# UY1: Magnetic-Field Energy In Inductor Establishing a current I in an inductor with inductance L requires an input of energy U.

An increasing current i in the inductor causes an emf $\epsilon$ between its terminals, and a corresponding potential difference $V_{ab}$ between the terminals of the source, with point a at higher potential than point b.

The source must be adding energy to the inductor, and the instantaneous power is:

\begin{aligned} \frac{dU}{dt} &= P \\ \frac{dU}{dt} &= iV_{ab} \\ \frac{dU}{dt} &= iL\frac{di}{dt} \\ dU &= Li \, di \end{aligned}

Note:

\begin{aligned} V_{ab} – L\frac{di}{dt} &= 0 \\ V_{ab} &= L \frac{di}{dt} \end{aligned}

The total energy supplied to the inductor while current increases from zero to a final value I is:

\begin{aligned} U &= L \int\limits_{0}^{I} i \, di \\ &= \frac{1}{2} L i^{2} \end{aligned}

When the current decreases from I to zero, the inductor acts as a source that supplies a total amount of energy:

$$U = \frac{1}{2} LI^{2}$$

to the external circuit.

An inductor carrying a current I has energy U stored in it. This energy is actually stored in the magnetic field within the coil, just as the energy of a capacitor is stored in the electric field between its plates.

Consider the ideal toroidal solenoid: \begin{aligned} B &= \frac{\mu_{0} Ni}{2 \pi r} \\ L &= \frac{\mu_{0}N^{2} A}{2 \pi r} \end{aligned}

The energy stored in the toroidal solenoid when the current is i is:

\begin{aligned} U &= \frac{1}{2} L i^{2} \\ &= \frac{1}{2} \frac{\mu_{0} N^{2}i^{2} A}{2 \pi r} \\ &= \frac{1}{2 \mu_{0}} \left( \frac{\mu_{0} N i}{2 \pi r} \right)^{2} 2 \pi rA \\ &= \frac{B^{2}}{2 \mu_{0}} \times 2 \pi rA \end{aligned}

The magnetic field and thus this energy are localized in the volume $V = 2 \pi rA$ enclosed by the windings. The magnetic field energy density:

\begin{aligned} u &= \frac{U}{2 \pi rA} \\ &= \frac{B^{2}}{2 \mu_{0}} \end{aligned}

When the material inside the toroid is no vacuum but a material with (constant) magnetic permeability $\mu = K_{m} \mu_{0}$,

$$\mu = \frac{B^{2}}{2 \mu}$$

Next: L-R-C Series Circuit

Previous: L-C Circuit

Back To Electromagnetism (UY1) 