Magnetic Circuits and Induction

The hysteresis loop is the closed B–H curve traced by a ferromagnetic material when the magnetising force H is taken through one complete cycle (+Hmax → 0 → −Hmax → 0 → +Hmax). Because B lags behind H, the curve for decreasing H does not retrace the rising curve, and the enclosed area represents energy lost as heat in every cycle.

The hysteresis loop

               B
               |      a (saturation)
           b   |  ____*
            *--|-'   /
           /   |    /
     c    /    |   /
  ---*---/-----+--/----*--- H
        /      |  /    f
       /   ____|_*
  d  *'        |  e
               |

• o–a: an unmagnetised sample is magnetised along the initial curve to saturation (point a).
• a–b: when H is reduced to zero, B does not fall to zero. The value ob is the residual flux density (retentivity).
• b–c: a reverse H is needed to make B zero. oc is the coercive force (coercivity).
• c–d: further reverse H gives reverse saturation at d.
• d–e–f–a: reducing H to zero gives the negative residual flux density oe, and positive H (of) is needed to bring B to zero; the loop closes at a.

Reason: the magnetic domains turn into line with H but do not all turn back when H is removed. Turning them back needs extra energy, which appears as heat.

Proof: loop area ∝ energy loss per cycle per unit volume

Take a ring of mean length $l$, area $A$, wound with $N$ turns carrying current $i$.

$$H = \frac{Ni}{l} \Rightarrow i = \frac{Hl}{N}$$

If the flux density changes by $dB$ in time $dt$, the induced emf is

$$e = N\frac{d\phi}{dt} = NA\frac{dB}{dt}$$

The source must supply power $ei$ against this emf. Energy supplied in time $dt$:

$$\begin{aligned} dW &= e\,i\,dt = NA\frac{dB}{dt}\cdot\frac{Hl}{N}\,dt \\ &= (A l)\, H\, dB \end{aligned}$$

Energy per unit volume (volume $= Al$): $dw = H\,dB$. Over one complete cycle:

$$w = \oint H\,dB \quad \text{J/m}^3 \text{ per cycle}$$

$\oint H\,dB$ is exactly the area enclosed by the B–H loop. Part of the energy given while B rises is returned while B falls; the difference, equal to the loop area, is lost as heat. Hence

$$W_h = (\text{area of loop}) \times \text{volume} \times f \quad \text{watts}$$

If the loop is drawn to scales of $x$ A/m per cm and $y$ T per cm, loop area (in J/m³) = measured area (cm²) × $xy$.

Practical result

Steinmetz found empirically $P_h = \eta B_{max}^{1.6} f V$. Machines therefore use silicon steel, whose loop is narrow (small area), to keep hysteresis loss low.

The centre limb and the right limb are two parallel paths between the same two junctions, so they have the same mmf drop. The left limb (with the coil) carries the total flux.

Assumed mean path lengths (from the figure): left path A = 25 + 15 + 25 = 65 cm, centre limb B = 15 cm, right path C = 25 + 15 + 25 = 65 cm. Area $a = 2 \times 2 = 4$ cm² $= 4\times10^{-4}$ m² throughout.

   +-----25-----+-----25-----+
   |            |            |
 [500T]  phi ^  | phi1  v    | phi2 v
   | A          | B (15)     | C
   +------------+------------+

Reluctances

$$\begin{aligned} S &= \frac{l}{\mu_0\mu_r a}, \quad \mu_0\mu_r a = 4\pi\times10^{-7}\times600\times4\times10^{-4} \\ S_A = S_C &= \frac{0.65}{3.016\times10^{-7}} = 2.155\times10^{6}\ \text{AT/Wb} \\ S_B &= \frac{0.15}{3.016\times10^{-7}} = 4.974\times10^{5}\ \text{AT/Wb} \end{aligned}$$

Fluxes

MMF across the parallel limbs (B and C):

$$\begin{aligned} F_{BC} &= \phi_2 S_C = 2\times10^{-3}\times2.155\times10^6 = 4310.4\ \text{AT} \\ \phi_1 &= \frac{F_{BC}}{S_B} = \frac{4310.4}{4.974\times10^5} = 8.667\ \text{mWb} \\ \phi &= \phi_1 + \phi_2 = 8.667 + 2 = 10.667\ \text{mWb} \end{aligned}$$

(Equivalently $\phi_1 = \phi_2 \times 65/15$ because the limbs differ only in length.)

Total mmf and current

$$\begin{aligned} NI &= \phi S_A + F_{BC} \\ &= 10.667\times10^{-3}\times2.155\times10^{6} + 4310.4 \\ &= 22989 + 4310 = 27299\ \text{AT} \\ I &= \frac{27299}{500} = 54.6\ \text{A} \end{aligned}$$

Answer: I ≈ 54.6 A.

(Note: with μr = 600 held constant the flux densities come out very high, e.g. 5 T in the right limb; a real cast-steel core would saturate, so the result is valid only for the linear data given.)

With leakage and fringing neglected, the same flux passes through the iron and the gap, and the area is the same. Iron path length is taken as $l_i = 0.8 - 0.005 = 0.795$ m.

Data: $\phi = 1$ mWb, $A = 120$ cm² $= 0.012$ m², $l_g = 5$ mm, $N = 220$, $\mu_r = 500$.

Flux density

$$B = \frac{\phi}{A} = \frac{1\times10^{-3}}{0.012} = 0.0833\ \text{T}$$

Ampere-turns for the air gap

$$\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{0.0833}{4\pi\times10^{-7}} = 66315\ \text{AT/m} \\ AT_g &= H_g l_g = 66315 \times 0.005 = 331.6\ \text{AT} \end{aligned}$$

Ampere-turns for the iron

$$\begin{aligned} H_i &= \frac{B}{\mu_0\mu_r} = \frac{66315}{500} = 132.6\ \text{AT/m} \\ AT_i &= H_i l_i = 132.6 \times 0.795 = 105.4\ \text{AT} \end{aligned}$$

Magnetising current

$$\begin{aligned} NI &= 331.6 + 105.4 = 437.0\ \text{AT} \\ I &= \frac{437.0}{220} = 1.986\ \text{A} \end{aligned}$$

Answer: magnetising current ≈ 1.99 A. (Taking the iron length as the full 0.8 m gives 1.989 A, practically the same.) Note that the 5 mm gap needs about three times the ampere-turns of the 0.795 m of iron.

For a sinusoidal supply voltage the core flux is forced to be sinusoidal; because the B–H (φ–i) curve is non-linear (saturation and hysteresis), the exciting current is non-sinusoidal. It is peaky and contains odd harmonics, mainly the 3rd.

Why the flux is sinusoidal

Neglecting winding resistance and leakage,

$$v = V_m \cos\omega t = N\frac{d\phi}{dt} \Rightarrow \phi = \frac{V_m}{N\omega}\sin\omega t = \phi_m \sin\omega t$$

So the flux follows the voltage waveform, whatever the core is like. The current needed for each value of flux is then read from the φ–i curve.

Shape of the exciting current (graphical method)

 phi-i curve            waveforms vs time
   phi                  phi (sine)   i0 (peaky)
    |    ___              |   _          ^
    |  /                  | /   \       / \
    | /                   |/     \  __./   \.__
 ---+------ i0         ---+-------\----------- t
   /|                             \_/
 _/ |

Steps: for each instant take φ from the sine wave, find the matching i on the φ–i curve, and plot i against time.

1. Saturation only (no hysteresis): near the peak of φ the core is saturated, so a small extra flux needs a large current. The current is peaked, symmetric about the flux peak, and in phase with φ. It contains the fundamental plus large 3rd, 5th, … harmonics.
2. With hysteresis: the rising and falling branches differ, so the current wave is also unsymmetrical and its zero crossings come before those of the flux. The current leads the flux slightly.

Expression and components

The exciting current can be written as a Fourier series of odd harmonics:

$$i_0(t) = I_1\sin(\omega t + \theta_1) + I_3\sin(3\omega t + \theta_3) + I_5\sin(5\omega t + \theta_5) + \dots$$

The fundamental component is split into:

• Magnetising current $I_m = I_0 \sin\phi_0$, in phase with φ (lags V by 90°); it sets up the flux.
• Core-loss current $I_c = I_0\cos\phi_0$, in phase with V; it supplies hysteresis and eddy losses, $P_c = V I_c$.

$$I_0 = \sqrt{I_m^2 + I_c^2}$$

Consequences

• The 3rd harmonic is typically 30–40% of the fundamental in a transformer worked near the knee.
• In three-phase banks, a delta winding or a neutral path is provided for these 3rd-harmonic currents; otherwise the flux and voltages become distorted.
• Because $I_0$ is only 2–5% of rated current, the distortion is usually neglected and an equivalent sine wave is used in the equivalent circuit.

Treat the core as a two-loop magnetic circuit. Assumption (polarities are not given): all three windings are connected so that their mmfs drive flux in the same sense through the centre limb, i.e. the outer coils push flux up the outer limbs and the centre coil pushes it down the centre limb. Each outer path has length $l_A = 50$ cm; the centre limb has iron $l_1 + l_2 = 30$ cm plus the 10 mm gap. Fringing is neglected.

   +----------+----------+
   |  phiA ^  |  phic v  |  phiB ^
 [N iA]       [N1 i1]  [N iB]
   |          |   g      |
   +----------+----------+

MMFs

$$F_A = 500\times20 = 10000\ \text{AT}, \quad F_B = 500\times15 = 7500\ \text{AT}, \quad F_1 = 1000\times10 = 10000\ \text{AT}$$

Reluctances ($A_c = 20\times10^{-4}$ m², $\mu_r = 4000$)

$$\begin{aligned} S_A = S_B &= \frac{0.5}{4\pi\times10^{-7}\times4000\times20\times10^{-4}} = 4.974\times10^{4}\ \text{AT/Wb} \\ S_{core,c} &= \frac{0.3}{4\pi\times10^{-7}\times4000\times20\times10^{-4}} = 2.984\times10^{4}\ \text{AT/Wb} \\ S_g &= \frac{0.01}{4\pi\times10^{-7}\times20\times10^{-4}} = 3.979\times10^{6}\ \text{AT/Wb} \\ S_c &= S_{core,c} + S_g = 4.009\times10^{6}\ \text{AT/Wb} \end{aligned}$$

Node equation

Let $U$ be the magnetic potential of the top junction relative to the bottom. Then $\phi_A = (F_A - U)/S_A$, $\phi_B = (F_B - U)/S_B$, $\phi_c = (U + F_1)/S_c$, and $\phi_c = \phi_A + \phi_B$:

$$\begin{aligned} U &= \frac{F_A/S_A + F_B/S_B - F_1/S_c}{1/S_A + 1/S_B + 1/S_c} = 8634\ \text{AT} \\ \phi_c &= \frac{8634 + 10000}{4.009\times10^{6}} = 4.648\times10^{-3}\ \text{Wb} \end{aligned}$$

(The outer limbs carry $\phi_A = 27.46$ mWb up and $\phi_B = -22.81$ mWb, i.e. 22.81 mWb down; their sum is 4.65 mWb.)

Flux density in the air gap

$$B_g = \frac{\phi_c}{A_c} = \frac{4.648\times10^{-3}}{20\times10^{-4}} = 2.32\ \text{T}$$

Answer: $B_g \approx 2.32$ T (for the assumed winding polarities, linear core, no fringing). Because the gap reluctance is about 80 times that of the iron, $B_g \approx \mu_0(\bar F_{outer} + F_1)/g$, so the result is mainly set by the gap.

A magnetic circuit is the closed path followed by magnetic flux, e.g. the iron core of a transformer with its air gaps. It behaves much like an electric circuit, with mmf driving flux against reluctance.

Similarities

Differences (in brief): flux does not actually flow and needs no energy to keep it up, while current does; $\mu$ changes with B (saturation), while R is nearly constant.

Work law (Ampere's circuital law)

The work done in moving a unit magnetic pole once round a closed path equals the ampere-turns enclosed:

$$\oint \vec H \cdot d\vec l = NI$$

For a path made of sections in which H is constant,

$$H_1 l_1 + H_2 l_2 + \dots = NI = F$$

So the total mmf equals the sum of the "magnetic drops" $Hl$ round the circuit, just as KVL says the emf equals the sum of IR drops.

Ohm's law for a magnetic circuit

Take a ring of mean length $l$, area $A$, permeability $\mu = \mu_0\mu_r$, with $N$ turns carrying $I$. From the work law $H = NI/l$. Then

$$\begin{aligned} B &= \mu H = \frac{\mu NI}{l} \\ \phi &= BA = \frac{\mu A NI}{l} = \frac{NI}{l/(\mu A)} \end{aligned}$$ $$\phi = \frac{F}{S}, \quad S = \frac{l}{\mu_0 \mu_r A}$$

i.e. flux = mmf / reluctance, the magnetic form of $I = V/R$.

Hysteresis is the lagging of flux density B behind the magnetising force H in a ferromagnetic material. Under ac excitation H goes through a full cycle every period, so the material traces a closed hysteresis loop each cycle and loses energy as heat.

Mechanism

• A ferromagnetic material is made of small domains, each fully magnetised. With no field they point randomly, so the net B is zero.
• As H rises, domains in line with H grow and others turn, so B rises steeply and then saturates when almost all are aligned.
• When H is reduced, many domains do not turn back; B stays higher than on the rising curve. Turning them back requires work, which is lost as heat. This is the cause of hysteresis.

Hysteresis loop with ac excitation

              B
              |   ___ a  (+Bmax)
          b  _|-'   /
            / |    /
     c     /  |   /    
  --*-----/---+--/---*--- H
         /    | /    f
        /  ___|'
   d  _/-'    |e
   (-Bmax)    |

With an ac current $i = I_m \sin\omega t$, H varies sinusoidally between $+H_m$ and $-H_m$:

1. H rises 0 → $+H_m$: B rises to $+B_m$ (point a).
2. H falls to 0: B falls only to residual flux density $B_r$ (point b), called retentivity.
3. H reversed to $-H_c$: B becomes zero (point c); $H_c$ is the coercive force.
4. H to $-H_m$: B reaches $-B_m$ (point d).
5. H back through 0 ($-B_r$, point e) and $+H_c$ (point f) to $+H_m$: the loop closes.

The loop is traced f times per second. For each cycle the energy lost per unit volume is $\oint H\,dB$, the area of the loop.

Hysteresis loss

$$P_h = \eta\, B_m^{1.6} f\, V \quad \text{W}$$

where $\eta$ is the Steinmetz coefficient (J/m³), $B_m$ the peak flux density, $f$ the frequency and $V$ the core volume. The index 1.6 varies from 1.5 to 2.5 for different materials.

Effects in machines

• Heats the core and lowers efficiency of transformers, motors and generators.
• Makes the exciting current non-sinusoidal and slightly leading the flux (core-loss component).
• Reduced by using soft magnetic materials (silicon steel, CRGO steel) with a narrow loop, and by keeping $B_m$ moderate.

The centre limb (with the coil) carries the total flux, which divides between the left and right outer paths. Each outer path contains an air gap, so the two outer paths are in parallel and both are in series with the centre limb.

Dimensions: limbs 7 cm wide, depth 7 cm → $A = 49$ cm² $= 4.9\times10^{-3}$ m². Windows 30 × 30 cm, so the centre-line spacing is 30 + 7 = 37 cm.

• Centre limb: $l_c = 37$ cm.
• Each outer path: top yoke 37 + outer limb 37 + bottom yoke 37 = 111 cm (minus its gap).
• Gap area with 5% fringing: $A_g = 1.05 \times 49 = 51.45$ cm².
• MMF: $F = 300 \times 10 = 3000$ AT.

Reluctances ($\mu_0\mu_r A = 4\pi\times10^{-7}\times4000\times4.9\times10^{-3} = 2.463\times10^{-5}$)

$$\begin{aligned} S_c &= \frac{0.37}{2.463\times10^{-5}} = 15022\ \text{AT/Wb} \\ S_{L,iron} &= \frac{1.1095}{2.463\times10^{-5}} = 45047, \quad S_{L,gap} = \frac{0.0005}{4\pi\times10^{-7}\times5.145\times10^{-3}} = 77335 \\ S_L &= 45047 + 77335 = 122381\ \text{AT/Wb} \\ S_{R,iron} &= \frac{1.1093}{2.463\times10^{-5}} = 45038, \quad S_{R,gap} = \frac{0.0007}{4\pi\times10^{-7}\times5.145\times10^{-3}} = 108269 \\ S_R &= 45038 + 108269 = 153307\ \text{AT/Wb} \end{aligned}$$

Total reluctance and fluxes

$$\begin{aligned} S_L \parallel S_R &= \frac{122381\times153307}{122381+153307} = 68055\ \text{AT/Wb} \\ S_{tot} &= 15022 + 68055 = 83077\ \text{AT/Wb} \\ \phi_c &= \frac{3000}{83077} = 0.03611\ \text{Wb} \\ F_{LR} &= \phi_c \times 68055 = 2457.6\ \text{AT} \\ \phi_L &= \frac{2457.6}{122381} = 0.02008\ \text{Wb}, \quad \phi_R = \frac{2457.6}{153307} = 0.01603\ \text{Wb} \end{aligned}$$

Flux density in the gaps

$$B_{gL} = \frac{0.02008}{5.145\times10^{-3}} = 3.90\ \text{T}, \quad B_{gR} = \frac{0.01603}{5.145\times10^{-3}} = 3.12\ \text{T}$$

Answer: $\phi_{middle} = 36.1$ mWb, $\phi_{left} = 20.1$ mWb, $\phi_{right} = 16.0$ mWb; $B_{gap,left} = 3.90$ T, $B_{gap,right} = 3.12$ T. (These densities are far above saturation of real iron; the result holds only for the constant μr = 4000 given.)

Flux is found from $\phi = NI/(S_i + S_g)$. Iron length is taken as $l_i = 80 - 0.2 = 79.8$ cm; gap area = core area (no fringing).

Data: $A = 16$ cm² $= 16\times10^{-4}$ m², $l_g = 2$ mm, $N = 1000$, $I = 6$ A, $\mu_r = 2000$.

MMF

$$F = NI = 1000 \times 6 = 6000\ \text{AT}$$

Reluctances

$$\begin{aligned} S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{0.798}{4\pi\times10^{-7}\times2000\times16\times10^{-4}} = 1.984\times10^{5}\ \text{AT/Wb} \\ S_g &= \frac{l_g}{\mu_0 A} = \frac{0.002}{4\pi\times10^{-7}\times16\times10^{-4}} = 9.947\times10^{5}\ \text{AT/Wb} \\ S &= S_i + S_g = 1.193\times10^{6}\ \text{AT/Wb} \end{aligned}$$

Flux and flux density

$$\begin{aligned} \phi &= \frac{F}{S} = \frac{6000}{1.193\times10^{6}} = 5.03\times10^{-3}\ \text{Wb} \\ B &= \frac{\phi}{A} = \frac{5.03\times10^{-3}}{16\times10^{-4}} = 3.14\ \text{T} \end{aligned}$$

Answer: φ ≈ 5.03 mWb, B ≈ 3.14 T in the core (and in the gap). The 2 mm gap has five times the reluctance of the whole iron path. (If the full 80 cm is used as iron length, φ = 5.02 mWb; the difference is negligible.)

By Faraday's law an emf is induced in a conductor whenever the flux linking it changes. If the change is caused by movement of the conductor (or field) it is a dynamically induced emf; if the conductor and field are both still and the flux itself changes with time, it is a statically induced emf.

Dynamically induced (motional) emf

A conductor of length $l$ moving with velocity $v$ at angle $\theta$ to a field of flux density $B$ cuts flux and has an emf

$$e = B\,l\,v\sin\theta \quad \text{V}$$

Direction is given by Fleming's right-hand rule.

Depends on:

• flux density B,
• active length of conductor l,
• speed v,
• angle θ between the motion and the field (maximum at 90°).

Example: the armature conductors of a dc generator or the stator conductors of an alternator; the emf of a dc generator is $E = \frac{\phi Z N P}{60A}$.

Statically induced emf

The conductor is stationary and the flux linking it changes. Two kinds:

(a) Self-induced emf: emf induced in a coil by the change of its own current.

$$e = -L\frac{di}{dt}, \quad L = \frac{N\phi}{I} = \frac{N^2}{S} = \frac{\mu_0\mu_r N^2 A}{l}$$

Depends on the number of turns, the reluctance (core material, length, area) and the rate of change of current. Example: the emf in a choke or the spark across a switch when an inductive circuit is opened.

(b) Mutually induced emf: emf induced in one coil by the change of current in a nearby coil.

$$e_2 = -M\frac{di_1}{dt}, \quad M = k\sqrt{L_1L_2}$$

Depends on the turns of both coils, the reluctance of the common path, the coupling coefficient $k$, and $di_1/dt$. Example: the transformer, where an alternating primary current induces an emf in the secondary: $E = 4.44 f N \phi_m$.

Comparison

The bar length becomes the mean length of the ring. Area is that of a 2 cm diameter circle.

$$A = \frac{\pi d^2}{4} = \frac{\pi (0.02)^2}{4} = 3.1416\times10^{-4}\ \text{m}^2$$ $$B = \frac{\phi}{A} = \frac{0.05\times10^{-3}}{3.1416\times10^{-4}} = 0.1592\ \text{T}$$

a) No air gap ($l_i = 0.30$ m)

$$\begin{aligned} H_i &= \frac{B}{\mu_0\mu_r} = \frac{0.1592}{4\pi\times10^{-7}\times4000} = 31.66\ \text{AT/m} \\ NI &= H_i l_i = 31.66\times0.30 = 9.50\ \text{AT} \\ I &= \frac{9.50}{600} = 0.0158\ \text{A} \end{aligned}$$

Answer (a): I ≈ 15.8 mA.

b) With a 4 mm air gap

The gap is cut from the same bar, so iron length $l_i = 0.30 - 0.004 = 0.296$ m.

$$\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{0.1592}{4\pi\times10^{-7}} = 126651\ \text{AT/m} \\ AT_g &= 126651\times0.004 = 506.6\ \text{AT} \\ AT_i &= 31.66\times0.296 = 9.37\ \text{AT} \\ NI &= 506.6 + 9.37 = 516.0\ \text{AT} \\ I &= \frac{516.0}{600} = 0.860\ \text{A} \end{aligned}$$

Answer (b): I ≈ 0.86 A.

A gap of only 4 mm raises the required current about 54 times, because the reluctance of air is 4000 times that of the same length of iron.

Both windings are on the same single-loop core, so their mmfs add (assuming they are wound to aid each other) and drive one flux through the core and gap in series. Gap area is taken equal to core area (fringing neglected).

i) Reluctances

$$\begin{aligned} R_c &= \frac{l_c}{\mu_0\mu_r A_c} = \frac{0.6}{4\pi\times10^{-7}\times2000\times1.8\times10^{-3}} = 1.326\times10^{5}\ \text{AT/Wb} \\ R_g &= \frac{g}{\mu_0 A_c} = \frac{2.3\times10^{-3}}{4\pi\times10^{-7}\times1.8\times10^{-3}} = 1.017\times10^{6}\ \text{AT/Wb} \end{aligned}$$

Total reluctance: $R = R_c + R_g = 1.149\times10^{6}$ AT/Wb.

ii) Flux in the air gap

$$\begin{aligned} F &= N_1 i_1 + N_2 i_2 = 83\times1.5 + 20\times1.25 = 124.5 + 25 = 149.5\ \text{AT} \\ \phi &= \frac{F}{R} = \frac{149.5}{1.149\times10^{6}} = 1.30\times10^{-4}\ \text{Wb} \end{aligned}$$

Gap flux density: $B_g = \phi/A_c = 1.30\times10^{-4}/1.8\times10^{-3} = 0.0723$ T.

Answer: $R_c = 1.33\times10^5$ AT/Wb, $R_g = 1.02\times10^6$ AT/Wb, $\phi = 0.130$ mWb (B = 0.072 T).

(If the two windings opposed each other, $F = 124.5 - 25 = 99.5$ AT and $\phi = 0.0866$ mWb.)

The steel path and the air gap are in series and carry the same flux; the area is the same throughout (fringing neglected).

Data: $\phi = 0.75\times10^{-4}$ Wb, $A = 1.5\times10^{-4}$ m², $l_s = 100\times10^{-3}$ m, $l_g = 2\times10^{-3}$ m, $\mu_r = 1424$, $N = 200$.

Flux density

$$B = \frac{\phi}{A} = \frac{0.75\times10^{-4}}{1.5\times10^{-4}} = 0.5\ \text{T}$$

Steel portion

$$\begin{aligned} H_s &= \frac{B}{\mu_0\mu_r} = \frac{0.5}{4\pi\times10^{-7}\times1424} = 279.4\ \text{AT/m} \\ H_s l_s &= 279.4\times0.1 = 27.9\ \text{AT} \end{aligned}$$

Air gap

$$\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{0.5}{4\pi\times10^{-7}} = 3.979\times10^{5}\ \text{AT/m} \\ H_g l_g &= 3.979\times10^{5}\times2\times10^{-3} = 795.8\ \text{AT} \end{aligned}$$

Current

$$\begin{aligned} NI &= 27.9 + 795.8 = 823.7\ \text{AT} \\ I &= \frac{823.7}{200} = 4.12\ \text{A} \end{aligned}$$

Answer: I ≈ 4.12 A. About 97% of the mmf is used by the 2 mm air gap.

Total mmf = gap ampere-turns + iron ampere-turns. Iron length = 30 − 0.1 = 29.9 cm. Same B in iron and gap (no leakage or fringing).

Data: $l = 0.30$ m, $A = 6$ cm² $= 6\times10^{-4}$ m², $N = 500$, $l_g = 1$ mm, $I = 4$ A, $B = 1$ T.

(i) Relative permeability

$$\begin{aligned} NI &= 500\times4 = 2000\ \text{AT} \\ AT_g &= \frac{B}{\mu_0} l_g = \frac{1}{4\pi\times10^{-7}}\times10^{-3} = 795.8\ \text{AT} \\ AT_i &= 2000 - 795.8 = 1204.2\ \text{AT} \\ H_i &= \frac{1204.2}{0.299} = 4027.5\ \text{AT/m} \\ \mu_r &= \frac{B}{\mu_0 H_i} = \frac{1}{4\pi\times10^{-7}\times4027.5} = 197.6 \end{aligned}$$

(ii) Inductance

$$\begin{aligned} \phi &= BA = 1\times6\times10^{-4} = 6\times10^{-4}\ \text{Wb} \\ L &= \frac{N\phi}{I} = \frac{500\times6\times10^{-4}}{4} = 0.075\ \text{H} \end{aligned}$$

Answer: μr ≈ 198, L = 0.075 H (75 mH). (Taking the iron length as the full 30 cm gives μr ≈ 198 as well.)

Assumption: the ring has a circular cross-section whose diameter equals the radial width, $(5.5 - 3.5)/2 = 1$ cm.

Dimensions

$$\begin{aligned} D_{mean} &= \frac{5.5 + 3.5}{2} = 4.5\ \text{cm} \\ l_{mean} &= \pi D_{mean} = \pi\times0.045 = 0.1414\ \text{m} \\ l_i &= 0.1414 - 0.002 = 0.1394\ \text{m} \\ A &= \frac{\pi(0.01)^2}{4} = 7.854\times10^{-5}\ \text{m}^2 \end{aligned}$$

Flux density

$$B = \frac{\phi}{A} = \frac{0.3\times10^{-3}}{7.854\times10^{-5}} = 3.82\ \text{T}$$

Ampere-turns

$$\begin{aligned} AT_g &= \frac{B}{\mu_0} l_g = \frac{3.82}{4\pi\times10^{-7}}\times0.002 = 3.0396\times10^{6}\times0.002 = 6079\ \text{AT} \\ AT_i &= \frac{B}{\mu_0\mu_r} l_i = \frac{3.0396\times10^{6}}{2000}\times0.1394 = 1519.8\times0.1394 = 211.8\ \text{AT} \\ NI &= 6079 + 211.8 = 6291\ \text{AT} \end{aligned}$$

Current

$$I = \frac{6291}{1000} = 6.29\ \text{A}$$

Answer: I ≈ 6.29 A (dc). (B = 3.82 T is beyond the saturation of real iron; the answer assumes μr stays 2000. If a square 1 cm × 1 cm section is assumed instead, A = 1 cm², B = 3 T and I ≈ 4.94 A.)

Magnetic hysteresis is the lagging of flux density B behind the magnetising force H when a ferromagnetic material is taken through a cycle of magnetisation. As a result the B–H curve forms a closed loop, and energy equal to its area is lost per cycle.

Hysteresis loop

              B
              |    __ a
          b  _|.-'  /
            / |    /
     c     /  |   /
  --*-----/---+--/---*--- H
         /    | /    f
        / ___.|'
     d -'     |e

ob = residual flux density (retentivity), oc = coercive force, a and d = positive and negative saturation.

Proof: area ∝ energy loss per unit volume

Ring: mean length $l$, area $A$, $N$ turns, current $i$. By Ampere's law $Hl = Ni$, so $i = Hl/N$. When B changes by $dB$ the induced emf is $e = NA\,dB/dt$. Energy supplied by the source in time $dt$:

$$\begin{aligned} dW &= e\,i\,dt = NA\frac{dB}{dt}\cdot\frac{Hl}{N}\,dt = (Al)\,H\,dB \end{aligned}$$

Per unit volume, $dw = H\,dB$. For one complete cycle

$$w = \oint H\,dB = \text{area of B–H loop} \quad (\text{J/m}^3)$$

The energy stored while B rises is only partly returned while B falls; the net amount, equal to the loop area, is converted into heat. Total hysteresis loss:

$$P_h = (\text{loop area}) \times V \times f \ \text{W} \approx \eta B_m^{1.6} f V$$

Hard and soft magnetic materials

Soft materials are chosen for ac machines because their small loop area gives low core loss; hard materials are chosen where the magnetism must be kept.

Flux density follows from $NI = H_i l_i + H_g l_g$ with the same B in the steel and in the gap.

Data: mean radius $r = 0.12$ m, section radius 1 cm, $l_g = 2$ mm, $N = 550$, $I = 3$ A, $\mu_r = 800$.

Dimensions

$$\begin{aligned} l &= 2\pi r = 2\pi\times0.12 = 0.7540\ \text{m} \\ l_i &= 0.7540 - 0.002 = 0.7520\ \text{m} \\ A &= \pi (0.01)^2 = 3.1416\times10^{-4}\ \text{m}^2 \end{aligned}$$

Flux density

$$\begin{aligned} NI &= \frac{B}{\mu_0}\left(\frac{l_i}{\mu_r} + l_g\right) \\ B &= \frac{\mu_0 NI}{\dfrac{l_i}{\mu_r} + l_g} = \frac{4\pi\times10^{-7}\times550\times3}{\dfrac{0.7520}{800} + 0.002} \\ &= \frac{2.0735\times10^{-3}}{9.40\times10^{-4} + 2\times10^{-3}} = \frac{2.0735\times10^{-3}}{2.940\times10^{-3}} \\ &= 0.705\ \text{T} \end{aligned}$$

Flux: $\phi = BA = 0.705\times3.1416\times10^{-4} = 2.22\times10^{-4}$ Wb.

Answer: B ≈ 0.705 T (flux ≈ 0.222 mWb).

A magnetic circuit is the closed path followed by magnetic flux. It is usually made of high-permeability iron (sometimes with air gaps) and is excited by a coil carrying current, e.g. the core of a transformer, relay or electrical machine.

     +-----------------+
     |    phi -->      |
   ==|== N turns       |  iron core,
   ==|== I             |  mean length l,
     |                 |  area A
     +-----------------+

Similarities between magnetic and electric circuits

Points of difference: flux does not truly flow, no energy is spent to maintain a constant flux (but $I^2R$ loss occurs with current), and $\mu$ is not constant because iron saturates, while R is nearly constant.

Ohm's law for a magnetic circuit

Take a uniform ring of mean length $l$ (m), area $A$ (m²), relative permeability $\mu_r$, wound with $N$ turns carrying $I$.

By Ampere's circuital law, $\oint H\,dl = NI$, so

$$H = \frac{NI}{l}$$

Then

$$\begin{aligned} B &= \mu_0\mu_r H = \frac{\mu_0\mu_r NI}{l} \\ \phi &= BA = \frac{\mu_0\mu_r A\, NI}{l} \\ \phi &= \frac{NI}{\dfrac{l}{\mu_0\mu_r A}} = \frac{F}{S} \end{aligned}$$

So flux = mmf ÷ reluctance, which is Ohm's law for the magnetic circuit ($I = E/R$). For a composite circuit (iron + air gap) the reluctances add: $\phi = NI/(S_i + S_g)$.

Eddy current loss is the $I^2R$ heat loss caused by circulating currents (eddy currents) induced in the body of a conducting magnetic core when the flux through it changes.

How it takes place with ac excitation

• With an ac supply the core flux alternates: $\phi = \phi_m \sin\omega t$.
• The iron core is itself a conductor. It can be regarded as many closed loops lying in planes perpendicular to the flux.
• By Faraday's law an emf $e = d\phi/dt$ is induced in each loop, so currents circulate in the core in closed paths, like whirlpools ("eddies").
• These currents flow through the resistance of the iron and produce heat $I^2R$. They also set up their own flux opposing the main flux (Lenz's law).

  Solid core              Laminated core
  +-----------+          +-+-+-+-+-+-+
  |  .-----.  |          |o|o|o|o|o|o|
  | (   o   ) |  phi     | | | | | | |  small loops,
  |  '-----'  | (out)    |o|o|o|o|o|o|  high resistance
  +-----------+          +-+-+-+-+-+-+
  large eddy loop        insulated sheets

Expression

$$P_e = K_e B_m^2 f^2 t^2 V \quad \text{W}$$

where $B_m$ = peak flux density, $f$ = frequency, $t$ = lamination thickness, $V$ = core volume, $K_e$ = a constant that falls as the resistivity of the material rises. Since the induced emf ∝ $fB_m t$ and loss ∝ emf², the loss depends on the square of each.

Methods of minimising eddy current loss

1. Laminating the core: build the core from thin sheets (0.25–0.5 mm) insulated from each other by varnish or oxide, with the sheets parallel to the flux. Each eddy path is confined to a thin sheet, its emf is small and its resistance high. Loss ∝ $t^2$, so halving thickness quarters the loss.
2. Using high-resistivity material: adding 3–4% silicon to steel raises its resistivity and lowers the eddy currents (and also reduces hysteresis loss).
3. Ferrite or powdered-iron cores for high-frequency use, where resistivity is very high.
4. Limiting $B_m$ and frequency in design, since loss ∝ $B_m^2 f^2$.

When an iron core is excited from an ac source, B and H vary cyclically. The ac magnetisation characteristic is the curve of peak flux density $B_m$ against peak magnetising force $H_m$ (or $\phi_m$ against $I_m$), obtained by joining the tips of the hysteresis loops traced at different excitation levels. It shows how the core magnetises under ac and gives its permeability and saturation.

Obtaining the characteristic

   B
   |             ___  saturation
   |         _.-' 
   |       /   (tips of loops)
   |     /:  
   |    / :  <- knee
   |   /  :
   |  /    linear region
   | / 
   |/  
 --+------------------- H
   |

1. A coil on the core is fed from a variable ac voltage at fixed frequency.
2. For a given voltage, the core goes round a hysteresis loop between $\pm B_m$ and $\pm H_m$. Since $V \approx 4.44 f N \phi_m$, $B_m$ is fixed by the voltage.
3. The voltage is raised step by step; each time a bigger loop is traced.
4. The curve through the tips $(H_m, B_m)$ of all the loops is the ac (normal) magnetisation curve.

Regions of the curve

• Initial region: at very low H the slope is small (domains move reversibly).
• Linear region: B rises almost in proportion to H; permeability is high and nearly constant. Machines and transformers are designed to work just below the knee.
• Knee: the slope falls as domains approach full alignment.
• Saturation region: further H gives very little increase in B; $\mu_r$ falls towards 1.

Features under ac excitation

• Hysteresis: each cycle traces a loop of area $\oint H\,dB$, giving loss $P_h = \eta B_m^{1.6}fV$.
• Eddy currents widen the dynamic loop further, and total core loss $P_c = P_h + P_e$ appears as the core-loss component of exciting current.
• Non-sinusoidal current: if the voltage (and so flux) is sinusoidal, the non-linear curve makes the magnetising current peaky with strong 3rd harmonic, especially near saturation.
• Apparent (ac) permeability $\mu = B_m/H_m$ varies with excitation level, being highest near the knee.

Use

The curve is used to choose the working flux density (about 1.2–1.7 T for silicon steel), to find the magnetising current for a given voltage, and to judge how much over-voltage a transformer can take before saturating.

Required mmf = flux × total reluctance; then $N = F/I$. Iron length is taken as $l_i = 1.25 - 0.005 = 1.245$ m; gap area = core area.

Data: $A = 0.02$ m², $l_g = 5$ mm, $\phi = 0.5$ Wb, $I = 2$ A, $\mu_r = 1000$.

Reluctances

$$\begin{aligned} S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{1.245}{4\pi\times10^{-7}\times1000\times0.02} = 4.954\times10^{4}\ \text{AT/Wb} \\ S_g &= \frac{l_g}{\mu_0 A} = \frac{0.005}{4\pi\times10^{-7}\times0.02} = 1.989\times10^{5}\ \text{AT/Wb} \\ S &= S_i + S_g = 2.485\times10^{5}\ \text{AT/Wb} \end{aligned}$$

MMF and turns

$$\begin{aligned} F &= \phi S = 0.5\times2.485\times10^{5} = 1.2424\times10^{5}\ \text{AT} \\ N &= \frac{F}{I} = \frac{1.2424\times10^{5}}{2} = 62120\ \text{turns} \end{aligned}$$

Answer: N ≈ 62,120 turns. (Using the full 1.25 m as iron length gives N ≈ 62,170.) Note: B = 0.5/0.02 = 25 T, which no iron can reach; the figure follows only from the constant μr given.

Retentivity and coercivity

• Retentivity (residual flux density, $B_r$): the flux density that remains in a magnetic material when the magnetising force H is reduced to zero after saturating it. It measures how much magnetism the material keeps.
• Coercivity (coercive force, $H_c$): the reverse magnetising force needed to reduce the residual flux density to zero. It measures how hard it is to demagnetise the material.

             B
             |   __ +Bsat
        Br  _|-'  /
           / |   /
   -Hc    /  |  /
  ---*---/---+-/----- H
        /    |/

Permanent magnets need high $B_r$ and high $H_c$; transformer cores need low $H_c$.

Proof that $W_h = \oint H\,dB$

Consider a ring of magnetic material with mean length $l$ (m), cross-sectional area $A$ (m²), uniformly wound with $N$ turns carrying current $i$.

1. By Ampere's circuital law the field intensity is

$$H l = N i \;\Rightarrow\; i = \frac{H l}{N}$$

2. If the flux density changes by $dB$ in time $dt$, the flux changes by $A\,dB$, and the emf induced in the coil is

$$e = N\frac{d\phi}{dt} = N A \frac{dB}{dt}$$

3. To keep the current flowing against this emf, the source supplies power $p = e\,i$. The energy supplied in time $dt$ is

$$\begin{aligned} dW &= e\, i\, dt \\ &= N A \frac{dB}{dt} \cdot \frac{Hl}{N}\, dt \\ &= A\,l\,H\,dB \end{aligned}$$

4. $Al$ is the volume of the core, so the energy per unit volume is $dw = H\,dB$.

5. Over a complete cycle of magnetisation (B going from $+B_m$ to $-B_m$ and back):

$$W_h = \oint H\,dB \quad \text{J/m}^3 \text{ per cycle}$$

During the parts of the cycle where H and dB have the same sign energy is taken from the source; where they have opposite signs energy is returned. The integral round the closed loop is the net energy absorbed, which equals the area of the hysteresis loop and is dissipated as heat.

Hence hysteresis power loss for a core of volume $V$ at frequency $f$:

$$P_h = f\,V \oint H\,dB \quad \text{W}$$

Required mmf = ampere-turns for the gap + ampere-turns for the iron, with the same flux density in both (no leakage or fringing). Iron length $l_i = 1.5 - 0.004 = 1.496$ m.

Data: $A = 50$ cm² $= 5\times10^{-3}$ m², $l_g = 4$ mm, $N = 500$, $\phi = 100$ mWb, $\mu_r = 2000$.

Flux density

$$B = \frac{\phi}{A} = \frac{0.1}{5\times10^{-3}} = 20\ \text{T}$$

Air gap

$$\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{20}{4\pi\times10^{-7}} = 1.5915\times10^{7}\ \text{AT/m} \\ AT_g &= H_g l_g = 1.5915\times10^{7}\times0.004 = 63662\ \text{AT} \end{aligned}$$

Iron

$$\begin{aligned} H_i &= \frac{H_g}{\mu_r} = \frac{1.5915\times10^{7}}{2000} = 7958\ \text{AT/m} \\ AT_i &= 7958\times1.496 = 11905\ \text{AT} \end{aligned}$$

Current

$$\begin{aligned} NI &= 63662 + 11905 = 75567\ \text{AT} \\ I &= \frac{75567}{500} = 151.1\ \text{A} \end{aligned}$$

Answer: I ≈ 151 A (dc). (B = 20 T is far above what iron can carry; the figure assumes μr stays constant at 2000 as given.)