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Block 19 — Magnetic Circuits and Inductance

Learning objectives

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

1. Warm-up (x min):
   1. ….
2. Core concepts & derivations (x min):
   1. …
3. Practice (x min): …
4. Wrap-up (x min): Summary box; common pitfalls checklist.

Conceptual overview

Core content

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 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

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 no stray field leaking out of the magnetic field conducting material.
3. 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 $l$

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

• a current-carrying coil
• a ferrite core
• an airgap (in picture (2) + (3) ).

These three parts will be investigated shortly:

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:

What do we know about this circuit?

1. The length of the average field line is $l$.
   The cross-sectional area shall be constant: $A = \rm const.$
   
   
2. magnetic voltage: \begin{align*} \theta = H \cdot l \tag{1} \end{align*}
3. magnetic flux: \begin{align*} \Phi = B \cdot A \tag{2} \end{align*}
4. 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:

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.

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 (Abbildung 6 (3)).

Electric Magnet with three Legs

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

With the reluctances:

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

Common pitfalls

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Exercises

Exercise E1 Magnetic Circuit
(written test, approx. 7 % of a 120-minute written test, SS2022)

The magnetic setup below shall be given. Draw the equivalent magnetic circuit to represent the setup fully. Name all the necessary magnetic resistances, fluxes, and voltages.
The components shall be designed in such a way, that the magnetic resistance is constant in it.
Formulas are not necessary.

Path

Watch for parts of the magnetic circuit, where the width and material are constant.
These parts represent the magnetic resistors which have to be calculated individually.
Be aware, that every junction creates a branch with a new resistor, like for an electrical circuit - there must be a node on each „diversion“. \begin{align*} R_{\rm m} = {{1}\over{\mu_0 \boldsymbol{\mu_{\rm r}}} }{\boldsymbol{l}\over{\boldsymbol{w}\cdot h}} \end{align*}

Result

Embedded resources