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

Two things have been neglected so far, and both come from the fact that air is a poor insulator for flux (the second failure of the analogy in Section 4.3.3). Figure 4.3.15 shows them on a gapped C-core. In the gap the flux tube spreads: the lines bulge out beyond the pole faces, and this is called fringing. And part of the flux of the coil short-cuts the window from limb to limb without going round the loop at all, and never reaches the gap: this is leakage.

**Figure 4.3.15.** Fringing and leakage in a gapped C-core (cross-section, winding on the left limb). The useful flux goes round the core and straight across the gap. The fringing flux bulges out of the gap on both sides. The leakage flux crosses the window from limb to limb and never reaches the gap.

Fringing makes the gap better than the formula says

Outside the gap the flux tube spreads, so the effective cross-section of the gap is larger than the pole area, and the gap reluctance is smaller than $l_{g} / (μ_{0} S)$ . A standard first correction simply adds $l_{g}$ to each transverse dimension of a rectangular pole $a \times b$ (Figure 4.3.16).

Fringing correction

S_{eff} = (a + l_{g}) (b + l_{g}), Z_{g} \approx \frac{l _{g}}{μ _{0} S _{eff}} (l_{g} ≪ a, b) . (4.3.27)

Fringing at a gap between two poles of cross-section a\times b. The flux tube widens outside the poles, and to first order the gap behaves as if its area were (a+l_g)(b+l_g). — **Figure 4.3.16.** Fringing at a gap between two poles of cross-section $a \times b$ . The flux tube widens outside the poles, and to first order the gap behaves as if its area were $(a + l_{g}) (b + l_{g})$ .

How large is the error? For a C-core with $a = b = 20 mm$ and $l_{g} = 2.0 mm$ ,

⟹ S_{eff} = (22 mm)^{2} = 4.84 c m^{2} = 1.21 S Z_{g} falls by 17 %,

which raises $Φ$ by about $20 %$ in a circuit dominated by the gap, far too much to ignore in a design. This is why the problems at the end of this section all say "neglect fringing" explicitly.

Switch fringing on at $S = 4$ cm² and $l_{g} = 2$ mm and see $Z_{g}$ fall by 17 %.

InteractiveLoading simulation…

Leakage makes the coil worse

Leakage is the other error, and it acts the other way. The flux that jumps the window without going round the loop is a reluctance $Z_{ℓ}$ in parallel with the rest of the circuit, so the coil carries more flux than the useful branch does:

Φ_{coil} = Φ_{useful} + Φ_{ℓ}, k_{ℓ} = \frac{Φ _{coil}}{Φ _{useful}} (1.1 to 1.2 typically) . (4.3.28)

The leakage coefficient $k_{ℓ}$ tells the designer how much larger the core section must be than the useful flux alone would require. Exercise 4.3.10 works through a circuit with $k_{ℓ} = 1.4$ .

In plain words

Fringing makes the gap better than the formula says, leakage makes the coil worse. Both grow when the gap is long compared with the pole size, which is one more reason to split a long gap into several short ones.

Quick check

A square pole of $10 mm$ side faces a $1 mm$ gap. By how much does the fringing correction change $Z_{g}$ ?

$S_{eff} = (11 mm)^{2} = 1.21 S$ , so $Z_{g}$ falls by $17 %$ (Exercise 4.3.9).

Application 4.3.1 · Reluctance and stepper motors

The machines of this application carry no magnet and no rotor winding. The rotor is a toothed piece of iron, and the only thing that changes as it turns is the reluctance seen by the excited stator phase. Figure 4.3.17 shows the commonest layout, a switched-reluctance machine with six stator poles and four rotor teeth. The two opposite poles of phase A carry one winding. When it is energised, the flux goes from one pole across the air gap into the nearest rotor tooth, through the rotor body, out of the opposite tooth, across the second gap into the opposite pole, and back through the stator yoke. With the teeth aligned under the poles the two gaps are short and the reluctance $Z$ of this path is small. With the teeth half-way between the poles the flux has to cross a wide stretch of air and $Z$ is large.

unaligned: Z large. (b) The two extreme positions. A 6/4 switched-reluctance machine. (a) The flux of the energised phase A crosses the two air gaps into the aligned rotor teeth and returns through the yoke. (b) Aligned teeth give a short gap and a small reluctance, unaligned teeth a long air path and a large reluctance. The rotor is pulled towards the aligned position. — **Figure 4.3.17.** unaligned: $Z$ large. (b) The two extreme positions. A $6/4$ switched-reluctance machine. (a) The flux of the energised phase A crosses the two air gaps into the aligned rotor teeth and returns through the yoke. (b) Aligned teeth give a short gap and a small reluctance, unaligned teeth a long air path and a large reluctance. The rotor is pulled towards the aligned position.

The torque. Let $ϑ$ be the rotor angle. The inductance of the excited phase is $L = N^{2} / Z (ϑ)$ (Section 4.2), largest in the aligned position, and the energy stored in the circuit is $W_{m} = \frac{1}{2} L i^{2}$ . If the rotor turns by $d ϑ$ while the current is held constant, the torque is the derivative of the stored energy at constant current (the magnetic counterpart of the constant-voltage force on a dielectric slab in Section 3.4):

T = \frac{\partial W _{m}}{\partial ϑ}_{i} = \frac{1}{2} i^{2} \frac{d L}{d ϑ} = - \frac{1}{2} \frac{N ^{2} i ^{2}}{Z ^{2}} \frac{d Z}{d ϑ} . (4.3.29)

The torque has the sign of $d L / d ϑ$ and the sign of $- d Z / d ϑ$ , whatever the sign of $i$ : the rotor is always pulled towards the position of minimum reluctance, exactly as a piece of iron is pulled into a coil. Energising the phases A, B and C in turn moves the position of minimum reluctance round the stator, and the rotor follows, by $3 0^{\circ}$ a step for the $6/4$ machine. With a constant current the torque is zero in the aligned position (stable) and in the unaligned position (unstable), and its maximum lies in between (Exercise 4.3.6).

Typical numbers. A small switched-reluctance drive with $L$ swinging between $30 mH$ (unaligned) and $70 mH$ (aligned) over $9 0^{\circ}$ of rotor angle, excited with $3 A$ , develops a peak torque of order $\frac{1}{2} i^{2} Δ L /Δ ϑ \approx \frac{1}{2} \times 9 \times 0.04/ (π /2) \approx 0.11 N m$ per phase, and with a sinusoidal $L (ϑ)$ of the same swing the maximum is $i^{2} L_{1} = 0.18 N m$ (Exercise 4.3.6). Since the torque scales with $i^{2}$ , doubling the current quadruples it, until the teeth saturate.

Where you meet it

The switched-reluctance machine ( $6/4$ stator and rotor poles, as in Figure 4.3.17) needs no rare-earth magnet: vacuum cleaners, washing-machine drives and some traction motors use it. The hybrid stepper adds a permanent magnet and fine rotor teeth to reach the familiar $1. 8^{\circ}$ step ( $200$ steps per revolution) of printers and machine tools.

Application 4.3.2 · Electromagnets and lifting magnets

A U-shaped core with a winding on its yoke, held above a steel plate, is the simplest electromagnet (Figure 4.3.18). The flux driven by $N I$ runs down one limb, crosses the air gap $l_{g}$ under the first pole face, passes along the plate, crosses the gap under the second pole face and returns through the second limb and the yoke: a series circuit with the iron of the magnet, the iron of the plate and two air gaps. The plate is pulled towards the poles, and we want the force.

A lifting electromagnet. The flux \Phi of the N-turn winding crosses the two air gaps of length l_g under the pole faces of area S and closes through the steel plate, which is pulled up with the force F. — **Figure 4.3.18.** A lifting electromagnet. The flux $Φ$ of the $N$ -turn winding crosses the two air gaps of length $l_{g}$ under the pole faces of area $S$ and closes through the steel plate, which is pulled up with the force $F$ .

The force on the plate. The energy stored in one gap is $w = B^{2} / (2 μ_{0})$ per unit volume (Section 4.2), that is $W = B^{2} S l_{g} / (2 μ_{0})$ in the gap of area $S$ and length $l_{g}$ . Pull the plate away by $d l_{g}$ at constant flux: $B = Φ/ S$ does not change, the volume of the gap grows by $S d l_{g}$ , and the energy added to the field, $d W = B^{2} S d l_{g} / (2 μ_{0})$ , is the work done against the magnetic force. Hence the pull on one pole face is

F = \frac{\partial W}{\partial l _{g}} = \frac{B ^{2} S}{2 μ _{0}} per gap, \frac{F}{S} = \frac{B ^{2}}{2 μ _{0}} . (4.3.30)

Magnetic pressure

A pole face of area $S$ across which the induction is $B$ pulls with the pressure $B^{2} / (2 μ_{0})$ . The U-magnet of Figure 4.3.18 has two gaps, so it lifts

F_{total} = 2 \times \frac{B ^{2} S}{2 μ _{0}} = \frac{B ^{2} S}{μ _{0}} .

Table 4.3.3. The magnetic pressure $B^{2} / (2 μ_{0})$ on a pole face.

$B$ (T)	$0.5$	$1.0$	$1.5$
$F / S$ (kPa)	$99.5$	$398$	$895$
$F / S$ ( $t/ m^{2}$ )	$10.1$	$40.6$	$91.3$

Typical numbers. Table 4.3.3 evaluates Eq. (4.3.30). The pressure goes as $B^{2}$ , and iron saturates near $2 T$ , so no electromagnet can pull much more than $1.6 MPa \approx 16 kg$ per $c m^{2}$ , a hard ceiling set by the material, not by the current. Scrap-yard magnets therefore get their tonnage from area, not from field: a $1.5 m$ disc at $1.0 T$ already gives about $70$ tonnes. The current, on the other hand, sets $B$ through the loop rule, in which the gaps dominate (Section 4.3.5): doubling $l_{g}$ (a rusty or uneven load) at fixed $N I$ roughly halves $B$ and quarters the force, which is why these magnets must sit flat on clean steel. Home exercise 4.3.3 and Exercise 4.3.8 put numbers to a $100 kg$ plate and to the $1/ l_{g}^{2}$ collapse of the force at fixed current.

Switch to 'Given I → F' and drag the plate away: the pull collapses roughly as $1/ l_{g}^{2}$ , while the iron takes only a small share.

InteractiveLoading simulation…

Common mistake · Forgetting the second gap

Writing $N I = H_{iron} l_{iron} + B l_{g} / μ_{0}$ for the U-magnet. The flux crosses the air twice, once under each pole, so the gap term is $2 B l_{g} / μ_{0}$ , and the force is the sum of the pulls on the two faces. The same "twice" occurs in Problem 4.3.1, where the flux crosses the gap between the two C-cores on both sides.

Quick check

A pole face of $10 c m^{2}$ carries $B = 1.11 T$ . What does it pull with, and how much mass do two such faces hold?

$F = B^{2} S / (2 μ_{0}) = 1.1 1^{2} \times 1 0^{- 3} / (2 \times 4 π \times 1 0^{- 7}) = 490 N$ per face, $981 N$ for two, i.e. $100 kg$ (Home exercise 4.3.3).

Table 4.3.4. The applications of this section.

Application	Physics used	Worked out in
Reluctance and stepper motors	$L = N^{2} / Z (ϑ)$ , $T = \frac{1}{2} i^{2} d L / d ϑ$	Exercise 4.3.6
Electromagnets and lifting magnets	two gaps in series, $F = B^{2} S / (2 μ_{0})$ per gap	Home exercise 4.3.3, Exercise 4.3.8

4.3.8 Fringing and leakage

Fringing makes the gap better than the formula says

Leakage makes the coil worse