Tutorial 7 - Magnetic Circuits

Learning Objectives

To use electrical circuits as a model for magnetic circuits.

To understand the similarities between electrical and magnetic circuits.

To understand the limitations of the analogy.

Key Questions

How are magnetic circuits similar to electrical circuits?

What is reluctance?

How are magnetic circuits different to electrical circuits?

Electrical Model for Magnetic Circuits

At A-level, we have compared electric fields with gravity fields, as the two are analogous.  This tutorial shows how magnetic fields can be modelled  using electrical principles.  A model in this case is a way of explaining difficult concepts in simple terms (and nothing to do with model trains or model boys and girls in the fashion industry).

 

Although magnetism has been a subject of study for many centuries, the association with electricity is more recent.  The detailed study of magnetism is quite complex.  Some physicists have used electrical circuits as models for magnetic circuits.  This can be useful.  Let's look at the two:

 

 

The electrical circuit has a source of voltage, called the electromotive force and given the physics code E (curly E).   The units are Volts (V).   The  magnetic circuit has a source of magnetism, which is shown as the current carrying coil of wire.  The current I flows through the coil of N turns.  The result of this is magnetomotive force, Fm, which is the product of the current and number of turns:

 

Fm = NI

 

The units for magnetomotive force are amp turns (At).  Since turns is not a unit with dimensions, we can correctly describe the unit as amps (A).

 

When a voltage is applied across a circuit, a current, I amps flows.  This is representative of the number of electrons flowing every second.

 

When a magnetic field is applied to a magnetic circuit, there is a magnetic flow, flux, of F Weber (Wb).  This is the number of field lines in the circuit.

 

In magnetic fields, the more commonly used quantity is flux density, B Tesla (T.  1 Weber per square metre (Wb m-2) = 1 T). 

 

This is equivalent of current density, J amp per square metre, where:

 

 

Reluctance

With an electric circuit there is resistance, R ohm (W) which is defined as the ratio between the voltage and the current:

 

In the same way there is a magnetic resistance called reluctance.  It is defined as the ratio of magnetomotive force to flux.  It is given the physics code S and has the units "per Henry" (H-1) or amp (turns) per Weber (A Wb-1).

We know that the magnetic field strength is given by:

We also know that:

F = BA

 

So we can write:

 

We can rewrite this equation as:

We know that:

and we can write:

If there is a magnetic material of relative permeability mr, we know that:

and the final relationship becomes:

Question 1

A toroidal (dough-nut shaped) piece of magnetic material has a relative permeability of 200.  It is 7.5 cm in radius and has a a circular cross section of radius 1.0 cm.  Calculate the reluctance.

Answer

Question 2

The coil providing the magnetic field has 200 turns.   The magnetic flux is 0.14 T.

(a)  Calculate the magnetic field strength.

(b) Calculate the current.

Answer

 

We can work out the reluctance if we know the area, the length, the number of turns, and the current.  All of these are easily measured.  We know that for a material of relative permeability mr:

Then we can work out the flux:

F = BA

 

From that we can work out the reluctance:

 

Question 3

A current of 2.0 A flows through a coil that has 150 turns.  The coil is made on a former of  square section 2.0 cm 2.0 cm, and 10 cm long.  In the coil there is a material of relative permeability 150.  Calculate the reluctance.

Answer

 

Ferromagnetic materials tend to have a low value of reluctance.  Reluctance has applications in transformers and some types of induction motors.

 

Permeance

This is a term that is used in magnetic circuits, and is analogous to conductance.  We know from basic electricity that conductance, physics code G, is the reciprocal to resistance and has the units Siemens (S) (or mho):

 

The permeance, Physics code P, is the reciprocal to reluctance:

 

Question 4

What is the permeance of the coil in Question 3?

Answer

 

 

Absolute Permeability

We have come across the permeability of free space m0 and the relative permeability, mr

The product of these two quantities gives as the absolute permeability, m.  The units for m are Henry per metre (H m-1)

 

m = m0mr

 

So we can rewrite our equation for reluctance as:

 

We have already compared magnetic reluctance with electrical resistance.  In electrical circuits the resistance depends on:

  • The length of the conductor;

  • The area of the conductor;

  • The material of the conductor.

The material has a property of resistivity, which is defined as the resistance of a conductor of unit cross-sectional area and unit length.  The physics code is r (rho, a Greek letter 'r')  and the units are Ohm metres.  The formula for resistance in terms of resistivity is:

In the same way, we can say that the reluctance of a magnetic material depends on:

  • The length;

  • The area;

  • The magnetic properties.

 

Both are directly proportional to the length and inversely proportional to the area.  The constant of proportionality in resistance is r, while the constant of proportionality in reluctance is 1/m.

 

The electrical analogy for absolute permeability, m, is conductivity, s.

 

 

 

Series Magnetic Circuits

Like resistors,  series reluctances add up:

Consider a magnetic circuit like this:

The magnetomotive force from the current is:

Fm = NI

 

This results in flux F Wb which flows around a magnetic material of absolute permeability m H m-1 which has length l m and an area A m2 throughout.  It crosses an air-gap that is x m wide, and also has the same area A m2.

 

The electric circuit model looks like this:

 

We know that for the block of magnetic material that:

 

For the air gap, the absolute permeability is the same as the permeability of free space, since the relative permeability is 1.0 for air.  So we can write down that S2 is given by:

 

The total reluctance is given by:

Question 5

The magnetic circuit above is made of a magnetic material of relative permeability 175.  It has a square cross section 2.5 cm 2.5 cm, and the average length is 25 cm.  There is a air gap cut into it that is 1.0 mm in length.  Calculate:

(a) the reluctance of the material;

(b) the reluctance of the air gap;

(c) the total reluctance.

Answer

 

Flux in the Magnetic Circuit

We can work out F:

F = BA

 

and:

These two combine to give:

 

Like electricity in which the current is the same all the way round, we can say that the flux is the same all the way round in the circuit.

 

Magnetomotive Forces

In an electric circuit we know that we can work out the voltage across a resistor using V = IR.  The same is true for magnetic circuits:

 

Fm = SF

 

We know that the magnetomotive force is the product between the current and the number of turns:

 

Fm = NI

 

So we can write:

NI = SF

 

Question 6

The magnetic circuit in Question 5 has a magnetic source that consists of 300 turns that carry a current of 0.75 A. 

(a) Work out the magnetomotive force.

(b) Work out the flux.

Answer

 

In a series electrical circuit the resistances add up to the total resistance, and the voltages add up to the battery voltage.  In series magnetic circuits, the reluctances add up to a total reluctance and the magnetomotive forces add up to the total magnetomotive force.

 

Question 7

Calculate the magnetomotive force across each reluctance in Question 5 using your answer to Question 6 (b).  Is your answer consistent with your answer to 6 (a)?

Answer

 

Your answer to Question 7 shows that series magnetic circuits seem to be consistent with Kirchhoff's Second Law (Kirchhoff II) in electrical circuits.

 

 

Parallel Magnetic Circuits

These can be modelled like parallel electrical circuits.  Consider this circuit:

This circuit consists of a block of material of relative permeability mr.  A coil of N turns carries a current of I A and produces a magnetomotive force of NI A turns.  There two branches of the same material.  Each branch has length l1 m and l2 m respectively and has areas A1 m2 and A2 m2 respectively.  The branches carry flux F1 and F2 respectively.  (The lengths have been omitted from the diagram for clarity and are shown below.)

 

This can be modelled as a parallel electrical circuit as shown below:

In a parallel electrical circuit the voltage is the same across each branch.  Therefore the magnetomotive force across each reluctance is the same.

 

In a parallel electrical circuit, the currents in the branches add up the total current, according to Kirchhoff I.  In a parallel magnetic circuit, the fluxes add up:

In a parallel electrical circuit, the reciprocals of the resistances add up.  Likewise, in a parallel magnetic circuit the reciprocals of the reluctances add up:

 

We will keep the magnetic material the same, so m (=m0mr) will be the same in each loop.  We can vary the length and the area for each loop to give reluctances:

and

 

 

Question 8

The diagram shows a piece of magnetic material of relative permeability 130.

 

The dimensions are shown on the diagram.  Calculate:

(a) The Magnetomotive force;

(b) The reluctance for Loop 1;

(c) The reluctance for Loop 2;

(d) The total reluctance;

(e) The total flux;

(f) The flux in Loops 1 and 2.

(g) Are the fluxes consistent?  Explain your answer.

 

Answer

 

Of course the material does not have to be the same in each loop.  It is more likely than not that the materials will be the same?

 

Question 9

How would the formula for reluctance of loop 1 change if the relative permeability were ms instead of mr?

Answer

 

Magnetic Fields and Electric Fields

Consider two plates, in which the positive plate is at +V V and the negative plate is at 0 V.  The plates are d m apart:

 

Electric Field Strength, E V m-1 in a uniform field is given by:

 

For a magnetic field, a coil of N turns carrying a current of I A will give a magnetomotive force of NI  amp-turns (A).  The magnetic field length is l m as shown:

 

The magnetic field strength H A m-1 is given by:

 

The electric field is electromotive force per unit length, while the magnetic field is magnetomotive force per unit length.

 

 

 

Differences between Electric Circuits and Magnetic Circuits

Not every analogy is perfect, of course.  This is certainly true when we use electric circuits as a model for magnetic circuits.  Let's look at this in more detail:

 

Energy

In an electric circuit, energy is transferred by a flow of charge (current).  (Volts are joules per coulomb).   Energy is needed all the time a current a flows.  Flux carries no energy.  Flux arises due to the alignment of domains.  There are no flux carriers that move.  Energy is used to make a magnetic field,  and is released when the magnetic field collapses.  No energy is needed to maintain magnetic flux.

 

Flow

Electrical charge carriers (electrons mainly) flow in a circuit.  Flux is not carried by flux carriers.

 

Insulation

Electrical circuits need insulating materials.  These include PVC, rubber, and many other materials.  Electric currents cannot pass through insulating materials.  Magnetic fields cannot be contained by insulating materials.  Some highly magnetic materials can be used to contain a magnetic field, but some lines of flux will still escape.

 

Resistance

Resistance in electrical circuits is independent of current density.  It can be changed if the temperature changes.  In magnetic circuits, the reluctance depends on the flux density.

 

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