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HSC PHYSICS ENRICHMENT
MOTORS AND GENRATORS
WORKBOOK - ANSWERS
1
C
V
E
d
plates must be conductors
2
E
m = 10-30 kg
q = +6×10-12 C
F  ma
a = 7.0×1021 m.s-2.
F  Eq
30
21
F m a 10  7  10 
E 

N.C-1  1.2  103 N.C-1
12
q
q
 6  10 
3
B = B A cos (B, A constant)
B
(0,0)
t
initial magnetic
flux is zero
I
4
v
B
right hand screw rule:
magnetic field into
page
University of Sydney
F
eright hand palm rule:
motion of electron is
directly towards the wire
Centre of continuing Education
1
5
Force on a conductor in a magnetic field
F  B I L sin    90o
F  B L I  (slope of line) I  m I
m 0.7 / 0.3
B 
T  1.1 T
L
2.2
m  BL
6
Right hand palm rule – initial force on wire is into page and because of symmetry – the
wire will rotate.
7 C
B
v
v
v  v sin 

B

v  v cos 
F  B q v sin   B q v
8
B = 0.55 T
 = 90o
I=3A
L = (0.42 + 0.42)1/2 m = 0.5657 m
F
F  B I L  (0.55)(3)(0.5657) N  0.93 N
B
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9
m = 2.456 g = 2.45610-3 kg L = 120 mm = 0.12 m
balance reading F = 4.89610-3 N B = ? T
I = 0.451 A
g = 9.80 m.s-2
Balance reading (up) = weight (down) + magnetic force (down – right hand rule)
F mg BI L
B
10
F  m g 4.896  103  (2.456  103 )

T  0.0451 T
IL
 0.451 0.12
right hand palm rule  electron travels in a circular orbit
v = 2.0107 m.s-1 |q| = 1.60210-19 C m = 9.1110-31 kg
R = ? m T = ? s f = ? Hz
Magnetic force = centripetal force
m v2
mv
Bqv 
R
 1.38  104
R
Bq
circumference
f 
2 R  v T
T
m
2 R
 4.34  1011 s
v
1
 2.31  1010 Hz
T
11
(A)
(B)
Wires carrying current in opposite directions  repulsive force between wires 
top wire will jump upwards.
L = 1.00 m m = 0.01 kg d = 100 mm = 0.100 m g = 9.8 m.s-2
I1 = I2 = I = ? A

k o
o = 4   10-7 T.m.A-1
k = 210-7 N.A-2
2
F = mg current must be sufficient to lift the tube
F
I I
k 1 2
L
d
(C)
I
Fd
mgd

 221 A
kL
kL
very large current
Transmission lines close together could experience large forces of attraction or
repulsion.
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12
I1 = 80 A I2 = ? A d = 200 mm = 0.200 m m/L = 0.12 g.m-1 = 0.1210-3 kg.m-1

k o
k = 210-7 N.A-2
o = 4   10-7 T.m.A-1
2
Force between current carrying conductors
F
I I
mg
k 1 2 
 m / L g
L
d
L
13
(A)
(B)
I2 
m / L g d
k I1
 14.7 A
Three wires carrying a current in the same direction will attract each other.
Three wires one of which carrying a current in an opposite direction, can not
attract or repel each other – there must be attraction and repulsion.
14
Wires carrying a current in opposite directions repel – in this case in different parts of the
wire have currents in opposite directions  all parts of the wire repel other parts  wire
will form a circle.
15 C
Use right hand palm rule.
16
n = 15
 = 0.15 N.m
I = 0.5 A
L = 100 mm = 0.1 m
torque on a current carrying coil
  n B I A cos   n B I L2
B
A = L2
=0
B=?T
force into page

2
nI L
B
 2.0 T
I
I
force out of
page
17
Torque on a coil is
  n B I A cos 
Turning effect can be increased by increasing the values of
n number of windings in coil
B strength of magnetic field
I current through coil
More coils at different orientations, iron core for coil, radial magnetic field
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18
Torque on a coil is
  n B I A cos 
When the coil is in the position shown  = 90o cos90o = 0
Net torque on coil is zero – coil will not turn
19
Current through the coil experience a torque due to the magnetic field. A spring applies
an opposing torque on the coil as in is compressed. Torques cancel out at a particular
deflection of the pointer. If the current through the coil is increased, needle has a greater
deflection due to increased torque due to interaction of current with magnetic field and
spring compressed more also gives greater torque at the greater deflection of the pointer.
20
Force on a current element in a magnetic field
F  B I L sin 
 angle between magnetic field and current element
F
F
force directed out of page due
to component of current at
right to magnetic field
diagram X
diagram Y
FX = B I L
magnitude of F is smaller than in diagram X
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FY = B I L sin < B I L
5
21
(A)
As current increased – balance
reading decreases – attractive force
between wires – currents must be in
same direction.
0.549
data 1
linear
y = - 5.8e-005*x + 0.55
0.5485
balance reading (kg)
0.548
(B)
(C)
mass of copper m = ? kg
0.5475
0.547
When current is zero, balance
reading is equal to equal to mass of
copper
m = 0.5487 kg
0.5465
0.546
0
5
10
15
20
current (A)
25
30
35
40
(D)
o
o = 4   10-7 T.m.A-1
2
When I1 = 0 A mbalance = 0.55 kg
I1 = 5.6 A mbalance = 0.5485 kg
L = 2.6 m
I2 = 50 A
k
magnetic force F = (0.5487 – 0.5485)g
k = 210-7 N.A-2
g = 9.8 m.s-2
Force between current carrying conductors
F
I I
k 1 2
L
d
d
k I1 I 2 L
 0.07 m
F
22
(A)
Right hand palm rule – coil would rotate clockwise – prevent rotation – weight
added at X
(B)
torque at X
X = m g d
m = 0.15 kg
X = 0.163 N.m
(C)
g = 9.8 m.s-2
d = (0.222 / 2) m = 0.111 m
motor stopped: torque at X = torque on coil in magnetic field
n=?
B = 0.1 T
 X   coil  n B I A
I=1A
n
X
A = 0.01 m2
BI A
 163
23
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n = 100 turns A = 0.05 m2
B1 = 5.0 T B2 = 0 T
induced emf  = ? V
d
d ( B A)
B
  N B  N
 N A
 100 V
dt
dt
t
24 B
emf  
B = 5.0 T
t = 0.25 s.
B
dB
dt

A
magneticflux B  B A cos 
thickness not so important
25 A
Lenz’s Law – an induced emf is always in a
direction that opposes the original change in
magnetic flux that caused it. Rotation clockwise
reduces flux – induced emf produces a current to
increase magnetic field to maintain the original
flux – induced B in same direction as original B
– right hand screw rule  induced current
clockwise as in A
26
  N
dB
dt
emf
time
27
copper with split R and plastic P fall faster than the copper ring R.
Copper ring – induced emf – induced currents – Lenz’s Law – motion opposed – ring
falls slower.
28
Falling magnet – changing magnetic flux in copper block – induced emf – induced
currents – Lenz’s Law – opposed motion – falling motion of magnetic opposed.
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If block cooled slowing motion greater – at lower temperature resistance of copper
smaller – large induced currents – large force to oppose motion.
29
First investigation – induced emf setup but no induced currents (no complete circuit) –
solenoid stays stationary.
Second investigation – induced emf – induced currents (complete circuit) – induced
currents set up a north pole at the end near the approaching magnet to oppose its motion –
the north poles will repel – solenoid moves away from magnet.
30
low motor speed
* small back emf
* small induced currents
* current from battery only slightly reduced
current
motor speed
large motor speed
* large back emf
* large induced currents
* current from battery
greatly reduced
31
*
*
*
*
*
*
Back emf opposes the supply emf .
Motor turns - torque due to the force on a conducting wire in magnetic field –
motor does not spin faster and faster – why?
Armature turns - magnetic flux through coil changes – induced emf in the coil –
this back emf opposes the motion (Lenz’s Law).
The greater the speed of the motor the greater the back emf – speed of motor
increases until back emf equals supply emf.
The greater the mechanical load, the slower the motor rotates – lower the induced
emf – the greater the current through the motor windings (coil) – if motor stops –
zero induced emf – max current through motor windings – motor heats up and
windings maybe damaged.
Need large resistance in windings at start to reduce the otherwise large current.
32
n = 500 L1 = 0.2 m
max = ? N.n
L2 = 0.1 m
A = L1 L2 = 0.02 m2
B = 0.001 T
I=4A
  n B I A cos 
 max  n B I A  0.040 N.m
Max torque occurs when plane of coil is parallel to the magnetic field lines.
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33
When motor is running, a back emf is induced, reducing the current. The faster the motor
spins the greater the induced emf and the greater the reduction in current. When the
motor is prevented from rotating, the induced emf = 0 and the max current flows through
motor.
34
The conductor (wheel) is moving through a magnetic field – induced emf in metal wheel
– induced currents (eddy currents) – produce a magnetic field to oppose the rotation of
the wheel. The closer the magnets are to the wheel the greater the opposition to the
motion. Therefore, the levels can be changed by varying the spacing between the wheel
and magnets.
35
Replace the coil with a metal strip – induced currents in the metal strip - forces to oppose
the motion – pendulum will stop move quickly.
36
Electromagnetic induction – creation of an emf in a conductor when there is relative
motion between the conductor and magnetic field or a changing magnetic flux.
Expt: moved a magnet through a coil attached to a sensitive galvanometer – observed a
deflection of the galvanometer’s needle indicating an induced emf which produced an
electrical current.
Applications of induced currents: induction cooktops – changing magnetic field in
cooktop induces eddy currents in the saucepan placed on the cooktop – eddy currents
have a heating effect.
Observation: construction of an induction motor – conductor carrying an electric current
in a magnetic field experiences a force.
Application of induced currents and force on a conductor in a magnetic field – induction
motor is an ac machine in which torque is produced by the interaction of a rotating
magnetic field produced by the stator and currents induced in rotor.
The information gathered about induction was reliable because the physical principles do
account for the operation of an ac induction motor and an induction cooktop.
The best responses made a judgment about the reliability of information gathered both from first-hand
data and/or second-hand data. Many candidates understood the general principles of assessing
reliability of information but had difficulty relating these principles to the identified applications of
induction.
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37
coil of wire
v
S
N
magnet
current
meter
38
Lenz’s Law (19th century):
The direction of the induced current (generated by changing magnetic flux) or induced
emf is such that it produces a magnetic field that opposes the original change in the
magnetic flux .
eB
force on
“+q”
electrons:
electrons will
move the this
end X
Electrons are free to move in a conductor – when the conductor is moved upward through
the magnetic field the free electrons experience a force out of the paper and hence
accumulate at point X – an emf is developed across the ends of the conductor with X(-)
and Y(+). If a compete circuit is setup, the induce emf will cause a current to flow. A
current through a resistance gives a heating effect.
39
Direction of induced current determined by Lenz’s Law – induced current opposes the
change in magnetic flux
clockwise current
attraction
force
anti-clockwise current
repulsion
switch
closed
switch
opened
40
Apply right hand palm rule to the motion of an electron in the wires at P and Q.
For P e- moving up the page ; +q moving down the page; magnetic field across the page
to the right  force on electron out of page  current into page
For Q current out of page.
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41
LCD = LEF = 12 mm = 1210-3 m LDE = LCF = 6.0 mm = 610-3 m
Rotation rate = 20 rpm (revolutions per minute)
Number of turns of coil N = 100
Total resistance of coil Rcoil = 10 
Resistance of external load connected between A and B Rload = 125 
 = 20 rpm = (20)(2) / (60) rad.s-1 = 2.09 rad.s-1
  2 f
f 

 0.333 Hz
2
T  3.00 s
1
2
magnetic flux (T.m )
Rtotal = Rcoil + Rload
B  B A cos( )  B A cos(t )
d
   N B   N B A sin(t )
x 10
-4
0.5
rotation rate 2
0
-0.5
-1
0
30
60
90
120
150
180 210 240
angle (deg)
270
300
330
360
30
60
90
120
150
180 210 240
angle (deg)
270
300
330
360
60
90
120
150
180 210 240
angle (deg)
270
300
330
360
0.04
dt
  N B A sin( )

 N B A sin( )
I

Rtotal
Rtotal
As rotation rate increase 
increase in induced emf 
increase in current 
greater force opposing rotation of
coil  counter torque (solid curve)
emf (V)
0.02
0
-0.02
-0.04
4
0
x 10
-4
I (A)
2
0
-2
-4
0
30
MOTOR
angle of coil  = 0
E
F
I
into
page
coil swings through
angle of coil  = 180o
half-turn
E
D
D
F
F
out of
page
F
out of
page
C
B-
C
B
F
into
page
I
F
B-
A+
A+
coil can not spin freely – every turn half-turn direction of toque changes direction
For the motor to spin freely – slip ring commutator replaced by split ring
commutator – the polarity to connections A and B reversed every half-turn –
torque does not change direction.
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42
Transformers can be used to transform AC voltages up or down. By transmitting at high
voltages, energy losses to heating of the wires can be reduced. The voltage from the
transmission lines can be step-down for use in homes.
43
Edison’s tactics were often less than honest – he lobbied politicians to pass laws
prohibiting the use of AC for voltages > 250 V – conducted experiments electrocuting
hundreds of animals to warn the public about the dangers of AC – agitated against the
introduction of AC – involved in the designs of an AC electric chair to show how
potentially dangerous AC was - he seemed to mistrust AC.
44
Chicago World Fair – May 1893 – 100 00s lights – 27 M people.
Niagara Falls – 1891 – power form the falls – AC generators developed – transmission
over long distances.
Publicly visible projects – people could see their success – demonstrated the efficiencies
of AC power transmission.
The final outcome did not depend on these projects – the deciding factor in the debate
was the use of transformers so that electrical energy could be transmitted over long
distances without a significant loss of power.
DC cannot be transformed easily. Any distribution system using DC would be more
expensive than AC: more generators required, different generators for different voltages,
separate supply lines for different voltage, higher transmission costs, difficult to transmit
over long distances.
45
AC generator
direction of
rotation
ac generator
split ring
single
DC commutator
generator and winding
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46
right hand palm rule
electrons move toward centre of spinning disk
current through globe is from Y to X
47
B  B A cos( )  B A cos(t )
d
   N B   N B A sin(t )
dt
The generators rotates twice as fast – emf has double the amplitude and frequency.
48
(A)
(B)
(C)
Brush – connects rotating coil with external circuit.
Generator P – slips rings – AC generator
Generator Q – split rings – DC generator
AC generators – transformers – step up or step down voltages
49
N = 500
A = 0.125 m2
f = 50 Hz
B = 1.5 T
max = ? V
T=?s
B  B A cos( )  B A cos(t )
d
   N B   N B A sin(t )  2 f N B A sin(2 f t )
dt
T
1
 0.02 s
f
3
x 10
 max  2 f N B A  3.0  104 V
4
2
DC
emf (V)
1
0
-1
-2
-3
0
University of Sydney
0.005
0.01
0.015
0.02 0.025
time (s)
Centre of continuing Education
0.03
0.035
0.04
13
50
max = 90 V
f = 50 Hz A= 200 cm2 = 20010-4 m2
max(N=1) = ? V
N=?
 max  2 f N B A
51
B = 2.39 T
max(N=1) = 15 V
N=6
B
magnetic flux and induced emf are 90o out of phase
P
Q
B
0
max
emf max
0
R
S
0
max
max
0
T
0
max
52
N S VS 6000


 25
N P VP
240
53
VP N P I S


VS N S I P
step down transformer VS  VP  N S  N P  I S  I P
Large currents can flow in secondary coils of a step transformer – need large diameter
wires to reduce resistance – reduce heating effects (I2R)
54
VS = (24)(1.5) V = 36 V
VP N P

VS N S
NS  NP
University of Sydney
VP = 240 V
NP = 640
NS = ?
VS (640)(36)

 96
VP
240
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55
battery
VP
opening & closing switch
` small
induced emf large
VS
t
close
switch
56
open
switch
B
57
To step up voltage to very large values – better to have transmission at high voltages to
reduced power losses.
58
Many appliances in the home can only operate at low DC voltages eg computer.
59
VS = 12 V
VP = 3.6 V
NP = 60
NS = ?
step up transformer
VP N P

VS N S
NS  NP
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VS (60)(12)

 200
VP
3.6
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60
Pplant = 120 kW = 120103 W
L = 10 km = 104 m R/L = 4.010-5 .m-1 R = 0.40 
P=VI I=V/R
Current draw from plant I = P / V
P = Pplant V1 = 240 V V2 = 24 kW = 24103 W
I1 = P = / V1 = 500 A
I2 = P / V2 = 5 A
Power losses in transmission lines Ploss = I2R
Ploss1 = 105 W
Ploss2 = 10 W
% Plosses = 100  Ploss / Pplant
% Plosses1 = 83 %
%Plosses2 = 0.01
61
(A) step down transformer: voltage transmission line = (30)(11000) V = 330 000 V
(B) power losses: current passing through resistance P = I 2 R
P = power loss, I = current carried by transmission line
R = resistance of transmission lines
power losses: eddy currents in transformers – thermal losses
(C) Superconductors below their critical temperature have zero resistance, R = 0.
P = I 2 R R = 0  P = 0 reduce dramatically thermal power losses
62
(A)
A transmission wires - conductors.
Ceramic insulating stacks – in dry air
static electricity can jump about 10 mm for
every 10 kV – for 330 kV transmission –
sparking could occur if spacing between
wires was less than about 330 mm – they
also protect against lightning strikes
(B)
Electrical energy transmitted at high
voltages – low currents.
(C)
DC was first used.
(D)
Use of transformers to transmit at high
voltages.
(E)
Lightning strike protection
Shielded conductors – two non-current
carrying wires at the very top of the tower
Earth cable – runs from top of tower
down into metal conductors in the ground
Insulation chains – saucer shaped stacks of ceramic material (very good
insulators) that attach wires to the tower.
Metal tower – tower connected to the Earth.
Distance – distance between towers ~ 150 m – enough to protect each tower from
neighbouring towers.
(F)
Ceramic insulating stacks
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63
Some claim that living near high voltage power lines is causing a variety of health
problems – cancers – leukemia – however, no conclusive scientific results to link health
risks and close proximity to high voltage power lines. Is there a danger?
64
Simple form of an induction motor.
Conductor (free electrons) in changing magnetic field due to motion of magnetic –
changing magnetic flux – induced eddy currents – induced magnetic field to oppose the
motion of the magnet – S pole induced at top of disk –S pole of disk attracted to N pole
of magnet – disk spins changing the magnet.
65
AC induction motor; rotor and stator (see notes).
66
AC generator rotated faster – large emf and greater frequency of ac voltage.
67
Westinghouse – arguments in favour of AC generation.
~ 1886
Nikola Tesla – transformer
Transformer – step up and step down voltages
68
Faster the rotation of a generator – greater induced currents – forces on induced currents
due to interaction with magnetic field – oppose rotational motion of generator – Lenz’s
Law.
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69
4.3
balance
data 1
linear
4.2
4.1
N
balance FB (N)
B
S
I
y
F
magnetic field in + x direction
current in – z direction
force on current element – y direction
y = 0.2*x + 3.7
4
3.9
3.8
x
z
3.7
3.6
0
0.5
1
1.5
current I (A)
2
2.5
Length of current element L = 6510-3 m
Straight line fits the data
FB = 0.20 I + 3.70
I=0
slope = 0.20 N.A-1
FB = 3.7 N (weight of coil)
intercept = 0.370 N
mcoil = (3.7/9.8) kg = 0.38 kg
Magnetic force on current element F = BL I
slope = BL
I = 5.0 A
B = slope / L = 3.1 T
FB = 4.70 N
I = -2.0 A current in opposite direction – magnetic force on conductor up reducing
balance reading
Fmagnetic = BIL = (3.1)(2)( 6510-3) N = 0.40 N
balance reading = (3.70 – 0.40) N = 3.30 N
If the magnetic field strength was increases by a factor of two – the magnetic force
increases by a factor of two – the straight line would have twice the slope.
Increasing the number of turns by N = 4000, the magnetic force would increase by a
factor N therefore the magnetic field strength would have to decrease by a factor of N
Bnew = Boriginal / N = (3.1/ 4000) T = 7.810-4 T
70
All answers are correct - a magnetic field exists around any wire carrying a current.
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71
B,(B,
A constant
BB =
cos
A constant) [T.m2]
 BBAAcos

MAGNETIC
Magnetic flux FLUX
[T.A2]
B
A

B
B
B
 = 0 cos = 1 B = B A
B
B
B
 = 90 cos = 0 B = 0
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MAGNETIC FLUX
All answers are correct
B = B A cos (B, A constant) [T.m2]
73
B  B A cos( )  B A cos(t )
d
   N B   N B A sin(t )  2 f N B A sin(2 f t )
dt
All answers are correct
74
right hand palm rule
magnetic field at N pole directed down into the ground (into page)
electrons moving in direction of plane – current in opposite direction (down the
page)
force on electrons directed form P to Q (across the page from left to right)
P positive Q negative R neutral S neutral
75
1 to 3 are correct but not 4 – high resistance wire would limit the current - reduce the
force in turning the coil in the magnetic field F = B I L
76
Right hand palm rule
top surface of magnet is N pole – magnetic field point up out the top surface
current is from X to Y
could also have magnetic fields and currents reversed.
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77
Lenz’s Law – induced currents in moving train oppose its motion
Loss of KE in moving train – energy associated with induced currents – energy
dissipated as thermal energy (I2R) - Conservation of energy.
78 D
Picture - train braking system – current through coil – magnetic field around coil –
induced currents in track oppose motion of train.
Conservation of energy - loss of KE in moving train – energy associated with induced
currents – energy dissipated as thermal energy (I2R).
I
79
L
B

F = B I L sin
I = 5.20 A
B = 0.516 T
L = 0.745
 = 60.6o
F = 1.74 N
right hand palm rule
F into page
Lsin
80
Major impacts AC generators and transformers
Transmission of high voltage and low currents over large distances.
Step up and step down voltages – many appliances use different voltages.
Lifestyle changes (“we now live in an electric world”)
Environment – coal powered electrical generators – green house gases.
81
rotation direction
rotation direction
I
+
-
DC generator
- energy input to rotate coil
- induced emf
- induced current
magnetic flux decreasing
induced current to increase flux
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I
+
DC motor
- energy input from battery
- current in magnetic field  force
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82
(A)
(B)
(C)
transformers – step up & step down voltages
Eddy currents – thermal energy losses – I2R
IP = 10.0 A VP = 1.10104 V IS = 185 A VS = 240 V
P=VI
PP = VP IP = 1.1105 W
PS = VS IS = 4.4103 W
4
Ploss = PR – PS = 6.5610 W
Eloss = Ploss t = 1.88109 J
t = 8 h = 2.88104 s
(D)
VP = 1.10104 V VS = 240 V
VP / VS = NP / NS
NP = 1.2104
NS = ?
NS = NP (VS / VP) = 2.62102
83
(A)
attraction – wires in same direction
(B) (C)
F
I I
k 1 2
L
d
F
F
force increases as
distance decreases
force decreases as
length decreases
d
L
(D)
Passes a current through a wire – observed the deflection.
84
F
I I
k 1 2
L
d
Wires in opposite directions – repulsion
k = 210-7 N.A-2
I1 = I2 = 2.23 A
L = 35510-3 m
d = 1810-3 m F = ? N
F = 2.010-5 N
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85
d = 5.75010-3 m V = 1222 V
q = e = 1.60210-19 C
v = 3.30106 m.s-1
(A) E = V / d = 2.13105 V.m-1
(B) FE = E q = 3.4110-14 N
(C) FB = FE = B q v
B = FE / q v = 6.410-2 T
(D)
+
E
FE
B
-
v
FB
right hand palm rule:
magnetic field into page
86
DC generator
direction of
rotation
single commutator and winding
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87
v = 250 m.s-1
FE = F
E = 200 V.m-1
+
B=?T
Eq = Bqv
FB
B
E
+
v
FE
B = E / v = 0.80 T
-
right hand palm rule:
electric field down the page
– top plate positive
88
Conservation of energy: energy input = energy output
power input (primary) = power output (secondary)
Changing magnetic flux in primary = Changing magnetic flux in secondary
d
 d 
VP  N P  B   N P B
dt
 dt  P
VP N P

VS N S
Step down transformer VP > VS
d
 d 
VS  N S  B   N S B
dt
 dt  S
IP NS

IS NP
PP  PS
NP > NS
IS > IP
89
Wires from coil not connected to split ring commutator.
Magnets too far from coils.
Zero net torque would be applied to the two coil windings.
90
B  B A

   B  B L v
I
t
t
X
The induced emf and current are proportional to v.
The rod moves to the right and then to the left at the
same speed – the induced emfs will have opposite
polarities and the currents opposite directions but with
the same magnitudes.
Right hand palm rule
rod moving to right
current –
anticlockwise
rod moving to the left current – clockwise
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A
B
F
v
I
D
G
C
Y
23
91
N
induced
current
Iinduced
N
92
e-
e- circular motion in magnetic field
The magnetic force on e- is always at
right angles to the direction – no work
is done on the e- - no change in its KE
or velocity
93
rotation of loop about axis of rotation
loop:
n turns each carrying current I
area A = L1 L2
B uniform magnetic field
L2
F into page
F out of page
magnetic force F produces a
torque on current loop
I
 net  
Fmagnetic  F  B (nI ) L2  n B I L2
L1
  net  F  L1 / 2  F  L1 / 2  F L1
 net   max  n B I L2 L1  n B I A  n I B
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B
top views of rotating coil:
forces are only on conducting
wires that are at perpendicular to
page
uniform magnetic field

F
Fcos
L1
F
I
F
rotation
axis

I
=0
 = max
F
  F L1 cos 
 = 90o
Fcos
cos(90o) = 0
=0

F  B (n I ) L2
  n B I A cos   n B I A sin 
The forces F does not change in
magnitude or direction as coil rotates.
 angle between magnetic field lines and
conductors length L1
 angle between magnetic field lines and
normal to conducting loop


(A)
Rotation
F
I
N
S
F
brushes
Commutator
split ring
(rotates with coil)
(B)
n = 900
B = 0.48 T
L2 = 40010-3 m
L1 = 16010-3 m
I=?A
F = B (n I) L2 = 2.6103 N
(C)
max = 4.1102 N.m
(D)
 = 30o
(E)
 = 0  = 90o
 = n B I A cos = 3.6102 N.m
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plane of coil at right angles to direction of magnetic field
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94
All currents in the same direction – each wire attracts each other.
 F / L 
 F / L 12
k I1 I 2
d
kI I
 1 2
d12
 F / L 23 
k I2 I3
d 23
 F / L 12  I 2   d 23 

 0.7  1
 F / L 23  I 3   d12 
Forces between wires 2 and 3 greater than between 1 and 2 – wire 2 move towards wire 3.
95
B
FB
magnetic force opposes the motion of the
metal plate – Lenz’s Law
v
induced
currents
Magnetic braking
96
Torque on a current carrying coil in a DC motor is not constant – it varies sinusoidally.
97
Step up or step down voltages.
Vp = 12.0 V IP = 0.045 A
6.3 V IS = 0.25 mA
VS =
P=VI
PP = VP IP = 0.54 W
PS
= VS IS = 0.41 W
Ploss = PR – PS = 0.13 W
Real transformers have insulating
laminations to reduce eddy currents.
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Laminations reduce induced eddy
currents - reduce energy losses due to
Joule heating I2 R
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98
magnet enters
current from BA
magnet exits
current from AB
Lenz’s law
99
AC generator
slip
rings
brushes
terminals to connect load
axle
100
N x4
4
f x3
3.5
B x2
3
emf
2.5
2
1.5
1
0.5
0
0
5
10
time
15
20
101
By using a coil rather than a loop the magnetic flux through the whole circuit is increased
in proportion to the number of turns. Thus the change in flux during each revolution is
increased and so is the magnitude of the induced emf.
102
The rotating coil represents a current carrying conductor moving in a magnetic field. This
conductor experiences a force that always opposes the motion - Lenz’s law. The greater
the speed of rotation the greater the induced current and the greater the opposing torque.
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103
1.5
DC generator – split ring
commutator
Commutator
wires not et twisted
as coils rotate
1
emf
AC generator
transformers to change
voltages
total
output
single coil
outputs
0.5
0
0
5
10
time
15
20
104
Generator – spinning magnet produces a charging flux through coils – induced emf –
induced current.
Simple to construct – no need for commutator and brushes.
Lots of energy wasted in spinning “heavy” magnet.
105
(A)
Deflection on the ammeter reading only on closing and opening the switch –
needle deflected in opposite directions on opening and closing – changing magnetic flux
– induced emf – induced current.
(B)
With AC power supply – magnetic flux various continuously – continuous AC
induced emf - AC current. DC ammeter would not respond to the AC – need an AC
meter.
106
Electron moves in a circular path of radius R.
m = 9.1110-31 kg q = 1.60210-19 C v = 2.0107 m.s-1
R=?m T=?s
B = 0.010 T
From Newton’s Second Law
Net force = Centripetal force = Magnetic force
m a = m v2/R = B q v
R = m v / q B = 1.110-2 m
For one complete orbit: distance traveled = circumference of a circle
2 R = v T R = m v / q B
T = 2 m / q B = 3.610-9 s
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107
Step down transformer
NS < NP VS < VP
IS > IP
AC input – AC magnetic flux through iron core of transformer – AC induced emf – AC
induced currents – I2 R heating of transformer core.
PS < PP because of thermal energy losses in transformer core – real transformers have
insulating laminations to reduce these losses.
108
A
Laminations reduce induced eddy
currents - reduce energy losses due to
Joule heating I2 R
109
VS = 1.5 V
VP = 12 V
NP = 2400
VP N P

VS N S
NS  NP
IP NS

IS NP
IS  IP
110
R/L = 6.2×10-4 .m-1
NS = ?
IP = 10 mA
= IS ? mA
VS
 300
VP
NP
 8 mA
NS
L = 104 m
R = 6.2  I = 50 A P = ? W
P = I 2 R = 1.6×104 W
111
B
112
A
113
D
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114
AC induction motors - avoid problems with wear, electrical arcing (sparking) across split
ring commutators of DC generators.
AC induction motors - Simple & rugged construction; low cost and minimum
maintenance; high reliability; efficient; no extra starting motor.
AC induction motors – torque on wire loops on a rotating armature (rotor) due to
induced currents in an ac magnetic field produced in stationary coils (stator).
Stator – groups of coils together with a steel core form an electromagnet
Rotor – most common type squirrel cage – stacks of steel laminations with evenly spaced
conductor bars around the circumference - induced current from alternating magnetic
field produced by stator – torque on rotor produces rotation from induced currents in
rotor interacting with magnetic field of stator.
115
An AC only motor in which torque is produced by the interaction of a rotating magnetic field
in the stator with induced magnetic fields of the induced current in the rotor.
116
As the magnet is rotated – changing magnetic flux through metal disk – induced currents
– oppose motion of disk – S pole induced at top of disk – disk attracted to magnet – disk
rotates.
This is not an efficient induction motor – lots of energy need to rotate “heavy” magnet.
117
Induced currents in copper sheet oppose the motion – Lenz’s Law.
118
No induced currents – no changing magnetic flux produced.
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