Tutorial 6 A - Magnetic Energy

Leaning Objectives

To learn about how energy is used in magnetic fields.

To learn what inductance is.

To link inductance, and energy in magnetic fields.

Key Questions

How is energy used to make a magnetic field?

What is inductance?

How are energy and inductance linked?

Energy in a Magnetic Field

It takes energy to make a magnetic field.   Let's look how:

You will remember about the domain model of magnetism.  In an unmagnetised piece of magnetic material, the domains are all randomly oriented.  When a magnetic field is applied, the domains line up.  In the picture we can see the effect on a single domain.  The domain acts as a couple, which turns through an angle q radians.  Since there is movement with a force, work is done (a basic principle in Physics).  The job of work done is given by:

W = tq

where:

W = work done (J)

t = torque (N m)

q = angle (rad)

 

Whenever a job of work is done, energy is transferred.  The amount of energy we need to align the single domain is not a large amount, but when there are huge numbers of domains, the energy needed starts to mount up.  To make sense of this, we need to use the concept of Energy Density, which is defined as the energy transferred per unit volume.  You will have come across energy per unit volume when studying Young's Modulus.  The Young's Modulus is the gradient of the stress strain graph.  The elastic strain energy per unit volume is area under the graph.

 

The physics code for energy density is either U or h (the strange looking symbol is 'eta' a Greek letter 'long e', ē) and the units are joules per cubic metre (J m-3).  It depends on what source you are using.

 

The energy per unit volume is for a magnetic field is given by this equation:

where

U = energy per unit volume ( J m-3)

B = magnetic flux density (T)

m0 = permeability of free space = 4 p 10-7 H m-1

 

If a magnetic material of relative permeability mr is incorporated into the magnetic field, the equation becomes:

Question 1

Calculate the energy per unit volume of a magnetic field of flux density 0.45 T

Answer

Question 2

A magnetic field has an energy density of 25 000 J m-3

What is the flux density, if a magnetic material of relative permeability of 100 is used?

Answer

 

On its own, this is not very meaningful.  But let us apply it to a solenoid.

 

Energy Density in a Solenoid

Consider a solenoid of N turns, of length l m, which is carrying a current of I A.  It will produce a field of flux density B T, according to the relationship:

Since N/l gives the number of turns per metre, n, we can rewrite the equation as:

B = m0nI

 

Now we can substitute this into the equation:

and this gives us:

Cancelling gives:

 

Question 3

A solenoid consists of 200 turns and is 5 cm long.  It is carrying a current of 2.0 A.  Calculate the energy density.

Answer

 

If we introduce a magnetic material or relative permittivity into the core of the of the solenoid, the relationship is:

 

This may be confusing to start with because at first sight it would appear that the energy density would be reduced by adding a magnetic material into the solenoid.  However the answer lies in the magnetic field flux density:

So if we substitute we get:

and we can do some cancelling to give:

This means that when a core of magnetic material of relative permeability mr is added, both the flux density and the energy density increase by mr times.

 

Question 4

A solenoid consists of 300 turns and is 5 cm long.  It is carrying a current of 1.5 A.   It has a core of relative permeability 150.

Calculate the energy density.

Answer

 

It is worth noting that a constant magnetic field requires no energy to maintain it.  This is how permanent magnets can retain their magnetic properties.  If a current is flowing in a solenoid to make the magnetic field, heat is given off from the coil because of the resistance of the coil.  In this case, the power is given by:

P = I2R

 

 

Inductance

Any coil of wire will make a magnetic field.  It takes energy to make a magnetic field, but no energy to maintain it.  This is linked with the inductance of the coil.  The inductance is defined as:

 

the property of an electric conductor or circuit that causes an electromotive force to be generated by a change in the current flowing.

 

Electromotive force is defined as:

the energy per unit charge supplied to a circuit.

 

This has important implications for the design of circuits in which there are inductive components.  Building a magnetic field takes energy out of the circuit.  When the current is turned off, the magnetic field is no longer maintained and the collapse of the magnetic field causes energy to be released back into the circuit.  The energy per unit charge is a potential difference, or voltage.  This voltage can be quite high and can wreck components in the circuit.

 

So let's have a look at what happens when we connect a solenoid to a battery and turn it on:

 

We need to use a data-logger, because the effect we see is very short-lived, less than a second.  We certainly could not measure it with a stopwatch!  We will look at this in more detail in a later tutorial.  If we plot a graph of voltage (or current) against time, we see an exponential rise in the voltage and the current.  This is because the current changes by a constant fraction in each interval of time.  The graphs look like this:

Assuming we have a voltage source with negligible internal resistance, we can say that the source voltage is constant.  Therefore some of the voltage has to be taken away form the source voltage.  It is not a voltage that we can measure directly.  We have to subtract the measured voltage from the source voltage to get the voltage from the inductance.  So this graph shows how the induced voltage falls:

 

 

The induced voltage is shown as EMF due to the inductance.  We can sum the graph up with this expression:

 

measured voltage (V) = supply voltage (V) - EMF (V)

 

In Physics code:

V = V0 - E

 

The curly E (E) is the physics code for electromotive force (EMF).  Electromotive force is not a force at all; it's energy per unit charge.  Perhaps it was named that by early physicists who got a belt off a high voltage battery that threw them across the room. 

 

 

The electromotive force opposes the supply voltage, so is given a negative sign.  It depends on the rate of change of the current.  This is written:

 

The constant of proportionality is the inductance, L, the unit of which is Henry (H).  EMF is in volt (V), and the rate of change of current, dI/dt is in amp per second (A s-1).  We can measure the rate of change of current by taking the gradient of the current-time graph.

 

If we work out the rate of change of the current for a number of different points, we get a straight line going through the origin, so the proportionality is direct:

Notice that I have deliberately referred to the EMF as the reverse voltage.  This avoids the need for the minus sign and makes the argument simpler.

 

The equation itself is called Faraday's Law, and the minus sign is due to Lenz's Law which states that the polarity of the induced emf is such that the current induced makes a magnetic field that opposes the change.  The equation again is:

Question 5

An inductor of inductance 0.14 H is connected to a 12 V supply of negligible internal resistance.  The exponential rise of the measured voltage is studied and at a certain time, it is found that the measured voltage is 2.5 V.  Calculate the rate of change of the current at this time.

Answer

 

Linking Flux, EMF, and Inductance

We have seen how when a magnetic field is generated, there is an EMF that opposes the supply voltage.  It is dependent on the rate of change of the current.  Since currents cause magnetic fields in the first place, it is reasonable to say that the EMF also results from the change in magnetic flux.  We can write:

But that is not the whole story, for the number of turns of wire in the coil is also important.  Lenz's Law applies, so there is a minus sign in the equation to show that the induced EMF is in the opposite direction to the applied voltage.  The equation becomes:

The term NF is called the flux linkage, and has the units Weber turns.  Turns is dimensionless, so the unit in unit analysis is Weber (Wb).  Therefore the EMF is the rate of change of flux linkage.

 

We know that:

F = BA

 

So we can rewrite the equation as:

The area is constant.  Only the flux density changes.  We can work out flux density for a solenoid with a core of relative permeability mr using:

 

The terms m0, mr, and n (= N/l) are all constant for a particular solenoid, so only the current changes.  So we can write:

And that tidies up to:

Question 6

A solenoid has a diameter of 2.0 cm and a length of 5.0 cm.  It has 150 turns and the core has a relative permeability of 120.  Calculate the EMF induced when the rate of change of current is 60.0 A s-1.  If the solenoid is connected to a 12.0 V power supply, calculate the voltage that would be measured by a data-logger.

Answer

 

From this, it does not take a genius to find an expression for inductance that takes into account the properties of the solenoid:

 

where:

L = inductance (H);

A = area (m2);

m0 = 4p 10-7 H m-1;

mr = relative permeability;

N = number of turns;

l = length (m).

 

Question 7

A solenoid has a diameter of 2.0 cm and a length of 5.0 cm.  It has 150 turns and the core has a relative permeability of 120.

What is the inductance?

Answer

 

Energy and Inductance

Power is defined as the rate of using energy.  We know it from GCSE as:

Power  (W) = energy (J) time (s)

Or in physics code as:

We also know that:

Power (W) = voltage (V) current (A)

 

P = VI

 

The equation above we have used for a steady voltage and current.  But what happens if the voltage and current are changing as would happen when an inductor is connected to a source of voltage? We can say that the power at any instant is the product of the voltage at that instant and the current at that instant, which in Physics code is written as:

 

p = vi

 

Consider a pure inductor (which has zero resistance) of inductance L which is connected to a voltage source of negligible internal resistance.  We know that the instantaneous voltage, v, is given by:

 

In this case I have used the physics code i for the instantaneous current.  The instantaneous power is given by:

 

Since energy = power time, we can write a calculus function:

And this gives us:

 

This is a useful result as it gives us the energy contained by the solenoid.

 

Question 8

The solenoid in Question 7 is now carrying a steady current of 4.5 A.  Show that the energy in the magnetic field is about 0.2 J.

Answer

 

The energy may not sound a lot, but consider what happens if the energy is released in a very short time, for example when the current is switched off.  The magnetic field collapses releasing the 0.2 J energy.  If this process takes 10 ms, we can easily see that the power is 20 W.  There is also a reverse voltage spike.

 

Question 9

The solenoid above is carrying a current of 4.5 A, as in Question 8.  When the current is turned off, it falls to zero in 10 ms.  

What is the value of the voltage induced?

Answer

 

The answer you worked out is quite tame, but could still damage a delicate electronic component.  However if you were to use a solenoid that was 0.2 H, the voltage would be ten times higher, which would quite easily wreck electronic components.  This can be prevented by using a reverse biased diode in parallel to the solenoid.  The idea is shown on the circuit below:

The data-logger sensors are now protected from a large reverse voltage spike.

 

Magnetic Energy Density in a Solenoid

We know that the energy in a solenoid is given by:

and that the inductance of a solenoid is given by:

So we can combine the two expressions to give us:

 

The energy density is defined as the energy per unit volume, U.  Volume = area length so we can write:

The area terms obligingly cancel out:

 

 

Question 10

A solenoid has a diameter of 1.5 cm and a length of 5.0 cm.  It has 200 turns and core of magnetic material that has a relative permeability of 130.  It has a current of 4.5 A flowing in it.  Calculate the energy density.  State the correct unit.

Answer

 

This equation can then take us back to the first equation that we had for the energy density:

We know that

and that n= N/l.  We can rearrange the equation for current, I:

We can substitute this into our previous relationship:

And this will give us:

We can do a lot of cancelling to give us:

Which is where we started from...

 

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