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

Almost every useful magnetic circuit has an air gap: a motor needs one so that the rotor can turn, a lifting magnet has one under each pole, a recording head writes through one, and an energy-storing inductor is given one on purpose. The gap is always short. Why does it matter so much?

Take the gapped core of Figure 4.3.7a, with an iron path of length $l_{c}$ and relative permeability $μ_{r}$ , and a gap of length $l_{g}$ . Core and gap carry the same flux, so their reluctances add, Eq. (4.3.17), and their ratio is

Z_{c} = \frac{l _{c}}{μ _{0} μ _{r} S}, Z_{g} = \frac{l _{g}}{μ _{0} S}, \frac{Z _{g}}{Z _{c}} = μ_{r} \frac{l _{g}}{l _{c}} . (4.3.19)

The ratio of the lengths is multiplied by $μ_{r}$ , which is in the thousands.

When the gap takes over

The gap and the iron have equal reluctance when

l_{g} = \frac{l _{c}}{μ _{r}}, (4.3.20)

that is, a millimetre of air is worth $μ_{r}$ millimetres of iron. For $μ_{r} = 4000$ and $l_{c} = 40 cm$ this happens at $l_{g} = 0.1 mm$ .

(a) C-core with a gap l_g. — (a) C-core with a gap $l_{g}$ .

(b) \Phi against the gap length at fixed NI. Why a millimetre of air beats half a metre of iron. (a) The gap is in series with the iron. (b) Flux of a silicon-steel C-core (l_c=0.40\,\mathrm{m}, S=4.0\,\mathrm{cm^2}, \mu_r=4000, F_m=NI=250\,\mathrm{At}) as the gap is opened: Z_c=1.99\times10^{5}\,\mathrm{H^{-1}}, and one millimetre of air adds Z_g=1.99\times10^{6}\,\mathrm{H^{-1}}, ten times the whole iron path. At l_g=0.1\,\mathrm{mm} the flux is already halved. — **Figure 4.3.7.** (b) $Φ$ against the gap length at fixed $N I$ . Why a millimetre of air beats half a metre of iron. (a) The gap is in series with the iron. (b) Flux of a silicon-steel C-core ( $l_{c} = 0.40 m$ , $S = 4.0 c m^{2}$ , $μ_{r} = 4000$ , $F_{m} = N I = 250 At$ ) as the gap is opened: $Z_{c} = 1.99 \times 1 0^{5} H^{- 1}$ , and one millimetre of air adds $Z_{g} = 1.99 \times 1 0^{6} H^{- 1}$ , ten times the whole iron path. At $l_{g} = 0.1 mm$ the flux is already halved.

Figure 4.3.7b puts numbers to this. The iron path of a silicon-steel C-core, $40 cm$ long with $μ_{r} = 4000$ and $S = 4 c m^{2}$ , has $Z_{c} = 0.40/ (4 π \times 1 0^{- 7} \times 4000 \times 4 \times 1 0^{- 4}) = 1.99 \times 1 0^{5} H^{- 1}$ . One millimetre of air in the same cross-section has $Z_{g} = 1 0^{- 3} / (4 π \times 1 0^{- 7} \times 4 \times 1 0^{- 4}) = 1.99 \times 1 0^{6} H^{- 1}$ , ten times as much. With $N I = 250 At$ the flux falls from $1257 μ Wb$ without a gap to about $114 μ Wb$ with the millimetre gap. Two practical consequences follow.

In a gapped circuit one may often neglect the iron altogether to first order: the flux is $Φ \approx μ_{0} S N I / l_{g}$ and the gap field is $H_{g} \approx N I / l_{g}$ . Problem 4.3.1 and Home exercise 4.3.3 show the gap taking three quarters of the mmf.
A gap one does not want hurts just as much. A $0.1 mm$ burr or a grain of dust in a butt joint of a laminated core halves the flux of the example, and a transformer core is stacked with interleaved laminations precisely to avoid such joints.

In plain words

Flux "sees" a millimetre of air as $μ_{r}$ millimetres of iron. In most gapped circuits the gap sets the flux and the iron is a short circuit, which is both a nuisance (unwanted gaps) and a design tool (wanted gaps, Section 4.3.7).

Quick check

A ring of $μ_{r} = 2000$ and mean length $30 cm$ is cut open by $1 mm$ . By what factor does the flux fall at constant current?

$Z_{g} / Z_{c} = μ_{r} l_{g} / l_{c} = 2000 \times 0.001/0.30 = 6.7$ , so $Z$ grows by the factor $1 + 6.7 = 7.7$ and the flux falls by $7.7$ . Home exercise 4.3.1 gives the full numbers.

4.3.5 The air gap