Tutorial 6 B - Hysteresis

Leaning Objectives

Distinguish between magnetic field strength and magnetic flux density.

To link the quantities with energy by a graphical method.

To consider the simple Steinmetz equation.

Key Questions

What is the difference between magnetic field strength and magnetic flux density?

How can we demonstrate the energy per unit volume?

What is hysteresis?

How can we work it out?

Energy in a Magnetic Field by Graphical Method

In A-level Physics, we tend to be rather slack and refer to the flux density as "magnetic field strength".  However, in these notes I have been careful to refer to magnetic flux density, because there is another separate quantity called magnetic field strength or magnetising force.  You will see this in many articles and textbooks at university level.  So what is the difference?

We know the flux density, B, which is defined as the flux (field lines) that pass through a unit area.  It is a vector quantity and is the quantity that is most widely used for practical physics and engineering.

The magnetic field strength or magnetic field intensity, H, which is defined as the force experienced by a unit North pole placed in a magnetic field.  It is a vector quantity.  It represents the force being generated by the source, i.e. the current.   When generated magnetic fields pass through magnetic materials, there is a contribution by the material itself.  So it's common to make the distinction between the the magnetic flux density and the magnetic field strength.

The initial driver for a magnetic field is an electric current.  The current gives a magnetomotive force which is defined as the product between the current and the number of turns.  It has the Physics code Fm and it's given by a simple formula:

Fm = NI

The units are amp-turns.  Turns are not a dimensional unit, so in unit analysis, the units are amps (A).

Like electromotive force, magnetomotive force NOT a force at all; it's a current.

It is measured in ampere-turns, NOT Newton.

The magnetic field strength is the result of the magnetomotive force. The formal definition is as above, but it can also be described as the magnetomotive force per unit length, or the product between the current and the number of turns per unit length.  The Physics code is H, and the formula is:

The unit is ampere per metre (A m-1).  The term l for length is called the length of the flux path.

The magnetic field strength, H, is related to flux density, B, by this simple equation:

If a magnetic material of relative permeability mr is added to make a core, the equation becomes:

 A current of 2.5 A is passed through a coil of diameter 14 mm.  The coil has 500 turns.  Calculate: (a) the magnetomotive force; (b) magnetic field strength; (c) the flux density.

If we plot a graph of magnetic flux density against magnetic field strength, we get a straight line through the origin, as the two quantities are are directly proportional.

The gradient is the permeability of free space, m0, which has the value 4p × 10-7 H m-1.  The  basic SI units  for m0 (H m-1) are kg m s-2 A-2.

If we increase the magnetic field strength by a tiny amount dH, the magnetic flux density increases by a corresponding tiny change in flux density, dB.  The shaded area is approximately a rectangle if the increase dH is sufficiently small.  The area is:

A = (H × dB)

The area represents energy per unit volume.  We can demonstrate this by unit analysis.

In basic SI units the Tesla (T)  is kg s-2 A-1.    So if we multiply the units for dB by H (A m-1),  we get kg s-2 A-1 × A m-1 which give kg m-1 s-2.  The SI basic units for energy (J) are kg m2 s-2.  So if we divide the energy by the volume, we get:

kg m2 s-2 ÷ m3 = kg m-1 s-2

We can add up all the little rectangles to get the total area under the graph:

So the total energy per unit volume is the area of the triangle:

We know that:

Combining the two equations, we get:

Alternatively we we can say:

And we can therefore write:

Which is where we were before...

The term H for magnetic field strength appears in textbooks and is useful for considering hysteresis which we will do in the next section.

Hysteresis

We need to put a certain amount of energy into a magnetic material to magnetise it.  This is to line up the domains.  Once the domains are lined up, no further energy is required to maintain the magnetic field.  The coil will get warm because there is an ohmic resistance.

In some materials the domains remain lined up.  In this case we have a permanent magnet.  To remove the magnetism (i.e. make the domains go random, we have to apply alternating magnetic fields or heat the magnet up to its Curie Temperature.  This is the temperature at which the magnet loses its magnetic properties.  For cobalt, it is 1400 K.

For other materials, the domains spring back to their randomly oriented state as soon as the current is turned off.  The change in the magnetic field induces an EMF, or a reverse voltage spike.  However we do not get that energy back that we put in when we release the magnetic field and the domains return to a random state  We get slightly less energy back as the field collapses.   Some energy is lost and this lost energy is called hysteresis.

Suppose we have a completely demagnetised material that releases its magnetism when the current stops flowing and start to apply a current with a view to lining up all the domains.  We get a graph like this:

When the applied magnetic field strength is low, there is a straight line region which shows proportionality between B and H.  Above the limit of proportionality, most of the domains are lined up, and the increase in B is no longer linear.  Eventually all the domains are lined up and there can be no increase in flux density.  This is saturation.

Now let's reduce the current, and then reverse it.  The graph looks like this:

As the current (hence the magnetic field strength) starts to fall at A, the flux density does not retrace the red line to O.  Instead it follows the line AB.  This is because some of the domains remain lined up.  This remaining flux density is called remanence

A reverse magnetic field has to be applied to make domains realign in the opposite direction, so that the overall flux density is zero.  This is shown by OC.  The graph follows the line BC.  The reverse magnetic field OC is called the coercive force or coercivity.

As the reverse field is increased, shown by line CD more and more domains line up in the opposite direction until all are aligned, and we have saturation in the opposite direction at D.

 Explain the shape of the hysteresis curve from D to A

Remanence has the physics code Mr and the units are Tesla, T.  Typical value for ferrite is 0.35 T.  For a neodymium magnet (a typical permanent magnet) the value for remanence is about 1.3 T.

Typical values for coercive force are:  Permalloy (another material for making permanent magnets) 79 A m-1; neodymium 900 A m-1.

A graph like the one above is often called a hysteresis loop.

Describing Hysteresis Losses

When the magnetic field is reversed, the domains are reversed and a job of work needs to be done.  This takes energy.  We can tell that because the material that is being magnetised starts to get warm.  Clearly we cannot get rid of hysteresis entirely, but we can try to minimise it as far as we can.

This is a  hysteresis graph of a material like hard steel::

 What are the magnetic properties of hard steel?

For soft steel, the remanence is high, but the coercivity is low:

Soft steel means magnetically soft, so it loses its magnetism rapidly.  It is not soft to touch.  It is hard and heavy.  If you drop some soft steel on your foot, you will know about it.

Ferrite is a magnetic material made from oxides of iron, cobalt, and nickel, along with oxides of magnesium, aluminium, and manganese.  The material is a ceramic.  You can see ferrite beads on signal cables for computers.  One is shown among the tangle of cables behind a PC.

Its hysteresis loop is very small.  It is shown on the graph below:

Quantifying Hysteresis Losses

Consider a ring made of a magnetic material of which the circumference is l m (so the radius is l/2p) and its cross-sectional area is A m2.  It has N turns of wire.  Experimental work using a magnetometer gives a hysteresis loop like this:

The magnetic field strength, H, is increased by a tiny amount, dH A m-1 in a time of dt s.  The instantaneous current that gives rise to the magnetic field strength OF is i A.   We know that:

So we can write an expression for the instantaneous current as:

We also know that the emf due to the change is:

The power = volts × amps, so we can write an expression for instantaneous power:

The little bit of energy, e, supplied in the time period dt seconds is power × time:

The dt terms obligingly cancel out to give us:

We now bring in the expression for the instantaneous current, i:

to give us:

And we can tidy up the expression to give:

Now the term lA is the volume of the ring.  The area of the strip = HOF × dB.  So we can say:

energy  = area of the shaded strip × volume

Therefore the area of the shaded strip is energy ÷ volume.

If we do this from O to A, we have to add up all the little strips.  The energy per unit volume is the area of the shape OABF.  If we do it for a whole cycle, the energy per unit volume is the area of the whole loop.  If the function that describes the loop were simple, calculus integration would give us the area (i.e. energy per unit volume).  However this hysteresis loop does not lend itself to that.  Instead we need to estimate the area by counting the squares on the graph paper.  If more than half the square is occupied by part of the shape, it is counted in.  If less than half is occupied, the square is ignored.

Once we have counted the squares (tedious), we need to work out the area, using the scale.  If 1 cm vertically is y Tesla, and 1 cm horizontally is x amps per metre, then the total area A cm2 gives the hysteresis loss in joules per cubic metre.

Hysteresis loss (J m-3) = A cm2 × x A m-1 cm-1 × y T cm-1

Of course we can count millimetre squares (even more tedious), which will reduce the uncertainty.

 Worked example The area of a hysteresis loop is 1250 mm2.  The horizontal scale is 1 mm = 30 A m-1, and the vertical scale is 1 mm = 0.010 T.  Determine the hysteresis loss per cycle.  Give you answer to an appropriate number of significant figures.  Give the correct units. Answer Hysteresis loss = 30 A m-1 mm-1 × 0.010 T mm-1 × 1250 mm2 = 375 J m-3 = 380 J m-3 (2 s.f. as data are to 2 s.f.)

 The area of a hysteresis loop is found to be 2000 mm2.  1 cm on the horizontal scale represents 400 A m-1 and 1 cm on the vertical axis represents 0.50 T.  Determine the hysteresis loss in one cycle.

The Significance of Hysteresis

With a pure direct current a certain amount of energy is lost to build up the magnetic field.  Once the magnetic field is established, no energy is needed to maintain the magnetic field, so energy losses are from the resistance of the coil.  We know this from:

P = I2 R

If the direct current is a rectified AC the loss is slightly more, as shown in the graph:

The real significance of hysteresis is when we use an alternating current.  Consider a coil that gives a hysteresis loop that has an area A.  We know that the hysteresis loss is given by:

Hysteresis loss (J m-3) = A cm2 × x A m-1 cm-1 × y T cm-1

This the loss per cycle.  So if we f cycles per second (Hertz), we can say that the hysteresis loss per second is:

Hysteresis loss per second (J m-3 s-1) = A cm2 × x A m-1 cm-1 × y T cm-1 × f Hz

Now energy (J) × frequency (Hz) = power (W).  This means that the hysteresis loss per second is actually power per unit volume.

To work the actual power loss, we can modify our equation to:

Power loss (W) = A cm2 × x A m-1 cm-1 × y T cm-1 × f Hz × V m3

 Use your answer to Question 14 to work out the power lost per cubic metre, if the supply has a frequency of 50 Hz.  The coil has a circular cross section with a diameter of 5.0 cm and a length of 10 cm.  Calculate the power loss.

In our study of hysteresis, we have considered the maximum possible hysteresis, which happens when the magnetic material is taken up to saturation.  However if we increase the magnetic field strength such that the resulting flux density is half that of saturation, we find that the hysteresis loss is rather less than half.

So far we have relied on counting squares and scaling up to work out the hysteresis loss per unit volume.

Steinmetz Equation

The American mathematician and electrical engineer C P Steinmetz (1865 - 1923) studied hysteresis losses in transformers and quantified them using the Steinmetz equation.

Suppose our hysteresis loss per cycle is given the physics code Qh J m-3.  If the maximum flux density is Bmax T, the Steinmetz theory states that:

The hysteresis loss (J m-3) per cycle depends on the magnetic material using a constant kh.   The constant kh is called the hysteresis coefficient.   So we can write:

We can then work out the power loss per cubic metre (W m-3), which we will give the physics code Ph.

If we know the volume, V m3, we get an expression for the power loss, Pm (W):

This relationship works for simple sinusoidal AC waveforms.  For a more complex waveform, a modified version is used.  The table shows some typical values for the hysteresis coefficient for typical magnetic materials:

 Material Hysteresis Coefficient Cast iron 27.63 to 40.2 Sheet iron 10.05 Cast steel 7.54 to 30.14 Hard cast steel 63 to 70.34 Silicon steel (4.8% Si) 1.91 Hard tungsten steel 145.7 Mild steel 7.54 to 22.61 Nickel 32.66 to 100.5 Permalloy 0.25

Data from

Clearly we want to keep hysteresis losses to a minimum, so it makes sense to use materials that have a low hysteresis coefficient in devices like motors and transformers.

 Worked Example Work out the power loss in a magnetic material that has a hysteresis coefficient of 10.5.  The maximum flux density at saturation is 3.5 T, and the material is magnetised by an alternating sinusoidal current of 50 Hz.  The volume of the same is 2.0 × 10-3 m-3. Answer Pm = 10.5 × 50 Hz × 2.0 × 10-4 m-3 × (3.5 T)1.6 = 7.8 W

 A coil has a circular cross section with a diameter of 6.0 cm and a length of 15 cm.  It has a core of hard cast steel of hysteresis coefficient 65.46.  The maximum flux density that the core can take is 4.3 T before saturation occurs.  Calculate the power loss if the coil is connected to a 50 Hz supply.  Give your answer to an appropriate number of significant figures.

In the equation:

the index (power to) is 1.6.  It can vary from 1.6 to 3.0.  Many sources give it as 2.