Magnetic Circuits Theory

eee-1109 EEE-1109 Basic Electrical Engineering

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Part 1: Permeability, Field Strength, Potential, and Flux

6.1. Absolute and Relative Permeabilities of a Medium

The phenomena of magnetism and electromagnetism are dependent upon a certain property of the medium called its permeability. Every medium is supposed to possess two permeabilities: (i) absolute permeability ($\mu$) and (ii) relative permeability (${\mu }_r$).

For measuring relative permeability, vacuum or free space is chosen as the reference medium. It is allotted an absolute permeability of ${\mu }_0=4\pi \times 10^{-7}$ henry/metre. Obviously, relative permeability of vacuum with reference to itself is unity. Hence, for free space, absolute permeability ${\mu }_0=4\pi \times 10^{-7}$ H/m relative permeability ${\mu }_r=1$.

Now, take any medium other than vacuum. If its relative permeability, as compared to vacuum is ${\mu }_r$, then its absolute permeability is $\mu ={\mu }_0{\mu }_r$, H/m.

Magnetic Field Strength (H)

Magnetic field strength at any point within a magnetic field is numerically equally to the force experienced by an N-pole of one weber placed at that point. Hence, unit of $H$ is N/Wb.

Suppose, it is required to find the field intensity at a point $A$ distant $r$ metres from a pole of $m$ webers. Imagine a similar pole of one weber placed at point $A$. The force experienced by this pole is $F=\frac{m\times 1}{4\pi {\mu }_0{r}^2}\text{ N}$ $\therefore H=\frac{m}{4\pi {\mu }_0{r}^2}\text{ N/Wb (or A/m)*** or oersted.}$

Magnetic Potential

The magnetic potential at any point within a magnetic field is measured by the work done in shifting a N-pole of one weber from infinity to that point against the force of the magnetic field. It is given by $M=\frac{m}{4\pi {\mu }_{0}r}\text{ J/Wb}$

Flux per Unit Pole

A unit N-pole is supposed to radiate out a flux of one weber. Its symbol is $\Phi$. Therefore, the flux coming out of a N-pole of $m$ weber is given by $\Phi =m\text{ Wb}$

Flux Density (B)

It is given by the flux passing per unit area through a plane at right angles to the flux. It is usually designated by the capital letter $B$ and is measured in weber/meter${}^2$. It is a Vector Quantity. If $\Phi$ Wb is the total magnetic flux passing normally through an area of $A$ m${}^2$, then $B=\Phi /A{\text{ Wb/m}}^2\text{ or tesla (T)}$

Intensity of Magnetisation (I)

It may be defined as the induced pole strength developed per unit area of the bar. Also, it is the magnetic moment developed per unit volume of the bar. Let $m=\text{pole strength induced in the bar in Wb}$ $A={\text{face or pole area of the bar in m}}^2$ Then $I=m/A{\text{ Wb/m}}^2$

If $l$ is the magnetic length of the bar, then the product $(m\times l)$ is known as its magnetic moment $M$. $\therefore I=\frac{m}{A}=\frac{m\times l}{A\times l}=\frac{M}{V}=\text{magnetic moment/volume}$

Susceptibility (K)

Susceptibility is defined as the ratio of intensity of magnetisation $I$ to the magnetising force $H$. $\therefore K=I/H\text{ henry/metre.}$

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Part 2: Ampere’s Law and Magnetic Circuit Fundamentals

Ampere’s Work Law or Ampere’s Circuital Law

The law states that m.m.f.* (magnetomotive force corresponding to e.m.f. i.e. electromotive force of electric field) around a closed path is equal to the current enclosed by the path. Mathematically, $\oint \vec{H}\cdot d\vec{s}=I\text{ amperes}$ where $\vec{H}$ is the vector representing magnetic field strength in dot product with vector $d\vec{s}$ of the enclosing path $S$ around current $I$ ampere and that is why line integral ($\oint$) of dot product $\vec{H}\cdot d\vec{s}$ is taken.

Work law is very comprehensive and is applicable to all magnetic fields whatever the shape of enclosing path e.g. (a) and (b) in Fig. 6.11. Since path $c$ does not enclose the conductor, the m.m.f. around it is zero.

The above work Law is used for obtaining the value of the magnetomotive force around simple idealized circuits like (i) a long straight current-carrying conductor and (ii) a long solenoid.

(i) Magnetomotive Force around a Long Straight Conductor

In Fig. 6.12 is shown a straight conductor which is assumed to extend to infinity in either direction. Let it carry a current of $I$ amperes upwards. The magnetic field consists of circular lines of force having their plane perpendicular to the conductor and their centres at the centre of the conductor.

Suppose that the field strength at point $C$ distant $r$ metres from the centre of the conductor is $H$. Then, it means that if a unit N-pole is placed at $C$, it will experience a force of $H$ newtons. The direction of this force would be tangential to the circular line of force passing through $C$. If this unit N-pole is moved once round the conductor against this force, then work done i.e. m.m.f. = force $\times$ distance = $I$ i.e. $H\times 2\pi r$ joules = Amperes or $H=\frac{I}{2\pi r}$ $=\oint \vec{H}\cdot d\vec{s}\text{ Joules = Amperes }=I$

Obviously, if there are $N$ conductors (as shown in Fig. 6.13), then $H=\frac{NI}{2\pi r}\text{ A/m or Oersted}$ and $B={\mu }_0\frac{NI}{2\pi r}{\text{ Wb/m}}^2\text{ tesla ...in air}$ $={\mu }_0{\mu }_r\frac{NI}{2\pi r}{\text{ Wb/m}}^2\text{ tesla ...in a medium}$

Magnetic Circuit

It may be defined as the route or path which is followed by magnetic flux. The law of magnetic circuit are quite similar to (but not the same as) those of the electric circuit.

Consider a solenoid or a toroidal iron ring having a magnetic path of $l$ metre, area of cross section $A$ m${}^2$ and a coil of $N$ turns carrying $I$ amperes wound anywhere on it as in Fig. 6.25. Then, as seen from Art. 6.15, field strength inside the solenoid is $H=\frac{NI}{l}\text{ AT/m}$ Now $B={\mu }_0{\mu }_{r}H=\frac{{\mu }_0{\mu }_{r}NI}{l}{\text{ Wb/m}}^2$ Total flux produce $\Phi =B\times A=\frac{{\mu }_0{\mu }_{r}ANI}{l}\text{ Wb}$ $\therefore \Phi =\frac{NI}{l/{\mu }_0{\mu }_{r}A}\text{ Wb}$

The numerator '$NI$' which produces magnetization in the magnetic circuit is known as magnetomotive force (m.m.f.). Obviously, its unit is ampere-turn (AT)*. It is analogous to e.m.f. in an electric circuit.

The denominator $\frac{l}{{\mu }_0{\mu }_{r}A}$ is called the reluctance of the circuit and is analogous to resistance in electric circuits. $\therefore \text{flux}=\frac{\text{m.m.f.}}{\text{reluctance}}\text{or}\Phi =\frac{F}{S}$

Sometimes, the above equation is called the “Ohm’s Law of Magnetic Circuit” because it resembles a similar expression in electric circuits i.e. $\text{current}=\frac{\text{e.m.f.}}{\text{resistance}}\text{or}I=\frac{V}{R}$

Definitions Concerning Magnetic Circuit

1. Magnetomotive force (m.m.f.). It drives or tends to drive flux through a magnetic circuit and corresponds to electromotive force (e.m.f.) in an electric circuit. M.M.F. is equal to the work done in joules in carrying a unit magnetic pole once through the entire magnetic circuit. It is measured in ampere-turns.

2. Ampere-turns (AT). It is the unit of magnetometre force (m.m.f.) and is given by the product of number of turns of a magnetic circuit and the current in amperes in those turns.

3. Reluctance. It is the name given to that property of a material which opposes the creation of magnetic flux in it. It, in fact, measures the opposition offered to the passage of magnetic flux through a material and is analogous to resistance in an electric circuit even in form. Its unit is AT/Wb.* $\text{reluctance}=\frac{l}{\mu A}=\frac{l}{{\mu }_0{\mu }_{r}A};\text{resistance}=\rho \frac{l}{A}=\frac{l}{\sigma A}$

4. Permeance. It is reciprocal of reluctance and implies the ease or readiness with which magnetic flux is developed. It is analogous to conductance in electric circuits. It is measured in terms of Wb/AT or henry.

5. Reluctivity. It is specific reluctance and corresponds to resistivity which is ‘specific resistance’.

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Part 3: Circuit Analysis and Comparison

Composite Series Magnetic Circuit

In Fig. 6.26 is shown a composite series magnetic circuit consisting of three different magnetic materials of different permeabilities and lengths and one air gap (${\mu }_r=1$). Each path will have its own reluctance. The total reluctance is the sum of individual reluctances as they are joined in series.

$\therefore \text{total reluctance }=\sum \frac{l}{{\mu }_0{\mu }_{r}A}$ $=\frac{l_1}{{\mu }_0{\mu }_{r_1}{A}_1}+\frac{l_2}{{\mu }_0{\mu }_{r_2}{A}_2}+\frac{l_3}{{\mu }_0{\mu }_{r_3}{A}_3}+\frac{l_a}{{\mu }_0{\mu }_{r_a}{A}_g}$

$\therefore \text{flux }\Phi =\frac{\text{m.m.f.}}{\sum \frac{l}{{\mu }_0{\mu }_{r}A}}$

How to Find Ampere-turns?

It has been shown in Art. 6.15 that $H=NI/l\text{ AT/m}$ or $NI=H\times l$. $\therefore \text{ampere-turns }\text{AT}=H\times l$

Hence, following procedure should be adopted for calculating the total ampere turns of a composite magnetic path. (i) Find $H$ for each portion of the composite circuit. For air, $H=B/{\mu }_0$, otherwise $H=B/{\mu }_0{\mu }_r$. (ii) Find ampere-turns for each path separately by using the relation $\text{AT}=H\times l$. (iii) Add up these ampere-turns to get the total ampere-turns for the entire circuit.

Comparison Between Magnetic and Electric Circuits

SIMILARITIES

Magnetic Circuit	Electric Circuit
1. Flux = $\frac{\text{m.m.f.}}{\text{reluctance}}$	1. Current = $\frac{\text{e.m.f.}}{\text{resistance}}$
2. M.M.F. (ampere-turns)	2. E.M.F. (volts)
3. Flux $\Phi$ (webers)	3. Current $I$ (amperes)
4. Flux density $B$ (Wb/m${}^2$)	4. Current density (A/m${}^2$)
5. Reluctance $S=\frac{l}{\mu A}=(\frac{l}{{\mu }_0{\mu }_{r}A})$	5. resistance $R=\rho \frac{l}{A}=\frac{l}{\sigma A}$
6. Permeance (= 1/reluctance)	6. Conductance (= 1/resistance)
7. Reluctivity	7. Resistivity
8. Permeability (= 1/reluctivity)	8. Conductivity (= 1/resistivity)
9. Total m.m.f. = $\Phi S_1+\Phi S_2+\Phi S_3+...$	9. Total e.m.f. = $I{R}_1+I{R}_2+I{R}_3+...$

DIFFERENCES

1. Strictly speaking, flux does not actually ‘flow’ in the sense in which an electric current flows.
2. If temperature is kept constant, then resistance of an electric circuit is constant and is independent of the current strength (or current density). On the other hand, the reluctance of a magnetic circuit does depend on flux (and hence flux density) established in it. It is so because $\mu$ (which equals the slope of $B/H$ curve) is not constant even for a given material as it depends on the flux density $B$. Value of $\mu$ is large for low value of $B$ and vice-versa. Hence, reluctance is small ($S=l/\mu A$) for small values of $B$ and large for large values of $B$.
3. Flow of current in an electric circuit involves continuous expenditure of energy but in a magnetic circuit, energy is needed only creating the flux initially but not for maintaining it.

Parallel Magnetic Circuits

Fig. 6.29 (a) shown a parallel magnetic circuit consisting of two parallel magnetic paths $ACB$ and $ADB$ acted upon by the same m.m.f. Each magnetic path has an average length of $2(l_1+l_2)$. The flux produced by the coil wound on the central core is divided equally at point $A$ between the two outer parallel paths. The reluctance offered by the two parallel paths is = half the reluctance of each path.

Fig. 6.29 (b) shows the equivalent electrical circuit where resistance offered to the voltage source is $=R||R=R/2$.

Series-Parallel Magnetic Circuits

Such a circuit is shown in Fig. 6.30 (a). It shows two parallel magnetic circuits $ACB$ and $ACD$ connected across the common magnetic path $AB$ which contains an air-gap of length $l_g$. As usual, the flux $\Phi$ in the common core is divided equally at point $A$ between the two parallel paths which have equal reluctance.

The reluctance of the path $AB$ consists of (i) air gap reluctance and (ii) the reluctance of the central core which comparatively negligible. Hence, the reluctance of the central core $AB$ equals only the air-gap reluctance across which are connected two equal parallel reluctances. Hence, the m.m.f. required for this circuit would be the sum of (i) that required for the air-gap and (ii) that required for either of two paths (not both) as illustrated in Ex. 6.19, 6.20 and 6.21.

The equivalent electrical circuit is shown in Fig. 6.30 (b) where the total resistance offered to the voltage source is $=R_1+R||R=R_1+R/2$.

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