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Block 18 — 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.

Flux

The magnetic flux is a measurement of the amount of magnetic field lines through a given surface area, as seen in Abbildung 3. The magnetic flux is the amount of magnetic field lines cutting through a surface area defined by the surface vector $\vec{A}$. If the angle between the $\vec{A} = A\cdot \vec{n}$ and magnetic field vector $\vec{B}$ is parallel or antiparallel, as shown in the diagram, the absolute value of the magnetic flux is the highest possible value given the values of the area and the magnetic field.

This definition leads to a magnetic flux similar to the electric flux studied earlier:

\begin{align*} \Phi_{\rm m} = \iint_A \vec{B} \cdot {\rm d} \vec{A} \end{align*}

Abbildung 4 depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume.

Since the magnetic field is a source-free vortex field, the flux over a closed area is always zero:

\begin{align*} \Phi_{\rm m} = {\rlap{\Large \rlap{\int} \int} \, \LARGE \circ}_{A} \vec{B} \cdot {\rm d} \vec{A} = 0 \end{align*}

By this, it can be shown that any open surface bounded by the circuit in question can be used to evaluate $\Phi_{\rm m}$. For example, $\Phi_{\rm m}$ is the same for the various surfaces $S$, $S_1$, $S_2$ of the figure.

The SI unit for magnetic flux is the $\rm Weber$ (Wb), \begin{align*} [\Phi_{\rm m}] = [B] \cdot [A] = 1 ~\rm T \cdot m^2 = 1 ~ Wb \end{align*}

If the $B$ field is homogenous and $B$ is perpendicular to the surface $S$, then the fomula simplifies to:

\begin{align*} \Phi_{\rm m} = B \cdot A \end{align*}

Based on this definition, the magnetic field unit is occasionally expressed as Weber per square meter ($\rm Wb/m^2$) instead of teslas. In many practical applications, the circuit of interest consists of a number $N$ of tightly wound turns (similar to Abbildung 5). Each turn experiences the same magnetic flux $\Phi_{\rm m}$. Therefore, the net magnetic flux through the circuits is $N$ times the flux through one turn, and Faraday’s law is written as

Linked Flux

When looking at the magnetic field in a coil multiple windings capture the passing flux, see Abbildung 8 (a). It can also be interpreted in such a way that the flux is going through the closed surface of the circuit multiple times (in picture (b)).

The linked flux $\Psi$ is defined as the resulting flux given by the sum of the partial fluxes of the closed circuit.

\begin{align*} \boxed{ \Psi = \sum_{i=1}^n \Phi_{i} } \end{align*}

For the case of a coil with $N$ windings, this leads to:

\begin{align*} \boxed{ \Psi = N \cdot \Phi } \end{align*}

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 2: 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

Common pitfalls

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Exercises

Exercise E4 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