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foundation31 min studyLesson 9 in Physics

Electric and magnetic fields as local physics

Replace action-at-a-distance sketches with fields, the Lorentz force, and a first account of field energy and waves.

Before you begin
On this page

The question for today

What you’ll understand

  • Distinguish a field from the force it produces on a chosen charge.
  • Predict the direction and energy effect of magnetic force.
  • Explain why fields are dynamical carriers of energy rather than decorative arrows.

A field assigns a physical quantity to every position and time. Temperature is a scalar-field example; the electric field is a vector field. This description lets a charge respond to its local environment while electromagnetic disturbances propagate through space. The field viewpoint is essential preparation for quantum field theory, where the fields themselves become the central dynamical objects.

A test charge reveals a field without defining its source

An electric field tells you the electric force per unit positive test charge at a point. A negative charge feels the opposite force. The test charge must be small enough that its influence does not substantially rearrange the source configuration. Field arrows visualize direction and relative strength; they are not tracks that a massive charged particle must follow while inertia is ignored.

For an isolated stationary point charge in vacuum, the electric field is radial and its magnitude decreases with squared distance. Multiple electric fields add within classical vacuum electromagnetism. This point-charge expression becomes singular at the source and is not a microscopic model of the source's internal structure. A softened diagram may avoid a visual singularity, but that modification must not be presented as Coulomb's exact law.

E(r)=14πε0qr2r^,FE=qtestE\mathbf E(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{\mathbf r},\qquad \mathbf F_E=q_{\mathrm{test}}\mathbf E
The first formula is the electrostatic vacuum field of a point source at the origin, for r > 0. Electric field has units N/C or V/m. The source charge q and test charge qtest play different roles.↔ Scroll wide equations horizontally.

Magnetic force depends on motion

A magnetic field acts on a moving point charge through a cross product. The force is perpendicular to both the velocity and magnetic field, and reverses with the charge sign. A velocity parallel to the magnetic field gives zero magnetic force even when both vectors are nonzero. Use a right-hand rule for a positive charge, then reverse the result for a negative charge.

Because magnetic force is perpendicular to instantaneous velocity, it does no work on the point charge. A uniform magnetic field can bend a trajectory without changing speed. Electric fields can change kinetic energy because their force can have a component along the motion. Time-dependent electromagnetic situations still require the full coupled fields; a changing magnetic field can induce an electric field that does perform work.

F=q(E+v×B),dKdt=F⋅v=qE⋅v\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B),\qquad \frac{dK}{dt}=\mathbf F\cdot\mathbf v=q\mathbf E\cdot\mathbf v
This is the Lorentz force on a point charge in prescribed fields, with radiation reaction neglected. B has units tesla. The magnetic contribution to instantaneous power vanishes because (v × B) · v = 0.↔ Scroll wide equations horizontally.

Fields have their own dynamics and energy

Maxwell's equations relate electric and magnetic fields to charge and current and to each other's time variation. In vacuum they permit waves traveling at a speed fixed by the electromagnetic constants. A change in a source does not update the distant field instantaneously. Static formulas are useful limiting descriptions, not permission to ignore causal propagation when sources change rapidly.

Electromagnetic fields store energy and carry momentum, so conservation laws must include them when needed. A pure particle-only force-pair account can be incomplete during field propagation. This is the bridge from mechanics to field theory: infinitely many spatial degrees of freedom have local dynamics and can exchange energy with matter. Quantization later changes their allowed states; it does not erase classical locality or bookkeeping.

c=1ε0μ0,u=12ε0∣E∣2+12μ0∣B∣2c=\frac{1}{\sqrt{\varepsilon_0\mu_0}},\qquad u=\frac12\varepsilon_0|\mathbf E|^2+\frac{1}{2\mu_0}|\mathbf B|^2
These are the vacuum wave speed and classical electromagnetic energy density. u is measured in J/m³. Material media require constitutive relations and potentially additional energy accounting; the vacuum formula should not be transplanted without qualification.↔ Scroll wide equations horizontally.

Work it through

A magnetic bend without a speed increase

A positive particle of charge 2.0×10−6 C2.0\times10^{-6}\,\mathrm{C} and mass 1.0×10−6 kg1.0\times10^{-6}\,\mathrm{kg} moves at 300 m s−1300\,\mathrm{m\,s^{-1}} along +x through a uniform 0.50 T0.50\,\mathrm{T} magnetic field along +z. There is no electric field. Find the initial force and orbit radius.

  1. The vectors are perpendicular, so ∣F∣=qvB=(2.0×10−6)(300)(0.50)=3.0×10−4 N|F|=qvB=(2.0\times10^{-6})(300)(0.50)=3.0\times10^{-4}\,\mathrm{N}. Since x^×z^=−y^\hat x\times\hat z=-\hat y, a positive charge initially accelerates toward −y.

  2. Magnetic force supplies the inward force for circular motion: mv2/r=qvBmv^2/r=qvB. Cancel one speed factor to obtain r=mv/(qB)r=mv/(qB), valid here because the initial velocity is entirely perpendicular to B.

  3. Insert the values: r=(1.0×10−6)(300)/[(2.0×10−6)(0.50)]=300 mr=(1.0\times10^{-6})(300)/[(2.0\times10^{-6})(0.50)]=300\,\mathrm{m}. The force remains perpendicular to velocity, so its work and the kinetic-energy change are zero.

The result

The initial force is 3.0×10−4 N3.0\times10^{-4}\,\mathrm{N} toward −y, and the ideal circular orbit has radius 300 m300\,\mathrm{m}. The field bends the path without increasing speed.

The living notebook

Explore the model

Interactive lab

Interactive lab · fields

Separate a source pattern from circulation

Change source strength q and rotation Ω in a softened two-dimensional vector field: q r/(r² + ε) plus Ω(−y, x). The first term supplies a source-like pattern and the second circulation. This normalized illustration is not a Maxwell solver, a Coulomb field in three dimensions, or a charged-particle trajectory.

Separate a source pattern from circulationThe softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Current readings: Fx at (1, 0): 0.833; Fy at (1, 0): 0.5; Field magnitude there: 0.972. All parameters are adjustable with the labeled controls below.−3.2−1.601.63.2−1.5−0.7500.751.5xyq = 1
Local field directionSource strength
Fx at (1, 0)
0.833
Fy at (1, 0)
0.5
Field magnitude there
0.972
t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA field diagram assigns vectors to locations; its arrows alone do not specify matter's motion.

Pause & reason

Check your understanding

A point charge moves in a static magnetic field with no electric field. What does the magnetic force do to its kinetic energy in the ideal Lorentz-force model?

Your turn

Try it yourself

A uniform electric field of 200 N C−1200\,\mathrm{N\,C^{-1}} points along +x. A charge of −3.0 μC-3.0\,\mu\mathrm{C} moves 0.50 m0.50\,\mathrm{m} along +x. Find the electric force and work.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Keep the negative charge sign in both F = qE and W = qEΔx.

Compare with the solution

The force is Fx=(−3.0×10−6)(200)=−6.0×10−4 NF_x=(-3.0\times10^{-6})(200)=-6.0\times10^{-4}\,\mathrm{N}, toward −x. Its work is W=FxΔx=−3.0×10−4 JW=F_x\Delta x=-3.0\times10^{-4}\,\mathrm{J}. If the electric force is the only force doing work, kinetic energy decreases by that amount. The stated displacement must be dynamically possible for the particle's initial energy.

Keep the idea

Take it with you

  • A field is a local assignment; force also depends on the responding object's charge and motion.
  • Magnetic force changes a point charge's direction without directly doing work.
  • Fields carry energy and momentum and obey dynamical propagation laws.

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