Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Since moving charges apply forces to each other, work has to be done to establish a set of currents in order: while a current is being built up, the changing flux induces a voltage that opposes the build-up (Faraday's law, Chapter 5), and the source has to push against it. This means there is an energy stored in the system to keep it in order. It is called the stored magnetostatic energy, and it sits in the field, exactly as the electrostatic energy $\frac{1}{2} \int E \cdot D d V$ of Section 3.4 does.

Magnetostatic energy

W_{m} = \frac{1}{2} \int_{V} H \cdot B d V = \frac{μ}{2} \int_{V} H^{2} d V [J], (4.2.25)

the integral extending over all the space where there is a field. The magnetostatic energy density is

w_{m} = \frac{1}{2} H \cdot B = \frac{μ}{2} H^{2} = \frac{B ^{2}}{2 μ} [J/ m^{3}] . (4.2.26)

For a single circuit the same energy reads

W_{m} = \frac{1}{2} L_{11} I_{1}^{2}, (4.2.27)

and for two coupled circuits

W_{m} = \frac{1}{2} L_{11} I_{1}^{2} + L_{12} I_{1} I_{2} + \frac{1}{2} L_{22} I_{2}^{2} . (4.2.28)

In plain words

Every cubic metre of field stores $\frac{1}{2} H B$ joules. Add it all up and you get $\frac{1}{2} L I^{2}$ , the energy a circuit designer attributes to the inductor. Both views are the same number: Home exercise 4.2.3 checks it for a toroid, where the field is confined to the core and the two integrals are easy.

The field form, Eq. (4.2.25), tells us where the energy is. At the same $B$ , the density $B^{2} / (2 μ)$ is $μ_{r}$ times smaller in iron than in air. In a gapped core (Figure 4.2.11) practically all the energy sits in the gap, which is why an inductor meant to store energy, as in a switch-mode power supply, always has a gap (Worked example 4.2.1), and why the force on the armature of an electromagnet is governed by the gap field alone (Exercise 4.2.8). The two-circuit form, Eq. (4.2.28), must be positive for any pair of currents, and this requires $L_{12}^{2} \leq L_{11} L_{22}$ , a fact used in Application 4.2.2.

Move $I_{2}$ and find the minimum of $W_{m}$ : it stays positive for any $k < 1$ .

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Quick check

A $64 mH$ inductor carries $0.22 A$ . How much energy does it store? Where is it, if the core is a closed ferrite toroid, and where if the toroid has a $1 mm$ gap?

$W_{m} = \frac{1}{2} \times 0.064 \times 0.2 2^{2} = 1.5 mJ$ . In the closed core it is spread through the ferrite. In the gapped core the gap, with $μ_{r} = 2000$ times the energy density of the ferrite at the same $B$ , holds the energy almost entirely, although it is only $1 mm$ of the $315 mm$ path (Worked example 4.2.1).

Application 4.2.2 · Transformers and the coupling coefficient

A transformer is two coils, the primary with $N_{1}$ turns and the secondary with $N_{2}$ , wound on a common core of high $μ_{r}$ (Figure 4.2.12). Not all of the flux $Φ_{11}$ made by winding 1 reaches winding 2: the rest closes through the air around the winding and is called leakage. The fraction that does link is measured by the coupling coefficient

Coupling coefficient

k = \frac{L _{12}}{L _{11} L _{22}}, 0 \leq k \leq 1. (4.2.29)

A transformer on a laminated core. The flux \Phi_{11} of the primary splits into the part \Phi_{12} that runs round the core and links the secondary (mutual flux) and the part that closes through the window (leakage flux). — **Figure 4.2.12.** A transformer on a laminated core. The flux $Φ_{11}$ of the primary splits into the part $Φ_{12}$ that runs round the core and links the secondary (mutual flux) and the part that closes through the window (leakage flux).

$L_{12} = L_{21}$ always (reciprocity). On a closed high- $μ$ core the same flux threads both windings. If one ampere-turn makes the flux $P$ webers in the core (the permeance of the core, $μ S / l_{c}$ for the toroid of Eq. (4.2.22)), then $L_{12} \to N_{1} N_{2} P$ , $L_{11} \to N_{1}^{2} P$ , $L_{22} \to N_{2}^{2} P$ , so that $k \to 1$ and

$\frac{L _{11}}{L _{22}} = (\frac{N _{1}}{N _{2}})^{2} .$
A mains transformer reaches $k > 0.99$ . That is exactly why $V_{2} / V_{1} ≃ N_{2} / N_{1}$ : the ideal-transformer law of Chapter 5 is the $k \to 1$ limit, in which the same flux is linked by both windings and the induced voltage per turn is the same.
Leakage is not always a fault. An arc-welding transformer and a microwave-oven transformer are built with a magnetic shunt that deliberately diverts flux, so that the leakage inductance limits the current by itself, without any external resistor.
Positive energy for any $I_{1}$ , $I_{2}$ in Eq. (4.2.28) requires $L_{12}^{2} \leq L_{11} L_{22}$ , which is the proof that $k \leq 1$ . Exercise 4.2.10 shows that the minimum energy at fixed $I_{1}$ is $\frac{1}{2} L_{11} I_{1}^{2} (1 - k^{2})$ , which is positive unless $k = 1$ .

Worked example 4.2.2 puts numbers in: a secondary of $100$ turns wound over the middle quarter of a $400$ -turn air-cored solenoid has $L_{12} = L_{22} = 0.316 mH$ , $L_{11} = 1.26 mH$ and $k = 0.50$ , although every line of the primary crosses the secondary. The coupling is only one half because the short coil's own flux escapes along the rest of the long one.

Shorten the secondary and watch $k = l_{2} / l_{1}$ .

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Where you meet it

Every power supply, from the substation transformer to the charger of a phone. Measuring the series inductance of two coils connected aiding and opposing gives $L_{12} = (L_{aid} - L_{opp}) /4$ (Exercise 4.2.10), the practical way of finding $k$ .

Quick check

Two coils have $L_{11} = 10 mH$ , $L_{22} = 40 mH$ and $L_{12} = 12 mH$ . What is $k$ ? Could $L_{12}$ be $25 mH$ ?

$k = 12/ 10 \times 40 = 0.6$ . No: $L_{11} L_{22} = 20 mH$ is the largest possible mutual inductance, reached only for perfect coupling.

4.2.5 Magnetostatic energy