advanced31 min studyLesson 246 in Physics

Motional emf and moving circuits

Combine Lorentz force with the total flux derivative for a moving contour and reconcile mechanical work with electrical heating.

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What you’ll understand

  • Compute the directed motional contribution to a circuit emf.
  • Distinguish a moving-surface total derivative from a local time derivative.
  • Balance drag power and Joule heating for a sliding conductor.

A circuit can generate an emf in a magnetic field that is constant in laboratory time. The source of the emf is then motion of the conductor, rather than a nonzero local time derivative of B. Faraday's flux rule still works, but only if the derivative follows the changing circuit. Starting from the Lorentz force makes the sign and the mechanical energy supply explicit instead of treating magnetic flux as a mysterious source of free electrical work.

The force per charge contains conductor motion

At a point on a thin moving wire, let the conductor velocity be u. In the laboratory frame the tangential driving force per unit charge contains E plus u crossed with B. The carrier drift relative to an ideal thin wire lies along its tangent, so its magnetic cross product has no tangential component. Integrating the remaining tangential force around the oriented circuit gives the electromagnetic emf, measured in volts. Batteries can add separate non-electromagnetic driving terms.

This expression does not say that a magnetic force performs work on an individual charge, since that force remains perpendicular to its instantaneous laboratory velocity. Mechanical constraints, electric fields, and charge motion together transfer energy from the moving conductor to the circuit. In an open moving rod, separated charges build an electric field that can balance the magnetic force. A measured terminal voltage then depends on how the leads and measurement contour are arranged.

E=∮C(t)(E+u×B)⋅dℓ\mathcal E=\oint_{C(t)}(\mathbf E+\mathbf u\times\mathbf B)\cdot d\boldsymbol\ell
E is in V/m, conductor velocity u in m/s, B in T, and emf in V. C(t) follows an oriented thin circuit with well-defined segment velocities. The expression assumes nonrelativistic motion and excludes an independently specified battery emf.↔ Scroll wide equations horizontally.

The derivative must follow the surface

Choose the surface orientation by the right-hand rule about the circuit. Its flux can change because the field changes at fixed positions or because the boundary sweeps through a stationary field. The transport identity contains both contributions. Combining it with the local Maxwell–Faraday equation yields the negative total flux derivative. Replacing that derivative by only an integral of the local time derivative erases the motional term and incorrectly predicts no emf for a generator in a static field.

The moving-contour rule assumes a smoothly evolving contour and a consistent spanning surface with no magnetic monopoles. Sliding contacts can be handled piecewise when the conducting path and segment velocities are specified. A circuit that reconnects through brushes or a rotating disk can require more care: the geometrical boundary velocity and the material velocity need not coincide everywhere. Identify the actual force integral before applying an unexplained change-of-area shortcut.

ddt∫S(t)B⋅da=∫S(t)∂tB⋅da−∮C(t)(u×B)⋅dℓ,E=−dΦBdt\frac{d}{dt}\int_{S(t)}\mathbf B\cdot d\mathbf a=\int_{S(t)}\partial_t\mathbf B\cdot d\mathbf a-\oint_{C(t)}(\mathbf u\times\mathbf B)\cdot d\boldsymbol\ell,\qquad \mathcal E=-\frac{d\Phi_B}{dt}
The total derivative follows the moving surface, while ∂t holds laboratory position fixed. The surface normal and boundary direction are right-hand related. With these assumptions Maxwell–Faraday plus the transport identity equals the Lorentz-force emf.↔ Scroll wide equations horizontally.

Lenz's sign becomes a mechanical drag

For a rod sliding along parallel rails, the enclosed area grows with displacement. An induced current opposes the increasing imposed flux. Its magnetic force on the rod opposes the motion, so maintaining a fixed speed requires a pull. When self-inductance is negligible, the current equals emf divided by total resistance and the pulling power equals resistive heating. If inductance matters, some input power also changes stored magnetic energy during the transient.

Uniform translation of an entire rigid loop through a uniform stationary field gives a contrasting limit. Its area and orientation stay constant, and the opposite sides' motional contributions cancel around the closed path. A single moving segment may have a nonzero end-to-end force integral while the whole loop emf is zero. Keeping the full circuit in view resolves this apparent disagreement without changing either the Lorentz force or Faraday's law.

A sliding rod with signed current

A 0.40 m rod lies along y and slides along x at +2.0 m/s through B=0.50z^\mathbf B=0.50\hat{\mathbf z} T, closing a rectangular rail circuit on its left. The total resistance is 2.0 Ω. Find emf for the +z normal, current direction, and steady pulling power; neglect self-inductance.

  1. The positive contour is counterclockwise viewed from +z and runs along +y on the right-hand rod. Since u×B=−1.0y^\mathbf u\times\mathbf B=-1.0\hat{\mathbf y} V/m, its contour integral is −0.40 V. The same sign follows from increasing positive flux.

  2. Divide by resistance: the signed current is I=−0.40/2.0=−0.20I=-0.40/2.0=-0.20 A, meaning clockwise viewed from +z. In the rod the current runs along −y, whose cross product with +z produces force along −x.

  3. The drag magnitude is Bℓ∣I∣=(0.50)(0.40)(0.20)=0.040B\ell|I|=(0.50)(0.40)(0.20)=0.040 N. Maintaining 2.0 m/s requires pulling power 0.080 W. Independently, I2R=(0.20)2(2.0)=0.080I^2R=(0.20)^2(2.0)=0.080 W, verifying the energy balance.

The result

The emf is −0.40 V for the +z orientation, producing 0.20 A clockwise. The required pull supplies 0.080 W, all converted to heat in the stated steady resistive approximation.

Explore the model

Interactive lab

Interactive lab · vectors

Project a motional driving field

Reset and set both controls to Angle 90° and Magnitude 1. Treat y as a wire tangent and predict tangential projection +1. For a fresh comparison, Reset and set both controls to Angle −90° and Magnitude 1; predict −1. This illustrates how the sign of a known u×B driving vector enters a directed wire integral. The lab itself does not compute the cross product or move a circuit.

Project a motional driving fieldPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forReversing the driving vector reverses its signed tangential contribution. A vector projection is an ingredient of emf calculation, not a moving-contour simulation.

Check your understanding

A rigid closed loop translates through a uniform time-independent magnetic field while keeping its area and orientation fixed. What is its net motional emf?

Try it yourself

A rigid single-turn loop of area 0.030 m² rotates in a uniform 0.20 T field. Its oriented normal makes angle θ with the field, increasing at 10 rad/s. Find the signed emf when θ=60°.

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

Give me a hint

Use the flux ΦB=BAcos⁡θ\Phi_B=BA\cos\theta and take its total derivative while B and area remain fixed.

Compare with the solution

Differentiating gives E=BAθ˙sin⁡θ\mathcal E=BA\dot\theta\sin\theta. At the specified instant this is (0.20)(0.030)(10)sin⁡60∘=0.05196(0.20)(0.030)(10)\sin60^\circ=0.05196 V, or +52.0 mV for the contour tied to the chosen normal. The positive sign corresponds to decreasing positive flux. No local time dependence of the imposed laboratory B is required; the changing orientation supplies the total flux derivative. Reversing the angular velocity would reverse this emf.

Take it with you

  • Moving-wire emf includes the tangential Lorentz term u crossed with B.
  • The flux rule uses a total derivative following the moving contour.
  • Motional current produces drag, so a generator's electrical energy comes from mechanical work.

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