Magnetic Circuits and Induction

The exciting current is found from the total mmf, which is the sum of the mmf for the iron path and for the air gap, using the reluctance of each part.

Given: mean diameter $D = 30$ cm, rod diameter $d = 2.5$ cm, gap $l_g = 1$ mm, $N = 500$, $\Phi = 4$ mWb, $\mu_r = 800$.

Dimensions

$$\begin{aligned} l_{total} &= \pi D = \pi \times 0.30 = 0.9425\ \text{m} \\ l_i &= 0.9425 - 0.001 = 0.9415\ \text{m} \\ A &= \frac{\pi}{4}(0.025)^2 = 4.909\times10^{-4}\ \text{m}^2 \end{aligned}$$

Reluctances

$$\begin{aligned} S_i &= \frac{l_i}{\mu_0\mu_r A} = \frac{0.9415}{4\pi\times10^{-7}\times800\times4.909\times10^{-4}} = 1.908\times10^{6}\ \text{AT/Wb} \\ S_g &= \frac{l_g}{\mu_0 A} = \frac{0.001}{4\pi\times10^{-7}\times4.909\times10^{-4}} = 1.621\times10^{6}\ \text{AT/Wb} \end{aligned}$$

The two reluctances are in series (the same flux passes through both), so $S = S_i + S_g = 3.529\times10^{6}$ AT/Wb.

Mmf and current

$$\begin{aligned} NI &= \Phi S = 4\times10^{-3}\times3.529\times10^{6} = 14\,116\ \text{AT} \\ I &= \frac{14\,116}{500} = 28.23\ \text{A} \end{aligned}$$

The flux density is $B = \Phi/A = 8.15$ T, which is far above the saturation of real iron, so the given $\mu_r = 800$ is only an assumption of the problem.

Answer: $I \approx 28.2$ A

Given: mean diameter $D = 10$ cm, $A = 100$ mm$^2 = 10^{-4}$ m$^2$, $l_g = 2$ mm, $N = 600$, $B_g = 1$ T, $\mu_r = 4000$.

With no leakage or fringing, the flux density is the same (1 T) in iron and in the gap.

Path lengths

$$l_i = \pi D - l_g = 0.31416 - 0.002 = 0.31216\ \text{m}$$

Mmf for each part

$$\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.31216}{4\pi\times10^{-7}\times4000} = 62.10\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\ F &= 62.10 + 1591.5 = 1653.7\ \text{AT} \end{aligned}$$

Current

$$I = \frac{F}{N} = \frac{1653.7}{600} = 2.756\ \text{A}$$

The air gap needs about 96% of the mmf even though it is only 2 mm long.

Answer: $I \approx 2.76$ A

Given: mean length of magnetic circuit $= 10$ cm (taken to include the gap), $A = 100$ mm$^2 = 10^{-4}$ m$^2$, $l_g = 2$ mm, $N = 600$, $B_g = 1$ T, $\mu_r = 4000$.

Neglecting leakage and fringing, $B$ is 1 T in both iron and gap.

Path lengths

$$l_i = 0.100 - 0.002 = 0.098\ \text{m}$$

Mmf

$$\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.098}{4\pi\times10^{-7}\times4000} = 19.50\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\ F &= 1611.0\ \text{AT} \end{aligned}$$

Current

$$I = \frac{1611.0}{600} = 2.685\ \text{A}$$

If the 10 cm is taken as the iron length only (gap extra), $F_i = 19.9$ AT and $I = 2.686$ A, so the answer is the same to three figures.

Answer: $I \approx 2.69$ A

The hysteresis loop is the closed $B$–$H$ curve traced by a ferromagnetic material when $H$ is taken through a full cycle. It shows how the flux density lags behind the magnetising force, and its shape and area decide where a material can be used.

Significance of the hysteresis loop

      B
      |      ___---- Bs (saturation)
      |   .-'
   Br |--/---.
      | /     \
 -----+-+------+------ H
   -Hc|        Hc
      |

1. Retentivity ($B_r$): the flux density left when $H = 0$. A high $B_r$ is needed for permanent magnets.
2. Coercivity ($H_c$): the reverse field needed to bring $B$ to zero. A small $H_c$ gives a narrow loop (soft iron, silicon steel) for transformer and armature cores. A large $H_c$ gives a wide loop (hard steel, alnico) for permanent magnets.
3. Area of the loop: it equals the energy lost as heat per cycle per unit volume. A narrow loop means low hysteresis loss and high efficiency.
4. Saturation flux density ($B_s$): it fixes the highest flux a core can carry, so it limits the size of the machine.
5. Material selection: the loop shows at a glance whether the material suits an ac machine core (narrow loop) or a permanent magnet (wide loop).

Hysteresis loss depends on volume

Energy spent in one cycle per unit volume is the loop area:

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

This value depends only on the material and on $B_{max}$, not on the size of the core. A core of volume $V$ has $V$ times as much material taking part in the magnetisation reversal, so

$$\begin{aligned} \text{Energy per cycle} &= w_h V \\ P_h &= w_h V f = \eta B_{max}^{x} f V\ \ \text{W} \end{aligned}$$

where $\eta$ is the Steinmetz coefficient and $x \approx 1.6$ to $2$. Hence $P_h \propto V$: doubling the core volume at the same $B_{max}$ and $f$ doubles the hysteresis loss. This is why machine cores are made as small as the design allows.

Magnetic hysteresis is the lagging of the flux density $B$ behind the magnetising force $H$ in a ferromagnetic material. When $H$ is increased, reduced, reversed and brought back, the $B$–$H$ curve does not retrace itself but forms a closed loop.

      B
      |     ____ saturation
      |   /
   Br |--/--.
      | /    \
 -----+-+-----+------ H
     -Hc     Hc
      |  loop area = energy lost

Explanation: the material is made of small magnetic domains. When $H$ rises they rotate and align, and $B$ increases up to saturation. When $H$ falls to zero, some domains stay aligned, so a residual flux density $B_r$ (retentivity) remains. A reverse field $-H_c$ (coercive force) is needed to bring $B$ to zero. Overcoming this internal friction between domains costs energy, which appears as heat.

Energy per cycle equals loop area

Take a core of cross-section $A$, mean length $l$ and $N$ turns carrying current $i$.

$$\begin{aligned} e &= N\frac{d\phi}{dt} = NA\frac{dB}{dt} \\ H &= \frac{Ni}{l} \Rightarrow i = \frac{Hl}{N} \end{aligned}$$

Energy supplied by the source in time $dt$ is

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

Here $Al = V$ is the volume of the core. Energy per unit volume is therefore

$$\frac{dW}{V} = H\,dB$$

For one full cycle,

$$w = \oint H\,dB = \text{area of the } B\text{–}H\text{ loop}$$

In the rising part of the cycle, part of this energy is stored in the field and is returned when the field falls. The rest is not returned and is lost as heat, which is the area enclosed by the loop. So the energy lost per cycle per unit volume equals the loop area (in J/m$^3$). For frequency $f$ and volume $V$, $P_h = f V \times (\text{loop area})$.

Hysteresis loss is the power lost as heat in a magnetic core because its domains are repeatedly turned back and forth by an alternating field. Eddy current loss is the $I^2R$ loss caused by circulating currents induced in the core itself by the changing flux.

Expressions

$$P_h = \eta B_{max}^{1.6} f V\ \text{W}, \qquad P_e = k_e B_{max}^{2} f^{2} t^{2} V\ \text{W}$$

where $t$ is the lamination thickness and $V$ the core volume.

Origin

• Hysteresis: energy per cycle equals the area of the $B$–$H$ loop, so loss rises with frequency and flux density.
• Eddy current: the alternating flux induces emf in the solid core. This drives currents in closed loops inside the iron, which heat it.

Significance in machines

1. Together they form the core (iron) loss. It is present whenever the core is excited, even at no load, and it is almost constant from no load to full load.
2. It lowers efficiency and creates heat that must be removed, so it limits the rating.
3. It fixes the choice of core material, lamination thickness and the flux density used in the design.
4. In transformers it is the constant loss $P_i$ that sets the condition for maximum efficiency ($P_i = P_{cu}$).

Methods of reduction

Both losses grow with frequency because a higher frequency means the core is magnetised and demagnetised more times each second and the induced emf in the core is larger.

Hysteresis loss

The energy lost in one cycle per unit volume is the area $w_h$ of the $B$–$H$ loop. It depends on the material and $B_{max}$ only. The number of cycles per second is $f$, so

$$P_h = w_h V f = \eta B_{max}^{1.6} f V$$

The loss is directly proportional to $f$. At twice the frequency, the loop is traversed twice as often and the loss doubles (for the same $B_{max}$).

Eddy current loss

The flux in the core is $\phi = \phi_m \sin\omega t$. The emf induced in a lamination is proportional to the rate of change of flux:

$$e \propto \frac{d\phi}{dt} = \omega\phi_m\cos\omega t \Rightarrow E_{eddy} \propto f B_{max}$$

The eddy current is $I_e = E_{eddy}/R_{core}$, and the loss is $I_e^2 R$. So the loss varies as the square of the induced emf:

$$P_e \propto E_{eddy}^2 \propto f^{2} B_{max}^{2}$$

Eddy loss is therefore proportional to $f^2$. At twice the frequency the loss is four times larger.

Note: in a transformer on a fixed supply, $B_{max} \propto V/f$. Then $P_e$ is almost independent of $f$, while $P_h \propto f^{-0.6}$. The statement is therefore true for a given $B_{max}$.

Eddy current loss is the power wasted as heat in a magnetic core when the alternating flux induces emf in the iron itself and this emf circulates currents inside the core.

Cause: by Faraday's law, a changing flux induces an emf in any conducting path linked with it. A solid iron core has low resistance, so closed current loops (eddy currents) flow in planes at right angles to the flux, and they cause $I^2R$ heating.

  Solid core          Laminated core
  +--------+          +--+--+--+--+
  | (  )   |          |  |  |  |  |
  | (  )   |  flux    |  |  |  |  |
  +--------+          +--+--+--+--+
  large loops         thin insulated layers,
                      small loops

Expression:

$$P_e = k_e B_{max}^{2} f^{2} t^{2} V\ \text{W}$$

where $t$ is the thickness of a lamination, $f$ the frequency, $B_{max}$ the peak flux density and $V$ the volume. The loss is proportional to $t^2$ and $f^2$.

Effects: lower efficiency, heating of the core, and a demagnetising effect of the eddy currents on the main flux.

Reduction:

• Build the core from thin laminations (0.35 to 0.5 mm) insulated from each other by varnish.
• Add silicon to the steel to raise its resistivity.
• Use ferrite or powdered iron cores at high frequency.

Uses: eddy currents are useful in induction furnaces, eddy current brakes and energy meters.

Reluctance ($S$ or $\mathcal{R}$) is the opposition that a magnetic circuit offers to the setting up of flux. It is the ratio of mmf to flux, and its unit is AT/Wb (or H$^{-1}$). Permeance ($P$) is the reciprocal of reluctance. It is the ease with which flux is set up, and its unit is Wb/AT (henry).

Derivation for a uniform circuit

Take a core of mean length $l$, cross-section $A$ and permeability $\mu = \mu_0\mu_r$, carrying a coil of $N$ turns and current $I$.

$$\begin{aligned} \text{mmf } F &= NI \\ H &= \frac{NI}{l},\qquad B = \mu H = \frac{\mu NI}{l} \\ \Phi &= BA = \frac{\mu A\,NI}{l} = \frac{NI}{\dfrac{l}{\mu A}} \end{aligned}$$

Comparing with $\Phi = \dfrac{\text{mmf}}{\text{reluctance}}$:

$$S = \frac{l}{\mu A} = \frac{l}{\mu_0\mu_r A}, \qquad P = \frac{1}{S} = \frac{\mu A}{l}$$

For other types of circuit

• Air gap of length $l_g$: $S_g = \dfrac{l_g}{\mu_0 A_g}$.
• Series circuit (core of several sections, or core with a gap), the same flux flows through each part:

$$S = S_1 + S_2 + \dots,\qquad NI = \Phi(S_1 + S_2 + \dots)$$

• Parallel circuit (for example the legs of a three-limb core): the mmf across them is common, so

$$P = P_1 + P_2 + \dots \Rightarrow \frac{1}{S} = \frac{1}{S_1} + \frac{1}{S_2} + \dots$$

• Non-uniform section, in general, by adding small elements: $S = \displaystyle\int \frac{dl}{\mu A(l)}$.

Points to note: $S$ is small for iron (high $\mu_r$) and large for air, so even a short air gap contributes a large share of the total reluctance. Unlike electrical resistance, reluctance stores energy and is not a dissipating quantity.

A magnetic circuit is the closed path followed by magnetic flux, made up of ferromagnetic material (and possibly air gaps), in which the flux is set up by an mmf such as that of a current-carrying coil.

Similarities with an electric circuit

Differences: current flow dissipates $I^2R$ energy, but a constant flux needs no energy to maintain it. Resistance is nearly constant, but reluctance changes with $B$ because $\mu$ is not constant (saturation). Flux leaks out of the core (leakage, fringing), but current can be kept inside the conductor.

Ohm's law for the magnetic circuit

$$\begin{aligned} H &= \frac{NI}{l} \\ B &= \mu H = \frac{\mu NI}{l} \\ \Phi &= BA = \frac{\mu A}{l}\,NI = \frac{NI}{l/\mu A} \end{aligned}$$ $$\Phi = \frac{F}{S}\qquad \text{i.e. flux} = \frac{\text{mmf}}{\text{reluctance}}$$

This is Ohm's law for a magnetic circuit: the flux set up is directly proportional to the mmf and inversely proportional to the reluctance of the path.

Why soft magnetic material is used for cores

Cores of transformers and armatures carry alternating flux, so every cycle the material goes round its $B$–$H$ loop. Soft magnetic materials (silicon steel, soft iron, CRGO) are chosen because of the following reasons.

• They have a narrow hysteresis loop, so the hysteresis loss (loop area $\times f\times V$) is small.
• They have high permeability, so a large flux is set up by a small magnetising current.
• They have low coercivity and low retentivity, so they magnetise and demagnetise easily without keeping residual magnetism.
• Silicon raises their resistivity, which lowers eddy current loss when the core is laminated.

Faraday's laws of electromagnetic induction

1. First law: whenever the flux linked with a coil changes, an emf is induced in it. The emf lasts only while the flux is changing.
2. Second law: the magnitude of the induced emf is equal to the rate of change of flux linkages:

$$e = -N\frac{d\phi}{dt}$$

The negative sign is Lenz's law: the induced emf drives a current that opposes the change that produced it.

Ways of inducing emf in a coil

1. Statically induced emf: the coil is at rest and the flux linking it changes with time.
   • Self-induced: the coil's own current changes, $e = -L\,di/dt$.
   • Mutually induced: the flux of a neighbouring coil changes, $e = -M\,di/dt$ (transformer action).
2. Dynamically induced emf: the flux is constant, but the coil moves across the field (or the field moves across the conductor). For a conductor of length $l$ moving with velocity $v$ at right angles to field $B$,

$$e = Blv\ \text{volt}$$

This is the principle of generators. A combination of both is also possible, for example a moving coil in a time-varying field.

Faraday's laws of electromagnetic induction

1. First law: when the magnetic flux linked with a coil or conductor changes, an emf is induced in it. There is no emf while the flux is constant.
2. Second law: the magnitude of the induced emf equals the rate of change of flux linkages, $e = N\dfrac{d\phi}{dt}$. The direction is given by Lenz's law, which opposes the cause of the change, hence $e = -N\dfrac{d\phi}{dt}$.

Statically induced emf vs dynamically induced emf

A coil on an iron ring has a self-inductance fixed by its turns and the reluctance of the core. The emf induced then follows $e = -L\,di/dt$.

Given: $l = 30$ cm, $N = 600$, $d = 20$ mm, $\mu_r = 4000$, $i = 5\sin 314t$ A.

Inductance

$$\begin{aligned} A &= \frac{\pi}{4}(0.02)^2 = 3.142\times10^{-4}\ \text{m}^2 \\ L &= \frac{\mu_0\mu_r N^2 A}{l} = \frac{4\pi\times10^{-7}\times4000\times600^2\times3.142\times10^{-4}}{0.3} \\ &= 1.895\ \text{H} \end{aligned}$$

Induced emf

$$e = -L\frac{di}{dt} = -L\times5\times314\cos314t = -2975\cos314t\ \text{V}$$

So the peak emf is $E_m = L\omega I_m = 1.895\times314\times5 = 2975$ V.

Average value

The average of a sinusoid over a half cycle is $\dfrac{2}{\pi}$ of its peak (over a full cycle it is zero):

$$E_{avg} = \frac{2}{\pi}E_m = \frac{2}{\pi}\times2975 = 1894\ \text{V}$$

This is the same as $4fLI_m$ with $f = 314/2\pi \approx 50$ Hz.

Answer: $L \approx 1.895$ H; average emf $\approx 1894$ V (peak 2975 V)

The flux in the air gap is the net mmf of the three coils divided by the total reluctance of the core and gap.

Assumption: the current directions in the figure are not legible. I assume coils $a$ and $b$ magnetise in the same sense and coil $c$ opposes them. If the figure shows a different arrangement, only the net mmf changes and the method stays the same.

Dimensions: $l_g = 5$ mm, $l_i = 100 - 0.5 = 99.5$ cm, $A = 2\times2 = 4$ cm$^2$ $= 4\times10^{-4}$ m$^2$, $\mu_r = 1400$.

Net mmf

$$\begin{aligned} F_a &= 335\times1.6 = 536\ \text{AT} \\ F_b &= 600\times4 = 2400\ \text{AT} \\ F_c &= 600\times3 = 1800\ \text{AT} \\ F_{net} &= 536 + 2400 - 1800 = 1136\ \text{AT} \end{aligned}$$

Reluctance

$$\begin{aligned} S_i &= \frac{0.995}{4\pi\times10^{-7}\times1400\times4\times10^{-4}} = 1.414\times10^{6}\ \text{AT/Wb} \\ S_g &= \frac{0.005}{4\pi\times10^{-7}\times4\times10^{-4}} = 9.947\times10^{6}\ \text{AT/Wb} \\ S &= S_i + S_g = 1.1361\times10^{7}\ \text{AT/Wb} \end{aligned}$$

Flux

$$\Phi = \frac{F_{net}}{S} = \frac{1136}{1.1361\times10^{7}} = 1.0\times10^{-4}\ \text{Wb}$$

The flux density in the gap is $B = \Phi/A = 0.25$ T.

Answer: $\Phi_{gap} \approx 0.1$ mWb ($1.0\times10^{-4}$ Wb)

The coil is on the central limb, so the central flux divides equally between the two outer limbs, which are identical. This gives a parallel-series circuit, solved with reluctances.

Assumptions (the figure is incomplete):

• The core depth (thickness) is not shown, so I take $d = 5$ cm throughout. The result is inversely proportional to $d$.
• Mean paths are taken through the centre of each member. Central limb and each outer limb: $20 - 4 = 16$ cm. Top or bottom yoke between centre limb and outer limb: $15 + 3 + 2 = 20$ cm.
• The flux is taken as given, 0.75 Wb. The answer is proportional to flux, so for 0.75 mWb divide by 1000.

   +---+---------------+------+---------------+---+
   |   top yoke 4 cm                              |
   |  outer      window    centre    window outer |
   |  4 cm       15 cm     6 cm      15 cm   4 cm |
   |   bottom yoke 4 cm                           |
   +----------------------------------------------+

Areas

$$A_c = 0.06\times0.05 = 3\times10^{-3}\ \text{m}^2,\qquad A_o = A_y = 0.04\times0.05 = 2\times10^{-3}\ \text{m}^2$$

Reluctances with $\mu = 4\pi\times10^{-7}\times4000 = 5.027\times10^{-3}$ H/m:

$$\begin{aligned} S_c &= \frac{0.16}{5.027\times10^{-3}\times3\times10^{-3}} = 1.061\times10^{4}\ \text{AT/Wb} \\ S_o &= \frac{0.16}{5.027\times10^{-3}\times2\times10^{-3}} = 1.592\times10^{4}\ \text{AT/Wb} \\ S_y &= \frac{0.20}{5.027\times10^{-3}\times2\times10^{-3}} = 1.989\times10^{4}\ \text{AT/Wb} \end{aligned}$$

One side path (outer limb + top yoke + bottom yoke) has $S_{side} = 1.592\times10^4 + 2(1.989\times10^4) = 5.570\times10^{4}$. Flux in each side path is $\Phi_c/2 = 0.375$ Wb.

Mmf

$$\begin{aligned} NI &= \Phi_c S_c + \frac{\Phi_c}{2}S_{side} \\ &= 0.75\times1.061\times10^{4} + 0.375\times5.570\times10^{4} \\ &= 7957 + 20\,889 = 28\,847\ \text{AT} \end{aligned}$$

Answer: $NI \approx 2.88\times10^{4}$ AT for 0.75 Wb (about 28.8 AT if the flux is 0.75 mWb), with $d = 5$ cm assumed.

Given: $D = 100$ cm, $A = 10$ cm$^2 = 10^{-3}$ m$^2$, $N = 1000$, $\mu_r = 2000$, $I = 1$ A.

(i) Reluctance (no saw cut)

$$\begin{aligned} l &= \pi D = \pi\ \text{m} \\ S &= \frac{l}{\mu_0\mu_r A} = \frac{\pi}{4\pi\times10^{-7}\times2000\times10^{-3}} = 1.25\times10^{6}\ \text{AT/Wb} \end{aligned}$$

(ii) Flux with 1 A

$$\Phi = \frac{NI}{S} = \frac{1000\times1}{1.25\times10^{6}} = 8\times10^{-4}\ \text{Wb} = 0.8\ \text{mWb}$$

(iii) Flux with a 1 mm saw cut

The iron path becomes $l_i = \pi - 0.001 = 3.1406$ m, and the gap is $l_g = 1$ mm.

$$\begin{aligned} S_i &= \frac{3.1406}{4\pi\times10^{-7}\times2000\times10^{-3}} = 1.2496\times10^{6} \\ S_g &= \frac{0.001}{4\pi\times10^{-7}\times10^{-3}} = 7.958\times10^{5} \\ S_{total} &= 2.0454\times10^{6}\ \text{AT/Wb} \\ \Phi &= \frac{1000}{2.0454\times10^{6}} = 4.889\times10^{-4}\ \text{Wb} \end{aligned}$$

A gap of only 1 mm cuts the flux from 0.8 mWb to 0.489 mWb, because air has a much higher reluctance than iron.

Answer: (i) $1.25\times10^6$ AT/Wb; (ii) 0.8 mWb; (iii) 0.489 mWb

Given: diameter $D = 0.15$ m (taken as mean diameter), $A = 10^{-3}$ m$^2$, $l_g = 2$ mm, $N = 300$, $B_g = 1$ T, $\mu_r = 800$.

Flux and lengths

$$\begin{aligned} \Phi &= B_gA = 1\times10^{-3} = 1\ \text{mWb} \\ l_i &= \pi\times0.15 - 0.002 = 0.4692\ \text{m} \end{aligned}$$

Since the flux and area are the same in iron and gap (no fringing), $B_i = 1$ T.

Mmf

$$\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{1\times0.4692}{4\pi\times10^{-7}\times800} = 466.8\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\ F &= 466.8 + 1591.5 = 2058.3\ \text{AT} \end{aligned}$$

Exciting current

$$I = \frac{F}{N} = \frac{2058.3}{300} = 6.861\ \text{A}$$

Inductance

$$L = \frac{N\Phi}{I} = \frac{300\times10^{-3}}{6.861} = 0.0437\ \text{H}$$

Answer: $I \approx 6.86$ A; $L \approx 43.7$ mH

Given: $l = 1.2$ m, $A = 0.005$ m$^2$, $N = 900$, $I = 2$ A, $B = 1.2$ T.

(i) Mmf

$$F = NI = 900\times2 = 1800\ \text{AT}$$

(ii) Total flux

$$\Phi = BA = 1.2\times0.005 = 6\times10^{-3}\ \text{Wb} = 6\ \text{mWb}$$

(iii) Magnetic field strength

$$H = \frac{F}{l} = \frac{1800}{1.2} = 1500\ \text{AT/m}$$

(iv) Relative permeability

$$\mu_r = \frac{B}{\mu_0 H} = \frac{1.2}{4\pi\times10^{-7}\times1500} = 636.6$$

Answer: mmf = 1800 AT; $\Phi$ = 6 mWb; $H$ = 1500 AT/m; $\mu_r \approx 637$

Given: mean circumference $= 30$ cm, $A = 6$ cm$^2 = 6\times10^{-4}$ m$^2$, $N = 500$, $l_g = 1$ mm, $I = 4$ A, $B_g = 1$ T.

(i) Relative permeability of the steel

Total mmf: $F = NI = 500\times4 = 2000$ AT.

Mmf used by the gap:

$$F_g = \frac{B l_g}{\mu_0} = \frac{1\times0.001}{4\pi\times10^{-7}} = 795.8\ \text{AT}$$

Mmf left for the iron: $F_i = 2000 - 795.8 = 1204.2$ AT. The iron length is $l_i = 0.30 - 0.001 = 0.299$ m.

$$\begin{aligned} H_i &= \frac{1204.2}{0.299} = 4027.5\ \text{AT/m} \\ \mu_r &= \frac{B}{\mu_0H_i} = \frac{1}{4\pi\times10^{-7}\times4027.5} = 197.6 \end{aligned}$$

(ii) Inductance

$$\Phi = BA = 1\times6\times10^{-4} = 6\times10^{-4}\ \text{Wb}$$ $$L = \frac{N\Phi}{I} = \frac{500\times6\times10^{-4}}{4} = 0.075\ \text{H}$$

Answer: $\mu_r \approx 198$; $L = 75$ mH

The ring is a toroid with a circular cross-section. Its mean diameter and section size follow from the outer and inner diameters.

Given: outer diameter $= 5.5$ cm, inner diameter $= 3.5$ cm, $\mu_r = 2000$, $l_g = 2$ mm, $N = 1000$, $\Phi = 0.3$ mWb.

Geometry

The radial thickness is $5.5 - 3.5 = 2$ cm, which is the diameter of the circular section. So

$$\begin{aligned} D_{mean} &= \frac{5.5 + 3.5}{2} = 4.5\ \text{cm} \\ A &= \frac{\pi}{4}(0.02)^2 = 3.142\times10^{-4}\ \text{m}^2 \\ l_i &= \pi\times0.045 - 0.002 = 0.13937\ \text{m} \end{aligned}$$

Flux density

$$B = \frac{\Phi}{A} = \frac{0.3\times10^{-3}}{3.142\times10^{-4}} = 0.9549\ \text{T}$$

Mmf

$$\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.9549\times0.13937}{4\pi\times10^{-7}\times2000} = 52.95\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.9549\times0.002}{4\pi\times10^{-7}} = 1519.8\ \text{AT} \\ F &= 1572.8\ \text{AT} \end{aligned}$$

Current

$$I = \frac{F}{N} = \frac{1572.8}{1000} = 1.573\ \text{A}$$

Answer: $I \approx 1.57$ A

Given: $A = 5$ cm$^2 = 5\times10^{-4}$ m$^2$, total mean length $= 25$ cm including the 4 mm gap, $N = 500$, $\mu_r = 2000$, $I = 10$ A.

Lengths: $l_g = 4$ mm, $l_i = 0.25 - 0.004 = 0.246$ m.

Reluctances

$$\begin{aligned} S_i &= \frac{0.246}{4\pi\times10^{-7}\times2000\times5\times10^{-4}} = 1.958\times10^{5}\ \text{AT/Wb} \\ S_g &= \frac{0.004}{4\pi\times10^{-7}\times5\times10^{-4}} = 6.366\times10^{6}\ \text{AT/Wb} \\ S &= 6.562\times10^{6}\ \text{AT/Wb} \end{aligned}$$

Inductance

$$L = \frac{N^2}{S} = \frac{500^2}{6.562\times10^{6}} = 0.0381\ \text{H}$$

Flux for 10 A

$$\Phi = \frac{NI}{S} = \frac{500\times10}{6.562\times10^{6}} = 7.62\times10^{-4}\ \text{Wb}$$

The flux density is $B = \Phi/A = 1.52$ T.

Answer: $L \approx 38.1$ mH; $\Phi \approx 0.762$ mWb

With no air gap and a constant $\mu_r$ (saturation neglected), the flux density follows directly from the field strength.

Given: $l = 100$ cm, $A = 50$ mm$^2$ (not needed for $B$), $N = 500$, $I = 10$ A, $\mu_r = 2000$.

$$\begin{aligned} H &= \frac{NI}{l} = \frac{500\times10}{1.0} = 5000\ \text{AT/m} \\ B &= \mu_0\mu_r H = 4\pi\times10^{-7}\times2000\times5000 \\ &= 12.57\ \text{T} \end{aligned}$$

The corresponding flux is $\Phi = BA = 12.57\times50\times10^{-6} = 6.28\times10^{-4}$ Wb.

A real iron core would saturate near 1.8 to 2 T, so this value is a theoretical result for the given linear assumption.

Answer: $B \approx 12.57$ T

Given: mean diameter $= 15$ cm, $A = 10$ cm$^2 = 10^{-3}$ m$^2$, $N = 200$, $l_g = 2$ mm, $B = 1$ T, $\mu_r = 500$.

Lengths and flux

$$l_i = \pi\times0.15 - 0.002 = 0.4692\ \text{m},\qquad \Phi = BA = 10^{-3}\ \text{Wb}$$

Mmf

$$\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.4692}{4\pi\times10^{-7}\times500} = 746.8\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.002}{4\pi\times10^{-7}} = 1591.5\ \text{AT} \\ F &= 2338.4\ \text{AT} \end{aligned}$$

Exciting current

$$I = \frac{F}{N} = \frac{2338.4}{200} = 11.69\ \text{A}$$

Inductance

$$L = \frac{N\Phi}{I} = \frac{200\times10^{-3}}{11.69} = 0.0171\ \text{H}$$

Stored energy

$$W = \tfrac{1}{2}LI^2 = \tfrac{1}{2}\times0.0171\times11.69^2 = 1.169\ \text{J}$$

Check: $W = \tfrac12 F\Phi = \tfrac12\times2338.4\times10^{-3} = 1.169$ J. About 68% of this energy is stored in the 2 mm air gap.

Answer: $I \approx 11.69$ A; $L \approx 17.1$ mH; $W \approx 1.17$ J

Given: rod diameter $= 1$ cm, mean diameter of ring $= 10$ cm, $l_g = 1$ mm, $N = 2000$, $\Phi = 0.2$ mWb, $\mu_r = 150$.

Area, flux density, iron length

$$\begin{aligned} A &= \frac{\pi}{4}(0.01)^2 = 7.854\times10^{-5}\ \text{m}^2 \\ B &= \frac{\Phi}{A} = \frac{0.2\times10^{-3}}{7.854\times10^{-5}} = 2.546\ \text{T} \\ l_i &= \pi\times0.10 - 0.001 = 0.3132\ \text{m} \end{aligned}$$

Mmf

$$\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{2.546\times0.3132}{4\pi\times10^{-7}\times150} = 4230.6\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{2.546\times0.001}{4\pi\times10^{-7}} = 2026.4\ \text{AT} \\ F &= 6257.0\ \text{AT} \end{aligned}$$

Current

$$I = \frac{6257.0}{2000} = 3.129\ \text{A}$$

The flux density of 2.55 T is beyond the saturation level of real iron, so the given $\mu_r = 150$ is a data assumption of the problem.

Answer: $I \approx 3.13$ A

For a material given by a $B$–$H$ table, $H$ for the working $B$ is read (interpolated) from the table instead of using $\mu_r$.

Given: section diameter $= 3$ cm, mean circumference $= 80$ cm, $l_g = 1$ mm, $N = 600$, $\Phi_g = 0.75$ mWb.

Flux density (the same in steel and gap, leakage and fringing neglected):

$$\begin{aligned} A &= \frac{\pi}{4}(0.03)^2 = 7.069\times10^{-4}\ \text{m}^2 \\ B &= \frac{0.75\times10^{-3}}{7.069\times10^{-4}} = 1.061\ \text{T} \end{aligned}$$

$H$ for steel at 1.061 T: this lies between (800 AT/m, 0.9 T) and (1000 AT/m, 1.08 T).

$$H = 800 + 200\times\frac{1.061 - 0.9}{1.08 - 0.9} = 978.9\ \text{AT/m}$$

Mmf

$$\begin{aligned} l_i &= 0.80 - 0.001 = 0.799\ \text{m} \\ F_i &= H l_i = 978.9\times0.799 = 782.2\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{1.061\times0.001}{4\pi\times10^{-7}} = 844.3\ \text{AT} \\ F &= 1626.5\ \text{AT} \end{aligned}$$

Current

$$I = \frac{1626.5}{600} = 2.711\ \text{A}$$

Answer: $I \approx 2.71$ A