-energy theorem: the work done by the electric field equals the kinetic energy gained.
W = qEd = ½mv²
v = √(2qEd/m)
Find the radius in the magnetic field:
R = mv/(qB) = (m/qB)√(2qEd/m) = (1/B)√(2mEd/q)
This type of two-stage problem (electric acceleration followed by magnetic deflection) is a classic exam setup.
Quantity | Formula |
|---|---|
Lorentz force | F = qv × B |
Magnitude of F | F = qvBsinθ |
Cross product magnitude | |A × B| = |A||B|sinθ |
Cyclotron radius | R = mv/(qB) = p/(qB) |
Kinetic energy from E field | ½mv² = qEd |
Velocity after E-field acceleration | v = √(2qEd/m) |
Mass spectrometers use exactly this principle: ions are accelerated through a voltage, then deflected by a magnetic field. The radius of curvature depends on the mass-to-charge ratio, so different isotopes trace different paths and can be separated. Cyclotrons and synchrotrons in particle physics accelerate charged particles using magnetic fields to bend them in circles.
Students often forget that the magnetic force on a stationary charge is zero. If v = 0, then F = q(0)× B = 0. You must be moving to feel the magnetic force.
The magnetic force does no work. Students sometimes try to use it in work-energy calculations. If the speed does not change, no work is done, and the magnetic force is the reason.
The right-hand rule gives the direction of v × B, not the force on any charge. For a negative charge, you must flip the direction.
When the velocity is parallel to B, the force is zero (sin0° = 0). The particle drifts in a straight line through the field. Only the perpendicular component of velocity produces circular motion.
⚠️ The two-stage problem (E-field acceleration, then B-field deflection) is extremely common on exams. Know how to connect the two stages through the velocity.
⚠️ Be careful with signs: positive charges curve one way, negative charges curve the other.
⚠️ Expect conceptual questions about why the magnetic force does no work, and what happens to speed vs. direction.
⚠️ Left-handed students: the right-hand rule is about your right hand, not your dominant hand. Practise it.
1. True or false: A magnetic force can change a particle's speed.
A: False. It can only change direction.
2. Fill in the blank: The Lorentz force on a charge q moving with velocity v in a field B is F = ______.
A: qv × B.
3. True or false: If a charged particle moves parallel to a magnetic field, it experiences no magnetic force.
A: True. sinθ = sin0° = 0.
4. Fill in the blank: Doubling the magnetic field strength while keeping everything else constant causes the cyclotron radius to ______.
A: Halve (R = mv/(qB), so doubling B halves R).
Q: A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves at 3 × 10⁶ m/s perpendicular to a 0.5 T magnetic field. What is the radius of its circular path?
A: R = mv/(qB) = (1.67 × 10⁻²⁷)(3 × 10⁶) / [(1.6 × 10⁻¹⁹)(0.5)] = 5.01 × 10⁻²¹ / 8 × 10⁻²⁰ = 0.0626 m ≈ 6.3 cm.
Q: An electron and a proton enter the same magnetic field with the same speed. Which has the larger radius of curvature?
A: The proton. R = mv/(qB), and both have the same charge magnitude and speed, but the proton has a much larger mass, so its radius is larger.
Q: A particle is accelerated from rest through a potential difference V, then enters a magnetic field B. Derive an expression for the radius of its circular path in terms of m, q, V, and B.
A: From energy conservation: qV = ½mv², so v = √(2qV/m). Then R = mv/(qB) = m√(2qV/m)/(qB) = √(2mV/q)/B = (1/B)√(2mV/q).
The Lorentz force on a single charge scales up to the force on a current-carrying wire (next topic), since a current is just many charges moving together. The circular motion here connects to the concept of cyclotron frequency, relevant in later electromagnetic wave discussions. The cross product appears again in torque on current loops, the Biot-Savart law, and throughout electromagnetism.
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