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Ampere's law vs continuity equation

Short answer

Static Ampere's law ∇×H = J implies ∇·J = 0, contradicting continuity ∇·J = −∂ρ_v/∂t for time-varying fields. Maxwell corrected it by adding displacement current ∂D/∂t: ∇×H = J + ∂D/∂t, making it consistent.

How does ∇×H = J conflict with the continuity equation in time varying fields? How is this conflict rectified (corrected) in such fields?

The conflict

Ampere's law for static fields is ∇×H = J. Take the divergence of both sides. The divergence of a curl is always zero:

∇·(∇×H) = 0 = ∇·J

So Ampere's law demands ∇·J = 0. But the continuity equation (conservation of charge) states

∇·J = −∂ρ_v/∂t

For time-varying fields ∂ρ_v/∂t ≠ 0 (for example, charging a capacitor), so ∇·J ≠ 0. The two equations contradict each other. Ampere's law is valid only for steady (DC) currents.

The correction (Maxwell's displacement current)

Add an unknown term J_d to the right side: ∇×H = J + J_d. Then ∇·J + ∇·J_d = 0, so ∇·J_d = −∇·J = ∂ρ_v/∂t. Using Gauss's law ρ_v = ∇·D:

∇·J_d = ∂(∇·D)/∂t = ∇·(∂D/∂t)    →    J_d = ∂D/∂t

The corrected equation is

∇×H = J + ∂D/∂t

Now ∇·(J + ∂D/∂t) = 0, which agrees with continuity. The term ∂D/∂t is the displacement current density (A/m²). It flows in the dielectric between capacitor plates and completes the current loop, even though no real charges move across the gap. In integral form: ∮H·dl = I_conduction + I_displacement.

✓ Answer checked by Claude Opus 5.5 (high effort)

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