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advanced30 min studyLesson 247 in Physics

Faraday induction and flux linkage

Relate circulating electric fields to changing flux and distinguish single-turn flux, winding linkage, and a lumped circuit response.

Before you begin
On this page

The question for today

What you’ll understand

  • Compute fixed-loop emf from a changing flux with a consistent orientation.
  • Sum turn fluxes to obtain winding linkage without multiplying by turn count twice.
  • Explain how an induced electric field can exist outside the region of changing B.

Transformer induction can drive a stationary circuit: its electric field circulates even though the conductor is not moving. The field exists before a wire is inserted, so Faraday's law is more general than a circuit recipe. A coil samples that field through its complete winding path. Converting the field statement into a voltage or current requires careful flux linkage and an explicit circuit model, including self-inductance when the induced current changes appreciably.

A circulating electric field need not have a scalar voltage

For a fixed oriented loop, the circulation of E is minus the time derivative of its magnetic flux. Locally this means that the curl of E equals minus the local rate of change of B. An electric field with nonzero curl cannot be represented throughout that region by the gradient of one single-valued scalar potential. This is why an electrostatic voltage intuition fails when two measurement lead paths enclose different changing flux.

The sign follows from a jointly chosen contour and surface normal. An increasing positive flux produces a negative circulation, whereas a decreasing positive flux produces positive circulation. Calling both cases opposition to the field is imprecise: the induced current opposes the change of flux under the stated passive-circuit assumptions. A decreasing positive field can induce a positive magnetic contribution, supporting the original direction while slowing its decrease.

∮CE⋅dℓ=−ddt∫SB⋅da,∇×E=−∂tB\oint_C\mathbf E\cdot d\boldsymbol\ell=-\frac{d}{dt}\int_S\mathbf B\cdot d\mathbf a,\qquad \nabla\times\mathbf E=-\partial_t\mathbf B
C and S are fixed in the laboratory and right-hand oriented. E is in V/m, B in T, and flux in Wb, making its time derivative volts. A moving contour additionally requires the motional term taught previously.↔ Scroll wide equations horizontally.

Linkage is a sum, not an extra magnetic field

For many series-connected turns, add the signed flux of each turn to obtain linkage. Equal turn fluxes reduce this sum to turn count times flux; unequal placement or fringing requires the actual sum. Once linkage is defined this way, the emf is its negative time derivative with no further turn factor. Changing turn count changes how the winding samples the imposed field; it does not by itself multiply the imposed magnetic flux density at a point.

In a simple resistive pickup, the induced emf divided by resistance approximates current only when self-inductive transients and capacitive effects are negligible. More generally the imposed-flux emf drives an RL equation, whose self-linkage belongs on the response side. Counting the current's self-flux both as an independent imposed source and as an inductive voltage double counts its effect. An open pickup can show an emf while carrying negligible current, so emf alone does not establish magnetic back reaction.

λ=∑k=1NΦk,Eext=−dλextdt,LdIdt+RI=Eext\lambda=\sum_{k=1}^{N}\Phi_k,\qquad \mathcal E_{\mathrm{ext}}=-\frac{d\lambda_{\mathrm{ext}}}{dt},\qquad L\frac{dI}{dt}+RI=\mathcal E_{\mathrm{ext}}
λ is linkage in Wb, Φk is the signed flux of turn k, N is integer turn count, L is in H, and R in Ω. The last equation assumes one fixed linear winding, an imposed external source, and a lumped RL response with consistent positive current orientation.↔ Scroll wide equations horizontally.

A flux confined elsewhere can still induce circulation here

An ideal very long solenoid provides a clean example. Its changing axial field is confined approximately inside its radius, but concentric electric-field circles exist outside. For a loop larger than the solenoid, the enclosed changing flux stops increasing with loop radius. The circumference continues to increase, so the azimuthal electric field decreases inversely with radius. Outside, its local curl can vanish even while its circulation around the excluded flux region is nonzero.

There is no conflict with Stokes' theorem: a spanning disk crosses the interior region where the curl is nonzero. Restricting attention to the exterior leaves a multiply connected domain and prevents a globally single-valued electrostatic potential for that circulation. Real solenoid ends, return fields, and propagation modify the idealization. The example isolates induction geometry and topology, rather than predicting a spatially instantaneous signal from a rapidly switched finite coil.

Work it through

A decreasing field in a pickup coil

A stationary 200-turn coil has area 0.0015 m² per turn and normal +z. A uniform imposed field along +z falls from 0.050 T to 0.010 T in 0.050 s. Find its average emf and the charge through a 12 Ω circuit if self-inductance is negligible.

  1. The field change is −0.040 T, so the linkage change is Δλ=NAΔB=(200)(0.0015)(−0.040)=−0.012 Wb\Delta\lambda=NA\Delta B=(200)(0.0015)(-0.040)=-0.012\,\mathrm{Wb}. This already sums all turns and requires no second factor of 200.

  2. Faraday's sign gives average emf −Δλ/Δt=+0.012/0.050=+0.24-\Delta\lambda/\Delta t=+0.012/0.050=+0.24 V. For a linear ramp this emf is constant during the ramp. Its positive contour direction is counterclockwise viewed from +z.

  3. Neglecting self-inductance gives average current Iavg=0.24/12=0.020I_{\mathrm{avg}}=0.24/12=0.020 A and Q=IavgΔt=0.0010Q=I_{\mathrm{avg}}\Delta t=0.0010 C. Equivalently integrate the resistive equation to obtain Q=−Δλ/RQ=-\Delta\lambda/R, which checks the duration cancellation.

The result

The average emf is +0.24 V and the transferred charge is +1.0 mC in the positive winding direction. The positive induced field opposes the decrease, rather than opposing the original field direction.

The living notebook

Explore the model

Interactive lab

Interactive lab · fields

Circulation without a radial source

Reset and set both controls to Source strength q=0 and Rotation Ω=0.5. Predict a purely circulating field with Fy=0.5 and field magnitude 0.5 at (1,0) in the readouts. For a fresh comparison, Reset and set both controls to Source strength q=0 and Rotation Ω=1; predict Fy=1 and magnitude 1 there. Analytic circulation around a fixed centred circle doubles, although it is not a readout and displayed arrow lengths use nonlinear compression. This illustrates the electric circulation of a locally uniform changing axial field after suitable scaling, not the exterior field of a finite solenoid.

Circulation without a radial sourceThe 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 circulating field can have zero radial source strength. The softened planar illustration neither solves Maxwell's time evolution nor supplies a coil, linkage, current, or induced-voltage readout.

Pause & reason

Check your understanding

A fixed 50-turn coil has equal signed flux Φ through every turn and linkage λ=50Φ. What is its induced emf?

Your turn

Try it yourself

An ideal long solenoid of radius 0.050 m has uniform axial field increasing at +0.40 T/s. Find Eφ at radius 0.10 m outside it and the counterclockwise circulation there, ignoring end and propagation effects.

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

Give me a hint

For an exterior circle, the changing flux is the solenoid cross-sectional area times the field change, while the electric-field integral uses the larger observation circumference.

Compare with the solution

The enclosed flux rate is π(0.050)2(0.40)=0.0031416 Wb/s\pi(0.050)^2(0.40)=0.0031416\,\mathrm{Wb/s}. Thus 2πrEϕ=−0.00314162\pi r E_\phi=-0.0031416 V, giving Eϕ=−0.0050 V/mE_\phi=-0.0050\,\mathrm{V/m} at r=0.10 m. The circulation is −3.14 mV. The exterior B time derivative is zero in this idealization, yet the electric circulation is nonzero because the spanning disk contains the changing interior flux. Negative azimuthal direction means clockwise viewed from +z.

Keep the idea

Take it with you

  • Transformer induction is a field statement that exists even without a wire.
  • Differentiate the complete signed linkage once; do not multiply by turn count again.
  • An exterior region can have zero local electric curl but nonzero circulation around enclosed changing flux.

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