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An electrical circuit connects components that carry electric current and sustain electric potential difference differences. Its dynamics may couple electrical resistance, inductance and capacitance.

For motion to the right in $B_0{\mathbf{e}}_z$, $B_0>0$, the induced current is clockwise. Its own magnetic field points into the page inside the circuit, opposing the increase in external flux, as required by Lenz's law. As that current changes, its changing flux also produces an electromotive force opposing the change in current: the circuit has inductance.

Consequently the algebraic current law used in parts (i)-(iv) must be replaced by a dynamical circuit equation that includes the derivative of the self-flux. The induced current cannot in general adjust instantaneously to the bar velocity. Energy is temporarily stored in the self-field and can later return to the motion or be dissipated; the simple exponential braking law can change. Self-field forces and changes of inductance with circuit geometry may also matter. The magnetic self-field changes the transient dynamics but does not create energy: mechanical energy, magnetic-field energy and Joule heat must be included in the conservation balance. A quantitative answer would require the circuit's inductance and its dependence on $s$.

With no electrical resistance or applied electric potential difference, divide the equation by the inductance $L$ to get a simple harmonic oscillator with angular frequency ${\omega }_0=(LC{)}^{-1/2}$. Its general real solution isEach nonzero solution has period $2\pi \sqrt{LC}$. Here ${\omega }_0$ is an angular frequency; the ordinary frequency in cycles per unit time is ${\omega }_0/(2\pi )$.

Taking curls givesBoth have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop areFor clockwise current $I$, the net force is zero in $B_1$, while in $B_2$

Put $K=2\pi b_2{r}^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-g{t}^2/2$. In $B_2$, using counterclockwise current as positive,ThusWith $\Gamma =K^2/R$, the velocity tends to $-mg/\Gamma$ andAt terminal speed, $I^{2}R=mg|\dot{z}|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and ideal motion has no resistive terminal state (inductance must be retained).

An RLC circuit combines electrical resistance, inductance and capacitance. Stored energy can oscillate between magnetic and electric forms, while Joule heating dissipates it.