Induced electric fields

Hold a copper ring still and ramp the magnetic field through it. A current flows, so work is being done on the conduction electrons, so a force is acting on them. It cannot be the magnetic force. Before the current starts the charges are not drifting, and F = qv × B is zero for a charge at rest; worse, the magnetic force is always perpendicular to v and so can never do work. The only force law left that acts on a charge at rest is F = qE. Since emf is defined as work per unit charge taken once round the loop, ℰ = ∮ (F/q) ⋅ dl = ∮ E ⋅ dl, and Faraday's law becomes ∮ E ⋅ dl = −dΦ_B/dt. Notice what has dropped out of that statement: the ring. Nothing on either side refers to a conductor, so the field is there in empty space and the ring was only ever the detector.

Both kinds of electric field push charges with F = qE; they differ in one integral. For charges at rest, ∮ E ⋅ dl = 0 on every closed path. That zero is the licence for everything you did in electrostatics: the work between two points does not depend on the route, so a single number V can be attached to each point and the field recovered as Eₛ = −dV/ds along any direction s. An induced field has ∮ E ⋅ dl = −dΦ_B/dt, which is not zero, so the work from a to b depends on which way round you go, and no V exists. In a real apparatus the total field is the sum of the two, E = E_charge + E_induced; surface charges pile up on the conductors and supply the first part, and only the second part carries circulation. Testing which you have is a single question: take a charge once round a closed loop and ask whether it came back with more energy than it left with.

Take a long solenoid of radius R whose interior field is uniform and changing at dB/dt, with no field outside. The arrangement looks the same after any rotation about the axis and after any slide along it, so on a circle of radius r centred on the axis the induced field must be tangential and equal in magnitude at every point. The integral collapses: ∮ E ⋅ dl = E(2πr). Inside, the enclosed flux is πr²B, so E(2πr) = πr²(dB/dt) and E = (r/2)(dB/dt). Outside, the circle encloses only the coil's own cross-section, so E(2πr) = πR²(dB/dt) and E = (R²/2r)(dB/dt). With R = 5.0 cm and dB/dt = 2.0 T s⁻¹ the field is 50 mV m⁻¹ at the winding and 25 mV m⁻¹ out at r = 10 cm. The same integral, run over an evacuated doughnut, is what a betatron uses to accelerate electrons turn after turn.

The minus sign is a direction rule and needs a convention to bite. Choose a direction to walk round the path; curl the right hand that way and the thumb gives the positive normal, which fixes the sign of Φ_B. Then a positive dΦ_B/dt makes ∮ E ⋅ dl negative, meaning the induced field points mostly against your chosen walking direction. In practice it is quicker to run Lenz's law and then read the field off the current it would drive. Suppose B points out of the page inside the solenoid and is increasing. An induced current would oppose that increase by making flux into the page, so it would run clockwise, so the induced electric field is clockwise — everywhere on that circle, inside the coil and outside it alike. If B instead points out of the page and is decreasing, everything reverses and E runs anticlockwise.

Field lines of a static field begin on positive charge and end on negative charge; if you follow one far enough it terminates. The lines of an induced field terminate nowhere. Around a solenoid they are concentric circles, closing on themselves, and that is the geometric face of ∮ E ⋅ dl ≠ 0: you can travel one line forever and keep gaining energy. It also means the field is not manufactured at the point where you measure it. At r = 6.0 cm outside a 3.0 cm solenoid there is no magnetic field at all and nothing is changing there, yet E ≠ 0. What the law constrains is the circulation round a whole path against the flux through the surface it spans, so the strength of E at your point depends on the flux somewhere else. Locally, a changing B does not create E; globally, a changing flux forces a circulation.

Put two identical voltmeters on the same two points of a ring encircling a solenoid, one with its leads routed left of the coil and one routed right. They disagree, and neither is broken. Each meter reads the line integral of E along its own leads and through its own resistor, and those two paths enclose different flux. So a sentence such as the voltage across this part of the loop has no answer until the route is stated, and V_b − Vₐ is not a property of the two points. Kirchhoff's loop rule survives only if you stop treating the induced field as a potential drop and put it in as a source: go round the loop, add up iR for every resistance, and set the total equal to ℰ = −dΦ_B/dt. Keep the meter leads out of the region where the flux is changing and the old bookkeeping works again.