01 · Intermediate → advanced

Lorentz force

Define the operational electric and magnetic force on a moving point charge, derive the cross-product geometry, prove that a static magnetic field does no mechanical work, solve constant-B circular/helical motion and carry the result to the force on a current element. Distinguish a force law for given fields from source laws for computing those fields.

220–280 minIntermediate → advanced
In this lesson
  • State F=q(E+v×B) in SI units and distinguish the source field from the force on a test particle.
  • Determine cross-product direction and the effect of reversing charge, velocity or B independently.
  • Derive the magnetic-force magnitude |q|vB sinθ and show its vanishing for v parallel B.
  • Prove Fmag·v=0 and state when that does or does not imply conserved kinetic energy.
  • Derive uniform-B cyclotron angular frequency, radius, period and helical pitch in the nonrelativistic regime.
  • State the sign of the gyration sense without replacing vectors by unsigned scalars.
  • Derive magnetic force on a thin current element dF=I dℓ×B and its assumptions from moving carrier charges.
  • Distinguish static uniform B kinematics from acceleration due to electric fields and from collisions/material interactions.
  • Explain why a force law does not by itself determine a magnetic field generated by a current; introduce the later role of Biot–Savart and Ampère.
  • Solve vector and numeric force/trajectory exercises with direction, units, limiting cases and regime checks.
Force law versus source law

Electrostatics supplied a force qE on a test charge. A magnetic field contributes a velocity-dependent force perpendicular to the particle velocity. We first use a prescribed B to derive trajectories and conductor forces; the next lessons ask what stationary currents actually produce B. These are distinct questions.

Lorentz: signed q v×B directionA positive charge +q moves right along positive x in a magnetic field pointing into the page along negative z, shown by crosses. Its magnetic Lorentz force points up along positive y. A negative charge with the same velocity experiences the opposite magnetic force.+B : −z ⊗
FB : +y ↑
v : +x →
FB = q v×B
q > 0
Lorentz: signed q v×B direction

Coordinates: +x RIGHT, +y UP, +z OUT OF PAGE. For v=+v x̂ and B=−B ẑ, v×B=+vB ŷ; therefore q>0 gives FB=+qvB ŷ. A negative charge reverses FB. The magnetic force is perpendicular to v, and this diagram assumes an imposed B, not a specific current source.

01.1The Lorentz force and the unit tesla

For a classical point charge q with instantaneous velocity v in prescribed fields E(r,t) and B(r,t), the electromagnetic force is the empirical Lorentz force law:

F=q(E+v×B)=Felectric+Fmagnetic
Fmagnetic=qv×B
[B]=T=N·s/(C·m)=N/(A·m)

Here v is the particle's velocity in the observer's chosen inertial reference frame, and E/B are fields evaluated at its instantaneous position. The magnetic part vanishes for a stationary charge. The formula is a FORCE law given E and B, not a method to calculate B from sources. Field transformations between moving reference frames require special relativity and are not obtained by simply switching signs of a fixed B.

01.2Derive direction and magnitude from the cross product

If θ is the angle between v and B, then the magnitude of their cross product is vB sinθ; the sign of q changes the direction but not the magnitude:

|Fmagnetic|=|q|vB|sinθ|
v∥B ⇒ Fmagnetic=0;   v⊥B ⇒ |Fmagnetic|=|q|vB

For B=+B ẑ and v=+v x̂, x̂×ẑ=−ŷ: a POSITIVE charge initially bends toward −y while a negative one bends toward +y. Reversing B or reversing v independently reverses the magnetic force; reversing two of q, v, B returns its original direction. Never use a scalar force magnitude to decide trajectory orientation.

01.3Prove that the magnetic contribution does no mechanical work

Mechanical power delivered to a point charge is P=F·v. A cross product is orthogonal to both factors, hence

Pmagnetic=q(v×B)·v=0
dK/dt=F·v=qE·v   (electromagnetic force only)

For E=0 and no other forces, K stays constant: B can curve the motion but cannot change the speed. This instantaneous orthogonality remains true even if B itself varies with time; however a changing magnetic field may induce E, and E can do work. Collisions, radiation reaction and external forces can change particle kinetic energy, so the zero magnetic-power identity alone does not guarantee conservation in a more complete system.

01.4Derive circular motion for v perpendicular to uniform B

Take B=B ẑ constant in space and time, E=0, no collisions and v·B=0 initially. Since the force is always perpendicular to v with magnitude |q|vB, it supplies centripetal acceleration at constant speed. Newton's second law gives

mv²/r=|q|vB ⇒ r=mv/(|q|B)
ωc=v/r=|q|B/m
Tc=2π/ωc=2πm/(|q|B);   fc=|q|B/(2πm)

The positive scalar ωc does not encode circulation direction. For the above +ẑ field, a positive charge starting toward +x bends toward −y and rotates clockwise as seen from +z; a negative one rotates the opposite way. If v=0 or B=0 there is no circular orbit to which the radius formula can be applied.

01.5Helix: decompose the initial velocity

For arbitrary initial v write v=v∥+v⊥, with v∥=(v·b̂)b̂ and b̂=B/|B|. The parallel component experiences no Lorentz magnetic force and stays constant for the stated uniform static field; the perpendicular component rotates. The resulting trajectory is a helix:

rL=mv⊥/(|q|B)
phelix=v∥Tc=2πmv∥/(|q|B)

Here the pitch is a SIGNED axial displacement if v∥ is signed along the chosen b̂; its geometric length is |phelix|. An initially parallel velocity has rL=0 and travels straight. A time-dependent or spatially nonuniform B, electric acceleration, and collisions generally invalidate the simple constant-radius/constant-pitch helix.

01.6Worked particle estimate and regime check

For an electron with mass m≈9.11×10⁻³¹ kg, charge magnitude e≈1.60×10⁻¹⁹ C, B=0.100 T, v⊥=1.00×10⁶ m/s and v∥=2.00×10⁵ m/s,

rL≈5.69×10−5 m
fc≈2.80×10⁹ Hz;   Tc≈3.57×10−10 s
|phelix|≈7.14×10−5 m

The total speed ≈1.02×10⁶ m/s is about 0.0034c, consistent with the nonrelativistic approximation. The electron's circulation direction is OPPOSITE that of a positive charge with the same initial velocity and B. Frequencies and radii are metrics of the chosen particle/field, not intrinsic fixed properties of an electron alone.

01.7Derive force on a filamentary current element from moving carriers

Take a thin stationary wire segment of vector length dℓ oriented along conventional current. Suppose n carriers per volume with individual charge q, drift velocity vd, and cross-section A, with a locally uniform applied B. A segment of length dℓ contains nA dℓ carriers and their total magnetic force is

dF=(nA dℓ)q(vd×B)
I=nqA vd,along ⇒ dF=I dℓ×B
F=I∫wiredℓ×B

For electron carriers q<0, their drift direction is opposite conventional current, so qvd still points along the current direction. The magnetic carrier force is transmitted to the conductor lattice in steady state. The thin-wire approximation assumes a specified local current direction and field; for a spatially distributed current, the local force density is f=J×B.

01.8Worked wire-force example and current orientation

A straight wire segment of length L=0.050 m carries I=2.00 A in +x̂ through a uniform B=0.300 T in +ẑ. Because x̂×ẑ=−ŷ,

F=ILB(−ŷ)=(2.00)(0.050)(0.300)(−ŷ)
F=−3.00×10−2 ŷ N

If the wire becomes parallel to B, the force vanishes. Reversing current reverses the force. The result does not say that the entire wire accelerates freely: clamps and support forces can balance the magnetic force.

01.9Keep the source law separate from the force law

The Lorentz law predicts motion or mechanical force once E and B are KNOWN. It says nothing by itself about the field produced by a specified wire shape or current density. Biot–Savart will compute B for steady currents; Ampère's law will express its circulation and becomes a practical field calculator when sufficient source symmetry exists. Time-varying systems eventually require Maxwell's equations including displacement current.

given E,B ⇒ F=q(E+v×B)
given stationary J ⇒ B   (source problem, next chapters)

01.10Mastery exercises with solutions

  1. A positive charge q=+1.00 nC moves with v=+1000 x̂ m/s in B=+0.200 ẑ T, E=0. Compute its magnetic force vector and the force for negative q.
  2. Prove algebraically that Fmag·v=0 for arbitrary θ. Can a time-dependent magnetic field nevertheless be accompanied by acceleration of particle speed? Explain.
  3. A proton has v⊥=2.00×10⁵ m/s and B=0.500 T. Estimate its radius and cyclotron frequency using mp≈1.67×10⁻²⁷ kg and e≈1.60×10⁻¹⁹ C.
  4. For a particle with v∥=3.00×10⁵ m/s, B=0.100 T, mass and charge magnitude equal to an electron, find its helical pitch.
  5. A straight wire segment carries I=2.00 A along +x̂, length 0.050 m, in uniform B=0.300 ẑ T. Find F with sign and units; what happens on current reversal?
  6. Can the Lorentz force on a test charge alone give the magnetic field of a nonuniform current distribution? Name the additional information/law needed.
Worked solutions and checks

1. x̂×ẑ=−ŷ, so F=−2.00×10⁻⁷ ŷ N; a negative charge of equal magnitude gives +2.00×10⁻⁷ ŷ N.

2. (v×B)·v=0 by antisymmetry. If B varies, it may induce E, which can do work: dK/dt=qE·v.

3. r=mpv/(eB)≈4.18×10⁻³ m; f=eB/(2πmp)≈7.62×10⁶ Hz. Nonrelativistic speed assumption is satisfied.

4. Tc=2πme/(eB)≈3.57×10⁻¹⁰ s at B=0.1 T, so |pitch|=v∥Tc≈1.07×10⁻⁴ m. Here me means electron MASS, and e is elementary charge magnitude.

5. F=−0.0300 ŷ N; reversing current gives +0.0300 ŷ N.

6. No. The force law assumes E and B known; use the stationary current distribution and a field-source law such as Biot–Savart (or Ampère with sufficient symmetry).

Next: derive the Biot–Savart integral for steady currents and integrate it for a line, a loop and a finite straight segment without inferring field geometry from force arrows.