====== Block 19 — Magnetic Circuits and Inductance ======

===== Learning objectives =====
<callout>
After this 90-minute block, you can
  * ...
</callout>

===== Preparation at Home =====

Well, again 
  * read through the present chapter and write down anything you did not understand.
  * Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting). 

For checking your understanding please do the following exercises:
  * ...

===== 90-minute plan =====
  - Warm-up (x min): 
    - .... 
  - Core concepts & derivations (x min):
    - ...
  - Practice (x min): ...
  - Wrap-up (x min): Summary box; common pitfalls checklist.

===== Conceptual overview =====
<callout icon="fa fa-lightbulb-o" color="blue">
  - ...
</callout>

===== Core content =====

<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 have come with such an analogy and where it can be practically applied.

==== Basics for 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 are applied and focusing on the average field line are applied. \\
The average field line has the length of $l$

<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) + (3) ).

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

These three parts will be investigated shortly:

<WRAP group><WRAP column third>
=== 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 $\theta = N\cdot i$.

</WRAP><WRAP column third>
=== 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)

</WRAP><WRAP column third>
=== 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)).

</WRAP></WRAP>

We also assume that the magnetic flux $\Phi$ remains constant along the ferrite core and in the air gap, so $\Phi_{\rm airgap}=\Phi_{\rm core}=\rm const.$. \\ 


==== Reluctance - the magnetic Resistance ====

Let's have a look at the simplest situation: 

<WRAP> 
<imgcaption ImgNr01 |  the simples magnetic circutit> 
</imgcaption> 
{{drawio>MagElCircuitSimplest01.svg}} 
</WRAP>

What do we know about this circuit?

  - The length of the average field line is $l$. \\ The cross-sectional area shall be constant: $A = \rm const.$ \\ \\
  - magnetic voltage: \begin{align*} \theta = H \cdot l          \tag{1} \end{align*}
  - magnetic flux: \begin{align*} \Phi = B \cdot A               \tag{2} \end{align*}
  - relationship between the two fields: \begin{align*} B= \mu H \tag{3} \end{align*}

This can now be combined. Let us start with $(3)$ in $(2)$ and then take this result to divide $(1)$ by it:

\begin{align*} 
                 \Phi &= \mu H \cdot A \\ \\
{{\theta}\over{\Phi}} &= {{ H \cdot l}\over{\mu H \cdot A}}  \\
                      &= {{l}\over{ \mu \cdot A}} 
\end{align*}

Hmm.. what have we done here? We divided the voltage by the flux, similar to ${{U}\over{I}}$ and we got something only depending on the dimensions and material. \\
We might see some similarities here: 

\begin{align*} 
{{U}\over{I}} =  \rho \cdot {{l}\over{A}} = R  \quad |\text {for the electic circuit} \\ \\
\end{align*}

\begin{align*} 
\boxed{ {{\theta}\over{\Phi}} =  {{1}\over{\mu}} \cdot {{l}\over{A}} =R_m} \quad |\text {for the magnetic circuit} 
\end{align*}

The quantity $R_m$ is called **reluctance or magnetic resistance**. \\
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.

So, we got an equivalent magnetic circuit:  

{{drawio>LumpedMagnCircuitV01.svg}}


==== Applications of Flux and Reluctance ====


=== Core with Airgap ===

Another common situation is to have a air gap separating the iron core. \\
The width of air gaps are commonly given by $\delta$. \\
The flux in the air gap and the core is the same, but the permeability $\mu$ differs strongly. 

<WRAP> 
<imgcaption ImgNr37 | lumped Circuit Model for a simple magnetic Circuit> 
</imgcaption> 
{{drawio>LumpedMagnCircuit.svg}} 
</WRAP>

If it would be an electical circuit, we would get for the source voltage $U_S$

\begin{align*} 
U_S  &= U_1 &&+ &&U_2 \\
     &= R_1 \cdot I &&+ &&R_2 \cdot I \\
     &= \rho_1 {{l_1}\over{A}} \cdot I &&+ &&\rho_2 {{l_2}\over{A}} \cdot I

\end{align*}

The resulting formula for the magnetic voltage $\theta$ is similar:

\begin{align*} 
\theta &= \theta_1 &&+ &&\theta_2 \\
       &= R_{m,1} \cdot \Phi &&+ &&R_{m,2}   \cdot \Phi \\
       &= {{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
\end{align*}

Additionally, the magnetic voltage $\theta$ is given by: 
\begin{align*} 
\theta &= N \cdot I 
\end{align*}

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

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


=== Electric Magnet with three Legs ===

<WRAP> 
<imgcaption ImgNr47 | lumped Circuit Model for a complex magnetic Circuit> 
</imgcaption> 
{{drawio>LumpedMagnCircuitComplexV01.svg}} 
</WRAP>

Assuming that $A$ is constant, we get the following:

<WRAP> 
<imgcaption ImgNr57 | lumped Circuit Model for a complex magnetic Circuit> 
</imgcaption> 
{{drawio>LumpedMagnCircuitComplexSolutionV01.svg}} 
</WRAP>

With the reluctances: 

\begin{align*} 
R_{m,x} = {{1}\over{\mu_0 \mu_{\rm r,x}}}{{l_{\rm x}}\over{A}} 
\end{align*}


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

Sections with ...
  * constant flux $\Phi$
  * constant cross-sectional area $A$
  * constant materiel $\mu_r$
... can be subsumed to a lumped magnetic resistans (reluctance)!

</callout>

===== Common pitfalls =====
  * ...

===== Exercises =====

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{{page>electrical_engineering_and_electronics:task_n1kwu944m7jac3tf_with_calculation&nofooter}}
{{page>electrical_engineering_and_electronics:task_0j7accfimmemytq9_with_calculation&nofooter}}

<panel type="info" title="Exercise 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="Exercise 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="Exercise 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="Exercise 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="Exercise 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="Exercise 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="Exercise 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="Exercise 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>

<panel type="info" title="Exercise 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 $l=280 ~\rm mm$. 
At 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 into 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>

===== Embedded resources =====
<WRAP column half>
Explanation (video): ...
</WRAP>

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