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Chapter 4
CP3 1 FYSL
Chapter 4: Magnetic Field
4.1 Magnetic Field
L.O 4.1.1 Define magnetic field
Magnetic field is defined as the region around a magnet where a magnetic force can be
experienced.
Magnetic field has two poles, called north (N) and south (S). These magnetic poles are
always found in pairs whereas a single magnetic pole has never been found.
Like poles (N-N or S-S) repel each other.
Opposite poles (N-S) attract each other.
L.O 4.1.2 Identify magnetic field sources and sketch their magnetic field lines
Magnetic field lines are used to represent a magnetic field.
The characteristics of magnetic field lines:
The lines do not intersect one another
The lines form a closed loop: magnetic field lines leave the North-pole and enter the
South-pole.
The lines are closer together at the poles. (The number of lines per unit cross-
sectional area is proportional to the magnitude of the magnetic field.)
Two sets of magnetic field lines can be superimposed to form a resultant magnetic
field line.
Magnetic field can be represented by crosses or by dotted circles as shown in figures
below:
Magnetic field lines enter
the page perpendicularly
Magnetic field lines leave
the page perpendicularly
Chapter 4
CP3 2 FYSL
The pattern of the magnetic field lines can be determined by using two methods:
Using compass needles
Using sprinkling iron filings on paper
Magnetic field sources:
i. Bar magnet
One bar magnet Horseshoe or U magnet
Two bar magnets (unlike pole)
Two bar magnets (like pole)
Chapter 4
CP3 3 FYSL
ii. Current-carrying conductor
A stationary electric charge is surrounded by an electric field only.
When an electric charge moves, it is surrounded by an electric field and a
magnetic field. The motion of the electric charge produces the magnetic field.
A circular coil A long straightwire A solenoid
iii. Earth magnetic field
Example
Note that the Earth’s “North
Pole” is really a south magnetic
pole, as the north ends of
magnets are attracted to it.
N S
N S
N S
S N
Chapter 4
CP3 4 FYSL
4.2 Magnetic Field Produced by Current-Carrying Conductor
L.O 4.2.1 Use magnetic field equations
a) Long Straight Wire
View (a) from the top:
Magnitude of the magnetic field at any point
from the conductor (wire):
where
B = magnetic field strength / magnetic flux
density (T)
I = current in the wire (A)
r = perpendicularly distance of P from
the wire (m)
μo = constant of proportionality
known as the permeability of
free space (vacuum)
= 4π x 10-7
Henry per metre (H m-1
)
The direction of magnetic field around the
wire or coil can be determined by using the
right hand grip rule as shown in Figure(b).
Thumb – direction of current
Other fingers – direction of magnetic
field (clockwise OR anticlockwise)
r
IB
2
0
r B •
Chapter 4
CP3 5 FYSL
b) Circular Coil
View from the top:
(a)
(b)
Magnitude of the magnetic field at the centre
of the circular coil:
For ONE circular coil
For N circular coils
where
R = radius of the circular coil.
µ0 = permeability of free space
4π × 10-7
H m-1
I = current
N = number of coils (loops)
The direction of magnetic field around the
wire or coil can be determined by using the
right hand grip rule as shown in Figure(a).
Example of multiple circular loops:
B
B
Chapter 4
CP3 6 FYSL
c) Solenoid
View from the top:
(a)
The magnitude of magnetic field intensity at
the centre (mid-point/ inside) of N turn
solenoid is given by
Since , therefore it can be written as
The magnitude of magnetic field intensity at
the end of N turn solenoid is given by:
where
n = numbers of turns per unit length
µ0 = permeability of free space
4π × 10-7
H m-1
I = current
N = number of coils (loops)
The directions of the fields can be found by
viewing the current flows in the solenoid from
both end or applying the right hand grip rule
as shown in Figure (a).
Thumb – north pole
Other fingers – direction of current in
solenoid.
Chapter 4
CP3 7 FYSL
Example
Question Solution
Determine the magnetic field strength at
point X and Y from a long, straight wire
carrying a current of 5 A as shown below.
A circular coil having 400 turns of wire in air
has a radius of 6 cm and is in the plane of the
paper. What is the value of current must
exist in the coil to produce a flux density of
2 mT at its center?
An air-core solenoid with 2000 loops is 60
cm long and has a diameter of 2.0 cm. If a
current of 5.0 A is sent through it, what will
be the flux density within it?
A solenoid is constructed by winding 400
turns of wire on a 20 cm iron core. The
relative permeability of the iron is 13000.
What current is required to produce a
magnetic induction of 0.5 T in the center of
the solenoid?
Chapter 4
CP3 8 FYSL
Question Solution
Two straight parallel wires are 30 cm apart
and each carries a current of 20 A. Find the
magnitude and direction of the magnetic field
at a point in the plane of the wires that is 10
cm from one wire and 20 cm from the other
if the currents are
i. in the same direction,
ii. in the opposite direction.
Exercise
Question
Two long straight wires are oriented perpendicular to the page as shown in Figure below.
The current in one wire is I1 = 3.0 A pointing into the page and the current in the other wire is
I2 = 4.0 A pointing out of the page. Determine the magnitude and direction of the nett
magnetic field intensity at point P. (Given 0=4 x 10-7
H m-1
)
Answer: 8.93×10-6
T, 63.1° below +ve x-axis
A 2000 turns solenoid of length 40 cm and resistance 16 is connected to a 20 V supply.
Find the magnetic flux density at the end of the axis of the solenoid.
Answer: 3.95×10-3
T
Chapter 4
CP3 9 FYSL
4.3 Force on a Moving Charged Particle in a Uniform Magnetic Field
L.O 4.3.1 Use magnetic force equation
A stationary electric charge in a magnetic field will not experience any force. But if the
charge is moving with a velocity, v in a magnetic field, B then it will experience a
force. This force known as magnetic force.
The magnitude of the magnetic force can be calculated by using the equation below :
BvqF
sinqvBF
The direction of the magnetic force can be determined by usingright hand rule:
Example
Determine the direction of the magnetic force, exerted on a charge in each problems below:
where
q : magnitude of the charge
: angle between ⃗ and ⃗⃗
F
v
B
IMPORTANT!
Thumb indicates the direction
of the magnetic force exerted
on a positive charge.
If the charge is negative,
direction of the force is
opposite.
Chapter 4
CP3 10 FYSL
L.O 4.3.2 Describe circular motion of a charge in a uniform magnetic field
L.O 4.3.3 Use relationship FB = FC
Consider a charged particle moving in a uniform magnetic field with its velocity
perpendicular to the magnetic field.
As the particle enters the region, it will experience a magnetic force which the force is
perpendicular to the velocity of the particle. Hence the direction of its velocity changes
but the magnetic force remains perpendicular to the velocity.
Since the path is circle therefore the magnetic force FB contributes the centripetal force
Fc (net force) in this motion. Thus
CB FF
r
mvBqv
2
sin and θ = 90°
Bq
mvr
The period of the circular motion, T makes by the particle is given by
v
rT
2
Bq
mT
2
The frequency of the circular motion makes by the particle is given by
T
f1
m
Bqf
2
where m : mass of the charged particle
v : magnitude of the velocity
r : radius of the circular path q : magnitude of the charged particle
Chapter 4
CP3 11 FYSL
Example
Question Solution
A charge q1 = 25.0 μC moves with a speed of
4.5 x 103 m s
-1 perpendicularly to a uniform
magnetic field. The charge experiences a
magnetic force of 7.31 x 10-3
N. A second
charge q2 = 5.00 μC travels at an angle of 40.0 o
with respect to the same magnetic field and
experiences a 1.90 x 10 -3
N force. Determine
a. The magnitude of the magnetic field
b. The speed of q2.
An electron at point A in figure above has a
speed v of 2.50 106 m s
-1. Determine the
magnitude and direction of the magnetic field
that will cause the electron to follow the
semicircular path from A to B.
(Given e = 1.601019
C and me= 9.111031
kg)
Exercise
Question
Calculate the magnitude of the force on a proton travelling 3.0 x107 m s
-1 in the uniform
magnetic flux density of 1.5 Wb m-2
, if
a. the velocity of the proton is perpendicular to the magnetic field.
b. the velocity of the proton makes an angle 50 with the magnetic field.
(Given the charge of the proton is 1.60 x 10-19
C)
Answer: 7.2×10-12
N, 5.5×10-12
N
An electron is moving in a magnetic field. At a particular instant, the speed of the electron is
3.0 x 106 m s
-1. The magnitude of the magnetic force on the electron is 5.0 x 10
-13 N and the
angle between the velocity of the electron and the magnetic force is 30. Calculate the
magnitude of the magnetic field. Given e = 1.601019
C.
Answer: 1.2 T
A proton is moving with velocity 3 x 10 5 m s
-1 vertically across a magnetic field 0.02 T.
Calculate
a. kinetic energy of the proton
b. the magnetic force exerted on the proton
c. the radius of the circular path of the proton. (mp = 1.67 x 10 -27
kg)
Answer: 7.52 x 10-17
J , 9.6 x 10 -16
N, 0.16 m
Chapter 4
CP3 12 FYSL
4.4 Force on a Current Carrying Conductor in a Uniform Magnetic Field
L.O 4.4.1 Use magnetic force equation
When a current-carrying conductor is placed in a magnetic field B, thus a magnetic force
will acts on that conductor.
The magnitude of the magnetic force exerts on the current-carrying conductor is given by
BlIF
sinIlBF
The direction of the magnetic force can be determined by using right hand rule:
One tesla is defined as the magnetic flux density of a field in which a force of 1 newton
acts on a 1 metre length of a conductor which carrying a current of 1 ampere and is
perpendicular to the field.
Example
Determine the direction of the magnetic force, exerted on a conductor carrying current, I in
each problems below.
where l : length of the conductor
: angle between l and ⃗⃗
F
l
B
Chapter 4
CP3 13 FYSL
Example
Question Solution
A wire of 20 cm long is placed perpendicular to
the magnetic field of 0.40 Wb m-2
.
a. Calculate the magnitude of the force on the
wire when a current 12 A is flowing.
b. For the same current in (a), determine the
magnitude of the force on the wire when its
length is extended to 30 cm.
c. If the force on the 20 cm wire above is
60 x 10-2
N and the current flows is12 A,
find the magnitude of magnetic field was
supplied.
A square coil of wire containing a single turn is
placed in a uniform 0.25 T magnetic field. Each
side has a length of 0.32 m, and the current in
the coil is 12 A. Determine the magnitude of the
magnetic force on each of the four sides.
Exercise
Question
A straight wire with a length of 0.65 m and mass of 75 g is placed in a uniform magnetic field
of 1.62 T. If the current flowing in the wire is perpendicular to the magnetic field, calculate
the current required to balance the wire? (g = 9.81 ms-2
)
Answer: 0.70 A
A wire of length 0.655 m carries a current of 21.0 A. In the presence of a 0.470 T magnetic
field, the wire experiences a force of 5.46 N. What is the angle (less than 90o) between the
wire and the magnetic field?
Answer: 57.62°
Chapter 4
CP3 14 FYSL
4.5 Forces between Two Parallel Current-Carrying Conductors
L.O 4.5.1 Derive magnetic force per unit length of two parallel current-carrying
conductors
L.O 4.5.2 Use magnetic force per unit length equation
Consider two identical straight conductors X and Y carrying currents I1 and I2 with length l are
placed parallel to each other as shown in figure below. The conductors are in vacuum and their
separation is d.
Force exerted on conductor X Force exerted on conductor Y
The magnitude of the magnetic flux density,
B2 at point Q on conductor X due to the
current in conductor Y is given by:
Conductor X carries a current I1 and in the
magnetic field B2 then conductor X will
experiences a magnetic force, F21.
The magnitude of F21 is given by
The magnitude of the magnetic flux density,
B1 at point P on conductor Y due to the
current in conductor X is given by
Conductor Y carries a current I2 and in the
magnetic field B1 then conductor Y will
experiences a magnetic force, F12.
The magnitude of F12 is given by
d
IB
2
101
d
IB
2
202
View from the top
sin1221 lIBF and 90
90sin2
120
21 lId
IF
lId
IF 1
2021
2
Direction: out of the page
Direction: towards
conductor Y
Direction: into the page
sin2112 lIBF and 90
90sin2
210
12 lId
IF
lId
IF 2
1012
2
Direction: towards
conductor X
d
LIIFFF
2
2102112
Conclusion: Type of the force is attractive force
Chapter 4
CP3 15 FYSL
If the direction of current in conductor Y is change to upside down as shown in figure
The magnitude of F12 and F21 can be determined by using equations above and its direction by
applying right hand rule.
Conclusion: Type of the force is repulsive force
L.O 4.5.3 Define one ampere
If two long, straight, parallel conductors, 1.0 m apart in vacuum carry equal 1.0 A currents
hence the force per unit length that each conductor exerts on the other is
The ampere (one ampere) is defined as the current that flowing in each of two parallel
conductors which are 1.0 metre apart in vacuum, would produce a force per unit length
between the conductors of 2.0 x 10-7
N m-1
.
Example
Question Solution
Two long straight parallel wires are placed 0.25
m apart in a vacuum. Each wire carries a current
of 2.4 A in the same direction.
a. Sketch a labelled diagram to show clearly
the direction of the force on each wire.
b. Calculate the force per unit length between
the wires.
c. If the current in one of the wires is reduced
to 0.64 A, calculate the current needed in
the second wire to maintain the same force
per unit length between the wires as in (b).
(Given µ0 = 4π × 10-7
T m A-1
)
View from the top
d
II
L
F
2
210).π(
).)(.)(πx(
L
F
012
0101104 7
17 100.2 mNxL
F
Chapter 4
CP3 16 FYSL
4.6 Torque on a Coil
L.O 4.6.1 Use torque equation
Torque is the tendency of a force to rotate an object about an axis.
For a coil of N turns, the magnitude of the torque is given by
BANI
sinNIAB
Top view Side view
where τ : torque on the coil
A : area of the coil
B : magnetic flux density I : current flows in the coil
θ : angle between vector area, A (the normal to plane of the coil) and B
N : number of turns (coils)
Chapter 4
CP3 17 FYSL
L.O 4.6.2 Explain the working principles of a moving coil galvanometer
The galvanometer is the main component in analog meters for measuring current and
voltage.
It consists of a magnet, a coil of wire, a spring, a pointer and a calibrated scale.
The coil of wire contains many turns and is wrapped around a soft iron cylinder.
The coil is pivoted in a radial magnetic field, so that no matter what position it turns, the
plane of the coil is always parallel to the magnetic field.
The basic operation of the galvanometer uses the fact that a torque acts on a current loop in
the presence of a magnetic field.
When there is a current in the coil, the coil rotates in response to the torque (τ = NIAB )
applied by the magnet.
This causes the pointer (attached to the coil) to move in relation to the scale.
The torque experienced by the coil is proportional to the current in it; the larger the current,
the greater the torque and the more the coil rotates before the spring tightens enough to stop
the rotation.
Hence, the deflection of the pointer attached to the coil is proportional to the current.
The coil stops rotating when this torque is balanced by the restoring torque of the spring.
From this equation, the current I can be calculated by measuring the angle θ.
This working principles of a moving coil galvanometer also used in voltmeter (multiplier),
ammeter (shunt), ohmmeter and multimeter.
Structure of a moving-coil galvanometer
coil theof anglerotation :θ radianin
constant) (torsional spring theofconstant stiffness theis where
torquerestoringforce magnetic todue Torque s
k
NAB
kI
kNIAB
Chapter 4
CP3 18 FYSL
Example
Question Solution
A 20 turns rectangular coil with sides 6.0 cm
x 4.0 cm is placed vertically in a uniform
horizontal magnetic field of magnitude 1.0 T.
If the current flows in the coil is 5.0 A,
determine the torque acting on the coil when
the plane of the coil is
a. perpendicular to the field,
b. parallel to the field,
c. at 60 to the field.
Exercise
Question
A rectangular coil of 10 cm x 4.0 cm in a galvanometer has 50 turns and a magnetic flux
density of 5.0 x 10-2
T. The resistance of the coil is 40 and a potential difference of 12 V is
applied across the galvanometer, calculate the maximum torque on the coil.
Answer: 3.0 x 10-3
N m.
A moving coil meter has a 50 turns coil measuring 1.0 cm by 2.0 cm. It is held in a radial
magnetic field of flux density 0.15 T and its suspension has a torsional constant of 3.0 × 10-6
N m rad-1
. Determine the current is required to give a deflection of 0.5 rad.
Answer: 1.0 × 10-3
A
Calculate the magnetic flux density required to give a coil of 100 turns a torque of 0.5 Nm
when its plane is parallel to the field. The dimension of each turn is 84 cm2 , and the current
is 9.0 A.
Answer: 66.1 mT
Chapter 4
CP3 19 FYSL
4.7 Motion of Charged Particle in Magnetic Field and Electric Field
L.O 4.7.1 Explain the motion of a charged particle in both magnetic field and electric
field
L.O 4.7.2 Derive and use velocity equation in a velocity selector
Velocity Selector
Velocity selector consists of a pair of parallel metal plates across which an electric field
can be applied in uniform magnetic field.
Consider a positive charged particle with mass m, charge q and speed v enters a region of
space where the electric and magnetic fields are perpendicular to the particle’s velocity
and to each other as shown in figure above.
The charged particle will experiences the electric force FE is downwards with magnitude
qE and the magnetic force FB is upwards with magnitude Bqv as shown in figure above.
If the particle travels in a straight line with constant velocity hence the electric and
magnetic forces are equal in magnitude (dynamic equilibrium). Therefore
Only particles with velocity equal to E/B can pass through without being deflected by
the fields. Equation above also works for electron or other negative charged particles.
Mass Spectrometer
EB FF
qEqvB 90sin
B
Ev
When the charged particle entering
the region consists of magnetic field
only, the particle will make a
semicircular path of radius r.
CB FF
r
mvBqv
2
and B
Ev
2rB
E
m
q
Chapter 4
CP3 20 FYSL
Example
Question Solution
A velocity selector is to be constructed to
select ions (positive) moving to the right at
6.0 km s-1
. The electric field is 300 Vm-1
upwards. What should be the magnitude and
direction of the magnetic field?
An ion enters a uniform magnetic field B.
The path of the ion is a spiral as shown in the
figure.
a) Determine whether the ion is positively
or negatively charged.
b) Give reasons for the shape of the path.
Exercise
Question
An electron moving at a steady speed of 0.50106 m s
1 passes between two flat, parallel
metal plates 2.0 cm apart with a potential difference of 100 V between them. The electron is
kept travelling in a straight line perpendicular to the electric field between the plates by
applying a magnetic field perpendicular to the electron’s path and to the electric field.
Calculate :
a. the intensity of the electric field.
b. the magnetic flux density needed.
Answer : 0.50104 V m
1; 0.010 T
An electron with kinetic energy of 8.01016
J passes perpendicular through a uniform
magnetic field of 0.40103
T. It is found to follow a circular path. Calculate
a. the radius of the circular path.
b. the time required for the electron to complete one revolution.
(Given e = 1.761011
C kg-1
, me = 9.111031
kg)
Answer: 0.595 m, 8.92×10-8
s
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