Faraday induction
Separate transformer emf from motional emf with an explicit oriented-loop convention: derive ∮E·dl=−dΦB/dt for a fixed contour, apply Lenz's law as a consequence of the signed flux rule, calculate induced azimuthal E for a changing axial B, derive the moving-rod emf from qv×B, reconcile both mechanisms via a moving-loop flux rule and account for mechanical versus Joule power.
- Define a signed magnetic flux through an oriented surface and the induced emf around its consistently oriented boundary.
- State Faraday's integral and local curl equations for a fixed contour and justify the minus sign with work/energy and Lenz's rule.
- Explain why an induced electric field can have nonzero circulation and cannot be described globally by a scalar electrostatic potential alone.
- Derive the azimuthal E inside/outside an ideal circular region with spatially uniform time-varying axial B.
- Compute the sign and magnitude of induced emf/current in a stationary resistance loop under a specified dB/dt.
- Use qv×B to derive motional emf of a rod and distinguish the charge-force from mechanical energy supplied by the external mover.
- State the moving-material-loop flux rule with local boundary velocity and its assumptions and correct orientation.
- Derive induced-current magnetic drag and P_mechanical=I²R for a simple moving rod/rail circuit.
- Distinguish fixed-contour transformer emf from motional emf and avoid double-counting their contributions.
- Solve six worked sign/geometry/units exercises and identify quasistatic and finite-geometry limits.
A changing magnetic flux and a moving conductor can both produce electromotive force, but they are not the same local mechanism. We first orient a FIXED loop and derive the circulation of the induced electric field. Then a moving rail/rod example makes the magnetic Lorentz term and mechanical work explicit. Only after identifying which contour moves do we use the generalized flux rule.
With the surface normal +z, increasing Bz produces dΦB/dt>0 and ∮C E·dℓ=−dΦB/dt<0: induced E is CLOCKWISE viewed from +z. A closed conductor can carry induced clockwise current. This is Faraday's law; a prescribed steady current producing H is the separate Ampère case.
0101.1 — Sign conventions: flux and electromotive force
Choose a stationary oriented surface S and its boundary contour C. The right-hand rule links positive traversal of C to the normal n̂ of S. Signed magnetic flux is ΦB=∫S B·n̂dS. Electric emf around that FIXED contour is the line integral of the electric field along its oriented tangent, with unit volt.
Reversing both n̂ and traversal reverses ΦB and the measured oriented emf, preserving the law. A negative flux is not an intrinsically negative magnetic-field magnitude; it reports orientation relative to the chosen surface.
0201.2 — Faraday's integral and curl equations for a fixed loop
Experimental electromagnetic induction and Maxwell–Faraday theory give an oriented electric circulation equal to the NEGATIVE rate of magnetic flux change through any fixed spanning surface. By Stokes' theorem, for a stationary surface and sufficiently regular fields the same law in local form is ∇×E=−∂B/∂t.
An induced E is not, in general, an electrostatic conservative field: if dΦB/dt is nonzero, ∮E·dℓ is nonzero and no globally single-valued electrostatic V alone can describe E on that loop. In a potential formulation E=−∇V−∂A/∂t; the time-dependent A contribution carries the circulation.
0301.3 — Lenz's rule follows from the SIGNED flux equation
Suppose a loop lies in the xy plane with normal +z; positive traversal is counterclockwise when viewed from +z. Let a uniform prescribed +z magnetic field increase. Then dΦB/dt>0, so ε_fixed<0: the induced electric circulation is clockwise. In a simple resistive loop the induced current is clockwise and its own field points toward −z, opposing the imposed INCREASE in positive flux.
If the imposed +z flux decreases instead, the emf and induced current reverse. The induced field does not always oppose the original B; it opposes the CHANGE of linked flux under the declared orientation and passive circuit assumptions.
0401.4 — Azimuthal induced E around a changing axial magnetic field
As an idealized symmetric source, take B=B(t) ẑ uniform within a cylinder r<a and negligible outside, with no dependence on azimuth. Choose a stationary concentric circular contour of radius r and positive normal +z. Rotational symmetry makes induced E=Eφ(r)φ̂ and its tangent component constant on that contour.
For r<a, ΦB=πr²B(t); for r>a, ΦB≈πa²B(t). Faraday gives Eφ(2πr)=−dΦB/dt, hence the piecewise result. The exterior E can be nonzero even where the prescribed time-dependent B is negligible; electric circulation encloses the changing flux. This sharply idealized spatial boundary is not a universal field of an arbitrary real solenoid.
0501.5 — Stationary resistive-loop numerical example
A stationary circular loop of radius a=0.100 m has normal +z and resistance R=2.00 Ω. A spatially uniform axial field over its area increases at dB/dt=+0.500 T/s. Its signed emf is ε=−πa²dB/dt≈−0.0157 V (negative relative to counterclockwise traversal).
If self-induction, capacitive effects and radiation are negligible on the variation timescale, the induced signed current is I≈ε/R≈−7.85 mA, clockwise. Joule power is I²R≈1.23×10⁻⁴ W. The induced-current reaction must be included if the external field source is not ideal/stiff.
0601.6 — Motional emf: a sliding rod in a static B
With u=+u x̂ and B=+B ẑ, the force on positive carriers is q(u×B)=−quB ŷ. The induced current goes DOWN the moving right rod and CLOCKWISE around the loop; its magnetic force is F=−IℓB x̂, opposing the motion. The loop has total resistance R. All directions and the signed emf are consistent with the +z-normal, counterclockwise-positive convention in this chapter.
A conducting rod of length ℓ lies along +y between two horizontal rails and moves with speed u along +x in a uniform static B=+B ẑ. Positive carriers in the rod experience q(u×B)=−quB ŷ: charge is driven toward the BOTTOM end. Orient the rectangular loop normal +z and traverse it counterclockwise; the moving rod is traversed bottom to top (+y).
For a conductor element moving at local velocity u, its motional contribution is ∫(u×B)·dℓ. On the rod it is −Bℓu; stationary rails contribute no motional term. Thus the induced loop current in a passive circuit is clockwise and runs top-to-bottom through the rod. Its magnetic force I dℓ×B opposes the +x motion, as required by mechanical-energy conservation.
0701.7 — Generalized flux rule for a moving MATERIAL contour
When C=C(t) moves/deforms, one must not insert d/dt of its moving flux into the FIXED-loop expression ∮E·dℓ alone. The emf of a thin material conducting contour moving with local boundary velocity u(r,t), within the standard nonrelativistic moving-conductor model, is ∮C(t)[E+u×B]·dℓ.
With consistent orientation, smooth moving surface and applicable flux-transport identity, this total equals −d/dt∫S(t)B·dS. Here the derivative includes both changing B and changing area/contour. It is not valid to add a separate −dΦ/dt to the transformer and motional terms; that would DOUBLE-COUNT the same induction.
0801.8 — Mechanical work and magnetic drag on the moving rod
For the sliding rod, changing loop area is A(t)=ℓx(t), so dΦB/dt=Bℓu. The moving-loop rule yields ε=−Bℓu, matching the direct u×B force on carriers. In the resistance-dominated quasistatic limit the current magnitude is I=Bℓu/R.
The magnetic force magnitude on the rod is Fmag=IℓB and points opposite u; an external motor must supply Fext=IℓB in +x. Its mechanical power Fext u=(Bℓu)²/R=I²R equals Joule heating in the ideal steady-speed circuit. The magnetic force on a point charge remains orthogonal to that charge's instantaneous velocity; the motor supplies the rod/circuit energy by moving the conductor against magnetic drag.
0901.9 — Regime and sign checklist
For a long straight steady current I>0 out of the page, the right-hand rule gives B counterclockwise (upward at the right-hand side of the wire). In vacuum, B=μ0H, and the oriented contour obeys ∮C H·dℓ=Ienc. This is Ampère's law, NOT Faraday induction.
Before integrating, choose and label normal n̂ and positive contour. Declare fixed versus moving boundary, identify the actual field and circuit, then use ∮E·dℓ for FIXED contours or ∮(E+u×B)·dℓ for moving material circuits. Compute sign with cross products and flux derivatives, not an unsigned mnemonic.
Use e.g. B=0.200 T, ℓ=0.100 m, u=2.00 m/s and R=2.00 Ω for the sliding rod: |ε|=0.0400 V, I=0.0200 A, |Fdrag|=4.00×10⁻⁴ N and Pext=8.00×10⁻⁴ W=I²R. Distributed circuit/wave propagation, appreciable self-inductance or relativistic u requires a richer model.
01.10 — Six exercises with worked solutions
- A stationary +z-oriented circular loop of radius 0.100 m encloses Bz increasing at 0.500 T/s; compute its signed emf and the direction of current through a passive resistor.
- For the same loop, if Bz instead decreases at 0.500 T/s, determine emf and induced-field direction.
- For an ideal changing-B cylindrical region a=0.100 m with dB/dt=+0.500 T/s, compute Eφ at r=0.0500 m and at r=0.250 m.
- A +y rod ℓ=0.100 m moves +x at 2.00 m/s through B=+0.200 ẑ T, with a 2.00 Ω rail circuit. Determine the signed CCW-positive emf and current direction through the rod.
- For the rod circuit calculate magnetic drag, external mechanical power at fixed speed, and Joule heating. Verify equality.
- State the correct flux rule for a moving material loop and explain in one sentence why ∮E·dℓ alone is insufficient and why adding an extra −dΦ/dt double-counts.
Worked solutions
1. ε=−π(0.100)²(0.500)≈−15.7 mV. Negative relative to CCW means clockwise electric circulation and current; induced B points −z.
2. ε≈+15.7 mV and circulation/current are CCW (viewed from +z), producing +z flux to oppose the decrease.
3. Inside Eφ=−(r/2)dB/dt=−0.0125 V/m. Outside Eφ=−a²/(2r)dB/dt=−0.0100 V/m. Negative means clockwise tangential direction relative to +z.
4. u×B=−uBŷ and dℓ on the rod is +ŷ under CCW traversal, so ε=−Bℓu=−0.0400 V. Current is clockwise, top-to-bottom on the rod; magnitude 0.0200 A.
5. Fdrag=−IℓB x̂=−4.00×10⁻⁴ x̂ N. Pext=|Fdrag|u=8.00×10⁻⁴ W=I²R=(0.0200)²×2.00 W.
6. ε=∮C(t)(E+u×B)·dℓ=−dΦB[S(t)]/dt under the material-loop assumptions. The moving magnetic term is missing from ∮E alone; the flux derivative already includes both mechanisms.
Chapter 02 derives the displacement-current correction to Ampère from charge continuity and the charging-capacitor surface paradox, then prepares the complete Maxwell equations.