1: Direct current electric motors

motors and

generators

Table 6.1.1  Parts of a DC motor

Part



Description and role



Coils



Many loops of wire that carry a direct current. These wires experience

a force (due to the motor effect) that causes the motor to turn when

current is flowing.

This is the part of a motor that contains the main current-carrying

coils or windings. For DC motors this is the rotor, but for AC motors it

is usually the stator.

In a DC motor, the rotor consists of coils of wire wound around

a laminated iron frame. The frame is attached to an axle or shaft that

allows it to rotate. The iron frame is laminated to reduce heating and

losses due to eddy currents. The iron itself acts to intensify the

magnetic field running through the coils.

Stationary permanent magnets (electromagnets in large DC motors)

that provide an external magnetic field around the coils. Permanent

magnets are curved to maximise the amount of time the sides of the

coil are travelling perpendicular to the magnetic field, to maintain

maximum torque.

A device with metal semicircular contacts that reverse the direction of

the current flowing in each coil at every half rotation. This reversal of

the current makes continuing rotation possible in a DC motor.

Conducting contacts (generally graphite or metal) that connect the

commutator to the DC power source.



Armature



Rotor



Stator



Commutator split-ring



Commutator brushes



Describe the main features

of a DC electric motor and

the role of each feature.

Identify that the required

magnetic fields in DC motors

can be produced either by

current-carrying coils or

permanent magnets.



Brushless DC motors



T



he designs of many modern DC motors are more complex

than the simple example we study in this chapter. The

brushless DC motor has numerous significant advantages, and

is used commonly in computer cooling fans (Figure 6.1.2). The

cylindrical permanent magnets in the rotor have alternating

poles around the circumference. Small electronics switch the

direction of the current in the stator coils. The magnetic field

from the stator coils rotates and this causes the rotor to rotate.



S



N

S

N



Figure 6.1.2 The brushless DC motor of a computer cooling fan



Now that we have seen the basic structure of a DC motor, we need to turn our

attention to understanding and explaining its operation. In section 4.3 we

encountered the motor effect, but now we will apply it to rotating a currentcarrying coil. Figure 6.1.1 shows that motors contain many coils of wire. For

simplicity we will begin by looking at a single coil and how it would act as a motor.



Let’s torque about rotating coils

A torque is the turning effect (or turning moment) of a force. The force

that causes an electric motor to turn is the motor effect, so we must see how to

calculate a torque before we can begin to perform calculations for motors.

The idea of torque is illustrated in Figure 6.1.3a, which shows a force F due to

a person sitting on a see-saw. In this case the force is acting perpendicular to the

The magnitude (τ) of the torque involved is calculated using

see-saw.

the equation:

τ = Fd



Define torque as the turning

moment of a force using: 

τ = Fd



115



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Motors: magnetic

fields make the

world go around



a



pivot



F



d



where torque in newton metres (Nm) is given by the distance d  from the pivot

If the force is not acting

in metres multiplied by the force F in newtons.

at right angles to the see-saw (Figure 6.1.3b), it is the perpendicular component

F⊥ of the force that is used. As F⊥ = F cos θ, the formula for torque becomes:

τ = F⊥d = F cos θd



b

d



F θ



F⊥



F

cos θ = ⊥

F



θ



Notice that the angle θ between F⊥ and F is the same as the angle θ the

see-saw makes with the horizontal. We will use this fact again soon. Note that

you could also use τ = Fd⊥.



Worked example

QUESTION



Figure 6.1.3 A force acting on the end of

a see-saw exerts a torque.

(a) Force at right-angles to

the see-saw and (b) force

angled to the see-saw.



Identify that the motor effect is

due to the force acting on a

current-carrying conductor in

a magnetic field.



a



axis of rotation



B



Y



X



FYZ



FWX



l



Z



W



Calculate the torque exerted on the see-saw in Figure 6.1.3b if a force of 980 N was acting

at a distance of 4.0 m from the pivot at an angle θ of 30°.



SOLUTION



τ = F⊥d = F cos θd





= 980 × cos 30 × 4.0







= 3390 Nm



Quantifying torque on a coil

When a current-carrying loop (or coil) is placed in a magnetic field, it is the

force we know as the motor effect that applies a torque on the coil. Figure 6.1.4a

If we know the

shows a current-carrying loop within a magnetic field.

direction of the current in the coil, we can use the right-hand palm rule to give

us the direction of the forces on each side of the coil. Recall from our previous

study of the motor effect that charges moving parallel to the magnetic field do

not experience a force. This explains why only the forces FWX and FYZ act on the

coil between points W and X, and Y and Z respectively, when it is in the position

shown in Figure 6.1.4a. During the coil’s rotation, forces are experienced by sides

XY and WZ, but they essentially cancel each other out and the coil is not free

to move in these directions. Now let’s apply our new understanding of torque to

this coil.

The forces FWX and FYZ apply a torque to the coil, and at the moment

shown in Figure 6.1.4a the magnitude of each torque (τ) is given by:



A



τ = F ⊥d



b

FWX



θ F⊥



WX



d



rotation



axis of rotation



θ

d



YZ



FYZ



Figure 6.1.4 (a) The current-carrying coil is

able to rotate about the axis of

rotation (dashed line). (b) The

same coil a short time later has

rotated slightly, as seen from

point A along the axis of rotation.

116



We will use the wire between points W and X as our example and consider

the situation in Figure 6.1.4b in which the coil has rotated. We must now

account for the fact that force FWX is not acting perpendicular to the coil, so we

use the formula seen previously:

τ = F⊥d = FWX cos θd

As the force FWX is the motor effect, we recall the equation:

FWX = BIl sin θ

Note that θ in this equation is the angle we met in section 4.3. It is the

angle of a current-carrying wire to the magnetic field through which it passes.

This is not the same angle as in the cos θ term of the previous equation. Since the



motors and

generators

two sides of the coil that experience a force (wx and yz in Figure 6.1.4) are

always perpendicular to the field, θ = 90°. This makes sin θ = 1 and therefore:

FWX = BIl

Combining this equation with τ = FWX cos θd (using the other θ!) gives:

τ = BIl cos θd

where l is the length W to X (WX) and d is half the length of X to Y (XY/2).

Now we consider the whole coil. The two forces FWX and FYZ are equal and

both make the coil rotate in the same direction. The total torque on the coil is:

τtotal = 2BIl cos θd

The product 2ld is the area of the coil A. This gives:

τ = BIA cos θ

Motors have more than one coil, so we make one final adjustment to this

expression, giving:

τ = nBIA cos θ

where n is the number of turns in the coil.

This formula quantifies the

torque (turning effect) on a rotating coil in newton metres (Nm). It is important

to note that the angle θ in this equation is the angle of the coil relative to the

magnetic field, as shown in Figure 6.1.4b.



Solve problems and analyse

information about simple

motors using:  

τ = n BIA cos θ



Worked example

QUESTION



axis of

rotation



The coil in Figure 6.1.5 contains 50 turns and is carrying a 6.0 × 10–2 A current in a

5.0 × 10–3 T magnetic field. Calculate the torque on the coil if the coil has rotated 30°

relative to the field.



= 4.5 × 10–5 Nm



S



cm



θ



I



N



cm







5



6



SOLUTION

Using τ = n BIA cos θ and converting the lengths to metres:

τ = 50 × 5.0 × 10–3 × 6.0 x 10–2 × (0.06 × 0.05) × cos 30



F

B



F



Figure 6.1.5



Operation of a simple DC motor

Let’s follow the rotation of a coil within a simple DC motor to understand how

motors work. We will analyse the situation shown in Figure 6.1.7. In Figure

6.1.7a the coil of a DC motor is connected to a battery via a split-ring

commutator at A. At the moment shown in Figure 6.1.7a, the two curved metal

contacts of the commutator split ring direct the conventional current around

the coil through points W, X, Y and Z in turn. Figure 6.1.7b shows the coil in

Figure 6.1.7a as seen from the point A along the axis of rotation. The coil is

seen end-on, and we can see the direction in which the current flows along sides

These sides experience a force F that imposes a torque τ

WX and YZ.

on the coil, causing it to turn. The magnitude of the torque on the coil at

positions 1–5 is shown on the graph in Figure 6.1.7c. Let’s now follow this

coil through half a rotation and see what happens to the current and torque

on the coil.

Recall that the torque τ on the coil in Figure 6.1.7 is given by the equation:



PRACTICAL

EXPERIENCES

Activity 6.2



Activity Manual, Page

48



Describe the forces

experienced by a currentcarrying loop in a magnetic

field and describe the net

result of the forces.



τ = nBIA cos θ

117



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Motors: magnetic

fields make the

world go around

In this example the values of all the variables on the right-hand side of the

The variation in the cos θ

equation remain constant, except for the angle θ.

term accounts for the variation in the magnitude of the perpendicular force F⊥

experienced by the sides WX and YZ. This force determines the amount of torque

acting on the coil, and therefore the torque varies accordingly as the coil rotates.

• At position 1 in Figure 6.1.7b, the perpendicular force F⊥ on the coil is at a

maximum and therefore the torque is at a maximum. This occurs when θ is

zero and cos θ is 1.

• At position 2, the coil has rotated 45° with respect to the magnetic field. The

magnitude of the perpendicular force (F⊥) acting on the coil is less than at

position 1. This means that the torque acting on the coil has reduced. At this

point, cos θ is approximately 0.7. Note that real motors have curved magnets

in the stator to ensure that each coil maintains θ = 0 as long as possible. This

provides maximum torque for the longest time possible during each rotation.

In the example in Figure 6.1.7, the magnets in the stator are not curved, so

only at positions 1 and 5 is θ = 0.

• At position 3, the perpendicular component of the force F⊥ has dropped to

zero. Consequently, there is no torque acting on the coil at this instant. Here θ

is 90° and cos θ is 0, so torque is zero. At this point the brushes and split ring

within the commutator have broken contact momentarily and no current

flows in the coil. The current needs to be reversed in the coil at this point, so

that the motor continues to turn for the next half rotation.

• Between position 3 and 4, the commutator has reconnected the coil and

reversed the direction of the current flowing in the coil (now flowing through

points Z, Y, X and W in turn). The opposite sides of the split-ring

commutator are now connected to the terminals of the battery. If the current

had remained in the original direction once the motor rotated past point 3,

the forces on the coil would have been in the opposite direction to the initial

rotation. This would stop further rotation.

• As the motor moves through position 4, the perpendicular component of the

force F⊥ is increasing. This means the torque is increasing and will reach a



Try this!

Model a simple motor



T



o get a better understanding of

the forces on a coil as it rotates,

make a model of a simple motor.

You need some coat-hanger wire,

a few bamboo skewers, Blu-Tack,

two small magnets, a cork and

paper. Construct a coil as shown in

Figure 6.1.6. Here the green

skewer indicates the magnetic

field. The red skewer points on the

magnets represent the direction of

the force due to the motor effect.

Work with a friend and model the

forces on a coil as it rotates.



Figure 6.1.6 A simple model of

an electric motor

b

axis of

rotation



a

X

F



Y



1



I



F



YZ



WX



F



F



B



S



c

Position



Z



2



F



N



θ = 0°



F⊥



WX



W



Torque

0 max.



F



F⊥



θ



YZ

F



θ = 45°



WX



+ –



3



A



Figure 6.1.7 (a) A simplified model of a DC motor.

(b) The coil as seen from point A in part a,

showing forces and current over a half

rotation. (c) Torque on the coil over a

half rotation

118



θ = 90°



YZ

F



4



F⊥

YZ

F



5



YZ



F



WX

F



F⊥



θ = 45°



WX

F



θ = 90°



Time



motors and

generators

maximum again at position 5. At position 5 the coil has performed half a

rotation and the pattern we have seen from positions 1 to 5 will repeat until

the coil is back to its original orientation at position 1.



PRACTICAL

EXPERIENCES

Activity 6.1



PHYSICS FEATURE



Activity Manual, Page

42



Galvanometers



T



he galvanometer was developed in the 1800s to measure the

relative strength and direction of electrical currents. Its

descendant is the analogue ammeter (Figure 6.1.8), which is

calibrated in units of amperes and its basic structure is shown in

Figure 6.1.9. In an ammeter, the coil is connected in parallel to a

low-resistance wire within the ammeter. When a current is passed

into the ammeter, only a small amount of the total current flows

through the coil and this is proportional to the total current.

When a current passes through the coil shown in Figure 6.1.9,

it experiences a force due to the motor effect. This force exerts a

torque on the coil, causing it to rotate around the pivot. A spring

provides resistance to the torque, and the needle comes to rest

when the torque on the coil equals the torque from the spring. The

torque on the coil is proportional to the current passing through

the coil. Therefore the amount the coil and needle move indicates

the size of the current.

To ensure that the needle moves (or deflects) by the same

amount with each ampere of current being measured, the magnets

are curved around the coil and iron core. This ensures that the

magnetic field is always perpendicular to the flow of current along

the sides of the coil. A uniform maximum torque will then be

experienced by the coil through its whole range of movement.

This means that the scale you read on the ammeter can be uniform

(i.e. the scale has the same-sized divisions throughout).









Identify data sources, gather and process

information to qualitatively describe the

application of the motor effect in:

• the galvanometer

• the loudspeaker.



Figure 6.1.8 A ‘moving coil’ ammeter



pointer



force



permanent magnet



spring



coil



N



pivot



moving

coil



S



soft iron

core

magnetic field



Figure 6.1.9 Cross-section of a ‘moving coil’

galvanometer



Checkpoint 6.1

1

2

3



Explain how each part of a DC motor contributes to its operation.

Define torque and explain how torque varies during the rotation of a DC motor.

Compare the features of a DC motor and a galvanometer.

119



6



Motors: magnetic

fields make the

world go around

B



C



D



Current (A)



A



motor current

0

0



Time (ms)



Figure 6.2.1 Graph of net current for a DC

motor averaged over many cycles



Account for Lenz’s Law in

terms of conservation of energy

and relate it to the production

of back emf in motors.



6.2 Back emf and DC electric motors

We now know that when a current is applied to the coil of a motor, the coil

experiences a force. This force exerts a torque on the coil and the rotor begins

to rotate. If we continue to apply the same current to the coil, the net force on

the coil continues increasing the motor’s speed (recall Newton’s second law).

This may make you wonder why a motor doesn’t just keep speeding up. We

know from experience that they don’t because our toys with motors don’t

accelerate forever.

The main reason DC motors reach a maximum operating speed is that a

Back emf is a

back emf is generated in the coil as the motor rotates.

potential difference across the terminals of the motor created by the changing

magnetic flux passing through the wire coils within the motor (see section 5.1).

Look back at Figure 6.1.7 for a moment. At position 1 the magnetic flux

ΦB through the coil is zero. At position 2 in Figure 6.1.7b, the magnetic field

is pointing right to left through the coil and the magnetic flux is increasing.

This tells us that there should be an emf induced that would produce a

magnetic field pointing from left to right to reduce the increasing magnetic

flux. This field would be produced by a current flowing through Z, Y, X and

W in turn. We can see from Figure 6.1.7a that this would be a current that

opposes the one generated by the supply emf (through W, X, Y and Z in

turn). The result of this would be that the net current in the coil would be less

than the supply emf could generate. Another way to think about it is to recall

the equation for Faraday’s law:

ε = n(∆ΦB/∆t)



Try this!

a back emf

Connect a DC motor to a

battery and place an ammeter

in the circuit to measure the

current flowing through the

motor. Predict how the amount

of current flowing will change

if you apply a significant load

on the motor. Try it! How did

you go?



Explain that, in electric

motors, back emf opposes

the supply emf.



120



We can see that this changing flux ∆ΦB over time ∆t would generate a

potential difference or emf ε.

This potential difference (back emf ) is in the opposite direction to

the applied potential difference (supply emf) that causes the rotor to turn. As

the speed of the motor increases the back emf increases, as ∆ΦB/∆t increases.

Eventually this potential difference cancels out most of the applied potential

difference and virtually no current flows through the coil. At this point there

is no net torque acting on the rotor and it turns at a constant speed.

Let us analyse this situation using Figure 6.2.1. At time A on the graph,

the motor is connected to a DC power source. The applied potential

difference causes a current (blue line) to flow in the motor’s coil and this

builds quickly to its maximum value. Once the coil starts turning, a back emf

is generated due to the changing flux within the coil. This back emf is in the

opposite direction to the applied potential difference and therefore reduces

the net potential difference, which in turn reduces the current flowing in the

coil as the speed of rotation increases.

At time B, the motor has reached its maximum rotational speed. Here

most of the supply emf has been cancelled out by the back emf. If the motor

has no load attached to it, only a small current continues to flow. This residual

current is required to overcome any friction within the motor and any voltage

drop due to losses such as resistive heating. There is no net torque on the coil

between times B and C and the motor operates at a constant speed.



motors and

generators

At time C, a large load is applied to the motor, such as the motor turning

a wheel to move a toy car. The motor slows down quickly under this load and

the amount of back emf is reduced. This means that the applied potential

difference is greater and therefore a larger current flows through the coil.

At time D, a larger current continues to flow through the coils. If the

motor is not designed to handle the resistive heating produced by this larger

current, the motor may ‘burn out’. Burn out occurs when insulation melts at

high temperature, and may cause other components to melt. The motor will

then cease to operate efficiently.



Back emf

measures

motor speed



I



Checkpoint 6.2

1

2

3



Define the term back emf.

Describe the relationship between back emf and supply emf during

the operation of a motor.

Analyse the production of back emf in terms of Lenz’s law.



n some motors back emf is

used to measure their speed.

The input voltage is turned off

for a short amount of time and

the back emf is measured. Since

the back emf is proportional to

the speed of the motor, a

calibration can be applied and

the speed can be calculated.



6.3 Alternating current electric

motors

Many large appliances in your home contain motors that use single-phase

240 V alternating current (AC). Single-phase AC was illustrated in Figure 4.1.2

as a blue curve, showing a current changing direction many times a second. To

connect an appliance to the mains power, you insert a plug (Figure 6.3.1) into

the wall socket. This allows single-phase AC to be supplied by the active pin and

the circuit to be completed by the neutral pin.

Some industrial motors require more power and torque than can be

supplied by single-phase AC. These motors are typically supplied with threephase 415 V AC by a plug with four or five pins. Figure 6.3.2 shows the AC

signal available for three-phase equipment. This involves three alternating

currents that can be applied to different coils within a motor at any one time.



active



neutral



earth



Current



+

0



Time



–



Figure 6.3.1 A typical plug for an electrical

appliance in Australia



Figure 6.3.2 Each of the three AC in

three-phase power is 120°

out of phase with the others.



121