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5. Magnetic Circuits

• Gyrator–capacitor model

In the previous chapters we got accustomed to the magnetic field. During this path some similarities from the magnetic field to the electric circuit appeared (see Abbildung 1).

In this chapter we will investigate, how far we come with such an analogy and where it can by practically applied.

5.1 Linear magnetic Circuits

For the upcoming calculations the following assumptions are made

1. The relationship between $B$ and $H$ is linear: $B = \mu \cdot H$
   This is a good estimation when the magnetic field strength lays well below saturation
2. There is not stray field leaking out of the magnetic-field conducting material.
3. The fields inside of airgaps is homogenious. This is true for small airgaps.

One can calulate a lot of simple magnetic circuits, when this assumptions and focusing on the average field line are applied.

Two simple magnetic circuits are shown in Abbildung 3: They consist of

• a current-carrying coil
• a ferrite core
• an airgap (in picture (2) ).

This three parts will be investigated shortly:

Current-carrying Coil

For the magnetic circuit, the coil is parameterized only by:

• its nuber of windings $N$ and
• the passing current $i$.

This parameters lead to the magnetic voltage $\theta = N\cdot i$.

Ferrite Core

• The core is assumed to be made of ferromagnetic material.
• Therefore, the relative permeability in the core is much larger than in air ($\mu_{r,core}>>1$).
• The ferrite core is also filling the inside of the current-carrying coil.
• The ferrite core conducts the magnetic flux around the magnetic circuit (and by this: also to the airgap)

Airgap

• The airgap interrupts the ferrite core.
• The width of the airgap is small compared to the dimensions of the cross section of the ferrite core.
• The field in the airgap can be used to generate (mechanical) effects within the airgap.
  An example for this can be the force onto a permanent magnet (see Abbildung 3 (3)).

With the above-mentioned assumtions the magnetic flux $\Phi$ must remain constant along the ferrite core, so $\Phi_{core}=const.$.
Since the magnetic field lines to neither show sources nor sinks, also the flux passing over to the airgap must be $\Phi_{airgap}=\Phi_{core}=const.$ This can also be seen in Abbildung 4 (1) ).
A different view onto this is the closed surface $\vec{A}$ (Abbildung 4 (2)): Based on the examination in Recap of magnetic Field we know that the flux into the volume must be equal the flux out of the volume, or $\Phi_m = \iint_{A} \vec{B} \cdot d \vec{A} = 0$.

The relationship $B=\mu \cdot H$, and $\mu_{core}\gg\mu_{airgap}$ lead to the fact that the ${H}$-Field must be much stronger within the airgap (Abbildung 4 (3)).

Magnetic Cirtuit in a Formula

Therefore, the following formula is given \begin{align*} \Phi_m &= &\iint_A &\vec{B} \cdot d \vec{A} &= &const. \\ &= & &B \cdot A &= &const. \\ &= & &B_{core} \cdot A_{core} &= &B_{airgap} \cdot A_{airgap} = const. \\ \end{align*}

The assumptions, that there is that the field inside of airgap is homogenious and there is no strayfield lead to the fact, that $A_{core} = A_{airgap}$.
Therefore: \begin{align*} B_{core} &= &B_{airgap} = B \\ \mu_0 \mu_{r,core} H_{core} &= &\mu_0 \mu_{r,airgap} H_{airgap} = {{\Phi}\over{A}} \tag{5.2.1} \end{align*}

Beside this, the magnetic field strength $H$ along one field line is directly given by: \begin{align*} \theta = N \cdot i &= \int_s \vec{H} d\vec{s} = \\ &= \int_{core} \vec{H} d\vec{s} + \int_{airgap} \vec{H} d\vec{s} \\ \end{align*}

With the assumtion of a linear and homogeneous $B$-Field and the width $\delta$ of the airgap, this leads to: \begin{align*} \theta = {H}_{core} l_{core} + {H}_{airgap} \delta \tag{5.2.2} \end{align*}

With the prevoius formula $5.2.1$, this gets to: \begin{align*} \theta &= {{B}\over{\mu_0 \mu_{r,core}}} l_{core} + {{B}\over{\mu_0 \mu_{r,airgap}}} \delta \\ &= {{\Phi \cdot l_{core}}\over{A \cdot \mu_0 \mu_{r,core}}} + {{\Phi \cdot \delta}\over{A \cdot \mu_0 \mu_{r,airgap}}} \\ &= {{1}\over{\mu_0 \mu_{r,core}}}{{l_{core}}\over{A}} \cdot \Phi + {{1}\over{\mu_0 \mu_{r,airgap}}}{{\delta}\over{A}} \cdot \Phi \tag{5.2.3} \end{align*}

Comparing the formula $5.2.3$ with the ohmic resistance and resistivity of two resistors in series, shows something interesting: \begin{align*} u &= &R_1 \cdot &i &+ &R_2 \cdot i \\ &= &\rho {{l_1}\over{A_1}} \cdot &i &+ &\rho {{l_2}\over{A_2}} \cdot i \end{align*}

This leads to:

• The magnetic voltage $\theta$ acts like the electic voltage $u$,
  the magnetic flux $\Phi$ like the current $i$.
• The linear relationship $\theta = f(\Phi)$ is also called Hopkinson's Law.

Application and Limitations of the Circuit Interpretation

The results are only allowed as first approximation. The following table shows the contrast between the electric and magnetic field:

5.2 Non-linear magnetic Circuits

not included in the present script

5.3 Mutual Induction and Coupling

Situation: Two coils $1$ and $2$ near by each other.
Which effect do the coils have onto each other?

Effect of Coils onto each other

1. Windings $N_1$ of coil $1$ gets feed with current $i_1$.
2. Coil $1$ generates change in flux $\Phi_{11}$
3. Coil $2$ gets passed by part of the flux ($\Phi_{21}$)
4. Stray flux $\Phi_{S1}$ takes not part in coupling

• $\rightarrow$ Coils are magnetically coupled:
  Flux $\Phi_{11}$ of coil $1$ gets divided into flux $\Phi_{21}$ in coil $2$ and stray flux $\Phi_{S1}$ not passing coil $2$:
  
  $\Phi_{11} = \Phi_{21} + \Phi_{S1}$
  
  
• Induced voltage in coil $2$:
  
  $ u_{ind,2} = -{{d}\over{dt}}\Psi_{21} $
  $\quad\quad = -N_2 \cdot {{d}\over{dt}}\Phi_{21} $
  
  
• Similar effect on coil $1$ due to a current $i_2$ through coil $2$:
  
  $\Phi_{22} = \Phi_{12} + \Phi_{S2}$
  
  $ u_{ind,1} = -N_1 \cdot {{d}\over{dt}}\Phi_{12} $

Linked Fluxes

For the single coil we got the relationship between the linked flux $\Psi$ and the current $i$ as: $\Psi = L \cdot i$.
Now the coils also are interacting with each other. This must also be reflected in the relationship $\Psi_1 = f(i_1, i_2)$, $\Psi_2 = f(i_1, i_2)$: \begin{align*} \Psi_1 &= &L_{11} &\cdot i_1 &+ M_{12} &\cdot i_2 \\ \Psi_2 &= &L_{22} &\cdot i_2 &+ M_{21} &\cdot i_1 \\ \end{align*}

With

• $L_{11}$ and $L_{22}$ as the self induction
• $M_{12}$ and $M_{21}$ as the mutual induction

The formula can also be described as: \begin{align*} \left( \begin{array}{c} \Psi_1 \\ \Psi_2 \end{array} \right) = \left( \begin{array}{c} L_{11} & M_{12} \\ M_{21} & L_{22} \end{array} \right) \cdot \left( \begin{array}{c} i_1 \\ i_2 \end{array} \right) \end{align*}

The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$?

Magnetic Circuit with 2 Sources

In order to get the self induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see Abbildung 10).
Ther the stray flux of the previous situation is only located in the middle leg. This also means, that there is no stray flux outside of the iron core.

The Abbildung 10 shows the fluxes on each parts. The filled black circles nearby the windings mark the direction of winding:
When there is one current for each winding ingoing into the marked pin, the fluxes will get added up positively.

In order to get $L_{11}$ and $L_{22}$, we look back to the inductivity $L$ of a long coil with the length $l$.
This was given in the chapter Self-Induction as

\begin{align*} L &= {{\Psi}\over{i}} \\ L &= \mu_0 \mu_r \cdot N^2 \cdot {{A }\over {l}} \\ \end{align*}

In this case, $\Psi$ was the flux through the coil generated from the coil itself - here $L_{11}$ and $L_{22}$.
Comparing this formula with the magnetic resistance $R_m = {{1}\over{\mu_0 \mu_{r}}}{{l}\over{A}}$ of the coil, one can conclude:

\begin{align} \boxed{L = {{N^2 }\over {R_{m,coil}}}} \end{align}

Based on this formula, the self inductions $L_{11}$ and $L_{22}$ can be written with the magnetic resistances of the coils as:

\begin{align*} L_{11} &= {{N_1^2 }\over {R_{m,11}}} \\ L_{22} &= {{N_2^2 }\over {R_{m,22}}} \end{align*}

In order to get the effect of the mutual induction, a coupling coefficient $k$ is introduced. $k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$):

\begin{align*} k_{21} = {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*}

When $k_{21}=100\%$, there is no flux in the middle leg but only in the second coil.
For $k_{21}=0\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling.
The resulting total coupling $k$ is given as

\begin{align*} k = \sqrt{k_{12}\cdot k_{21}} \\ \end{align*}

The mutual induction $M_{21}$ can be calculated as the fraction of the linked flux $\Psi_{11}$ from coil $1$ acting on the current $i_2$ of coil $2$:

\begin{align*} M_{21} &= {{\Psi_{21}}\over{i_2}} \\ &= {{N_2 \cdot \Phi_{21}}\over{i_2}} \\ &= {{N_2 \cdot k_{21} \cdot \Phi_{11}}\over{i_2}} \\ &= {{N_2}\over{N_1}}k_{21} \cdot {{\Psi_{11}}\over{i_2}} \\ &= {{N_2}\over{N_1}}k_{21} \cdot L_{11} \\ &= {{N_2}\over{N_1}}k_{21} \cdot {{N_1^2 }\over {R_{m,11}}} \\ &= k_{21} \cdot {{N_1 \cdot N_2 }\over {R_{m,11}}} \\ \end{align*}

\begin{align*} M_{21} &= {{\Psi_{21}}\over{i_2}} \\ &= {k_{21} \cdot {\Psi_{11}}\over{i_2}} \\ &= {k_{21} \cdot N_1 \cdot {\Phi_{11}}\over{i_2}} \\ \end{align*}

\begin{align*} L \cdot i = \Psi_{11} &= N_1 \cdot \Phi_{11} \\ \end{align*}

With the

The formula can also be described as: \begin{align*} \left( \begin{array}{c} \Psi_1 \\ \Psi_2 \end{array} \right) = \left( \begin{array}{c} L_{11} & M_{12} \\ M_{21} & L_{22} \end{array} \right) \cdot \left( \begin{array}{c} i_1 \\ i_2 \end{array} \right) \end{align*}

5.4 Magnetic Energy

Tasks