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Transcript
```Physics
for CCEA
A2 Level
Colourpoint
Educational
Rewarding Learning
Pat Carson and Roy White
Contents
Unit 4 (A2 1): Momentum, Thermal Physics, Circular Motion,
Oscillations, Atomic and Nuclear Physics
4.1
Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
4.2
Thermal Physics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4.3
Uniform Circular Motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.4
Simple Harmonic Motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4.5
The Nucleus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.6
Nuclear Decay. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.7
Nuclear Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
4.8
Nuclear Fission. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.9
Nuclear Fusion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
Unit 5 (A2 2): Fields and their Applications
5.1
Force Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.2
Gravitational Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
5.3
Electric Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5.4
Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
5.5
Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
5.6
Deflection of Charged Particles in Electric and Magnetic Fields. . . . . . . . . . 163
5.7
Particle Accelerators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
5.8
Fundamental Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
Unit 6 (A2 3): Practical Techniques
6.1
Measuring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
6.2
Precision, Accuracy and Errors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
6.3
Analysis and Interpretation of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
Unit 7: Synoptic Assessment
7.1
Synoptic Assessment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
3
42a4.4
a = –Harmonic
ω x
Simple
Motion
2
a = – ω2x
42a
Students should be 42b
able to:
€
ω = 2πf
4.4.1 Define simple harmonic motion using
the equation a = 42b
– ω 2 x where ω = 2πf ;
42a
€
4.4.2 Perform calculations
using x = A cos ωt ;
42c
€
4.4.3 Demonstrate an understanding of s.h.m.
measuring velocity
x = Afrom
cos the
ωt
ω =42c
2πf
42b graphs to include
€
€
gradient of a displacement time graph;
42d
x
€
4.4.4 Know and be able to use the terms free vibrations, forced vibrations, resonance and
x
x = 42d
A cos ωt
42c
damping in this context;
€
€
2
4.4.5 Understand 42e
the concepts
ofωlight damping, overdamping and critical damping;
€
42d
x
42e
4.4.6 Describe mechanical examples of resonance
€ and damping; €
42f
€
ω2
F = ma = – mω 2 x
2
ω 42f point will oscillate
42ea €
= –itsmω 2 x
Any object that is initially displaced slightly from
stable equilibrium
€
equilibrium position. It will, in general, experience
a restoring force that is proportional to its
–ω 2
42g
€
displacement from the equilibrium
point.
42f
F = 42g
ma = – mω 2 x–ω 2
€ It consists of a €
A simple pendulum is a system that behaves in this way.
mass, called the bob,
2
attached to a length of42h
string.
–mω
€
Pendulum at rest
Pendulum oscillating
When hanging vertically the
2
2
–ω
42g
–mω
42h
resultant force on the bob is
€
€
zero, the tension in the string is
1
2π
f =
=
€
equal and opposite to 42i
the weight
T 42hω
Tension
1
2π
–mω 2
of the bob. When the pendulum
in string
€
42i €
f =
=
T
ω
bob is moved a little to one side
and released it oscillates
θ = ωt
42j
1
2π
€
Weight of 42i €
backwards and forwards. The
f =
=
the bob
θ = ωt
42j
T
ω
resultant force always acts
This component
of the bob’s
€
weight
provides a restoring
towards the equilibrium position
x
force
towards the equilibrium point.
€
Then
:
cos
θ
=
cos
ωt
=
so the bob will vibrate around
θ =OPωt
42j
this point.
€
Then : cos θ x = cos ωt =
€
O
A
42kωt
But, since
x OP = OA = am
gives : x = A cos
Defining Simple
x
= Rearranging
– ω 2(SHM)
42aHarmonic aMotion
€
Then : cos θ = cos ωt =
OP gives : x = A c
ω = 2πfmotion if
Rearranging
A particle moves42b
with€simple harmonic
42kωt from a fixed
But,point
sinceand
OP = OA = amplitude, A, then : co
1 its acceleration42b
is proportional
to=its
displacement
x2πf
= A cos
42L
ω
€
2 the direction 42c
of the€
acceleration
point.
x =isAalways
cos ωttowards that fixedRearranging
gives : xx == A
A cos
cos ωt
ωt
42L
€
€
The defining equation for simple harmonic motion is:
42m
ω x 0 sin ωt
x = vA =
cos–ωt
42c
2
€
€
x
a
=
–
ω
x
42d
v = – ω x 0 sin ωt
x = 42m
A cos ωt
42L
€
€
€ in ms–2
where a is the acceleration
2
A cos ωt
– ω2x = – ω
x is athe= displacement
42d 42n
in
m
ω = 2πf
€ €
2
–2
€
in s
ω is a constant
42e €
42m
v = 42n
– ω x 0 sin ωt a = – ω 2 x = – ω 2 A cos ω
€
€
2
ω
42e
x = A cos ωt€ €
2
€
49
F = ma = – mω
42f
a = – ω 2 x = – ω 2 A cos ωt
42n x
€
€
€
F = ma = – mω 2 x
42f
x
€
42a 42k
42a
42b
42c
42d
a = –But,
ω 2 xsince OP = OA = amplitude, A, then : cos θ =
42c
€
x = A cos ωt
Physics for ccea a2 level • UNIT 4
42d
x
€
The minus indicates that the acceleration and displacement
are in opposite directions.
Newton’s 2nd law tells that a resultant force F is needed to produce this acceleration.
2
ω force and displacement.
42e € acceleration,
The graphs below illustrate these relationships between
a = – ω2x
42a
a = – ω2x
42a
42f
€
a = – ω2x
42a
42b
Acceleration, a
ω = 2πf
42g
€Gradient is –ω2 +ω2 Α
€
ω = 2πf
42b
€
x = A cos ωt
42c
42h
€
€
x = A cos ωt
42c
€
+Α
Displacement, x
x
42d
€
42i € –Α
x
42d
–Α
€
42b
42e €
42f
42g
€
F = ma = – mω 2 x
€
–ωis2–mω2
42c
€
–mω 2
42d
€
ω = 2πf
Force, F
+mω2 Α
x = A cos ωt
x
+Α
1
2π
f =
=
42e € T
ωω 2
Displacement, x
ω2
ω2
42e €
42j
FΑ
= ma = – mω x
–ω
F = ma = – mω 2 x
42f
€
2
€
€
–mω2 Α
–ω 2
42g
€ : cos θ = cos ωt =
Then
2
–ω 2
€ 42gGradient is –ω 2
€
F = ma = – mω 2 x
θ = ωt
42f
€
x
OP
42k
2
But,
since isOP–mω
= OA
= amplitude, A, then : c
42h
€
Rearranging gives : x = A cos ωt
–mω 2
42h
2
The
amplitude
displacement
of the motion is denoted by the letter A.
€ 42hor maximum–mω
1
2π
€
f = velocity
=
€
In simple harmonic motion the object’s displacement from the42i
equilibrium
position,
T
ω and
x =that
A cos
ωt each of these
42L
acceleration all vary with1 time. 2π
To derive the appropriate
equations
describe
42i €quantitiesfit =is necessary
=
1to return
2π to circular€motion.
physical
42i €
T f =ω =
T
ω
θ = ωt
42j
42m
v =€ – ω x 0 sin ωt
Consider a point P moving in circle of radius A with €
P
θ = ωt θ = ωt
42j velocity
angular
x
€ 42j ω€.
€
Then : cos θ = cos ωt =
The projection of P onto the diameter XY is the 42n
point
a = – ω 2 x = –Aω 2 A cos ωt
OP
R. As P moves in a circle starting at Y then to X and €
x
42k
But, since OP = OA = amplitude,
€ to Y,€the
Then
: cos
θ =: cos
cos
=cos ωt = x
finally back
point
R
moves
along
Then
θωt=the
OP
OP
diameter from Y to X and back to Y.
€ X
x x x R gives :Y x = A cos ωt
O Rearranging
42k
But,
since
OP
=
OA
=
amplitude,
A,
then
:
cos
θ
=
The point R42k
moves along the
diameter
with=simple
But,
since OP
OA = amplitude, A, then : cos θ =
A
A
harmonic motion.
Rearranging
gives : xgives
= A: cos
Rearranging
x =ωtA cos ωt42L
x = A cos ωt
€
42L
50
42m
42n
€
x =ωtA cos ωt
42L x = A cos
€
42m
€
42m v = – ωvx 0=sin– ω
ωtx 0 sin ωt
€
42n
2 2
2
x =
– ωωt
A cos ωt
ωω
42n a = – ωa2 x= = – –
A cos
€
€
v = – ω x 0 sin ωt
a = – ω 2 x = – ω 2 A cos ωt
42c
42d
€42c
42g€
€
x = A cos ωt
x –ω
= 2A cos ωt
4.4 simple harmonic motion
x
€42d
The periodic
T is the time it takes R to complete one oscillation. This is of course the same as
2
x –mωtime
42h€
€ it takes P to complete one circle. As with all vibrations, the frequency, f, is the number of
vibrations made per second.
ω2
42e €
42e €
42i €
ω2
1
2π
f =
=
F = ma = –Tmω 2 xω
42f
€
F = ma = – mω 2 x where T is measured in s
42f
€
ω is measured in rad s–1
θ = ωt
42j
€ f is measured in Hz
–ω 2
42g
€
2
–ω
42g
€
x R from the
displacement of the point
€ The maximum
2 Then : cos θ = cos ωt =
–mω
42h
centre of oscillation O is called the amplitude.
In this
OP
€
2
–mω
42h
case
it
equals
the
of
the
circle
OY
or
OP
and is
x
42k€ denoted But,
since
OP
=
OA
=
amplitude,
A,
then
:
cos
θ
=
by A. The displacement x of R from O varies
A
1
2π
with time.
Rearranging
gives
:
x
=
A
cos
ωt
42i €
f =
= 1
Tf =at time
ω =t =2π0 R is at Y, i.e. at extremity of the
42i € Suppose
X
T
ω
42j
oscillation.
42L
€42j
x
42m
42n
€42L
€42m
€
Y
x = A cos ωt
θ = ωt further at time t, P has moved through an
€ Suppose
angle θ = ωt . At this time the displacement of R from
€ O is x.
42m
v = – ωA sin ωt
€
x
€
Then : cos θ = cos ωt =
OP= x
€
Then : cos θ = cos ωt
OP
x
a = – ω 2 x = – ω 2 A cos ωt
42n
42k
€ But, since OP = OA = amplitude, A, then : cos θ = A
x
42k
But, since OP = OA = amplitude, A, then : cos θ =
A
Rearranging gives : x = A cos ωt
Rearranging gives : x = A cos ωt
€ This is the displacement equation for SHM.
42L
R
θ
O
P
x = A cos ωt
x = A cos ωt
v = – ωA sin ωtvelocity and acceleration during Simple Harmonic Motion
Displacement,
v = – ω x 0 sin ωt
€ In the diagram below
the object R moves along the line YX with simple harmonic motion. The
centre of oscillation
is the point O. At time t = 0 the object is at Y and moving to towards O.
a = – ω 2 x = – ω 2 A cos ωt
The graphs
theωt
displacement x, (represented by the vector OR), the velocity v
€42n
– ω 2how
A cos
a = overleaf
– ω 2 x =show
€ and the acceleration a of the object vary with time t. The time to complete one oscillation is
the period T.
€
€
Direction of motion of R
X
O
Y
R
51
t=T
Displacement
: cos= θ –=mω
cosx ωt =
FThen
= ma
42f €
€
OP
= 2πf
F
=
ma = – mω 2 x
42f
x
x = since
A cosOP
ωt
42c
€ = OA Physics
42k €
But,
=
amplitude,
A, a2
thenlevel
: cos•θ UNIT
= 4
for ccea
2
A
–ω
42g
€
Rearranging gives : x = A
cos ωt
= A cos ωt The displacement
T
T
3T
The
–ω 2
42gx varies sinusoidally.
t=0
x
42d
t=_
t=_
t=_
€
4
2
4
€ value at any instant
is
given
by
–mω 2
42h
€
x = A cos ωt
42L €
–mω 2
42h
ω2
42e €
€
where A is the amplitude of the oscillation
1
2π
42i
fv == – ω=x sin ωt
42m €
0ω
T
€
2
1
2π
F
=
ma
=
42f
=
€– mω x f =
the displacement-time
graph,
we
€ Note that given42i
ω
can obtain the velocity at any instantT by drawing
2
2
x = –atωthe
Apoint
cos ωt
– ωtangent
42n
the2appropriate
and finding
x θa =2 ωt
= ma42j
= –€mω
–ω
€
θ = ωt
42j
€
2
x
€
Then2: cos θ = cos ωt =
–mω
42h
OP
€
x
€
Then : cos θ = cos ωt =
x
But, since OP = OA = amplitude, A, then : cosOP
θ =
2 42k
ω
A
x
1
2π
42k
But,
since
OP
=
OA
=
amplitude,
: cos θ 3T
=
42i €
f =
= gives : x = A cos ωt
Rearranging
T A, then_
T
_
_
t
=
0
t=
t=
t=
A
T any instant
ω
The velocity at
is equal to the
4
2
4
Rearranging
gives : x = A cos ωt
graph at that
1
2π gradient of the displacement-time
=
instant.
While
the
general
equation
for
v
is
not
x = A cos ωt
T 42L ω
ωt CCEA specification, it turns out
42j € requiredθby= the
€
to be:
x = A cos ωt
42L
€
= ωt 42m
v = – ωA sin ωt
x
€
€
Then : cos θ = cos ωt =
The velocity has
a maximum value
at the
42m
ωt
v = of
–
OPωA sin
€
instant the object
through
the centre of
2
x
x– ω 2passes
xOP
= =– ω
ωt
42n
42kθ €= oscillation.
OAA =cosamplitude,
A, then : cos θ =
en : cos
cos ωt aBut,
== since
A
OP
xx = – ω 2 A cos ωt
– ω 2ωt
42n gives : x a= =A cos
Rearranging
, since OP = OA = amplitude,€A, then : cos θ =
A
€
arranging gives : x = A cos ωt
x = A cos ωt
42L
€
€
Time
Velocity
t=T
instant.
= – ω x 0 sin ωt
42n €
a = – ω 2 x = – ω 2 A cos ωt
The acceleration has a maximum value of
at the
ωtinstant the object reaches the
= – ω x = – ω 2 A cos
€ extremities of its oscillation. The acceleration is
zero when the object reaches the centre of the
oscillation.
t=0
T
t=_
4
T
t=_
2
_
t = 3T
4
t=T
Acceleration
= A cos ωt The acceleration at any instant is equal to the
42m
v = – ω x 0 sin ωt
€ gradient of the velocity-time
graph at that
Time
2
The minus sign tells us that the acceleration is
always in the opposite direction to the
displacement from O.
52
Time
x
But, since OP = OA = amplitude, A, then : cos θ =
1
2π A
θf == ωt
42j
42i4.4€
= motion
€simple harmonic
Rearranging gives : x = A cos ωt
T
ω
42k
The specification makes it clear that candidates are expected to know how xto perform calculations
Then
: cos θ = cos ωt =
θ
= ωt
42j € for displacement
is based on time tOP
= 0 occurring at the
based x = A cos ωt . This expression
€
€ point of maximum positive displacement.
For the sake of completeness, the equations below show
x
42k
OA
= these
amplitude,
A, then
: cos θ =
how the displacement and acceleration
changeBut,
withsince
time OP
and =also
how
equations
would
A
x moving in the
42m
ω x 0t sin
changevif =
the–time
= 0 ωt
occurred at €
the equilibrium
position
with
the
particle
Then
:
cos
θ
=
cos
ωt
=
Rearranging
gives
:
x
=
A
cos
ωt
€
direction of positive displacement. There is no requirement to know the equation
showing how
OP
velocity changes with time.
x
42L
42n
€
43a
€
43b
€
42k
cos ωt
a = – ω 2 x = – ω 2 A 42L
€
x = A sin ωt
But, since OP = OA = amplitude, A, then : cos θ =
A
x = A cos ωt
Rearranging gives : x = A cos ωt
Based on time t = 0 at x = A
42m 43a v = – ωx x=0 sin
ωt ωt
A sin
€
x = A cos ωt
42L
Displacement,
ωt = – ωx22€
x x = A sin ωt
a = – ω 22 A sin
43a
– 2ωA2 sin
A cos
ωt ωt
= – ω2x
43b a = – ωa2 x= =– ω
€
€
42m
v = – ω x sin ωt
€ a = – ω 2 A sin ωt0 = – ω 2 x
Acceleration,
42na
43c €
43a
43b
43d
43c
43e
43f
43d
43e
43g
43f
x = A cos ωt 43b
€
22
2
x 22 == AAsin
cos
ωt2 ωt
x = A cos ωt
€ 43c €
2 cos ωt
a = – ω 22x = –2 ω 2 A
42n
Velocity and Displacement
€ x = A cos ωt x = A cos ωt
€
2 sin 43c
2
va =
=the––velocity
ωA
ωt
A
sin
ωt
=
– ωdisplacement
x x 2 = A2 from
ω
How
does
varies
with
the centre of oscillation?
cos2 ωt
€
2
2
2
2
2 2
2
2
v
=
ω
A
sin
ωt
= – ωA
ωt time.
The equations below show how the€displacement and vvelocity
varysin
with
43d €
2
2 2
2
x = A cos ωt
v = – ωA sin vωt = ω A sin ωt
€
2
2
2 2 43d €
2 2
θ= +A2cos
θ ωt
= 1
sin
x
cos
v 2 = ω 2 A2 sin 2 ωt
€
The relationship between sine and cosine
43etells us that sin 2θ + cos 2θ = 1 . So,
€
22
v = – ωA sin
ωt
sin43e
ωt + cos 22 ωtsin
=2θ1 + cos 2θ = 1
€
€
2 2
2
v 222 A
=22 ω
ω
sinA22 ωtsin+ ωt
ω 22 A22 cos22 ωt = ω 22 A22
sin 2 ωt + cos 2 ωt = 1
43f €
2
2
2
2
2
2 2 A 2 sin 2 ωt
Substituting for x and v this into the above gives: sinω
ωt + cos 2 ωt+ =ω 1A cos ωt = ω A
€
2 43f
22 2
+22 cos
θ ω
= 22 1A22
vsin
+θ ω
x 22 =
ω 2 A2 sin 2 ωt + ω 2 A2 cos2 ωt = ω 2 A2
€
€
So : v 22 = ω 22 A22 – ω 22 x 22 and hence :
v 2 + ω 2 x 2 = ω 2 A2
€
2
2
2
2 2
22 2 ωt
22 + cos 2 ωt 2= 1
sin
–
x
v
=
±
ω
A
43g
(
)
v
+
ω 2 x 2 = So
ω 2 :Av2 = ω A – ω x and hence :
€
€
2
2
2 2
2
ω 2 A2 sin 2 ωt +
ω 2 A2 cos2 ωtSo=: ω
v 2 A= ω 2 Av2 =– ω
and
43g
–hence
x2 :
± ωx A
(
43h
43g
43i
43j
43h
43k
43i
43L
43j
)
– x ) i.e. to left or to the right or up or
= ± ω or( A
The ± indicates
the ωt
velocity can bevpositive
negative
v = – ωthat
A sin
2 2
2
2
€ down, inv 2other
words
it
refers
to
direction.
The
equation
above is not required by the
+ω x = ω A
€ specification, but it is so useful that it is43h
v
=
ω Asake
sin of
ωtcompleteness.
reproduced here for–the
€:
So : v 2 =22 ω 2 A2 – ω 2 x 2 and hence
ωt
a = – ω A cos43h
v = – ω A sin ωt
€
v = ± ω ( A2 – x 2 )€
a = – ω 2 A cos ωt
43i
€
±ωA
a = – ω 2 A cos ωt
43i
€
€
v = – ω A sin ωt
±ωA
43j
€
€
2
ω 2A
±ωA
43j
€
€
2
a = – ω A cos ωt
ω 2A
43k
€
€
ω 2A
v = ± ω ( A22 43k
– x 22 )
€
€
±ωA
€
v = ± ω ( A2 – x 2 )
43L €
2
2
53
v 43g
= – ω A sin ωtSo : v = ω A – ω x and hence :
43h
x
Then : cos θ = cos ωt €
=
2
2
2 ±2ω ( A – x )
= – ω A sin ωt
43h
v 2 +OP
ω 2 x 2 =v ω=43i
A
x 2v)ωt
v
=
±=ω– (ωA2 2A–cos
a
v = a2
– ωlevel
A sin •ωtUNIT €
43hPhysics
€ for ccea
€
4
2
x :
€ = OA
2 2 x 2 and hence
v 2 = ωa 2 =A
–ω ω
43gOP
A: cos
43i So
42k
But, since
= :amplitude,
A, –then
cos ωt
θ =
€
A
2
2
v
=
–
ω
A
sin
ωt
43h
a = – ω 2 A cos ωt
43i
–
x
v
=
±
ω
A
2
Summary of EquationsRearranging
for Displacement,
Velocity
and
Acceleration
(
)
±ωA
43j
gives : xa == A
cos
ωt
v
=
–
ω
A
sin
ωt
43h
€
A
cos
ωt
–
ω
43i
€
€
€ €
±ωA
43j
Displacement
Acceleration
€ Velocity
2
Variation
a ωt
=43k
– ω A cosa ωt
43i
ω 2–Aω€2 A cos ωt±ωA
x = A43j
cos ωt
42L with
v =€ – ω A sin
43h
=43j
43i
±ωA
€
time €
€
€€
ω 2A
43k
€At fixed point,
At extreme displacement,
Maximum value
Amplitude = A
2
2 = ±ωA
43j
43k
Max
velocity
maximum
acceleration =2 ω A2
2
42m
v = 43k
–43i
ω x 0 sin ωt ωa A=€ – ω A cos
±ωA
43j ωt
€
v
=
±
ω
A
–
x )
43L
(
€
€
€
€€
At fixed point,
At extreme displacement,
2
2
Minimum value
v = ± ω ( AAt fixed
– x )point, acceleration = o
43Lminimum
€
displacement = o
velocity = 0
2
ω A2
v = ± ω ( A2 – x 2 )
±ωA
43j
A
cos
ωtω A243k
a = –43L
ω 2 x = – ω 2 43k
42n with
ω 2 A43L € 2
Variation
€
v
=
±
–
x
€
€
x
a
=
–
ω
43m
(
) €
€
€
displacement
a = – ω2x
43m
€2
2
v = ± ω ( A2 v– x=43m
43L
ω A€ 2
43k
) ± ω 2π(A2 – x 2a) = – ω 2 x
43L
€
€
€
a = –ω x
43m
Other Equations
43n €
ω =€
= 2πf
€
T
2π
Angular velocity (sometimes called angular
€
43nfrequency):
ω =
= 2πf
1
1
2
2
T =2
=
= 0.2 seconds
T
2π
v = ± ω ( A43m
–=x –) ω 244a
43L €
x
a
43m
f
5
a =43n
– ω€x
2π
ω =
= 2πf
€
43n €
ω =
= 2πf €
ω = 2πf = 2π × 5 = 10π
T = 31.4 rad s–1
1
1
T
Worked Examples
T = €=
= 0.2 seconds
44a
f
5
2
2π
2
2
2π
a€=€ – ω x ω ω= = 2πf 44b
43m
a == ω31.4
Example 1
== 2π
× 5 = 10π
s–1 ) × 5 ×10 –2 = 5π 2 = 49
43n
2πf
€
€
43n
ω
=
=
2πf
€
T
A steel strip is clamped at one end and made to vibrate at a frequency of 5 Hz T
and
€ with an
€
2
2
2
amplitude of 50 mm.
x =–2A=cos
50 cos
44b
a = ω x 44c
= (10π ) × 5 ×10
5πωt
== 49.3
mm(10π
s–2 × 0.04 ) = 15
€
€
2π
1
1
43n
ω
= 2πfω.
(a) (i) Calculate the period,
and
velocity,
T T=€
= the =angular
0.2 =
seconds
44a
f
5
T
€
= A
ωt mm
44c
x €
= A cos ωt = 50 cos (10π × x0.04
= 15.5
) cos
1
1
2π × 5 = 10π =€ 31.4 rad s–1
€
Solution T =
= ω ==2πf0.2= seconds
33 = 50 cos 10πt
44a
1f
15
33
T =
=
= 20.2 seconds
–1
cos 10πt =
= 0.660
x 44d
= A–2cos ωt
2
2
44a
ω44b= 2πf
= 5a 2π
× x5 ==(10π
10π
=
31.4
f
= ω
×
5
×10
=
5π
=
49.3
mm
s
)
50
€
€
€
33
=
50
cos
10πt
10πt = cos–1 0.660 = 0.850
ω the
= 2πf
= 2π × 5at=maximum
10π = 31.4
(ii) Calculate
acceleration
displacement.
33
cos 10πt =
= 0.660
0.850
44d
= 0.027 s
t =
44c
x = A cos
10π × 0.04
= 15.5 mm 50–2
2 ωt = 50 cos
2
–2 (
2 )
10π
€
44b
a
=
ω
x
=
10π
×
5
×10
=
5π
=
49.3
mm
s
–1
(
)
Solution
10πt = cos 0.660 = 0.850
1€ 1
2
–2
2
T =
=
= 0.2 seconds 2
44b
a
=
ω
x
=
10π
×
5
×10
=
5π
=
49.3
mm
s–2 = 0.027 s
0.850
(
)
5
(b) f€
At time
t = 0, the displacement
x = isA+50
cos ωtmm. Calculate
t =
x = A cos ωt
–1
10π
€
44a
44b
44c
€
€
€
44d
€
44e
54
ω = 2πf = 2π × 5 = 10π€= 31.4 rad s 33 = 50 cos 10πt
€
at ωt
time=t 50
= 0.04
–33 = 50 cos 10πt
44c (i) the displacement
x = A cos
cos s.10π × 0.04 = 15.5 mm
33
€
33
cos 10πt =
= 0.660
44d
cos 10πt =
= – 0.660
44e
= A cos ωt
2 x = A
2
ωt
50mm
cos
50s–2 10π × 0.04 = 15.5 xmm
a44c
= ω 2 xSolution
= (10π ) ×
5 ×10 –2 cos
= 5π
==49.3
50
€
€
–33 = 50 cos 10πt
–1
10πt = cos–1 0.660 = 0.850
cos – 0.660 = 2.292
(ii) the times during the first period at which the distance from the fixed point10πt
is 33= mm.
x = A cos ωt 0.850
33
cos 10πt =
= – 0.660
2.292
44e s
= 0.027
t =mm
= 0.073 s
t =
cos
ωt = 50 cos (10π × 0.04 ) = 15.5
50
x = A€
Solution
10π
33
50cos
cosωt
10πt
x == A
10π
–1
10πt = cos – 0.660
= 2.292
€quarter cycle,
In the 1st
In the 3rd quarter cycle,
In the 4th quarter cycle,
In the 2nd quarter cycle,
33 = 50
33cos 10πt
the displacement
is –33
the displacement is –33
2.292
=x =0.660
44d x = A cos ωtcos 10πt =
A cos ωt
= 0.073
s0.027 half
t =half a period
t =exactly
+ 0.1
mm exactly
mm
a
10π
€
50 –33 = 50 cos 10πt
33
€ in
(0.1 s) after 44f
the time
period (0.1 s) after the
33
=
50
cos
10πt
t
=
0.127
s
cos 10πt =
44d
–1= 0.660
the first quarter cycle.
time in the second
10πt = cos
33
50 0.660
33 = 0.850
quarter cycle.
cos 10πt =
= 0.660
cos 10πt =
= – 0.660
44e
t
=
0.027
+
0.1
–1
50
50 = 0.850
44f €
10πt
=
cos
0.660
0.850
t = 0.073 + 0.1
–1
s €
s = 2.292 t = 0.12744g
t = 10πt ==cos0.027
10πt = cos–1 0.660 = 0.850
– 0.660
t = 0.173 s
10π
0.850
0.850
2.292
= 0.027 s t =
t =
= s0.073 s
t == 0.027
10π
t = 0.073 + 0.1
10π
10π
44g €
2π
2π
t = 0.173 s €
T =
so : 1.5 =
x calculations
= A cos ωt above it is essential that the44h
Note that in the
calculator is setωto
ω
t = 0.027 + 0.1
x = A cos ωt
€
2π
44f
calculating
the cosines and inverse cosines. Rearranging : ω =
€ x when
== tA
= 4.19 rad s–1
50
cosωt
10πt
= cos
0.127
s
–33 = 50 cos 10πt –33
2π
2π
1.5
T =
so : 1.5 =
€
€
ω
ω
33
–33 = 50
33cos 10πt
44h
cos
10πt
=
=
–
0.660
cos
10πt
=
=
–
0.660
44e
2π
50
t 50
= 0.073 + 0.1
= 4.19 rad s–1
Rearranging : ω =
–1
33
44g €
44i
1.5 ±ωA = 4.19 × 0.018 = ± 0.075 ms
= 2.292
cos 10πt
= t = 0.173
=
–
0.660
44e10πt = cos–1 – 0.660
€
s
–1
10πt = cos
50 – 0.660 = 2.292
2.292
= 0.073 s
t =
–1
–1
10π
2.292
10πt = cos
– 0.660 = 2.292
(
(
)
)
5.5 Magnetic Fields
Students should be able to:
5.5.1 Explain the concept of a magnetic field;
F = BI that there is a force on a current-carrying conductor in a perpendicular
5.5.2 Understand
magnetic field and be able to predict the direction of the force;
5a
5a using the equation F = BI ;
5.5.3 Define magnetic flux density
5b
€
5a
F = BI
φ = BA
5.5.4 Define
the unit of magnetic flux density, the tesla;
5.5.5 Understand the concepts of magnetic flux and magnetic flux linkage;
5b
5c
€€
φ = BA
5b for magnetic flux, φ = BA , and flux linkage,
5.5.6 Recall and use the equations
€
Nφ = BAN ;
5.5.7 Define the unit of magnetic flux, the weber;
€
5d
5cexperimentally
Nφ = BAN
5.5.8 State, use and demonstrate
Lenz’s laws of
5c
Nφ =and
BAN
€
electromagnetic
induction;

5.5.9 Recall and calculate induced e.m.f. as rate of change of flux linkage with time;
€
€

€
I
Ns
V
F
= p for transformers;
B = and use5athe equation s =
5.5.11 Recall
Vp
Np
Is
I
5.5.12 Explain power losses in transformers and
of high voltage transmission of
FF
5e
B
=
electricity;
5e
B = I
I
€
F = BI
€
5d works;
5.5.10 Describe how a transformer
5d
5e
= 4.5 ×10 –2 × 1.25 × 0.25
5f
The space surrounding a magnet where a magnetic force is experienced is called a magnetic field.
= BI
BIof metals such as iron and steel
0.014is N
€€ objects
This magnetic=force
experienced by other magnets,
–2
and also by current carrying conductors.
The direction of a magnetic
field
a1.25
point× is0.25
taken as
4.5×10
×10
–2 at
== 4.5
×× 1.25
× 0.25
5f5f
the direction of the force that acts on a north pole placed at that
point.N
= 0.014
5g
€ shape B
The
of a magnetic field can be represented by a magnetic field lines (lines of magnetic force).
⊥
Arrows on the lines show the direction of the magnetic force. Since a north pole is repelled from
another north pole and attracted by a south pole the direction of a magnetic field is from North to
5g
B
€€
South.
5g
B⊥
= 0.014 N
5h
⊥
€
are commonly used in school laboratories to
=
produce a strong magnetic field. Two are arranged on a U-shaped
steel yoke as shown in the diagram. The diagram€also shows the
5h
B=
€
shape and direction of the magnetic5hfield that is created byBthis
=
€
F = B⊥ I
arrangement.
N
S

5i
B⊥ in
= the
0.2 central
× cos 40
The field lines
portion are parallel and uniformly
€
spaced this indicates
a
magnetic
field in this region
F = B
€ that is uniform,
= 0.153 T
F = B⊥⊥II
i.e. it has the same strength at all points in the region. The field
B⊥ = 0.2 × cos 40 
F = 0.153 × 0.5 × 0.3
× lines
cos 40
weakens at the edges as indicated by the increased spacingBbetween
of force.
⊥ = 0.2the
= 0.023 N
5j
€
–5
2.0 ×10 T
5i5i
= 0.153 T
= 0.153 T
F = 0.153 × 0.5 × 0.3
F = 0.153 × 0.5 × 0.3
= 0.023 N
= 0.023 N
141
Physics for ccea a2 level • UNIT 5
Electromagnetism
A conductor carrying an electric current is surrounded by a magnetic field. Different shaped
conductors produce magnetic fields of various shapes. The magnetic field lines due to the current
in a straight wire are circles concentric with the wire. The shape can be revealed by sprinkling iron
filings around the wire. The direction can be obtained by using the right hand screw rule. If a right
handed screw moves forward in the same direction as the current, then the direction of rotation of
the screw gives the direction of the magnetic field lines.
Current carrying wire
Iron filings
Current
Direction of magnetic field
The direction can also be obtained using the right hand grip rule, this is illustrated in the
diagram above.
Pretend to grab the wire with your right hand, your thumb pointing in the direction of the
current. Your fingers curl in the direction of the magnetic field.
The diagram below shows a view looking along the current carrying wire.
represents a current flowing into the page (away from you) and the symbol
The symbol
represents a current flowing out of the page (towards you). You should apply the right hand grip
rule to the diagrams below to confirm the magnetic field direction in each case.
Magnetix flux lines
Current carrying wire with
current direction into the page
142
Current carrying wire with
current direction out of the page
5.5 Magnetic Fields
Force on a Current in a Magnetic Field
When a current flows through a conductor in a magnetic field the conductor experiences a force.
This can be demonstrated using the apparatus shown. When a current is passed along the flexible
wire, the wire moves up. The force acts at right angles to both the current and the magnetic field
direction.
Upward force on the
current carrying wire
Force on the current
carrying conductor
F
N
Magnetic field
direction
North to South
S
B
I
Conventional current
direction, positive
to negative
The direction of the force (movement) of the wire is obtained from Fleming’s Left Hand rule.
Extend the thumb, first and second fingers of the left hand so that they are at right angles to each
as shown in the diagram above. The first finger represents the magnetic field direction. The
magnetic field direction is from the north pole (N) to the south pole (S). The second finger
represents the conventional current direction, from positive to negative. The thumb represents the
force acting on the current carrying wire.
This force arises because of the interaction of the two magnetic fields: the uniform field of the
permanent magnet and the field due to the current in the wire. The magnetic field lines of force are
vectors and the field lines due to two fields have to be combined vectorially. This interaction of
magnetic fields is illustrated below and the resultant field is some times called a catapult field. The
field lines resemble the stretched rubber band in a catapult, the conductor experiencing a force in
the direction shown.
S
N
Uniform field of the
permanent magnet
+
=
Magnetic field due to
the current which is
INTO the page
S
N
The resultant magnetic
field is sometimes known
as a catapult field
143
5a
5b
5c
5d
5e
5f
5g
5h
5i
5a
F = BI
Physics for ccea a2 level • UNIT 5
F = BI
When the current and €
the magnetic field are in the same direction or act at 180o to each other
5b
φ = BA
the force is zero. It is a maximum when the current and the magnetic field are perpendicular to
€ each other.
φ = BA
In the case when the magnetic field and the current are at right angles the force is given by:
€
F 5c
= BI
€
Nφ = BAN
where B is the magnetic flux density in tesla (T)
Nφ = BAN
I is the current in the conductor in amperes (A)
€
φ 5d
= BA
€

€
 is the length of the conductor in the magnetic field in metres (m)
F = BI flux density is used to indicate the strength of a magnetic field. Magnetic flux
The term magnetic
measured in €
units known as the
€ density is
F tesla, symbol T. The tesla is defined from the equation
Nφ
BAN
5e = Re-arranging
B =
shown above.
the equation
gives:
I
F
B φ= = BA
I
€

The magnetic flux density
the force per unit current carrying length. A current
€ is defined
F =asBI
€
€
carrying length is the product of the current and the length of the conductor in the magnetic field.
–2
€
= 4.5
× 1.25
× 0.25
5f
=A
BAN
of
and a length of 0.5
m is×10
a current
carrying
length of 1.0 A m.
€ A current
F Nφ
= 2.0
BI
=
0.014
N
€ A magnetic field
–2 density 1.0 T will exert a force of 1.0 N on a current carrying length of
flux
F of
== 4.5
×10
× 1.25 × 0.25
1.0 Am B
when
the
current
and magnetic field directions are perpendicular to each other.
I N
= 0.014
€

€
5g
B⊥
Worked Examples
€
F
€ Example
B⊥=1 BI
€
–2
= 4.5wire,
× 1.25 × 0.25
F×10of
A straight
B =
€length 60Bcm, carries a current of 1.25 A. Calculate the value of the
5h
=
I on
force that
acts
a length of 25 cm of the wire is placed at right angles to a
= 0.014
N this wire when
€
magnetic field of flux density 4.5×10–2 T.
B=
Solution
€€
B⊥F = BI
€
F = B⊥ I
Uniform magnetic
field of flux density
4.5 × 10-2 T acting
into the page

12.5 A
= 0.2 × cos 40
= 4.5 ×10 –2 × 1.25B×⊥ 0.25
F = B⊥ I
= 0.153 T
5i = 0.014 N
25 cm

B
=
0.2
×
cos
40
⊥
€
0.153 × 0.5 × 0.3
Only
B= 0.25 m of the wire is in Fthe=magnetic
=
0.153
T
field.
= 0.023 N
F
=
0.153
×
0.5
×
0.3
€ Application
of Fleming’s Left Hand Rule tells
B⊥
us that =
the0.023
force N
acts on the wire to the left.
€
F = B⊥ I
€
2.0 ×10 –5 T
5j
B⊥ 2= 0.2 × sin 50 
Example
€
B = 0.153
–5
€ A straight
T
wire
a current of 500 mA is placed in a magnetic field of flux density
T carrying
2.0 =×10
€
€€
144€
0.2 T.
length×of0.5
the×wire
F The
= 0.153
0.3 in the magnetic field is 30 cm and the wire makes an angle of
€
o
50 with the magnetic field direction. Calculate the force on the current carrying wire due
= 0.023 N
to the magnetic field.
F = B⊥ I
B⊥ = 0.2 × cos 40 
= –50.153
T T
2.0 ×10
F = 0.153 × 0.5 × 0.3
= 0.023 N
€
5g
€
B⊥
5.5 Magnetic Fields
€
5g
5h
= 4.5 ×10 –2 × 1.25 ×
= 0.014 N
5f
B⊥
€ SolutionB
5g
The magnetic flux density is resolved into a component perpendicular
to€the wire B⊥ and a
€
5hparallel to the wire, B= . Only the perpendicular component produces a force on
component
the wire..
=
€
F = B⊥ I
€ 50
B⊥ = 0.2 × sin
F =
= 0.153 T
B⊥ =
F
=
0.153
×
0.5
×
0.3
=
5i
= 0.023 N
F =
5h

5i
B€
=
B⊥ I
B=
50°
0.2 × cos 40 
0.153 T
0.153 × 0.5 × 0.3
= 0.023 N
5j
F = BI
€
B⊥
F = B⊥ I
B⊥ = 0.2 × cos 40 
= 0.153 T
F = 0.153 × 0.5 × 0.3
5i
€
2.0 ×10
Verification
of –5F T
= BI
= 0.023 N
€ can be2.0
The apparatus
used
how the force acting on a current carrying
T
×10to–5 investigate
5j shown below
conductor depends on the current flowing in the conductor and the length of the conductor in the
€
magnetic field.
€
Conventional
current
direction
5j
€
2.0 ×10 –5 T
Magnet
(south)
€
Magnet
(north)
Aluminium rod
(clamps not shown)
Electronic balance
An aluminium rod (e.g. the upright from a retort stand) is clamped horizontally above a sensitive
electronic balance. The rod is connected to a variable low voltage supply. An ammeter connected
in series with both will allow the current to be measured.
A permanent magnet is placed on the balance and the aluminium rod positioned so that it is
located in the centre of the magnetic field. The balance is set to read zero after the magnets have
been placed on it. When a current is then passed along the clamped aluminium rod the rod
experiences a force due to the interaction of the permanent magnetic field and the magnetic field
due to the current in the aluminium rod. In the case shown in the diagram, the force is upwards,
use Fleming’s Left Hand Rule to verify this.
Since the magnet exerts an upward force on the rod, then by Newton’s third law, the rod must
exert an equal but opposite force on the magnets. This downward force will cause the reading on
the electronic balance to increase.
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The current is varied with a single magnet in place. This ensures that
the length of the conductor in the magnetic field remains constant.
A graph of the force (y–axis) against current (x–axis) produces a
straight line through the origin. This is verification that the force on
the current carrying conductor in the magnetic field is directly
proportional to the current.
Force / N
Physics for ccea a2 level • UNIT 5
0
Current / A
The current is then fixed and a number of identical magnets are
placed side by side, this changes the length of the conductor in the
magnetic field. A graph of force (y–axis) against the length of
conductor in the magnetic field (x–axis) produces a straight line
through the origin. This is verification that the force on the current
carrying conductor in the magnetic field is directly proportional to
the length of the conductor in the magnetic field.
Force / N
0
0
0
Length of conductor in
the magnetic field / m
Exercise 33
1 The horizontal component of the Earth’s magnetic field has a flux density of 2.0×10–5 T. A
straight piece of wire XY 1.2 m long of mass 0.8 g is resting on a wooden bench so that it is
at right angles to the magnetic field direction. A current is passed through the wire which
just causes the wire to lift off the bench.
X
Y
(a) State the direction of the current in the wire.
(b) Calculate the current.
2 A single-turn rectangular wire loop ABCD hangs from a
sensitive balance so that its lower portion is in a region of
uniform magnetic field. The direction of the magnetic field
is at right angles to the plane of the loop. This arrangement
is shown in the diagram. The upper portion of the loop is
not in the magnetic field.
When a current passes round the loop in the direction
ABCD, the balance reading gets less.
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