Transcript 3.4 Electromagnetism

Background knowledge Magnetic field

Magnetic field can be produced by a magnet or a
current-carrying conductor.
Magnetic field lines

Magnetic field lines are used to show the
strength and direction of a magnetic field.
Neutral point
N
Note:
1.
2.
S
N
x
N
Field lines run from the N-pole round to the S-pole.
When field lines are closely-spaced, field is strong and vice-versa.
Example 1 (a)

In each of the following cases, draw the magnetic field
lines between the magnets and mark the positions of
the neutral points (if any).
S
N
Example 1 (b)

In each of the following cases, draw the magnetic field
lines between the magnets and mark the positions of
the neutral points (if any).
N
S
Example 1 (c)

In each of the following cases, draw the magnetic field
lines between the magnets and mark the positions of
the neutral points (if any).
S
S
Example 1 (d)

In each of the following cases, draw the magnetic
field lines between the magnets and mark the
positions of the neutral points (if any).
Earth’s magnetic field

The Earth has a weak magnetic
field. It is so weak that it does
not affect much the field patterns
of permanent magnet.

It is rather like the field around a
huge bar magnet.

The south pole of the ‘earth
magnet’ is actually in the north.
Earth’s magnetic field

It is convenient to resolve the
earth’s field strength B into
horizontal and vertical component,
BH and BV respectively.
BV
B
BH
a
N
ground
We have BH = Bcos a and BV = Bsin a.
Compass needles, whose motion is confined to a horizontal plane,
are affected by BH only.
Fleming’s left-hand rule
F
Magnetic force
current
B
I



F: force
B: magnetic field
I: current
Factors affecting magnetic force
F
Magnetic force
current
B
I



B: magnetic field
I: current
l: length of conductor in B-field
Current balance


magnetic force F
acting upwards
weight of the rider
is equal to the
magnetic force
when the frame
PQRS is balanced.
magnetic force and current

Record the
currents required
to balance different
number of identical
riders on PQ.

Results: magnetic
force (number of
riders)  current

i.e. F  I
magnetic force and length



Keep the current through PQ constant.
Record the number of riders required to balance the metal frame
when one, two and three pairs of equally strong magnadur
magnets are used as shown.
Results:
The magnetic force (number of riders)  length of conductor
(number of pairs of magnadur magnets)
i.e. F  l
magnetic force and magnetic field





Replace the magnadur magnets
by a coil of many turns (for
example 1100 turns) as shown
below.
Hence, the magnetic field is
provided from the coil and the
magnetic field can be increased
by increasing the coil current.
Keep the current through PQ
constant.
Record the coil currents required
to balance different number of
identical riders on PQ.
Results:
The magnetic force (number of riders)  magnetic field (coil
current)
i.e. F  B
Conclusion

The magnetic force is increased if

the current is increased,(F  I)
the magnetic field is increased, (F  B)
 there is a greater length of wire inside the
magnetic field. (F  l)

Precautions of using current
balance





Make sure the direction of the magnetic field is
perpendicular to the current-carrying arm.
Minimize the effect of the earth’s magnetic field by
aligning the current-carrying arm along the N-S
direction.
The set-up should be far from any current-carrying
conductors so as to avoid the effect of stray
magnetic fields.
To avoid overheating, the current should be switched
off as soon as measurements have been taken.
Shield the set-up from the disturbance of wind.
Magnetic flux density B


Definition:
Electric field E: force per unit charge
(E = F / Q)
 Gravitational field g: force per unit mass.
(g = F / m)
 Magnetic flux density B (magnetic field): force
acting per unit current length.
(B = F / Il
)
 In words: The magnetic flux density B is equal to the
force acting on a conductor of unit length and carrying
a unit current at right angles to the field.
 Unit of B: Tesla (T) or N A-1 m-1
Typical Values of the magnetic
flux density
Source
Magnetic field / T
Smallest value in a magnetically
shielded room
10-14
Interstellar space
10-10
Earth's magnetic field
5 x 10-5
Small bar magnet
0.01
Big electromagnet
1.5
Strong lab magnet
10
Surface of a nucleus
106
Magnetic force on a currentcarrying conductor


From definition: B = F/Il
F = BIl
I sin q
l
q

I
I cos q
B
If the conductor and field are not at right angles, but make an
angle q with one another, the expression becomes
F = BIl sin q.

Note that:
1. when q = 90o
2. when q = 0o
(conductor ⊥ field),
(conductor // field),
F = BIl.
F = 0.
Magnetic force on moving charge in
a magnetic field
conductor containing n particles
each having charge q

Current = Q / t
total charge in the conductor

time required

I 
nq
l
v

for all charges
nqv
l
to pass through
a section
Magnetic force on moving charge in
a magnetic field
conductor containing n particles
each having charge q

Magnetic force acting on each moving charge

total force acting on the conductor
n

If the conductor makes an angle q with the magnetic field,
 nqv 
B
 l sin q
BIl sin q
 l 
F 

n
n

F = Bqv sin q
Motion of a charged particle in a
magnetic field

1
Moving parallel to the field
+q

v
Magnetic
field B
When a charged particle moves parallel to the
magnetic field, i.e. q = 0o, there is no force acting on it.
Its velocity remains unchanged and its path is a
straight line.
2
Moving perpendicular to the field
uniform
magnetic
field B
Circular path of
motion of +q
+q
+q

Radius of circular path
uniform
magnetic
field B
2
Moving perpendicular to the field
uniform
magnetic
field B
Circular path of
motion of +q
+q
+q

Period of circular motion
uniform
magnetic
field B
2
Moving perpendicular to the field
uniform
magnetic
field B
Circular path of
motion of +q
+q
+q



uniform
magnetic
field B
Notes:
The magnetic force is always perpendicular to the motion of the
charged particle; no work is done by the magnetic field.
Kinetic energy of the charged particle remains constant.
Example 2


The path of an electron in a uniform magnetic field of flux density
0.01 T in a vacuum is a circle of radius 0.05 m. Given that the
charge and mass of an electron are -1.6 x 10-19 C and 9.1 x 10-31 kg
respectively. Find
(a) the speed and
(b) the period of its orbit.
Solution:
Current balance
d1
d2
rider
Magnetic force
BIl



mechanical
force
mg
To measure steady magnetic
field by using principle of
moment.
Clockwise moment = anticlockwise moment
mgd2 = BIld1
Example 3 Current balance
A rider of mass 0.084 g is required to balance the frame
when an arm PQ, of length 25 cm and carrying a current of
1.2 A, is inside and in series with a flat, wide solenoid as
shown below.
Find the magnetic flux density inside the solenoid.
Search coil and CRO



A search coil is only used in measuring a varying magnetic
field.
A typical search coil consists of 5000 turns of wire and an
external diameter d ≤ 1.5 cm so that it samples the field
over a small area.
The search coil is placed with its plane perpendicular to a
varying magnetic field.
Search coil and CRO



Due to the change of magnetic flux, an e.m.f., which is
induced in the search coil, can be measured by a CRO.
The induced e.m.f. E is proportional to the flux density B
and the frequency f of the varying magnetic field. i.e. E 
Bf
The earth’s magnetic field can be ignored because it is a
steady field.
Cathode-ray oscilloscope (CRO)

The CRO is a perfect voltmeter as its resistance is very
high. It can measure both d.c. and a.c. voltages and show
how they vary with time.
simulation
Measuring d.c. voltages


To measure a d.c. voltage, the time base is usually
switched off. Thus, it is the light spot which is
deflected.
Or the time base may be switched on to any high value
so that the horizontal trace is deflected. From the
deflection on the screen and the gain control setting, the
d.c. voltage is then calculated.
Y-gain
control:
2 V cm-1
Vd.c. = amount of deflection (cm) x Y-gain control setting (V cm-1)
=3x2=6V
Measuring a.c. voltages

To measure an a.c. voltage, the time base is usually
switched off. The waveform displayed becomes a vertical
trace so that the amplitude can be easily read on the screen.
The maximum voltage or peak voltage is then calculated
from the gain control setting. Note that the peak voltage
refers to half the length of the vertical trace.
Y-gain
control:
5 V cm-1
Time base
setting:
10 ms cm-1
Peak voltage =
Period =
Frequency =
Example 4 CRO waveform
The figure below shows a waveform on a screen.
(a)
If the controls on the CRO are set at 0.5 V cm-1
and 2 ms cm-1,
(i)
the peak voltage, and
(ii)
the frequency of the input signal.
Solution:
1 cm
1 cm
Example 4 CRO waveform
The figure below shows a waveform on a screen.
(b)
If the gain control is changed to 1 V cm-1, sketch
the trace on the figure.
Solution:
1 cm
1 cm
Example 4 CRO waveform
The figure below shows a waveform on a screen.
(c) If the time base control is changed to 5 ms cm-1,
sketch the trace on the figure.
Solution:
1 cm
1 cm
Time base



When Vx increases linearly with time from A to B, the spot
of light on the screen moves at a constant speed from the
left to right of the screen.
Then the spot of light flies back to the left quickly when Vx
suddenly drops from B to C.
The saw-tooth voltage Vx causes no vertical movement of
the spot of light..
Example 5 displaying a.c.
waveforms
Example 5 displaying a.c.
waveforms
Measuring of phase relationships
 The phase difference f of two p.ds (Vx and
Vy) can be observed on phase difference f
a double-beam CRO.
phase difference f



If a double-beam CRO is not available, the phase
difference can be found by applying the two p.ds of the
same frequency and amplitude to the X- and Y-plates
(time base off) simultaneously.
The phase difference can be determined from the trace on
the screen of the CRO as follows.
In general, the trace is an ellipse except when the f is 0o,
90o, 180o, 270o or 360o.
Comparing of frequencies

When two p.ds of different frequencies fx and fy are
applied to the X- and Y-plates (time base off), more
complex figures are obtained, know as Lissajous’ figures.
simulation
In any particular case, the frequency ratio can be found by
fy
fx

no. of loops touching
no. of loops touching
horizontal
verticall
line
line
In any particular case, the frequency ratio can be found by
fy
fx

no. of loops touching
no. of loops touching
horizontal
verticall
line
line
Magnetic field around a long
straight wire
1. The field lines are circles
around the wire.
2. The magnetic field is the
strongest close to the wire.
3. Increasing the current makes
the magnetic field stronger.
4. Reversing the current also
reverses the direction of field
lines, but the field pattern
remains unchanged.
Right hand grip rule
If the right hand grips the wire so that the thumb points
the same way as the current, the fingers curls the same
way as the field lines.
Experiment to show that B ∝ I
and B ∝ 1 / r



B ∝ I and B ∝ 1 / r
B ∝ I/r
B = m0I/(2pr) where I is the current and is the
permeability of free space (m0 = 4p x 10-7 T m A-1)
B = m0I/(2pr)
(m0 = 4p x 10-7 T m A-1)



Example 7
Two long wires X and Y each carries a current of 20 A
in the directions as shown in the figure. If the
distance between the wires is 10 mm, find the
magnitude and direction of the magnetic flux density
at
Q
P
(a) P
5 mm 5 mm 5 mm
I = 20 A
X
I = 20 A
Y
B = m0I/(2pr)
(m0 = 4p x 10-7 T m A-1)



Example 7
Two long wires X and Y each carries a current of 20 A
in the directions as shown in the figure. If the
distance between the wires is 10 mm, find the
magnitude and direction of the magnetic flux density
at
Q
P
(b) Q.
5 mm 5 mm 5 mm
I = 20 A
X
I = 20 A
Y
Magnetic field around a flat
coil


At the centre of the coil
 The field lines are straight and
at right angles to the plane of
the coil
Outside the coil,
 The field lines run in loops.
Experimental set-up
B 
m 0 NI
2r

By using the experimentalsetup shown, it is found that
the magnetic field is
1. directly proportional
to the current and
the number of turns,
and
2. inversely proportional
to the radius of the
coil.

Note: The magnetic field is
greatest at the centre.
Magnetic field due to a long solenoid
N
I

S
I
From the field pattern of the solenoid, it can be found that
1. inside the solenoid, the field lines are straight and
evenly-spaced. This indicates that the field is of
uniform strength.
2. outside the solenoid, the pattern is similar to that
around a bar magnet, with one end of the solenoid
behaving like a N-pole and the other end like a S-pole.
Right hand grip rule
If the right hand grips the solenoid so that the fingers curls
the same way as the current, the thumbs points to the north
pole of the solenoid.
Magnetic field due to a long solenoid
N
I

S
I
The magnetic field of the solenoid can be increased by
1. increasing the current,
2. increasing the number of turns on the coil.
Magnetic field due to a long solenoid
N
I



S
I
The magnetic field of the solenoid can be increased by
1. increasing the current,
2. increasing the number of turns on the coil.
In vacuum, the magnitude of the magnetic flux density B at a
point O on the axis near the centre of the solenoid of length l,
having N turns and carrying a current I is given by
B = m0NI/l or B = m0nI where n is the number of turns per unit
length (n = N/l)
Experiment to show that B ∝ N, B ∝ I and
B∝1/l
Magnetic field due to a long solenoid
N
I

S
I
Note: The magnetic field within a solenoid is
independent of the shape and the cross section
area of the solenoid.
Magnetic field due to a long solenoid


At the ends of solenoid
The magnetic field at the ends of the solenoid is weaker. It is half
that in the central region within the solenoid.
B'
1
2
B 
m 0 NI
2l

m 0 nI
2
B 
m 0 NI
2r
Current-carrying
conductor
B 
m 0 NI

2l
Position of
magnetic field
m 0 nI
2
B 
m0I
2p r
B 
m 0 NI
l
Magnetic flux
density (B)
 m 0 nI
Symbol
1. Long straight wire
Around the
wire
r = perpendicular distance
from wire
2. Circular coil
At the centre
N = number of turns
r = radius of coil
Inside
N = number of turns
l = length of solenoid
At the ends
n 
3. Solenoid
N
l
Force between currents
B1
F2
P

Magnetic field due to current through P
B1 

Q
m 0 I1
2p r
Magnetic force acting on Q
F 2  B1 I 2 l 
m 0 I1I 2l
2p r
Applying Fleming’s left hand rule, force acting on wire Q is towards wire P.
P
F1
Q
B2



Similarly, the force acting on wire P is towards wire Q.
Hence, the two wires attract each other.
By Newton’s third law, the magnetic force acting on P
= magnetic force acting on Q.
∴
F1 
m 0 I1I 2l
2p r
Summary: Unlike current repel,
like current attract


Parallel wires with current flowing in the same
direction, attract each other.
Parallel wires with current flowing in the opposite
direction, repel each other.
Summary: Unlike current repel,
like current attract
F1  F 2 


m 0 I1 I 2l
2p r
The force per unit length on each conductor  F 
l
When the current I1 = I2 = 1 A, and the separation
between the wires r = 1 m,
the force per unit length on the conductor
1
4p  10 11  2  10
7

2 p 1 
7
N m-1
F2
l

m 0 I1I 2
2p r
F 2  B1 I 2 l 
m 0 I1I 2l
Summary: Unlike current repel,
like current attract


2p r
Definition of the ampere
The ampere is constant current which, flowing in two
infinitely long, straight, parallel conductors of
negligible cross-section and placed in a vacuum 1
metre apart, produces between them a force equal
to 2 x 10-7 Newton per metre of their length.
Example 8

The figure below shows two horizontal wires, P and Q, 0.2 m
apart, carrying currents of 1.5 A and 3 A respectively.
.



(a)
(b)
(c)
Calculate the force per metre on wire Q.
State the direction of the force.
State the direction and magnitude of the force per metre on
wire P due to the current in wire Q
Moment and couple

Couple ─ consists of 2 equal and opposite parallel
forces whose lines of action do not coincide (重疊).
F
F
d
d/2
F
torque of couple = F x d/2 + F x d/2 = Fd
d/2
F
Torque on a rectangular currentcarrying coil in a uniform B-field
F’
F’

Torque = F(b sin q)
= NBIlb sin q
= NBAI sin q
Maximum and minimum torque on a
coil
q  90o
q  0o


The maximum torque is NBAI when the plane of coil is
parallel to the field (q = 90o).
The torque on the coil is zero when the plane of coil is
perpendicular to the field (q = 0o).
Example 9



Example 9
A square coil has sides of length 5 cm. The coil consist
of 20 turns of insulated wire carrying a current of 0.2 A.
The plane of the coil is at an angle 40o to a uniform
magnetic field of flux density of 25 mT. Calculate the
torque acting on the coil.
Solution:
q  40o
Moving-coil galvanometer



The moving-coil meter
contains a coil wound on an
aluminium former around a
soft-iron cylinder.
The coil is pivoted on
bearings between the poles
of a cylindrical magnet.
Current flows through the coil
via a pair of spiral springs
called hair springs.
Theory


When a current is passed
through a coil in a magnetic
field, the coil experiences a
torque. The coil rotates,
moving the pointer across the
scale.
The normal of plane of the coil
is always perpendicular to the
magnetic field, the torque on
the coil is given by
T = NBAI sin 90o = NBAI
Theory
The movement of the coil is
opposed by the hair
springs.
 The restoring torque (t)
exerted by the hair springs
to oppose the rotation is
given by
t = kq
where k is the torsion
constant of the hairsprings

Linear scale


The coil comes to rest when the magnetic turning
effect (torque) on the coil is balanced by the turning
effect (restoring torque) from the hair springs.
BANI = kq
Hence, I ∝ q; the galvanometer scale is linear
Sensitivity



The current sensitivity of a galvanometer is defined as the
deflection per unit current
i.e.
current sensitivity = q/I = BAN/k.
The voltage sensitivity of a galvanometer is defined as the
deflection per unit voltage
i.e.
voltage sensitivity = q/V = q/(IR) = BAN/(kR)
where R is the resistance of the coil.
The sensitivity can be increased, i.e. the coil deflects more for a
certain current or voltage, by
1.
using a stronger magnet (larger B),
2.
using weaker hair springs (smaller k).
3.
using a coil with larger area (larger A), and
4.
increasing the number of turns of the coil (larger N).