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

A winding of $N$ turns around a branch links the loop $N$ times, so the current on the right-hand side of the loop equation is $N I$ , not $I$ (Figure 4.3.4a). This product is the quantity that drives the flux, and it gets a name of its own.

Definition · Magnetomotive force

The magnetomotive force (mmf) of a winding of $N$ turns carrying the current $I$ is

F_{m} = N I [A], (4.3.15)

usually quoted in ampere-turns (At) to recall where it comes from. It is the circulation of $H$ around any closed path that threads the winding once.

With this, the single-loop core of Figure 4.3.4a obeys $Z Φ = N I$ , which looks exactly like Ohm's law $R I = E$ for the electric circuit of Figure 4.3.4b. The resemblance is not a coincidence: a resistor of length $l$ , cross-section $S$ and conductivity $σ$ has $R = l / (σ S)$ because $J = σ E$ and the voltage drop is $E l$ , while a branch of a core has $Z = l / (μ S)$ because $B = μ H$ and the magnetic potential drop is $H l$ . The current $I$ is the flux of $J$ , the magnetic flux $Φ$ is the flux of $B$ , and both are conserved at nodes. Table 4.3.1 lists the dictionary.

(a) Magnetic circuit: NI=Z\Phi, Z=l/(\mu S). — (a) Magnetic circuit: $N I = Z Φ$ , $Z = l / (μ S)$ .

(b) Electric circuit: \mathcal E=RI, R=l/(\sigma S). The magnetic circuit and its electric analogue. (a) A winding of N turns on a closed core drives the flux \Phi around the mean path. (b) A battery drives the current I through a resistor. The loop rule \sum Z\Phi=NI is formally Kirchhoff's voltage law. — **Figure 4.3.4.** (b) Electric circuit: $E = R I$ , $R = l / (σ S)$ . The magnetic circuit and its electric analogue. (a) A winding of $N$ turns on a closed core drives the flux $Φ$ around the mean path. (b) A battery drives the current $I$ through a resistor. The loop rule $\sum Z Φ = N I$ is formally Kirchhoff's voltage law.

Table 4.3.1. The dictionary between a magnetic circuit and a resistive electric circuit.

magnetic circuit	electric circuit
magnetomotive force $F_{m} = N I$ (A, "At")	electromotive force $E$ (V)
flux $Φ = \int B \cdot d S$ (Wb)	current $I = \int J \cdot d S$ (A)
reluctance $Z = l / (μ S)$ ( $H^{- 1}$ )	resistance $R = l / (σ S)$ ( $Ω$ )
permeability $μ$ , $B = μ H$	conductivity $σ$ , $J = σ E$
magnetic potential drop $H l = Z Φ$ (A)	voltage drop $E l = R I$ (V)
$\sum Z Φ = \sum N I$ around a loop	$\sum R I = \sum E$ around a loop
$\sum Φ = 0$ at a node	$\sum I = 0$ at a node

Ohm's law of the magnetic circuit

F_{m} = N I = Z Φ, Z = \frac{l}{μ S} . (4.3.16)

In plain words

Ampere-turns push flux through reluctance, as volts push current through resistance. Every rule you know for resistor networks (series, parallel, mesh and node equations) can be used, with the dictionary of Table 4.3.1.

Common mistake · Trusting the analogy too far

The analogy is formal, and it fails in two places. First, $σ_{Cu}$ is a constant, whereas $μ_{iron}$ depends on $B$ (Section 4.3.6): a magnetic circuit with iron in it is nonlinear, and $Z = l / (μ S)$ is only useful while the iron is far from saturation. Second, flux always leaks. The ratio $μ_{iron} / μ_{0}$ is $1 0^{3}$ to $1 0^{5}$ , whereas $σ_{Cu} / σ_{air}$ is about $1 0^{20}$ : air is a nearly perfect insulator for current but a rather poor insulator for flux, so a few per cent of the flux take short cuts outside the core (Section 4.3.8). There is no magnetic "wire" as good as a copper wire.

Quick check

A closed ring core of mean length $l = 20 cm$ , cross-section $S = 2 c m^{2}$ and $μ_{r} = 800$ carries $200$ turns with $I = 1 A$ . What are $Z$ , $Φ$ and $B$ ?

$Z = l / (μ_{0} μ_{r} S) = 0.20/ (4 π \times 1 0^{- 7} \times 800 \times 2 \times 1 0^{- 4}) = 9.95 \times 1 0^{5} H^{- 1}$ , $Φ = N I / Z = 200/ (9.95 \times 1 0^{5}) = 0.201 mWb$ and $B = Φ/ S = 1.0 T$ (Exercise 4.3.1 continues this calculation).

4.3.3 The electric analogue