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MC-A2 · Inductance & magnetic energy

Inductance and Magnetic Energy

To set up a current in a coil you have to do work against the back emf the coil produces, and that work does not vanish but collects in the magnetic field. Watch how the stored energy grows as the current rises, and read off W = ½LI².

How fast does the stored energy grow with current?

The blue line on the graph is the flux linkage line λ = LI, and the area of the gold triangle under it is the magnetic energy stored in the coil. Push the current with the slider and the triangle grows, widening its base and raising its height together. Below, first pick how many times larger the area becomes when the current is made 2 times larger, then check with the slider.

Current II = 1.5 A
Slide for continuous current. Triangle area = stored energy.
Stored magnetic energy
W = ½ L I² = 0.56 J
It is half the product of current and flux linkage. It grows not in proportion to the current but with its square, so 2 times the current gives 4 times the energy.
Current × 2: what does W become?
Misaligned

Inductance is the flux-linkage efficiency

The inductance L is the flux linkage produced per unit current: L = λ/I = NΦ/I. Solving with the magnetic circuit gives Φ = NI/R, so λ = NΦ = N²I/R, and therefore L = N²/R. It goes with the square of the number of turns because the turns count once in making the flux and once more in linking with it, and it is inversely proportional to the reluctance. Cutting an air gap into the core raises the reluctance and lowers the inductance, but it keeps L from dropping suddenly when the core saturates.

ObserveL = λI
Inductance is the flux linkage per unit current.
ChooseL = ?
Insert the magnetic circuit and L becomes N²/R.

Why energy grows as the square

The work needed to raise the current by a further di is i·dλ = Li·di. The larger the current, the larger the back emf to push against, so the same di takes more work; stacking it up from 0 to I gives the triangle area under the line, ½LI². The half in front comes from the average current being half of I while the current grew evenly from 0 to I. This calculation holds when L is constant; if the core saturates and the line bends, the area has to be measured again under that curve.

Fill indW = i ? = Li di
The small amount of work needed to raise the current a little more.
On your ownW = ?
The triangle area stacked up from 0 to I.

The same energy in current and inductance

Putting λ = LI into ½λI gives W = ½LI², and putting in I = λ/L instead gives W = ½λ²/L. The three formulas just write the same area with different pairs of current, flux linkage, and inductance. For the same current a larger inductance holds more energy, but for the same flux linkage it is the other way around: the smaller the inductance, the larger the energy. That is why, in a core with an air gap, the energy collects mostly in the gap, where the reluctance is large.

Back to the first screen

Making the current 2 times larger made the triangle area 4 times larger because the area grows its base (current) and its height (flux linkage = LI) together. Both are proportional to the current, so their product grows with the square of the current. That is why the stored energy is W = ½LI², and the half in front comes from averaging while the current grew evenly from 0 to I.

The magnetic energy: a coil of inductance L carrying current I stores W = ½LI² in its field. The inductance L = λ/I = N²/R is flux linkage per unit current. The same quantity is also W = ½λI = ½λ²/L. The work done raising the current pools in the field and flows back out as the current falls.
Once you hold this energy

½LI² is the bridge between circuits and machines. The leakage inductance of transformers, the torque of motors, and the energy an inverter handles as it switches all come down to trading this energy stored in coils. The ideal transformer of the next lesson is a device in which two coils share the same flux and move energy across with almost no loss (MC-A3).