Tutorial 4B - Putting Magnetic Fields to Work  (Linear Applications)

Learning Objectives

To learn about the action of a relay.

To understand the use of a solenoid

To calculate the force from a solenoid

To learn about the structure and action of a loudspeaker.

Key Questions

How does a relay work and what is it used for?

How is a solenoid actuator used?

How does a loudspeaker work?

Electromagnetic Relay

The devices we are going to look at in this tutorials are linear transducers.  They convert electrical energy into linear motion, i.e. motion in a straight line.  They are often used as actuators, which are devices that use an electrical current to control another system.

 

The relay at its simplest is an electromagnetic switch that allows a small current to turn on a big current.  There are many different kinds of relay, but here we will discuss a very simple relay.  It is shown below:

When the coil is energised, the magnetic material is magnetised.  The armature is attracted, causing a clockwise turning moment around the pivot.  This movement closes the contacts.  The spring is also stretched.

 

When the coil current is turned off, the stretched spring applies an anti-clockwise turning moment which pulls the armature back into its original position.  The contacts are pulled apart and the big current is turned off.

 

Consider a coil of N turns.  It has an area, A m2.  It carries a current I A.

 

Let's look at the magnet and armature in close up:

 

Notice that the magnetic field lines are concentrated more in the magnetic material.  There is an air-gap between the core and the armature.  The relative permeability of air is 1.0, so we have to be careful.  It can be shown that the actual force is given by:

The table shows the quantities:

Term Meaning Unit
F Force Newton (N)
I Current Amp (A)
m0 permeability of free space 4p 10-7 Henry per metre (H m-1)
A area metre2 (m-2)
N Coil turns turns
x Gap between the armature and the core metre (m)

 

 

It's easiest to do a stepwise calculation.  We will do a worked example:

 

Worked Example

A relay has a coil of 600 turns.  The coil has a core of circular cross-section with diameter 1.0 cm.  Between the core and the armature, there is an air gap of 1.5 mm. 

Calculate the force on the armature if a current of 0.20 A is flowing.

m0 = 4p 10-7 H m-1

Answer

Calculate the area:

A = p (5.0 10-3 m)2 = 7.85 10-5 m2

 

Now calculate the force:

F = [(600 0.20 A)2 4p 10-7 H m-1 7.85 10-5 m2] [2 (1.5 10-3 m)2] = 0.316 N = 0.32 N to 2 s.f.

 

 

Now try one for yourself

 

Question 11

A relay has a coil of 1200 turns. The diameter of the coil core is 6 millimetres and the air gap is 1.8 millimetres. The spring exerts a force on the armature of 0.15 newtons at the part of it opposite the air gap. What coil current will just operate the relay?

m0 = 4p 10-7 H m-1

Answer

 

The closer the armature is to the core of the coil, the greater the force, because:

This is true to a first approximation as the field is not truly radial.  In the case of a relay, the armature is within a couple of millimetres of the core.

 

Let's look at the moments about the pivot:

 

The armature, which has mass m, is acted on by a force of F N which we have looked at above.  Therefore there is a clockwise moment:

 

 

The strange looking symbol that looks like a gallows, G, is "Gamma", a Greek capital letter 'G'.  It is sometimes used as the physics code for moment. 

 

At a certain value of the current I A, the relay is just about to click over, but not quite.  The system is in balanced equilibrium.  The clockwise moment is opposed by an anticlockwise moment given by the spring.  By the principle of moments, we can write:

 

The notation F'  shows that this force has a different value to the force acting on the armature.

 

The spring obeys Hooke's Law, so we can write:

F' = ke

 

where k is the spring constant (N m-1) and e is the extension (m).

 

Therefore we can now write:

G = kes1

 

We can then go on to write:

We will see this relationship being used in the next worked example:

 

Worked example

A relay has a coil of 800 turns.  It has a core of magnetic material 12 mm in diameter.  The armature, of mass 2.0 g, is held in the open position by a light spring of spring constant 200 N m-1.  The centre of mass of the armature is 10 mm from the pivot.  The other dimensions are shown in the picture below:

Calculate the current that is needed to activate the relay.

m0 = 4p 10-7 H m-1

g = 9.81 N kg-1

Answer

We need to work out the area A:

 

A = p (6.0 10-3 m)2 = 1.13 10-4 m2

 

We also need to know the extension e.  This is worked out by proportion.  s1 = 0.25 s2

 

e = 0.25 1.8 10-3 m = 0.45 10-3 m

 

Work out the force from the spring:

F' = 200 N m-1 0.45 10-3 m = 0.090 N

 

The anticlockwise moment:

G = 0.090 N 5 10-3 m = 4.5 10-4 N m

 

The clockwise moment = (F 20 10-3 m + 2.0 10-3 kg 9.81 N kg-1 10 10-3 m) = 4.5 10-4 N m

 

F = [(4.5 10-4 N m) - (2.0 10-3 kg 9.81 N kg-1 10 10-3 m)] 20 10-3 m = (4.5 10-4 N m - 1.962 10-4 N m) 20 10-3 m

 

F = 0.0127 N

 

Now we can work out the current:

I2 = [2 0.0127 N (1.8 10-3 m)2] [8002 4p 10-7 H m-1 1.13 10-4 m2] = 9.05 10-4 A2

 

Don't forget to square-root it:

I = 0.030 A = 30 mA

 

This calculation has many steps in it.  Don't over-complicate things, otherwise errors creep in very easily.

 

The model here only applies if the armature is close to the electromagnet.   However if a larger "throw" is needed, an electromagnetic relay would not be used.  Instead a mechanism driven by an electric motor or pneumatic system would be used for such an application.

 

Make and Break

The make and break circuit is widely used in electric bells and buzzers.  These are still common, even if they have been superseded by electronic devices.

 

When the push-to-make switch is pressed, the coil is energised and magnetises the core.  The armature is attracted downwards and the contacts open, cutting off the current.  The armature is pulled back to its original position and the contacts close again.  The whole process starts again until the push to make switch is released.

 

A hammer can be attached to the armature, which in turn strikes a gong.  If you want to see an animation of this, click on the link:

https://animatedscience.co.uk/blog/electromagnetic-bell

Piezoelectric devices are more commonly used instead of mechanical buzzers of this type. 

Solenoid Actuators

A solenoid is simply a coil of wire wrapped around a former.  When a current flows through the solenoid, it makes a magnetic field, which can attract a piece of magnetic material.  The moving material is called the armature.  It moves into the solenoid.

 

Consider a solenoid of N turns, carrying a current of I amps that is acting on an armature of diameter d metres, which is x metres from the end of the solenoid.  The relationship that describes the force of attraction in this case is:

Of course we need to work out the area, A, which we do using:

 

Question 12

A solenoid of 300 turns is carrying a current of 1.5 A.  An armature of diameter 1.0 cm is 2.0 mm from the end of the solenoid.  What is the force that is acting on the armature?

m0 = 4p 10-7 H m-1

Answer

 

In the case above, we considered what would happen if the armature were outside the coil of the solenoid.  Let's think about what happens when the armature is in the coil like this:

At the right hand end, there is an end-stop made of a magnetic material.  Most solenoid actuators are actually made like this.  The relationship between the force and the current is the same as the previous case discussed above, i.e.,

From this relationship, we can also see that:

The distance between the end stop and the armature is sometimes called the throw

 

Question 13

What is the significance of the length of throw (the distance between the armature and the end stop) on the force?

Answer

 

In real solenoid, a spring is inserted into the solenoid like this:

This returns the armature to its original position when the current is turned off.  If a more powerful return spring is needed, this can be placed outside the coil.

 

The spring obeys Hooke's Law in compression, i.e.

F = kc

where F = (N), k = compression constant (N m-1), and c = compression (m).

 

Let's do a worked example:

 

Worked Example

A solenoid actuator consists of an armature of area 1.0 10-4 m2.  The solenoid coil consists of 500 turns.  Inside the coil there is a spring of compression spring constant 20 N m-1.  When a current flows, the armature moves 2.0 mm into the coil so that its final position is 2.0 mm from the end stop.  Calculate the current.

Answer

The force can be easily worked out:

F = kc = 20 N m-1 2.0 10-3 m = 0.040 N

 

Now to work out the current.  Formula to be rearranged:

Now out the numbers in:

I2 = [2 0.040 N (2.0 10-3 m)2] [5002 4p 10-7 H m-1 1.0 10-4 m2] = 0.0102 A2

 

I = 0.10 A

 

 

Question 14

A solenoid actuator consists of a coil of 600 turns wrapped around a former that has a cylindrical space in it of uniform diameter 12 mm.  The armature of the actuator fits so that it moves easily, but is a close fit.  It has a return spring of compression constant 150 N m-1 and an uncompressed length of 5 mm.

A current of 1.56 A flows through the coil.  Calculate how far the armature moves.

Answer

 

Some solenoid actuators have a second coil, like this:

The armature will move to the left if Coil 1 is turned on.  It will move to the right if Coil 2 is turned on.  Such an arrangement is common on point-work for a model railway.

 

 

For powerful solenoids, a large current is needed.  If the current is allowed to flow for a long time, the solenoid can get quite hot.  (I know - it happened on one of my model railways and the polystyrene casing of the motor melted.)   Therefore many solenoid control circuits have a way of reducing the current to hold the solenoid in place, since the current needed to hold the solenoid in place is less than what is needed to activate it.

 

In a model railway point motor, the point blades are held in place by a spring and do not need the solenoid to be on.  A charged capacitor gives a big current for a brief moment.  If the solenoid remains connected for some reason, the current falls way to almost zero, so the solenoid coils will not get hot.

 

How can we measure the force from a solenoid?

We can do this using simple physics laboratory equipment used in a school or college laboratory.

The solenoid return spring is removed and a brass spacer is put in to keep a constant separation, x.  The current is turned up to a maximum.  The armature is attracted to the right.  A force is then applied to the force meter, so that it reads, say, 10 N.  The current is then reduced and the current at which the armature is released is noted.  The restraining cord is there to stop the armature from flying across the lab when the armature is released.  Even so, it's a good idea to wear eye protection.

 

The results can be plotted on graphs which look like this:

                                                                                                Graph 1                                                                                                 Graph 2

  

                                                                                              

Graph 1 is the most obvious one to plot, force against current.  It is a parabola.   Graph 2 is of Force against Current2, which gives a straight line.

 

Question 15

What would the gradient represent?  What are the units?

Which constant could you use the gradient to work out?

Answer

 

If we plot a graph of the logarithms of the force against the logarithm of the current, we get this:

 

We get a straight line and the gradient is 2, indicating that F I2.

 

Maths Note

A logarithm is one number expressed as a power of a base number.  The most common base number is 10.  Another base number that is commonly used is e, where e is the exponential number 2.718...

 

Logarithms to the base 10 are written either log10 or lg.  Logs to the base e are written loge or ln.

 

For example log10 2 = 0.3010 (= 100.3010).

 

The logarithms of two numbers multiplied add up.  So 20 = 2 10.  Log10 20 = log10 10 + log10 2 = 1 + 0.3010 = 1.3010

If two numbers are divided, the logarithms subtract.

 

If a number is squared, its logarithm is multiplied by 2.  If the number is square-rooted, the logarithm is halved.

 

If we plot a log-log graph, the gradient of the line represents the power of the quantity relationship.  If a = kb2, where k is a constant, then log a against log b will give a gradient of 2.  There will be an intercept of value log k.

log a = 2logb + log k

 

 

Question 16

The graphs above were produced by data modelling with the formula:

 

The data used were:

Armature diameter - 12 mm;

Space thickness - 2.0 mm;

Number of turns = 300;

Permeability of free space = 4p 10-7 H m-1.

 

Show that the value of the y-axis intercept on the graph of log10 F against log10 I is about 0.2.

Answer

Loudspeaker

Almost every home or car has a loudspeaker.

 

This one is one of a stereo pair in a domestic hi-fi system.  The structure of a loudspeaker is like this:

The magnet is designed to have a field that is as close as possible to radial.

The coil obeys Fleming's Left Hand Rule so that if the current goes clockwise, the coil moves inwards.  If the current is anticlockwise, the motion of the coil is outwards.  Check it for yourself.  The movenebt of the coil is limited by the coil mount (sometimes called the spider).  The frame of the loudspeaker is sometimes called the basket.

 

A laboratory vibration generator is simply a loudspeaker without the cone:

 

 

The coil mount (spider) can be seen clearly.

 

The loudspeaker can be modelled as the coil being a mass and the coil mount being s spring.  The spring obeys Hooke's Law:

 

F = kx

where:

F = force (N);

k = spring constant (N m-1);

x = displacement (m);

 

We also know that the force is given by:

F = BIl

where:

B = magnetic flux density (T);

I = current (A);

l = length of wire in the field (m).

 

The length of the wire is worked out by multiplying the circumference of the coil by the number of turns.  So for N turns, on a coil of radius r m, the force is given by:

 

F = 2pBINr

 

To get the displacement of the coil we can combine the two formulae:

kx = 2pBINr

 

Question 17

The coil mount on a loudspeaker can be modelled as a spring of spring constant 50 N m-1.  The loudspeaker has a magnet of magnetic flux density of 0.15 T.  The coil has a radius of 15.0 mm and has 50 turns of copper wire.  The resistance is 4.00 ohms and the coil is connected to a DC supply of 2.0 V.

 

Calculate how far the coil will move.  Give your answer to an appropriate number of significant figures.

Answer

 

The trouble with this answer is that a loudspeaker is just serving as a physics curiosity.  With direct current, the loudspeaker has no use at all, although a DC offset is something that electronics engineers need to be aware of.  The DC offset can result in the characteristic switch on thump we get with a cheap amplifier.  More expensive audio amplifiers have methods of keeping the speakers disconnected until the electronics have settled down.

 

The loudspeaker is used with a continually varying electrical signal.  While this makes it useful, there are some problems that we need to be aware of.  The first is there is a certain amount of inductive reactance due to the interaction of a changing voltage (and current).  This combines with the resistance to make an impedance, which has a higher value that the resistance due to the wire.

 

The second problem is that the movement of the coil in the magnetic field makes the coil act as a generator.  We will look at this effect in another tutorial.  The coil generates back e.m.f. which makes the loudspeaker act as a microphone.  (If you speak into a loudspeaker, it will act as a microphone anyway.)

 

The third problem is that the speaker has a certain value of resonance.

 

We can work out the resonant frequency using a simple formula:

where:

f0 = resonant frequency (Hz)

k = spring constant (N m-1)

m = mass (kg).

 

In some text books, you may see the term compliance.  Compliance is the reciprocal of the spring constant and is measured in metres per Newton.

 

The resonance of the speaker is damped by the movement of the cone in air.

 

Question 18

A loudspeaker has a spring constant provided by the coil mount of 150 N m-1 and the mass of the coil is 2.4 g.

 

Calculate the resonant frequency.

Answer

 

The resonance will increase the impedance, as well as changing the quality of the sound.

The behaviour of loudspeakers and their enclosures is complex and well beyond the remit of these notes.  We will look at crossovers when we look at inductive reactance in a later tutorial.

 

Linear Motors

The linear motor is a rotary AC induction motor that has been slit down one side and flattened out.  So instead of rotary motion, the two parts move in a straight line relative to each other.  The concept was first thought about as early as 1840 by Charles Wheatstone, but the first attempts at making a practical motor happened from 1947 onwards.  The British engineer Eric Laithwaite (1921 - 1997) did a lot of work to develop the linear motor.  There several different types of linear motor, but we will look at a simple generic linear motor.

 

Linear motors like these are AC devices.  Most linear motors use 3-phase AC, enabling a greater deal of control than is possible with single phase AC.

 

 

In the picture above we can see the effect of flattening out the rotary motor into a linear form.  The rotor is often referred to as the reaction plate, although we will use rotor in this explanation.  The maximum force comes when the rotor is completely covering the stator.  If the 10 % of the rotor is covering the stator, then the force is 10 % of the maximum.

 

The maximum force can be worked out using the equation:

F = BIl

We can work out the flux density using the equation:

We have to be careful, because the current in the stator will be different to the current in the rotor.  We can combine the two formulae to give:

where Is and Ir are the stator and rotor currents respectively.

 

Since the n term is the number of turns per metre, we can write this term as:

where N is the number of turns and l is the length (m) respectively.

 

So we can put this into the equation to give:

It does not take a genius to see that the l terms cancel out, so we get:

 

While the stator current is easy to measure with an AC ammeter, the rotor current is not.  We can model the motor as a simple transformer, where the stator acts as the primary, and the stator is the secondary.  The transformer equation is:

Let's do a worked example:

 

Worked Example

A large linear motor consists of a stator made from a magnetic material of which the relative permeability is 150.  The stator has 200 turns and takes a current of 1.3 A.  The rotor is modelled as a secondary which has a single turn.  What is the maximum theoretic force available from this motor, if we model the motor as a simple transformer?

Answer

Work out the current in the rotor:

Ir = (200 1) 1.30 A = 260 A

 

F = 4p 10-7 H m-1 150 200 1.3 A 260 A = 13 N

 

This assumes that the transformer is perfect, which it will not be, because there is a gap between the rotor and the stator.  In reality the maximum force will be lower than this, because there is an air gap between the rotor and the stator.  This will reduce the efficiency of the transformer effect.

 

This idea is used in maglev trains

 

Question 19

A maglev train is to be held on a magnetic field using a linear motor arrangement.  It consists of a single carriage of mass 30 tonnes.  The stator has a total of 200 turns and the magnetic material has a relative permeability of 75.

It is estimated that the current induced in the rotor (reaction plate) is 10 000 A.  Calculate the stator current that will be required to keep the train levitated.

 

m0 = 4p 10-7 H m-1

g = 9.81 N kg-1

Answer

 

Linear motors are used in a number of applications, for example shuttling packages along fixed lengths of track.  This can be a more efficient than a conveyor belt.  They also can be used to move a patient underneath a hospital scanner.

 

The motors are usually three phase, in order to provide better control.

 

It was once hoped to use linear motors under high speed trains.  The advantage of this would be that the motor would have no moving parts.  The disadvantage is that a continuous rotor would be needed between the rails and this would be expensive.  Also the track is not always perfectly level (if you don't believe this, try riding on a Pacer train - you will find out very quickly), so there would be variation between the reaction plate and the stator.  This would reduce the efficiency of such a motor.

 

Now try the multiple choice quiz.

Self Test

 

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