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Lenz's law expresses the minus sign in Faraday's law of induction: the induced electromotive force drives a passive resistive loop's electric current in the direction whose own magnetic field opposes the imposed change in magnetic flux. It opposes the change of flux, not invariably the external magnetic field. Choose an oriented surface and its positive boundary direction first, find the sign of $d{\Phi }_{\mathrm{ext}}/dt$, and then use the right-hand rule to translate the signed electric current into a physical direction. ^[ref].

In the fixed-loop example in Faraday's law of induction, out-of-page magnetic flux increases at $0.0060Wb/s$. With counterclockwise positive, $I=-0.0030\mathrm{A}$ is clockwise, producing an in-page magnetic field through the loop. Reverse the ramp to $B(t)=0.40\mathrm{T}-(0.30T/s)t$ on the same one-second interval: now $\mathcal{E}=+0.0060\mathrm{V}$ and $I=+0.0030\mathrm{A}$. The counterclockwise electric current produces an out-of-page magnetic field, supporting the decreasing external magnetic flux. A steady out-of-page magnetic field with fixed loop geometry gives no induced electromotive force in either direction.

For the increasing-area example in motional electromotive force, clockwise electric current runs down the right-hand rod. Its magnetic force on a current-carrying wire points left, opposing the imposed motion to the right. Writing $I_{\mathrm{mag}}=B\ell v/R$ for the current magnitude, the power required to maintain constant speed isThus the sign agrees with conservation of energy: the external drive supplies the Joule heating. The magnetic forces do no work on individual charges; mechanical work by the drive supplies the electrical energy. This also explains the usefulness of electromagnetic braking of a sliding rod and the terminal speed of a falling conducting loop. ^[ref].

The sign of induced electromotive force alone does not determine an arbitrary circuit's instantaneous electric current. Retaining constant self-inductance $L$ gives $LdI/dt+RI=-d{\Phi }_{\mathrm{ext}}/dt$, so a pre-existing electric current can persist while the induced drive opposes its change. The distinction matters in the lossless limit of magnetic loop braking. The resistive toy examples use $L=0$ as an approximation, not a general consequence of Lenz's law.

In Cartesian coordinates the magnetic field is $(bx,by,-2bz)$. Its divergence is $b+b-2b=0$, and all components of its curl vanish. Thus it satisfies the static vacuum magnetic equations $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$, with no electric current in the region.

Fix clockwise orientation as viewed from above, looking down the positive $z$-axis. Its positive area normal is $-{\mathbf{e}}_z$. The loop's external magnetic flux is thereforeWith the opposite normal its flux has the opposite sign. On the circle $\mathbf{B}=ba{\mathbf{e}}_r-2bz{\mathbf{e}}_z$ and clockwise line element is $d\mathbf{\ell }=-a{\mathbf{e}}_{\phi }d\phi$. The Lorentz force element isThe radial terms cancel around the circle, soThis fixes the current/flux/force signs together.

Neglect self-inductance initially, so Faraday's law and Ohm's law give $RI=-K\dot{z}$. Including gravity then givesFor $R>0$ and $K\ne 0$, the terminal solution is $z=z_0-vt$ withThe general velocity relaxes exponentially to $-v$ on time scale $mR/K^2$. In the terminal state $I=Kv/R$, and the Joule-heating rate isThus all the lost gravitational potential energy becomes resistive heat when kinetic energy is constant.

Within the zero-inductance approximation $v\to 0$ as $R\to 0$; finite-speed motion would require an unbounded current. This limit is singular, so it is important to describe the physical ideal-conductor limit separately. With finite self-inductance $\mathcal{L}$, the circuit equation is $\mathcal{L}\dot{I}+RI=-K\dot{z}$. At $R=0$, total linked flux $\mathcal{L}I+Kz$ is conserved and there is no Joule heating. If $I=0$ initially at $z_0$, thenThis lossless limit of magnetic loop braking has undamped oscillations about $z_0-mg\mathcal{L}/K^2$, with magnetic energy storing and returning mechanical energy. It has no generic dissipative terminal descent. Thus the vanishing terminal speed is the formal resistive-model limit, not a justification for dropping inductance in a perfectly conducting moving loop.