Background knowledge Magnetic field
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Transcript of Background knowledge Magnetic field
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Background knowledge Magnetic field
Magnetic field can be produced by a magnet or a current-carrying conductor.
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Magnetic field lines
Magnetic field lines are used to show the strength and direction of a magnetic field.
N SN S
Note:1. Field lines run from the N-pole round to the S-pole.2. When field lines are closely-spaced, field is strong and vice-versa.
N NN N
Neutral point
x
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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
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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
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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
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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).
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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.
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Earth’s magnetic field
It is convenient to resolve the
earth’s field strength B into
horizontal and vertical component,
BH and BV respectively.
B
BH
BV
N
ground
We have BH = Bcos and BV = Bsin .Compass needles, whose motion is confined to a horizontal plane, are affected by BH only.
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Fleming’s left-hand rule
F: force B: magnetic field I: current
F
B
I
Magnetic forcecurrent
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Factors affecting magnetic force
B: magnetic field I: current l: length of conductor in B-field
F
B
I
Magnetic forcecurrent
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Current balance
magnetic force F acting upwards
weight of the rider is equal to the magnetic force when the frame PQRS is balanced.
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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
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magnetic force and length
Keep the current through PQ constant. Record the number of riders required to balance the metal fram
e 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
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magnetic force and magnetic field
Results: The magnetic force (number of riders) magnetic field (coil current) i.e. F B
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 bala
nce different number of identical riders on PQ.
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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)
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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.
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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
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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
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Magnetic force on a current-carrying conductor
From definition: B = F/Il F = BIl
If the conductor and field are not at right angles, but make an angle with one another, the expression becomes
F = BIl sin .
Note that:1. when = 90o (conductor field), ⊥ F = BIl.2. when = 0o (conductor // field), F = 0.
B
Il
I cos
I sin
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Magnetic force on moving charge in a magnetic field
Current = Q / t
conductor containing n particles each having charge q
sectionathroughpasstochargesallforrequiredtime
conductortheinchargetotal
l
nqv
vl
nqI
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Magnetic force on moving charge in a magnetic field
Magnetic force acting on each moving charge
If the conductor makes an angle with the magnetic field,
F = Bqv sin
conductor containing n particles each having charge q
n
conductortheonactingforcetotal
n
ll
nqvB
n
BIlF
sinsin
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Motion of a charged particle in a magnetic field
1 Moving parallel to the field
When a charged particle moves parallel to the magnetic field, i.e. = 0o, there is no force acting on it. Its velocity remains unchanged and its path is a
straight line.
vMagnetic field B+q
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2 Moving perpendicular to the field
Radius of circular path
Circular path of motion of +q
uniform magnetic field B
uniform magnetic field B
+q+q
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2 Moving perpendicular to the field
Period of circular motion
Circular path of motion of +q
uniform magnetic field B
uniform magnetic field B
+q+q
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2 Moving perpendicular to the field
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.
Circular path of motion of +q
uniform magnetic field B
uniform magnetic field B
+q+q
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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:
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Current balance
To measure steady magnetic field by using principle of moment.
Clockwise moment = anti-clockwise moment
mgd2 = BIld1
rider
Magnetic forceBIl
mechanical forcemg
d1 d2
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Example 3 Current balanceA 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.
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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.
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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.
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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.
Cathode-ray oscilloscope (CRO)
simulation
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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) = 3 x 2 = 6 V
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Measuring a.c. voltages To measure an a.c. voltage, the time base is usually switch
ed 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 =
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Example 4CRO waveformThe 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
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Example 4CRO waveformThe 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
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Example 4CRO waveformThe 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
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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..
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Example 5 displaying a.c. waveforms
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Example 5 displaying a.c. waveforms
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Measuring of phase relationships The phase difference of two p.ds (Vx and
Vy) can be observed on phase difference a double-beam CRO.
phase difference
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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.
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Comparing of frequencies When two p.ds of different frequencies fx and fy are applie
d to the X- and Y-plates (time base off), more complex figures are obtained, know as Lissajous’ figures.
simulation
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In any particular case, the frequency ratio can be found by
lineverticalltouchingloopsofno.
linehorizontaltouchingloopsofno.
x
y
f
f
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In any particular case, the frequency ratio can be found by
lineverticalltouchingloopsofno.
linehorizontaltouchingloopsofno.
x
y
f
f
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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.
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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.
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Experiment to show that B ∝ I and B ∝ 1 / r
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B ∝ I and B 1 /∝ r B ∝ I/r B = 0I/(2r) where I is the current and is the permeability of free
space (0 = 4 x 10-7 T m A-1)
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B = 0I/(2r) (0 = 4 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
(a) P P Q
X Y
I = 20 A I = 20 A
5 mm 5 mm 5 mm
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B = 0I/(2r) (0 = 4 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
(b) Q. P Q
X Y
I = 20 A I = 20 A
5 mm 5 mm 5 mm
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Magnetic field around a flat coil
At the centre of the coil The field lines are straight an
d at right angles to the plane of the coil
Outside the coil, The field lines run in loops.
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Experimental set-up
By using the experimental-setup shown, it is found that the magnetic field is 1. directly proportional
to the current and the number o
f turns, and 2. inversely proportional to the radius of the
coil.
Note: The magnetic field is greatest at the centre.
r
NIB
20
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Magnetic field due to a long solenoid
From the field pattern of the solenoid, it can be found that1. 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.
I I
NS
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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.
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Magnetic field due to a long solenoid
The magnetic field of the solenoid can be increased by
1. increasing the current,
2. increasing the number of turns on the coil.
I I
NS
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Magnetic field due to a long solenoid
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 = 0NI/l or B = 0nI where n is the number of turns per unit length (n = N/l)
I I
NS
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Experiment to show that B ∝ N, B ∝ I and B 1 /∝ l
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Magnetic field due to a long solenoid
Note: The magnetic field within a solenoid is independent of the shape and the cross section area of the solenoid.
I I
NS
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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.
222
1' 00 nI
l
NIBB
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r
IB
2
0r
NIB
20
nIl
NIB 0
0
l
Nn
2200 nI
l
NIB
N = number of turnsl = length of solenoid
Current-carrying conductor
Position of magnetic field
Magnetic flux density (B)
Symbol
1. Long straight wireAround the wire
r = perpendicular distance from wire
2. Circular coil At the centreN = number of turnsr = radius of coil
3. Solenoid
Inside
At the ends
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Force between currents
Magnetic field due to current through P
Magnetic force acting on Q
QP
B1
F2
r
IB
2
101
r
lIIlIBF
2210
212
Applying Fleming’s left hand rule, force acting on wire Q is towards wire P.
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B2
F1PQ
r
lIIF
2210
1
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.
∴
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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.
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Summary: Unlike current repel, like current attract
The force per unit length on each conductor
When the current I1 = I2 = 1 A, and the separation
between the wires r = 1 m,
the force per unit length on the conductor
r
lIIFF
2
21021
r
II
l
F
l
F
2
21021
77
10212
11104
N m-1
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Summary: Unlike current repel, like current attract
Definition of the ampere The ampere is constant current which, flowing in two infinit
ely 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.
r
lIIlIBF
2210
212
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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) Calculate the force per metre on wire Q. (b) State the direction of the force. (c) State the direction and magnitude of the force per metre
on wire P due to the current in wire Q
.
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Moment and couple
Couple ─ consists of 2 equal and opposite parallel forces whose lines of action do not coincide ( 重疊 ).
torque of couple = F x d/2 + F x d/2 = Fd
F
Fd/2 d/2
F
F
dd
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Torque on a rectangular current-carrying coil in a uniform B-field
Torque = F(b sin )= NBIlb sin = NBAI sin
F’
F’
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Maximum and minimum torque on a coil
The maximum torque is NBAI when the plane of coil is parallel to the field ( = 90o).
The torque on the coil is zero when the plane of coil is perpendicular to the field ( = 0o).
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Example 9
Example 9 A square coil has sides of length 5 cm. The coil consis
t 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:
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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.
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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
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Theory
The movement of the coil is opposed by the hair springs.
The restoring torque () exerted by the hair springs to oppose the rotation is given by
= k where k is the torsion
constant of the hairsprings
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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 = k Hence, I ∝ ; the galvanometer scale is linear
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Sensitivity
The current sensitivity of a galvanometer is defined as the deflection per unit current
i.e. current sensitivity = /I = BAN/k. The voltage sensitivity of a galvanometer is defined as the deflect
ion per unit voltage
i.e. voltage sensitivity = /V = /(IR) = BAN/(kR)
where R is the resistance of the coil. The sensitivity can be increased, i.e. the coil deflects more for a c
ertain 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).