Tutorial 4A - Putting Magnetic Fields to Work  (Motors and Meters)

Learning Objectives

To learn about the force acting on a coil.

To learn about applications of these concepts in motors and meters.

To learn how to read an analogue meter.

Key Questions

What happens when a coil is placed in a magnetic field?

How does an electric motor work?

How do analogue meters work?

How do we use them?

What goes on in a multimeter?

How does a watt-meter work?

Coil in a magnetic field

Electric motors use magnetic fields to convert electrical energy to rotational kinetic energy.  They are sometimes called rotational transducers.

We have looked at a number of phenomena involving magnetic fields.  We have done calculations.  However, unless there is a use, then what we have studied in previous tutorials is simply a Physics curiosity.  The good news is, of course, that there are many uses for magnetic forces.  Indeed they have had a central role in making our technological society what it is today.  Think about how many times you have put the motor effect to good use before you read these notes today.  The chances are that the hard disc drive that stores your web-browser that you are using to read this stuff has discs that are being spun at high speed by a tiny electric motor.

We are now going to look at how coils are turned in magnetic fields.

Consider a coil that is l metres in length, w metres wide, that is placed vertically in a magnetic field of flux density B Tesla.  It is free to turn on a shaft XY when it carries a current of I amps.  It is shown in the picture below:

Let us have a look at the set-up from above:

The force acts along the length, l, of the coil.  Since the top and bottom parts of the coil are parallel to the field, the force is zero.  The forces act about the shaft, XY, to form a couple.  We know that the force acting on the wire is given by:

F = BIl

We also know that the turning effect, or torque, acting around the shaft is given by:

t = Fw

The symbol t is a 'tau', a Greek lower case letter 't', and it's the physics code for torque.  Torque is measured in Newton metre (N m).

So if we substitute for the F term, the torque equation becomes:

Since length multiplied by width gives us the area, we can write:

The torque makes the coil turn through an angle q as shown below:

The force acting on the coil is no longer F, but F' ("F-prime") where:

F' = F cos q

So our torque equation is now modified to:

If we have N turns, each turn experiences the couple.  So we multiply the torque by the number of turns:

 A light coil of dimensions 3.0 × 5.0 cm with 500 turns is placed in a magnetic field of flux density 14 mT.  A steady current of 0.25 A flows through it.  Calculate: (a) The maximum torque; (b) The torque when the coil has turned through 27o.

This concept is very useful in electric motors and analogue meters.  We will look at the basic electric motor next.  More detailed discussion about electric motors is on the pages about electric motors.

The Electric Motor

The picture below shows a very simple electric motor.

This motor works on direct current.  The current is carried to the armature through carbon brushes and a split-ring commutator.  If we reverse the current, the direction of the rotation changes as well.  Using Fleming's Left Hand Rule:

The commutator acts as a change-over switch so that the current on the left hand side of the coil always moves towards us (and on the right hand side, away).  However when the coil is at this position, the force is zero.

In this case, we see that the armature is at 90 degrees, so that the current is at parallel to the field.

There is zero force, and the motor stalls.  However, if the rate of rotation is high enough the motor will keep on turning, because the momentum of the armature will ensure that the motor keeps on turning.

In real motors there are several coils.  The picture shows the components of a motor from a small power tool:

The armature itself is shown in the picture below:

The magnetic field of practical electric motors is made to be as close to radial as possible.  A ring-shaped magnet or magnets with a circular cut-out are used as in the picture below:

In a real motor, the magnet (or yoke) looks like this:

The idea of this is to keep the magnetic field as close as possible to 90o to the rotating coil.  Therefore the torque remains at a maximum.

The power of the motor can be easily worked out.  From A-level linear dynamics we learned that:

Power (W) = force (N) × speed (m s-1)

In Physics code:

P = Fv

As a motor rotates, many of the concepts of rotational dynamics apply.  In this case:

Power (W) = torque (N m) × angular velocity (rad s-1)

In Physics code:

P = tw

You will remember the idea of angular velocity from circular motion.  Angular velocity is the angle swept every second by a rotating object.  Linear speed is not that useful, as it depends on the radius.  However a rotating object sweeps the same angle every second.  We could use degrees per second as the unit, but it's much easier to use radians per second (rad s-1).  The physics code for angular velocity is w (omega, a Greek lower case letter 'long o').

Radians is a dimensionless unit.  So it takes no part in checking for unit consistency. In some books it is left out altogether and is just written as s-1.  In these notes I will include it for clarity.

The units for power (Watt) are the same as N m s -1 which are the same as N m rad s-1.

The rate of rotation is the number of revolutions per second that the motor turns at.  It is the same as frequency, is given the Physics code f, and is measured in Hertz (Hz).  1 Hz = 1 s-1

The angular velocity is worked out using the formula:

w = 2pf

Make sure that your calculator is set to radians, NOT degrees.  Many errors arise because of this.

Often the rate of rotation is given as revolutions per minute (rpm).  It is important that you convert the rpm to Hz by dividing by 60.

 Question 2 A motor takes a current I, and the armature is placed in a magnetic field for flux density B.  The armature has N turns and is of length l and width w, rotating at a frequency f. Show that the equations P = Fv and P = tw are consistent.  What is the final relationship between power and angular velocity?

From Question 2 you will have seen that the power depends on the angular velocity and you will have worked out that:

P = BANIw

 A motor has an armature of 200 turns of rectangular form 4 cm × 5 cm.  The armature is placed in a magnetic field of flux density 0.15 T and a current of 1.5 A is flowing through it.  The motor is turning at 2400 rpm.  Calculate the power.

The useful output of the motor will be rather less than this.  Frictional losses will occur in:

• The bearings;

• The brushes;

• The movement of cooling air by the fan.

Moving Coil Meters

Although most modern voltmeters and ammeters are digital, analogue meters are still widely used in the electrical engineering industry.  Although analogue meters are sometimes considered old hat, you can still buy one for as little as £3.50 (€4.50).  An analogue meter has a scale with a pointer.  People like analogue instruments.  Most cars still have them.  Although the most modern aeroplanes have touch-screen computer technology for the pilots, these still show images of analogue meters, rather than columns of numbers.  Why?  Because pilots get a better picture of the reading of the instrument and compare it with what it should be.  Humans, like all other animals, are analogue.

A digital instrument can read to several decimal places, while an analogue meter usually reads to no more than 2 decimal places.  For most purposes, that is quite adequate.

Many students think that digital meters with their direct readout are more accurate than analogue meters.  They may be more precise, but, unless they have been calibrated correctly, they may not be accurate.  Instruments with certified calibration are very expensive indeed, because it takes time to verify the calibration.

Many students have a lot of difficulty with analogue instruments, especially if scales need to be converted.  For example, a meter may have a scale that reads from 0 to 100.  If it is used to measure 5.0 V, the 100 mark means 5.0 V, while the 60 mark represents 3.0 V.

In this section, we will look at how moving coil (analogue) meters work, and how they can be used to measure voltage, current, and power.

I have used this meter since I got it as a teenager.  And it still works.

A moving coil meter works by using a very light coil connected to a needle that acts as a pointer.  The coil is placed over a stationary cylinder of magnetic material between the circular poles of a magnet.  This has the effect of making the field as close to radial as possible, ensuring a uniform torque.

The torque from the current is balanced by a torque in the opposite direction from a very fine hair spring.  There is one of these at each end of the coil.  It also acts as a conductor for the current going into the coil.  The picture below shows the idea in a real meter.

We know that the torque on the coil is given by:

t = BANI

There is a torque in the opposite direction from the hair spring.  We know that for a linear spring, Hooke's Law applies:

Force (N) = Spring constant (N m-1) × extension (m)

For a spring that's a flat coil spring, a similar rotational formula applies:

Torque (N m) = torsion constant (N m rad-1) × angle turned (rad)

In physics code, the formula is written:

t = kq

Remember that the angle is in radians.

It does not take a genius to equate the two relationships:

kq = BANI

 Worked Example A meter has a coil consisting of 500 turns of very thin copper wire on a very light rectangular former 1.0 cm × 2.0 cm.  It is held in place on a shaft where the friction is negligible.  Its motion is opposed by two very fine hair springs of which the torsion constant is 1.5 mN m rad-1 for each spring.  The magnetic field has a flux density of 0.35 T.    Calculate the angle through which the needle turns when a current of 50 mA flows. Answer Formula rearranged: q = (0.35 T × 2.0 × 10-4 m2 ×500 × 50 × 10-6 A) ÷  (2 × 1.5 × 10-6 N m rad-1)   There are two springs...   q = 0.58 rad (= 33.4 o)

Remember that if the angle is given in degrees, you need to convert to radians.

1 rad = (360 ÷ 2p)o » 57.3o

1o = (2p ÷ 360) rad

 A moving coil meter  has a coil of dimensions 1.5 cm × 2.0 cm.  The coil has 600 turns and is placed in a magnetic field that has a flux density of 0.25 T.  The coil is mounted by two springs of torsion constant 1.2 mN m rad-1 and is deflected by 40o when a current passes through it. Calculate the current flowing in the coil.

The currents that are used in moving coil meters are very small.  A common value for the full scale deflection of a moving coil meter in a college physics laboratory is 100 microamps.

Shunts and Multipliers

To make a meter useful, it needs to have different ranges.  A voltmeter that reads 5 V would not be much good for reading 15 V.  To make a meter read the right quantity, we use resistors in different ways:

• For an ammeter, a resistor is placed in parallel, and is called a shunt.

• For a voltmeter, a resistor is place in series and is called a multiplier.

In the pictures below, the meter part is shown as a galvanometer.   For the voltmeter, the circuit diagram is like this:

For the ammeter, the circuit is like this:

The two pictures show a typical college meter with its collection of multipliers and shunts:

Meter with shunts                                                                                                                                 Meter with multipliers

We will do some examples using some typical values.  The full scale deflection occurs with a current of 100 mA.  The resistance of the coil is 100 W.

 Calculate the voltage across the 100 W coil when a current of 100 mA flows.

 Worked Example A moving coil ammeter is to measure a current of 5.0 A at full scale deflection.  The meter current at full scale deflection is 100 mA and the coil resistance is 100 W. Calculate the resistance of the shunt. Answer Use Kirchhoff I to work out the current through the shunt.   Current in shunt = 5.0000 A - 1 × 10-4 A = 4.9999 A Voltage across the shunt = voltage across the coil = 0.01 V (see Question 5) Resistance of the shunt = V/I = 0.01 V ÷ 4.9999 A = 2.00004 × 10-3 W

A couple of points to note here:

• The resistance of shunts is very low, but NOT zero;

• The total resistance of an ammeter is slightly lower than the shunt resistance, as it's a parallel combination.  It is not quite zero.

Using Analogue Meters

The perfect voltmeter has an infinite resistance, so it takes no current at all.  The digital voltmeter has a very high resistance, about 107 ohms, so we can say that it is almost perfect.  However the analogue voltmeter has a resistance of between 20 kW and 400 kW, depending on what range we use.

The perfect ammeter has zero resistance.  In reality it has a very small resistance.  For most purposes, we can ignore the resistance.  The resistance of a digital ammeter is the same as the resistance of an analogue ammeter.

So what?  If we are going to use an analogue voltmeter with something that takes a relatively high current, there is little difference.  However, if we hare measuring the voltage across a resistor of 100 kW, then there is a considerable difference.  We will use that figure in our next worked example.

We model the analogue voltmeter with a resistance of 40 kW as a perfect voltmeter in parallel with a 40 kW resistor.

This is a model for problem-solving only.  It does NOT show the actual construction, which is a meter in series with a multiplier.

 Worked Example A current of 20 mA is flowing through a resistor of resistance of 100 kW.  (a) A digital voltmeter of infinite resistance is used to measure the voltage.  What is the voltage? (b) The battery of the digital voltmeter goes flat.  The only voltmeter available is an analogue meter of which the resistance is 40 kW.  It is known to be accurately calibrated.  What is the reading now? Answer (a) V = 20 × 10-6 A × 100 × 103 W = 2.0 V   (b) The circuit now becomes: Work out the parallel combination: Rtot = 28571 W   V = 20 × 10-6 A × 28571 W = 0.57 V

From this result we can see that the value of voltage is considerably different from the true voltage.  So analogue meters need to be used with care.

 An analogue voltmeter is known to be accurately calibrated.  A digital voltmeter of infinite resistance measures the voltage across a 150 kW resistor as 3.5 V, but when the analogue meter is used, the voltage is measured as 1.4 V.   Calculate the resistance of the meter.

The model for the ammeter with a resistance is a perfect ammeter in series with a very low value resistor.

 An accurately calibrated ammeter has a resistance of 400 W and is being used to measure a current of 100 mA from a 12.0 V supply through a resistor.  Calculate the resistance of the resistor.  Hence discuss the uncertainty in the reading of the current.

Analogue Multimeter

It is possible to buy separate voltmeters and ammeters, but a multipurpose instrument, a multimeter, is of far more use to the electrical engineer.   Analogue multimeters are still available. The instrument has ranges for current, voltage, and resistance.  The resistance meter is sometimes called an ohmmeter.   We will look at one now.  It has been opened up to show what happens inside:

The fried resistor still works.  I remember the smoke coming from the instrument, but I can't remember what I was doing when I fried it.  Yes, I put too much current through it!

The circuit below shows a very simple multimeter, with one range for voltage, one for current, and one for resistance.

In the ohmmeter circuit there is a battery in series with a variable resistor to adjust the zero point.  Note that the zero in the ohmmeter is when the maximum current is flowing, i.e. at full scale deflection.  The ohm scale goes down from 0 to infinity.  This is shown below:

This ohm scale can read up to 10 kW on the ×1 scale.

 On this multimeter, there is a ×100 range.  What is the maximum resistance that can be read from the scale?  Comment on the accuracy and precision of the reading of a resistor of 750 kW, assuming the instrument is accurately calibrated.

My experience of thirty years' teaching of A-level students is that they struggle with the reading of an analogue meter.  I am sure that many a tutor of a first year class in a university course would agree.  Actually it's not that difficult.  We will look at how to do it.  Suppose we want to measure a voltage of 8.3 V.

1.  We need to make sure that the negative lead is in the common terminal

2. We need to select the 10 V range.  If a meter with a 5 V range, the needle would go off the scale.  This meter uses probes rather than a switch.   This is shown in the picture:

3. Connect the meter to the voltage.

4. Read from the 0 - 10 scale.  Use the mirror behind the needle to avoid parallax errors.  The meter will read to 2 significant figures only.  This should be quite sufficient.

 The picture shows the scale of an analogue meter.   The meter is set to the range 0 - 50 mA.  What is the current?

The Watt Meter

The watt-meter is a combination of an ammeter and a voltmeter.  Unlike other analogue meters, the magnet in a watt-meter is an electromagnet, not a permanent magnet.  The higher the current, the larger the flux density.  The coil in the meter is connected to a multiplier and acts as the voltmeter.  This is shown in the circuit diagram below:

Consider a watt-meter that has an electromagnet of n turns per metre, which is carrying a current of I amps, and is wrapped around a magnetic material of relative permeability mr.   In Tutorial 3 we saw that:

Earlier in this tutorial we also saw that for the angle turned by a meter:

kq = BANI

Now we have to be careful here as the two currents are NOT the same.  If we wrote this:

kq = m0mrnANI2

We would be wrong.

So let's call the load current through the electromagnet I1, and the current through the moving coil I2.  Therefore the our relationship is now:

kq = m0mrnI1ANI2

The current through the moving coil is given by Ohm's Law:

where RM is the multiplier resistance and RC is the coil resistance.

We can now substitute into the equation:

Of course P = VI, and we have the I1 term (the current through the load) and the V term, the voltage across the load.  So it doesn't take a genius to write:

Now we can rearrange the expression:

There are lots of terms in this relationship:

 Term Meaning Unit P Power Watt (W) k torsion constant Newton metre per radian (N m rad-1) q Angle turned by the coil Radian (rad) RM Multiplier Resistance Ohm (W) RC Coil resistance Ohm (W) m0 permeability of free space 4p × 10-7 Henry per metre (H m-1) mr relative permeability no units n electromagnet turns per metre turns per metre (m-1) A area metre2 (m-2) N Coil turns turns

Let's do a worked example:

 Worked Example A watt-meter has a coil consisting of 100 turns of very thin copper wire on a very light rectangular former 1.0 cm × 2.0 cm.  It is held in place on a shaft where the friction is negligible.   Its resistance is 50 W.   Its motion is opposed by two very fine hair springs of which the torsion constant is 1.5 mN m rad-1 for each spring.  The coil is connected to the supply through a multiplier of which the resistance is 37 kW.   The magnetic field is formed by a coil consisting of 10 turns of thick wire in a length of 5 cm.  The core of the electromagnet has relative permeability of 200.   The meter turns though 30o when the load is connected to the supply.   Calculate the power. Answer Work out the area:  A = 0.010 m × 0.020 m = 2.0 × 10-4 m2. Work out the number of turns per metre: n = 10 ÷ 0.050 m = 200 m-1. Two springs, so k = 2 × 1.5 × 10-6 N m rad-1 = 3.0 × 10-6 N m rad-1 Angle q = (30 × 2p) ÷ 360 = 1/6 p rad (= 0.524 rad).   Now put all the numbers in:   The p terms obligingly cancel out in this case: P = 58 W

 A watt-meter has a coil consisting of 40 turns of very thin copper wire on a very light rectangular former 1.0 cm × 2.0 cm.  It is held in place on a shaft where the friction is negligible.   Its resistance is 200 W.   Its motion is opposed by two very fine hair springs of which the torsion constant is 1.5 mN m rad-1 for each spring.  The coil is connected to the supply through a multiplier of which the resistance is 50 kW.   The magnetic field is formed by a coil consisting of 30 turns in a length of 5 cm.  The core of the electromagnet has relative permeability of 200.   The load takes a power of 250 W   Calculate the angle through which the needle turns.