electrical_engineering_2:magnetic_circuits [MEXLE Wiki]

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====== 5 Magnetic Circuits ======

<callout> For this and the following chapter the online Book 'DC Electrical Circuit Analysis - A Practical Approach' is strongly recommended as a reference. In detail this is chapter [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|10.3 Magnetic Circuits]] </callout>

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 <imgref ImgNr01>).

<WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP>

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

===== 5.1 Linear Magnetic Circuits =====

For the upcoming calculations, the following assumptions are made

  - 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
  - There is no stray field leaking out of the magnetic field conducting material.
  - The fields inside of airgaps are homogeneous. This is true for small air gaps.

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

<WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP>

Two simple magnetic circuits are shown in <imgref ImgNr02>: They consist of

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

<WRAP> <imgcaption ImgNr02 | A simple magnetic Circuit> </imgcaption> {{drawio>SimpleMagnCircuit.svg}} </WRAP>

These three parts will be investigated shortly:

==== Current-carrying Coil ====

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

  * its number of windings $N$ and
  * the passing current $i$.

These 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 ($\mu_{\rm r,core}\gg 1$).
  * 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 air gap interrupts the ferrite core.
  * The width of the air gap is small compared to the dimensions of the cross-section of the ferrite core.
  * The field in the air gap can be used to generate (mechanical) effects within the air gap. \\ An example of this can be the force onto a permanent magnet (see <imgref ImgNr02> (3)).

With the above-mentioned assumptions the magnetic flux $\Phi$ must remain constant along the ferrite core, so $\Phi_{\rm core}=\rm const.$. \\ 
Since the magnetic field lines neither show sources nor sinks, also the flux passing over to the airgap must be $\Phi_{\rm airgap}=\Phi_{\rm core}=\rm const.$ This can also be seen in <imgref ImgNr04> (1) ). \\ 
A different view of this is the closed surface $\vec{A}$ (<imgref ImgNr04> (2)): Based on the examination in [[:electrical_engineering_2:the_time-dependent_magnetic_field#recap_of_magnetic_field|Recap of magnetic Field]] we know that the flux into the volume must be equal the flux out of the volume, or $\Phi_{\rm m} = \iint_{A} \vec{B} \cdot d \vec{A} = 0$.

The relationship $B=\mu \cdot H$, and $\mu_{\rm core}\gg\mu_{\rm airgap}$ lead to the fact that the ${H}$-Field must be much stronger within the airgap (<imgref ImgNr04> (3)).

<WRAP> <imgcaption ImgNr04 | B- and H-field along the ferrite core> </imgcaption> {{drawio>BHfieldFerriteCore.svg}} </WRAP>

==== Magnetic Circuit in a Formula ====

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

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

Besides 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_{\rm core} \vec{H} d\vec{s} + \int_{\rm 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}_{\rm core} l_{\rm core} + {H}_{\rm 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_{\rm r,core}}} l_{\rm core}                       + {{B}\over{\mu_0 \mu_{\rm r,airgap}}} \delta \\ 
       &= {{\Phi \cdot l_{\rm core}}\over{A \cdot \mu_0 \mu_{\rm r,core}}}      + {{\Phi \cdot \delta}\over{A \cdot \mu_0 \mu_{\rm r,airgap}}} \\ 
       &= {{1}\over{\mu_0 \mu_{\rm r,core}}}{{l_{\rm core}}\over{A}} \cdot \Phi + {{1}\over{\mu_0 \mu_{\rm 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 {{https://en.wikipedia.org/wiki/Magnetic_circuit#Hopkinson's_law|Hopkinson's Law}}.

<callout icon="fa fa-exclamation" color="red" title="Notice:">

  * Also for the magnetic circuit one can set up a lumped circuit model (see <imgref ImgNr07>).
  * Similar to Ohms law, there is a **magnetic resistance**  or **reluctance**: \\ \begin{align*} \boxed{R_{\rm m} = {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}} \end{align*}
  * The unit of $R_{\rm m}$ is $[R_{\rm m}]= [\theta]/[\Phi] = ~\rm 1 A / Vs = 1/H $
  * The length $l$ is given by the mean magnetic path length (= average field line length in the core).
  * Kirchhoff's laws (mesh rule and nodal rule) can also be applied:
      * The sum of the magentic fluxes $\Phi_i$ in into a node is: $\sum_i \Phi_i=0 $
      * The sum of the magnetic voltages $\theta_i$ along the average field line is: $\sum_i \theta_i=0 $
  * The application of the lumped circuit model is based on multiple assumptions. \\ In contrast to the simplification for the electric current and voltage the simplification for the flux and magnetic voltage is not as exact.

<WRAP> <imgcaption ImgNr07 | lumped Circuit Model for magnetic Circuits> </imgcaption> {{drawio>LumpedMagnCircuit.svg}} </WRAP>

</callout>

==== Application and Limitations of the Circuit Interpretation ====

<callout icon="fa fa-exclamation" color="red" title="Notice:"> "Recipe" for solving magnetic circuits:

  - Watch out for individual magnetic resistances in the circuits:
      - Separate the magnetic circuit into parts, where the permeability and area of the cross-section are constant. \\ For these parts $B$ and $H$ is constant. These parts will have individual magnetic resistances.
      - Each airgap also gets an individual magnetic resistance.
      - Calculate the magnetic resistance by the mean magnetic path length through the individual parts.
  - Calculate the magnetic circuit as a circuit.
      - Magnetic voltages depict voltage sources; magnetic flux depicts currents.
      - Use the known way to solve the circuit.

Be aware that the orientation of the current $I$ and the field strength $\vec{H}$ are also connected by the right-hand rule. </callout>

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

^Property        ^Electric Field                                                               ^Magnetic Field |
|Materials       |There are "pure" isolators, which are completely non-conductive.             |All materials have a permeability $>0$ |
|Sources         |The source is concentrated \\ (= there are field sources / electric charges) |The magnetic source (= coil) is distributed |
|Simplifications |The simplifications often work for good results \\ (small wire diameter, relatively constant resistivity) |The simplification is often too simple \\ (widespread beyond the mean magnetic path length, non-linearity of the permeability) |

<panel type="info" title="Task 5.1.1 Coil on a plastic Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A coil is set up onto a toroidal plastic ring ($\mu_{\rm r}=1$) with an average circumference of $l_R = 300 ~\rm mm$. 
The $N=400$ windings are evenly distributed along the circumference. 
The diameter on the cross-section of the plastic ring is $d = 10 ~\rm mm$. In the windings, a current of $I=500 ~\rm mA$ is flowing.

Calculate

  - the magnetic field strength $H$ in the middle of the ring cross-section.
  - the magnetic flux density $B$ in the middle of the ring cross-section.
  - the magnetic resistance $R_{\rm m}$ of the plastic ring.
  - the magnetic flux $\Phi$.

<button size="xs" type="link" collapse="Solution_5_1_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_1_1_Result" collapsed="true">

  - $H = 667 ~\rm {{A}\over{m}}$
  - $B = 0.84 ~\rm mT$
  - $R_m = 3 \cdot 10^9 ~\rm {{1}\over{H}}$
  - $\Phi = 66 ~\rm nVs$

</collapse>

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.2 magnetic Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions:

  - $l=35.8~\rm cm$, $d=1.90~\rm cm$
  - $l=11.1~\rm cm$, $d=1.50~\rm cm$

#@HiddenBegin_HTML~5_1_2s,Solution~@#

The magnetic resistance is given by:
\begin{align*}
\ R_{\rm m} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} 
\end{align*}

With
  * the area $ A = \left({{d}\over{2}}\right)^2 \cdot \pi $
  * the vacuum magnetic permeability $\mu_{0}=4\pi\cdot 10^{-7} ~\rm H/m$, and 
  * the relative permeability $\mu_{\rm r}=1$.

#@HiddenEnd_HTML~5_1_2s,Solution ~@#

#@HiddenBegin_HTML~5_1_2r,Result~@#
  - $1.00\cdot 10^9 ~\rm {{1}\over{H}}$
  - $0.50\cdot 10^9 ~\rm {{1}\over{H}}$
#@HiddenEnd_HTML~5_1_2r,Result~@#

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.3 magnetic Resistance of an airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Calculate the magnetic resistances of an airgap with the following dimensions:

  - $\delta=0.5~\rm mm$, $A=10.2~\rm cm^2$
  - $\delta=3.0~\rm mm$, $A=11.9~\rm cm^2$

<button size="xs" type="link" collapse="Solution_5_1_3_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_3_1_Result" collapsed="true">

  - $3.9\cdot 10^5 ~\rm {{1}\over{H}}$
  - $2.0\cdot 10^6 ~\rm {{1}\over{H}}$

</collapse>

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Calculate the magnetic voltage necessary to create a flux of $\Phi=0.5 ~\rm mVs$ in an airgap with the following dimensions:

  - $\delta=1.7~\rm mm$, $A=4.5~\rm cm^2$
  - $\delta=5.0~\rm mm$, $A=7.1~\rm cm^2$

<button size="xs" type="link" collapse="Solution_5_1_4_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_4_1_Result" collapsed="true">

  - $\theta = 1.5\cdot 10^3 ~\rm A$, or $1000$ windings with $1.5~\rm A$
  - $\theta = 2.8\cdot 10^3 ~\rm A$, or $1000$ windings with $2.8~\rm A$

</collapse>

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.5 Magnetic Flux"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Calculate the magnetic flux created on a magnetic resistance of $R_m = 2.5 \cdot 10^6 ~\rm {{1}\over{H}}$ with the following magnetic voltages:

  - $\theta = 35   ~\rm A$
  - $\theta = 950  ~\rm A$
  - $\theta = 2750 ~\rm A$

<button size="xs" type="link" collapse="Solution_5_1_5_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_5_1_Result" collapsed="true">

  - $\Phi =14  ~\rm µVs$
  - $\Phi =0.38~\rm mVs$
  - $\Phi =1.1 ~\rm mVs$

</collapse>

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.6 Two-parted ferrite Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A core shall consist of two parts as seen in <imgref ImgExNr08>. 
In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$.

The cross sections are $A_1=530 ~\rm mm^2$ and $A_2=460 ~\rm mm^2$. 
The mean magnetic path lengths are $l_1 = 200 ~\rm mm$ and $l_2 = 130 ~\rm mm$.

The air gaps on the coupling joint between both parts have the length $\delta=0.23 ~\rm mm$ each. 
The permeability of the ferrite is $\mu_r = 3000$. 
The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$

<WRAP> <imgcaption ImgExNr08 | Two-parted ferrite Core> </imgcaption> {{drawio>TwoPartedCoil.svg}} </WRAP>

  - Draw the lumped circuit of the magnetic system
  - Calculate all magnetic resistances $R_{\rm m,i}$
  - Calculate the flux in the circuit

<button size="xs" type="link" collapse="Solution_5_1_6_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_6_1_Result" collapsed="true">

  - -
  - magnetic resistances: $R_{\rm m,1} = 100 \cdot 10^3 ~\rm {{1}\over{H}}$, $R_{\rm m,1} = 75 \cdot 10^3 ~\rm {{1}\over{H}}$, $R_{{\rm m},\delta} = 400 \cdot 10^3 ~\rm {{1}\over{H}}$
  - magnetic flux: $\Phi = 0.80 ~\rm mVs$

</collapse>

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.7 Comparison with simplified Calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The magnetic circuit in <imgref ImgExNr09> passes a magnetic flux density of $0.4 ~\rm T$ given by an excitation current of $0.50 ~\rm A$ in $400$ windings. 
At position $\rm A-B$, an air gap will be inserted. After this, the same flux density will be reached with $3.70 ~\rm A$

<WRAP> <imgcaption ImgExNr09 | Example of a magnetic circuit> </imgcaption> {{drawio>ExMagncirc01.svg}} </WRAP>

  - Calculate the length of the airgap $\delta$ with the simplification $\mu_{\rm r} \gg 1$
  - Calculate the length of the airgap $\delta$ exactly with $\mu_{\rm r} = 1000$

<button size="xs" type="link" collapse="Solution_5_1_7_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_7_1_Result" collapsed="true">

  - $\delta = 4.02(12) ~\rm mm$
  - $\delta = 4.02(52) ~\rm mm$

</collapse>

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.8 Coil on a ferrite Core with airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The choke coil shown in <imgref ImgExNr10> shall be given, with a constant cross-section in all legs $l_0$, $l_1$, $l_2$. 
The number of windings shall be $N$ and the current through a single winding $I$.

<WRAP> <imgcaption ImgExNr10 | Example for a Choke Coil> </imgcaption> {{drawio>ChokeCoilEx1.svg}} </WRAP>

  - Draw the lumped circuit of the magnetic system
  - Calculate all magnetic resistances $R_{{\rm m},i}$
  - Calculate the partial fluxes in all the legs of the circuit

</WRAP></WRAP></panel>

===== 5.2 Non-linear magnetic Circuits =====

not included in the present script

===== 5.3 Mutual Induction and Coupling =====

Imagine charging your phone wirelessly by simply placing it on a charging pad. 
This seamless experience is made possible by the fascinating phenomenon of mutual induction and coupling between two coils.

This situation is depicted in <imgref ImgNr09>: 
When an alternating current flows through one coil (Coil $1$), it creates a time-varying magnetic field that induces a voltage in the nearby coil (Coil $2$), even though they are not physically connected. 
This mutual influence is governed by the principle of electromagnetic induction.

<WRAP center 35%> <imgcaption ImgNr09 | Mutual Induction of two Coils> </imgcaption> {{drawio>MutualInductionTwoCoils1.svg}} </WRAP>

The key factor determining the strength of mutual induction is the mutual inductance ($M$) between the coils. 
It quantifies the magnetic flux linkage and depends on factors like the number of turns, current, and relative orientation of the coils.

While geometric properties play a role, the fundamental principle can be described using electric properties alone, making mutual induction a versatile concept with numerous applications, including:

  * Wireless power transfer
  * Transformers
  * Inductive coupling in communication systems
  * Inductive sensors

As we explore this chapter, we'll delve into the mathematical models, equations, and practical considerations of mutual induction and coupling, unlocking a world of innovative technologies that shape our modern lives.
We explicitly try to answer the following questions:

  * Which effect do the coils have on each other?
  * Can we describe the effects with mainly electric properties (i.e. no geometric properties)


==== Effect of Coils on each other ====

  - Windings $N_1$ of coil $1$ gets feed with current $i_1$.
  - Coil $1$ generates change in flux $\Phi_{11}$
  - Coil $2$ gets passed by part of the flux ($\Phi_{21}$)
  - Stray flux $\Phi_{\rm S1}$ 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 $\Phi_{\rm S1}$ not passing coil $2$: \\  \\ $\Phi_{11} = \Phi_{21} + \Phi_{\rm S1}$ \\
  * Induced voltage in coil $2$: \\  \\ $ u_{\rm ind,2} = -{{\rm d}\over{{\rm d}t}}\Psi_{21} $ \\ $\quad\quad = -N_2 \cdot {{\rm d}\over{{\rm d}t}}\Phi_{21} $ \\
  * Similar effect on coil $1$ due to a current $i_2$ through coil $2$: \\  \\ $\Phi_{22} = \Phi_{12} + \Phi_{\rm S2}$ \\  \\ $ u_{\rm ind,1} = -N_1 \cdot {{\rm d}\over{{\rm d}t}}\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 interact 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 &= &\Psi_{11}        &+ \Psi_{12} \\ 
\Psi_2 &= &\Psi_{22}        &+ \Psi_{21} \\ \\ 

\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: 
{{drawio>VectorialformulaOfInduction.svg}}

The view of the magnetic flux is advantageous when effects like an acting Lorentz force is of interest. 
However, more often the coils couple electric circuits, like in a transformer or a wireless charger. 
Here, the effect on the circuits is of interest. This can be calculated with the induced electric voltages $u_{\rm ind,1}$ and $u_{\rm ind,2}$ in each circuit. 
They are given by the formula $u_{{\rm ind},x} = -{\rm d}\Psi_x /{\rm d}t$:

\begin{align*} 
\left( \begin{array}{c} u_1 \\ u_2 \end{array} \right) &= -{{\rm d}\over{{\rm d}t}} \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} {{\rm d}\over{{\rm d}t}} i_1 \\ 
{{\rm d}\over{{\rm d}t}}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 two Sources ====

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 <imgref ImgNr08>). \\ 
There 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.

<WRAP> <imgcaption ImgNr08 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils.svg}} </WRAP>

The <imgref ImgNr08> shows the fluxes on each part. The black dots near the windings mark the direction of the windings, and therefore the sign of the generated flux. \\ 
All the fluxes caused by currents flowing into the __marked pins__ are summed up __positively__ in the core. \\
When there is a current flowing into a __non-marked pin__, its flux has to be __subtracted__ from the others.

To get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\ 
This was given in the chapter [[:electrical_engineering_2:the_time-dependent_magnetic_field#self-induction1|Self-Induction]] as

\begin{align*} 
L &= {{\Psi}\over{i}} \\ 
L &= \mu_0 \mu_{\rm 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 $R_{\rm m} = {{1}\over{\mu_0 \mu_{\rm r}}} {{l}\over{A}}$ of the coil, one can conclude:

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

Important here is, that we assumed no stray field. 
When taking this into account for calculating the self-induction the magnetic resistance $R_{\rm m,coil}$ ends op to be the resistance of the magnetic circuit, seen by the magnetic voltage source!

\begin{align} \boxed{L = {{N^2 }\over {R_{\rm m1}}}} \end{align}

The magnetic resistance $R_{\rm m1}$ in the example <imgref ImgNr08> is $R_{\rm m1} = R_{\rm m,11} + (R_{\rm m,ss}||R_{\rm m,22})$. 
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_{\rm m1}}} \\ 
L_{22} &= {{N_2^2 }\over {R_{\rm m2}}} 
\end{align*}

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} = \pm {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*}

The sign of $k_{21}$ depends on the direction of $\Phi_{21}$ relative to $\Phi_{22}$! If the directions are the same, the positive sign applies, if the directions are opposite, the minus sign applies.

When $k_{21}=+100~\%$, there is no flux in the middle leg but only in the second coil and in the same direction as the flux that originates from the second coil. \\ 
When $k_{21}=-100~\%$, there is no flux in the middle leg but only in the second coil and in the opposite direction as the flux that originates from 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 mutual induction $M_{21}$ can be calculated as the fraction of the linked flux $\Psi_{11}$ in coil $2$ based on the current $i_1$ from the coil $1$ :

\begin{align*} 
M_{21} &= {{\Psi_{21}}                              \over{i_1}} \\ 
       &= {{N_2                   \cdot   \Phi_{21}}\over{i_1}} \\ 
       &= {{N_2 \cdot k_{21}      \cdot   \Phi_{11}}\over{i_1}} \\ 
       &= {{N_2}\over{N_1}}k_{21} \cdot {{\Psi_{11}}\over{i_1}} \\ 
       &= {{N_2}\over{N_1}}k_{21} \cdot L_{11} \\ 
       &= {{N_2}\over{N_1}}k_{21} \cdot {{N_1^2 }        \over {R_{\rm m1}}} \\ 
       &= k_{21}                  \cdot {{N_1 \cdot N_2 }\over {R_{\rm m1}}} \\ 
\end{align*}

Note, that also $M_{21}$ and $M_{12}$ can be either positive or negative, depending on the sign of the coupling coefficients.

The formula is finally: 
\begin{align*} 
\left( \begin{array}{c} \Psi_1                                                                     \\ \Psi_2                                                                     \end{array} \right) 
= 
\left( \begin{array}{c} {{N_1^2}\over{R_{\rm m1}}} & k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} \\ k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} & {{N_2^2}\over{R_{\rm m2}}} \end{array} \right) 
\cdot 
\left( \begin{array}{c} i_1                                                                        \\ i_2                                                                        \end{array} \right) 
\end{align*}

For most of the applications the induction matrix has to be symmetric((This can be derived from energy considerations, when only electric circuits are coupled without additional flow of mechanical energy. This is, for example  not the case for motors with a mechanical load.)). \\ Therefore, the following applies:

  * In General: the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$
  * For symmetric induction matrix: The mutual inductances are equal: $M_{12} = M_{21} = M$
  * The resulting **total coupling** $k$ is given as \begin{align*} k = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \end{align*}



<panel type="info" title="Task 5.3.1 Example for magnetic Circuit with two Sources"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The magnetical configuration in <imgref ExImgNr01> shall be given. \\ 
The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, the length $l=12 ~\rm cm$ and the number of windings $N_1 = 400$, $N_2=300$.

<WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP>

1.  Simplify the configuration into three magnetic resistors and 2 voltage sources. Draw the problem as an equivalent circuit

#@HiddenBegin_HTML~5311,Result~@#
<WRAP> <imgcaption ExImgNr11 | Equivalent Network> </imgcaption> {{drawio>CoreWithTwoCoils2network.svg}} </WRAP>
#@HiddenEnd_HTML~5311,Result~@#

2. Calculate all magnetic resistances. Additionally, calculate the magnetic resistances $R_{\rm m1}$  and $R_{\rm m2}$ seen from the magnetic voltage source $1$ and $2$.

#@HiddenBegin_HTML~5312,Path~@#

<WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP>

The magnetic resistance is summed up by looking at the circuit from the source $1$ (see <imgref ExImgNr13>): 
\begin{align*} 
R_{\rm m1} &= R_{\rm m,11} + R_{\rm m,ss} || R_{\rm m,22} \\ 
\end{align*}

where the parts are given as 
\begin{align*} 
R_{\rm m,11} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{3\cdot l}\over{A}} &&= 398 \cdot 10^{3} ~\rm {{1}\over{H}} \\ 
R_{\rm m,ss} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{1\cdot l}\over{A}} &&= 133 \cdot 10^{3} ~\rm {{1}\over{H}} \\ 
R_{\rm m,22} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{2\cdot l}\over{A}} &&= 265 \cdot 10^{3} ~\rm {{1}\over{H}} \\ 
\end{align*}

With the given geometry this leads to 
\begin{align*} 
R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\ 
           &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{3}} &&= 486 \cdot 10^{3} ~\rm {{1}\over{H}}\\ 
\end{align*}

Similarly, the magnetic resistance $R_{m2}$ is 
\begin{align*} 
R_{\rm m2} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{4}} &&= 365 \cdot 10^{3} ~\rm {{1}\over{H}}\\ 
\end{align*}

#@HiddenEnd_HTML~5312,Path~@#

3. Calculate the self-inductions $L_{11}$ and $L_{22}$

#@HiddenBegin_HTML~5313,Path~@#
For the self-induction the effect on the electrical circuit is relevant. That is why the number of windings has to be considered.
\begin{align*} 
L_{11} &= {{N_1^2}\over{R_{\rm m1}}}                    &= 329 ~\rm mH\\ \\ 
L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\ 
\end{align*}
#@HiddenEnd_HTML~5313,Path~@#

4. Calculate the coupling factors $k_{12}$ and $k_{21}$.

#@HiddenBegin_HTML~5314,Path~@#
<WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP>

The coupling factor $k_{21}$ is defined as "how much of the flux created by one coil ($\Phi_{11}$) crosses the other coil ($\Phi_{21}$) ":
\begin{align*} 
k_{21} &= {{\Phi_{21}}\over{\Phi_{11}}}
\end{align*}

For this, we look at the circuit considering only one coil ("magnetic voltage source"), see <imgref ExImgNr12>. Based on the image, we see that the flux $\Phi_{11}$ divides into $\Phi_{S1}$ and $\Phi_{21}$ based on the magnetic resistance $\Phi_{\rm m,SS}$ and $\Phi_{\rm m,22}$. 
In step 2, we have calculated that $R_{\rm m,22}$ is twice $R_{\rm m,SS}$.  So from the incoming flux, only $1/3$ reaches $R_{\rm m,22}$ and - by this - "crosses the other coil". 

Therefore, the coupling factor $k_{21}$ is: $k_{21}= 1/3$.

A similar approach leads to $k_{12}$ with $k_{12}= 1/4$.
#@HiddenEnd_HTML~5314,Path~@#

5. Calculate the mutual inductions $M_{12}$, and $M_{21}$,

#@HiddenBegin_HTML~5315,Path~@#
\begin{align*} 
M_{21} &= k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} &&= {{1}\over{3}}\cdot{{400 \cdot 300}\over{ 486 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\ 
M_{12} &= k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} &&= {{1}\over{4}}\cdot{{400 \cdot 300}\over{ 365 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\ 
\end{align*}
#@HiddenEnd_HTML~5315,Path~@#

</WRAP></WRAP></panel>

#@TaskTitle_HTML@# 5.3.2 Wireles Charging #@TaskText_HTML@#

For Electric vehicles sometimes wireless charging systems are employed. These use the principle of mutual inductance to transfer power from a charging pad on the ground to the vehicle's battery pack. \\
This system consists of two coils: a transmitter coil embedded in the charging pad and a receiver coil mounted on the underside of the vehicle.

  * The transmitter coil has a self-inductance of $L_{\rm T} = 200 ~\rm \mu H$. 
  * The receiver coil has a self-inductance of $L_{\rm R} = 150 ~\rm \mu H$. 
  * The mutual inductance between the coils at this distance is measured to be  $M = 20 ~\rm \mu H$ - when the vehicle is properly aligned over the charging pad.

1. Calculate the coupling coefficient $k$ between the transmitter and receiver coils when the vehicle is properly aligned over the charging pad.

#@HiddenBegin_HTML~53211,Path ~@#

The given self-inductances are $L_{\rm T} = L_{11}$, $L_{\rm R} = L_{22}$. \\
By this, the following formula can be applied:

\begin{align*}
M = k \cdot \sqrt{L_{\rm T} \cdot L_{\rm R}}
\end{align*}

Therefore, $k$ is given as:
\begin{align*}
k = {{M}\over{ \sqrt{ L_{\rm T} \cdot L_{\rm R} } }}
\end{align*}

#@HiddenEnd_HTML~53211,Path ~@#

2. If the vehicle is misaligned by 10 cm from the center of the charging pad, the mutual inductance drops to $M = 12 ~\rm \mu H$. Calculate the new coupling coefficient in this misaligned position.

#@TaskEnd_HTML@#



==== Effects in the electric Circuits ====

  * Whenever two coils are magnetically coupled, not only the self-induction $L$ but also the mutual induction $M$ applies.
  * Based on the currents $i_1$, $i_2$ in the two circuits, the induced voltages are given by:

\begin{align*} 
u_1 &= L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} &+ M      \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ 
u_2 &= M      \cdot {{{\rm d}i_1}\over{{\rm d}t}} &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ 
\end{align*}

It is important to consider the polarity of the fluxes for the calculation in circuits (see <imgref ImgNr11>). 
The **sign of the mutual induction**  is influenced by

  * the direction of the windings, and
  * the orientation/counting of the current in the circuit.

<WRAP center 50%> <imgcaption ImgNr11 | Polarity of Coupling> </imgcaption> {{drawio>DirectionOfCoupling.svg}} </WRAP>

=== positive Polarity ===

The polarity is positive when both currents either flow into or out of the dotted pin (see <imgref ImgNr12>).

<imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}}

In this case, the **mutual induction is positiv $(M>0)$**.

The formula of the shown circuitry is then: 
\begin{align*} 
u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} &+ M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ 
u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} &+ M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ 
\end{align*}

=== negative Polarity ===

The polarity is negative when only one current either flows into the dotted pin and the other one out of the dotted pin (see <imgref ImgNr13>).

<WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP>

In this case, the **mutual induction is negativ $(M<0)$***.

The formula of the shown circuitry is then: 
\begin{align*} 
u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ 
u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ 
\end{align*}

<panel type="info" title="Task 5.3.3 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$. 
On the core, there are two coils $N_1=500$ and $N_2=250$ wound. The currents on the coils are $I_1 = 250 ~\rm mA$ and $I_2=300 ~\rm mA$.

  - The coils shall pass the currents with positive polarity (see the image **A** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm A}$ in the coil?
  - The coils shall pass the currents with negative polarity (see the image **B** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm B}$ in the coil?

<WRAP> <imgcaption ImgEx14 | toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg.svg}} </WRAP>

#@HiddenBegin_HTML~5_3_2s,Solution~@#

The resulting flux can be derived from a superposition of the individual fluxes $\Phi_1(I_1)$ and $\Phi_2(I_2)$, or alternatively by summing the magnetic voltages in the loop ($\sum_x \theta_x = 0$).

**Step 1 - Draw an equivalent magnetic circuit**

Since there are no branches all of the core can be lumped to a single magnetic resistance (see <imgref ImgEx14circ>).
<WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP>

**Step 2 - Get the absolute values of the individual fluxes**

Hopkinson's Law can be used here as a starting point. \\
It connects the magnetic flux $\Phi$ and the magnetic voltage $\theta$ on the single magnetic resistor $R_\rm m$. \\
It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\

\begin{align*} 
\theta_x             &= R_{\rm m}                                  \cdot \Phi_x \\
N_x \cdot I_x        &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} \cdot \Phi_x \\
\rightarrow \Phi_x   &= \mu_0 \mu_{\rm r} {{A}\over{l}} \cdot N_x \cdot I_x 
                      = {{1}\over{R_{\rm m} }}          \cdot N_x \cdot I_x \\
\end{align*}

With the given values we get: $R_{\rm m} = 495 {\rm {kA}\over{Vs}}$

**Step 3 - Get the signs/directions of the fluxes**

The <imgref5_3_2_Solution> shows how to get the correct direction for every single flux by use of the right-hand rule. \\
The fluxes have to be added regarding these directions and the given direction of the flux in question.
<WRAP> <imgcaption 5_3_2_Solution| toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg_solution.svg}} </WRAP>

Therefore, the formulas are
\begin{align*} 
\Phi_{\rm A}   &= \Phi_{1} - \Phi_{2} \\
               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  -  N_2 \cdot I_2 \right) \\
               & = 0.25 ~{\rm mVs} - 0.15 ~{\rm mVs} \\
\Phi_{\rm B}   &= \Phi_{1} + \Phi_{2} \\
               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  +  N_2 \cdot I_2 \right) \\
               & = 0.25 ~{\rm mVs} + 0.15 ~{\rm mVs} 
\end{align*}

#@HiddenEnd_HTML~5_3_2s,Solution~@#


#@HiddenBegin_HTML~5_3_2r,Result~@#
  - $0.10 ~\rm mVs$
  - $0.40 ~\rm mVs$
#@HiddenEnd_HTML~5_3_2r,Result~@#

</WRAP></WRAP></panel>

===== 5.4 Magnetic Energy =====

The magnetic field of a coil stores magnetic energy. 
The energy transfer from the electric circuit to the magnetic field is also the cause of the "current dampening" effect of the inductor. 
The energetic turnover for charging an conductor from $i(t_0=0)=0$ to $i(t_1)=I$ is given by:

\begin{align*} W_m = \int_0^\infty u(t)\cdot i(t) {\rm d}t \end{align*}

With $u_L = L \cdot {\rm d}i/{\rm d}t$, this becomes:

\begin{align*} 
W_{\rm m} &=        \int_0^\infty L\cdot {{{\rm d}i}\over{{\rm d}t}}\cdot i(t) {\rm d}t \\ 
          &=        \int_0^\infty L\cdot                                  i(t) {\rm d}i \\ 
          &= L\cdot \int_0^\infty                                         i(t) {\rm d}i \\ 
          &= L\cdot \left[ {{1}\over{2}}i^2\right]_0^I \\ 

\boxed{W_m = {{1}\over{2}}L\cdot I^2 } 
\end{align*}





==== magnetic Energy of a magnetic Circuit ====

With this formula also the stored energy in a magnetic circuit can be calculated. For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit: \begin{align*} \boxed{W_{\rm m} = {{1}\over{2}} \Psi \cdot I = {{1}\over{2}} {{\Psi^2}\over{L}}= {{1}\over{2}}{{\Phi^2 }\over{N^2 \cdot L}} = {{1}\over{2}} \Phi^2 \cdot R_{\rm m} = {{1}\over{2}}{{\theta^2 }\over{R_{\rm m}}}} \end{align*}

==== magnetic Energy of a toroid Coil ====

The formula can also be used for calculating the stored energy of a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line. 
By this, the following formulas can be used: 
\begin{align*} 
\theta = H \cdot l = N \cdot I \\ 
\Phi   = B \cdot A 
\end{align*}

With the above-mentioned formulas of the magnetic circuit, we get: 
\begin{align*} 
W_{\rm m} &= {{1}\over{2}} \Psi             \cdot I \\ 
          &= {{1}\over{2}}N \cdot B \cdot A \cdot {{H \cdot l}\over{N}} \\ 
          &= {{1}\over{2}}B \cdot H \cdot A \cdot l \\ 
\end{align*} \begin{align*} 
\boxed{W_{\rm m} = {{1}\over{2}}B\cdot H \cdot V} 
\end{align*}

The **magnetic energy density**  $w_{\rm m}$ can therefore be calculated as: 
\begin{align*} 
w_{\rm m} &= {{W_{\rm m}}\over{V}} \\ 
          &= {{1}\over{2}} B \cdot H \\ 
\end{align*}

This formula is also true for other types of coils.

==== generalized magnetic Energy ====

The general term to find the magnetic energy (e.g. for inhomogeneous magnetic fields) is given by 
\begin{align*} 
W_{\rm m} &= \iiint_V{w_{\rm m}            {\rm d}V} \\ 
          &= \iiint_V{\vec{B}\cdot \vec{H} {\rm d}V} 
\end{align*}

==== Application of the magnetic Energy ====

The circuit shown in <imgref ImgNr14> shall now be investigated. 
The inductor shall be a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line.

<WRAP> <imgcaption ImgNr14 | Example Circuits for calculating the magnetic Energy> </imgcaption> {{drawio>CircuitMagnEnergy.svg}} </WRAP>

The Kirchhoff mesh law leads to:

\begin{align*} 
u_{\rm s} = u_R       + u_L \\ 
u_{\rm s} = R \cdot i + N {{{\rm d}\Phi}\over{{\rm d}t}} 
\end{align*}

Multiplying with $i$ and with $dt$ we get the principle of conservation of energy $dw = u \cdot i \cdot {\rm d}t$ for each small time step.

\begin{align*} 
u_{\rm s} \cdot i \cdot dt &= R \cdot i^2 \cdot {\rm d}t   & + N {{{\rm d}\Phi}\over{dt}} \cdot i \cdot {\rm d}t   & \\ 
                        dW &=                   {\rm d}W_R & +                                          {\rm d}W_m & 
\end{align*}

In this way, we get the magnetic energy as: 
\begin{align*} 
dW_{\rm m} &= N {{{\rm d}\Phi}\over{{\rm d}t}}          \cdot i \cdot {\rm d}t \\ 
 W_{\rm d} &=   \int                                                  {\rm d}W_{\rm m} \\ 
           &= N \int_0^t {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\ 
           &= N \int_0^                                {\Phi} i \cdot {\rm d}\Phi \\ 
\end{align*}

In a toroid coil with a given cross-section $A$ the flux change ${\rm d}\Phi$ can only be given as a change in the field $B$. 
Therefore, ${\rm d}\Phi = A \cdot {\rm d}B$. Additionally, we know that the magnetic voltage is given by $\theta(t) = N \cdot i = H(t) \cdot l$. 
Including this in the formula gives us:

\begin{align*} 
W_{\rm m} &= N \int_0^{B} I            \cdot A \cdot {\rm d}B \\ 
          &=   \int_0^{B} H(B) \cdot l \cdot A \cdot {\rm d}B \\ 
          &= V \int_0^{B} H(B)                 \cdot {\rm d}B \\ 
\end{align*}

We can conclude that the magnetic energy $W_{\rm m}$: 
$W_{\rm m}$ can be calculated from the $H$-$B$ curve by integrating the external magnetic field strength $H$ for each small step of the flux density ${\rm d}B$. 
This will be shown for the case of a linear magnetic behavior, a nonlinear behavior, and the situation with magnetic hysteresis shortly.

=== Circuit with linear magnetic Behavior ===

In <imgref ImgNr15> the situation for a magnetic material with a linear relationship between $B$ and $H$ is shown. 
Given by the maximum current $I_{\rm max}$ the maximum field strength $H_{\rm max}$ can be derived. 
In the circuit in <imgref ImgNr14>, the inductor will experience increasing and decreasing current. 
Therefore, the $B$-$H$-curve gets passed through positive and negative values of $H$ and $H$ along the line of $B=\mu H$.

<WRAP> <imgcaption ImgNr15 | H-B-Curve for linear material> </imgcaption> {{drawio>HBcurveLin.svg}} </WRAP>

The situation for integrating the area in the graph is also shown: 
For each step ${\rm d}B$ the corresponding value of the field strength $H$ has to be integrated. 
For $B_0=0$ to $B=B_{\rm max}$ the magnetic energy is

\begin{align*} 
W_{\rm m} &=               V\int_0^{B}              H(B) \cdot {\rm d}B \\ 
          &=               V\int_0^{B} {{B  }\over{\mu}} \cdot {\rm d}B \\ 
          &= {{1}\over{2}} V           {{B^2}\over{\mu}} \\ 
          &= {{1}\over{2}} V \cdot       B \cdot H \\ 
\end{align*}

This situation is a good approximation for air or non-magnetic materials. 
However, it does not work well for ferrite materials, since they show nonlinear behavior and hysteresis.

=== Circuit with nonlinear magnetic Behavior ===

In <imgref ImgNr15> the situation for a magnetic material with a nonlinear relationship between $B$ and $H$ is shown.

<WRAP> <imgcaption ImgNr16 | H-B-Curve for nonlinear material> </imgcaption> {{drawio>HBcurveNonLin.svg}} </WRAP>

In this case, the permeability $\mu_{\rm r}$ is not a constant but can be represented as a function: $\mu_{\rm r}= f(B)$. 
Here, the formula $W_{\rm m} = V\int_0^{B} H(B) \cdot {\rm d}B$ also applies - so the magnetic energy is again the area between the curve and the $B$-axis. 
As an example, the situation of the field strength $H(t_1)=H_1$ is shown. 
This shall be the field strength after magnetizing the ferrite material to $H_{\rm max}$ (yellow arrows) and then partly demagnetizing the material again (blue arrow). 
The magnetization corresponds to an energy intake to the magnetic field and the demagnetization to an energy outtake.

Moving along the $H$-$B$-curve, one can see, that the energy intake and outtake are the same when coming back to a start point. 
This means that the magnetization and demagnetization take place lossless in this example. 
This is a good approximation for magnetically soft materials, however, does not work for magnetically hard materials like a permanent magnet. 
Here, hysteresis also has to be considered.

=== Circuit with magnetic Hysteresis ===

<WRAP> <imgcaption ImgNr17 | H-B-Curve material with Hysteresis> </imgcaption> {{drawio>HBcurvehyst.svg}} </WRAP>

===== Tasks =====

<panel type="info" title="Task 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The <imgref ImgTask01> and <imgref ImgTask02> show a shaded pole motor of a commercial oven.

  * Find out how this motor works - explicitly: why is there a preferred direction of the motor?
  * In which direction does the shown motor run?

<WRAP group> <WRAP column half> <imgcaption ImgTask01 | Core of a shaded Pole Motor> {{:electrical_engineering_2:shadedpolemotor_core.jpg?400|shadedpolemotor_core.jpg}}</imgcaption>

</WRAP> <WRAP column half>

<imgcaption ImgTask02 | Setup of the full shaded Pole Motor> {{:electrical_engineering_2:shadedpolemotor.jpg?450|shadedpolemotor.jpg}}</imgcaption> </WRAP> </WRAP>

</WRAP></WRAP></panel>

<panel type="info" title="Task 5.1.10 Further exercise"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

The book [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|DC Electrical Circuit Analysis - A Practical Approach (Fiore)]] has some nice exercise for beginning in the topic of magnetic circuits.

</WRAP></WRAP></panel>

===== Further Information =====

An alternative interpretation of the magnetic circuits is the {{https://en.wikipedia.org/wiki/Gyrator–capacitor model|Gyrator–capacitor model}}. 
The big difference there is, that there the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$. 
This model can solve more questions, however, is a bit less intuitive based on this course and less commonly used compared to the {{https://en.wikipedia.org/wiki/Magnetic_circuit#Resistance–reluctance_model|Magnetic_circuit}}, which was also presented in this chapter.

==== Moving a Plate into an Air Gap ====

<WRAP> <imgcaption ImgNr81 | a magnetic circuit with a moving plate> {{electrical_engineering_2:plate_in_airgap_50_.gif}} {{electrical_engineering_2:stiftt_in_luftspalt_2.3.gif}} </imgcaption> </WRAP>

==== Switch Reluctance Motor ====

<WRAP> <imgcaption ImgNr98 | a switch reluctance Motor as an application for the magnetic resistance> {{:electrical_engineering_2:switched_reluctance_motor_rotation_animation.gif}}</imgcaption> </WRAP> Based on a [[https://commons.wikimedia.org/wiki/File:Switched_Reluctance_motor_rotation_animation.gif|wikimedia image]] from Hamidreza D ([[https://creativecommons.org/licenses/by-sa/4.0/deed.en|CC-SA 4.0]])

==== Resolver ====

<WRAP> <imgcaption ImgNr99 | Resolver> </imgcaption> {{drawio>Resolver.svg}} </WRAP>

<WRAP> <imgcaption ImgNr100 | Overview of the magnetic Formalism> </imgcaption> {{drawio>OverviewONMagnetics}} </WRAP>

xxxxxxxxxx
 
====== 5 Magnetic Circuits ======​<callout> For this and the following chapter the online Book 'DC Electrical Circuit Analysis - A Practical Approach' is strongly recommended as a reference. In detail this is chapter [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|10.3 Magnetic Circuits]] </callout>​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 <imgref ImgNr01>).​<WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP>​In this chapter, we will investigate, how far we come with such an analogy and where it can be practically applied.​===== 5.1 Linear Magnetic Circuits =====​For the upcoming calculations, the following assumptions are made​  - 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  - There is no stray field leaking out of the magnetic field conducting material.  - The fields inside of airgaps are homogeneous. This is true for small air gaps.​One can calculate a lot of simple magnetic circuits when these assumptions and focusing on the average field line are applied.​<WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP>​Two simple magnetic circuits are shown in <imgref ImgNr02>: They consist of​  * a current-carrying coil  * a ferrite core  * an airgap (in picture (2) ).​