Title: CHAPTER 19 : MAGNETIC FIELD (7 hours)
1CHAPTER 19 MAGNETIC FIELD (7 hours)
19.1 Magnetic field 19.2 Magnetic field
produced by current-carrying conductor 19.3
Force on a moving charged particle in a uniform
magnetic field 19.4 Force on a
current-carrying conductor in a uniform
magnetic field 19.5 Forces between two
parallel current-carrying conductors 19.6
Torque on a coil 19.7 Motion of charged
particle in magnetic field and electric
field
219.1 Magnetic Field (1 hour)
- Learning Outcomes
- At the end of this lesson, the students should be
able to - Define magnetic field.
- Identify magnetic field sources.
- Sketch the magnetic field lines.
3Magnetic field
19.1 Magnetic field
- is defined as a region surrounding a magnet
- or a conductor carrying current where a
- magnetic force is experienced.
- Magnets always have two poles
- a) North and south poles.
- b) Like poles repel and unlike poles attract.
Magnetic field lines
- A magnetic field can be represented by
- magnetic field lines (straight lines or curves).
- Arrows on the lines show the direction of the
- field the arrows point away from north poles
- and towards south poles.
419.1 Magnetic field
- A uniform field is represented by parallel
- lines. This means that the number of lines
- passing perpendicularly through unit area
- at all cross-sections in a magnetic field
- are the same as shown below.
519.1 Magnetic field
- A non-uniform field is represented by non-
- parallel lines. The number of magnetic field
- lines varies at different unit cross-sections
- as shown below.
A1
A2
619.1 Magnetic field
- Magnetic field lines do not intersect
- one another.
- The tangent to a curved field line at a
- point indicates the direction of the
- magnetic field at that point.
719.1 Magnetic field
- The number of lines per unit cross section
- area is an indication of the strength of
- the field. The number of lines per unit
- cross-sectional area is proportional to
- the magnitude of the magnetic field.
A1
A2
stronger field in A1
819.1 Magnetic field
- Magnetic field can also be represented by
- crosses or by dotted circles as shown
- below.
Magnetic field lines enter the page
perpendicularly
Magnetic field lines leave the page
perpendicularly
B into the page
B out of page
9Field Patterns
19.1 Magnetic field
- The magnetic field lines pattern can be
- obtained by using iron filings or a plotting
- compass.
the arrowhead of a compass needle is a north pole.
1019.1 Magnetic field
- The direction of the magnetic field at a point
- is defined as the direction of a compass
- needle points when placed at that point.
1119.1 Magnetic field
Field Sources
b. Horseshoe or U magnet
a. A bar magnet
c. Two bar magnets (unlike pole) - attractive
1219.1 Magnetic field
d. Two bar magnets (like poles) - repulsive
Neutral point (point where the resultant magnetic
force/field strength is zero)
1319.1 Magnetic field
e. A circular coil
f. A long straight wire
g. A solenoid
1419.1 Magnetic field
h. Earth Magnetic Field
1519.1 Magnetic field
Magnetic Flux, ?
- is a measure of the number of field lines
- that cross a surface area.
- is defined as the scalar product between
- the magnetic flux density, B and the vector
- of the surface area, A.
1619.1 Magnetic field
- scalar quantity.
- unit weber(Wb)/ tesla-meter squared(T.m2)
- 1 T.m2 1 Wb
- Consider a uniform magnetic field B
- passing through a surface area A as shown
- in figure below.
In Figure below, ? 0?
1719.1 Magnetic field
If ? 90?
Magnetic Flux Density, B
- is defined as the magnetic flux per unit
- area at right angles to the magnetic field.
1819.1 Magnetic field
- vector quantity and its direction follows the
direction of the magnetic field. - unit weber per metre squared (Wb m-2)
- or tesla (T).
19CHAPTER 19 MAGNETIC FIELD
19.1 Magnetic field 19.2 Magnetic field
produced by current-carrying
conductor 19.3 Force on a moving charged
particle in a uniform magnetic
field 19.4 Force on a current-carrying
conductor in a uniform magnetic
field 19.5 Forces between two parallel
current-carrying conductors 19.6 Torque on a
coil 19.7 Motion of charged particle in
magnetic field and electric field
2019.2 Magnetic field produced by current-carrying
conductor (1 hour)
- Learning outcomes
- At the end of this lesson, the students should be
able to - Apply magnetic field formula
- (a) for a long straight wire
- (b) for a circular coil
- (c) for a solenoid
2119.2 Magnetic field produced by current-carrying
conductor
Types of current-carrying conductor
2219.2 Magnetic field produced by current-carrying
conductor
1. A long straight wire
- B is a vector quantity.
- Magnitude
(4?? x 10 -7 H m-1)
View from the top
2319.2 Magnetic field produced by current-carrying
conductor
B magnetic field strength / 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) 4p x 10-7 Henry per metre (H
m-1)
r
P
Direction right-hand grip rule
out of the page
2419.2 Magnetic field produced by current-carrying
conductor
Example 1
Determine the magnetic field strength at point X
and Y from a long, straight wire carrying a
current of 5 A as shown below.
2519.2 Magnetic field produced by current-carrying
conductor
2. A circular coil
N
S
2619.2 Magnetic field produced by current-carrying
conductor
r
Magnetic field strength at the center given as
r radius of the coil (m)
For N loops / number of turns on the coil
2719.2 Magnetic field produced by current-carrying
conductor
Example 2
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 ?
2819.2 Magnetic field produced by current-carrying
conductor
3. A solenoid
2919.2 Magnetic field produced by current-carrying
conductor
Magnetic field strength at the center
L
number of turns per length
3019.2 Magnetic field produced by current-carrying
conductor
Example 3
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 ?
3119.2 Magnetic field produced by current-carrying
conductor
Example 4
1. 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 ?
3219.2 Magnetic field produced by current-carrying
conductor
Exercise
1. 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.
3319.2 Magnetic field produced by current-carrying
conductor
- 2. A student is provided with a 3.0 m long wire
with a current of 0.15 A flowing through it. What
is the strength of the magnetic field at the
centre of the wire if the wire is bent into a
circular coil of one turn ? ( B 1.97 x
10-7 T ) - 3. A circular coil has 15 turns and a diameter of
45.0 cm. If the magnetic field strength at the
centre of the coil is 8.0 x 10-4 T, find the
current flowing in the coil. ( I 19.1 A
) - ( µ0 4p x 10-7 Hm-1 )
34CHAPTER 19 MAGNETIC FIELD
19.1 Magnetic field 19.2 Magnetic field
produced by current-carrying conductor 19.3
Force on a moving charged particle in a uniform
magnetic field 19.4 Force on a
current-carrying conductor in a uniform
magnetic field 19.5 Forces between two
parallel current-carrying conductors 19.6
Torque on a coil 19.7 Motion of charged
particle in magnetic field and electric
field
3519.3 Force on a moving charged particle
in a uniform magnetic field (1 hour)
- Learning Outcomes
- At the end of this lesson, the students should be
able to - Use force,
- Describe circular motion of a charge in a uniform
magnetic field. - Use relationship
3619.3 Force on a moving charged particle in a
uniform magnetic field.
- A charge q moving with speed v at angle ? with
the direction of a uniform magnetic field of
magnitude B experiences a magnetic force of
magnitude,
Where ? angle between B and v
For electron , q e.
3719.3 Force on a moving charged particle in a
uniform magnetic field.
Direction of F
- Flemings right hand rule - negative charge
- Flemings left hand rule - positive charge
-
positive charge
Thumb direction of Force (F) First finger
direction of Magnetic field (B) Second finger
direction of Velocity (v)
3819.3 Force on a moving charged particle in a
uniform magnetic field.
Example 5
A.
B.
C.
D.
3919.3 Force on a moving charged particle in a
uniform magnetic field.
Example 7
- A charge q1 25.0 µC moves with a speed of
- 4.5 x 103 m/s 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 -
- The magnitude of the magnetic field and
- The speed of q2.
4019.3 Force on a moving charged particle in a
uniform magnetic field.
Solution 7
4119.3 Force on a moving charged particle in a
uniform magnetic field.
Exercise
1. Calculate the magnitude of the force on
a proton travelling 3.1 x 107 m s-1 in the
uniform magnetic flux density of 1.6 Wb m-2,if
(i) the velocity of the proton is
perpendicular to the magnetic field. (ii)
the velocity of the proton makes an angle 60?
with the magnetic field. (charge of the proton
1.60 x 10-19 C)
4219.3 Force on a moving charged particle in a
uniform magnetic field.
Circular Motion of a Charged Particle in a
Uniform Magnetic Field
- Consider a charged particle moving in a uniform
magnetic field with its velocity (v)
perpendicularly to the magnetic field (B). - As the particle enters the region, it will
experience a magnetic force (F) 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. - This magnetic force causes the particle to move
in a circle.
4319.3 Force on a moving charged particle in a
uniform magnetic field.
- The magnetic force provides the centripetal
force for the particle to move in circular motion.
r? m?
4419.3 Force on a moving charged particle in a
uniform magnetic field.
- The time for one rotation (period),
and
and
4519.3 Force on a moving charged particle in a
uniform magnetic field.
Exercise (DIY)
A proton is moving with velocity 3 x 10 5 m/s
vertically across a magnetic field 0.02 T. (mp
1.67 x 10 -27 kg) 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.
7.52 x 10-17 J , 9.6 x 10 -16 N, 0.16 m
4619.3 Force on a moving charged particle in a
uniform magnetic field.
Exercise
1. An electron is projected from left to right
into a magnetic field directed into the page. The
velocity of the electron is 2 x 10 6 ms-1 and
the magnetic flux density of the field is 3.0 T.
Find the magnitude and direction of the magnetic
force on the electron. (charge of electron
1.6 x 10-19 C) (9.6 x
10-13 N, downwards)
4719.3 Force on a moving charged particle in a
uniform magnetic field.
- 2. A proton with a mass of 1.67 x 10-27 kg is
moving in a circular orbit perpendicular to a
magnetic field. The angular velocity of the
proton is 1.96 x 104 rad s-1 . Determine - (i) the period of revolution,
- (ii) the magnetic field strength of the field.
- (charge of proton 1.6 x 10-19 C)
-
- (T 3.2 x 10-4 s , B 2.05 x 10-4 T)
48CHAPTER 19 MAGNETIC FIELD
19.1 Magnetic field 19.2 Magnetic field
produced by current-carrying conductor 19.3
Force on a moving charged particle in a uniform
magnetic field 19.4 Force on a
current-carrying conductor in a uniform
magnetic field 19.5 Forces between two
parallel current-carrying conductors 19.6
Torque on a coil 19.7 Motion of charged
particle in magnetic field and electric
field
4919.4 Force on a current-carrying conductor in a
uniform magnetic field (1 hour)
- Learning Outcomes
- At the end of this lesson, the students should be
able to - (i) Use force,
5019.4 Force on a current-carrying conductor in a
uniform magnetic field.
- When a current-carrying conductor is placed in a
magnetic field B, thus a magnetic force will act
on that conductor. - The magnitude of the magnetic force exerts on the
current-carrying conductor is given by - In vector form,
5119.4 Force on a current-carrying conductor in a
uniform magnetic field.
- Direction of F Flemings left hand rule.
Thumb direction of Force (F) First finger
direction of Magnetic field (B) Second finger
direction of Current (I)
5219.4 Force on a current-carrying conductor in a
uniform magnetic field.
5319.4 Force on a current-carrying conductor in a
uniform magnetic field.
Example 19.4.1
- Determine the direction of the magnetic
force, exerted on a conductor carrying current, I
in each problem below. - a. b.
b.
a.
5419.4 Force on a current-carrying conductor in a
uniform magnetic field.
Example 19.4.2
- 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?
5519.4 Force on a current-carrying conductor in a
uniform magnetic field.
Exercise
- 1. 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.
B
90o
0.96 N (top and bottom sides) 0 N (left and right
sides)
I
5619.4 Force on a current-carrying conductor in a
uniform magnetic field.
- 2. 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 ) ( I 0.70 A )
57CHAPTER 19 MAGNETIC FIELD
19.1 Magnetic field 19.2 Magnetic field
produced by current-carrying conductor 19.3
Force on a moving charged particle in a uniform
magnetic field 19.4 Force on a
current-carrying conductor in a uniform
magnetic field 19.5 Forces between two
parallel current-carrying
conductors 19.6 Torque on a coil 19.7 Motion
of charged particle in magnetic field and
electric field
5819.5 Force between twp parallel
current-carrying conductors(1 hour)
- Learning Outcomes
- At the end of this lesson, the students should be
able to - Derive force per unit length of two parallel
current-carrying conductors. - Use force per unit length,
- Define one ampere.
5919.5 Forces between two parallel current-
carrying conductors.
- Consider two identical straight conductors X and
Y carrying currents I1 and I2 with length L are
placed parallel to each other as shown below.
- The conductors are in vacuum and their separation
is d.
6019.5 Forces between two parallel current-
carrying conductors.
- The magnitude of the magnetic flux density, B1 at
point P on conductor Y due to the current in
conductor X is given by
into the page
- Conductor Y carries a current I2 and in the
magnetic field B1 then conductor Y experiences
a magnetic force, F12.
6119.5 Forces between two parallel current-
carrying conductors.
to the left (towards X)
to the right (towards Y)
Attractive force
6219.5 Forces between two parallel current-
carrying conductors.
- If the direction of current in conductor Y is
changed to upside down as shown below
Repulsive force
- The currents are in the same direction
- (2 conductors attract each other)
- The currents are in the opposite direction
- (2 conductors repel each other)
6319.5 Forces between two parallel current-
carrying conductors.
Rearrange,
If I1 I2 1 A and d 1 m , then
Definition of 1 Ampere
6419.5 Forces between two parallel current-
carrying conductors.
Definition of 1 Ampere
One ampere is defined as the constant current
that, when it is flowing in each of two
infinitely long, straight, parallel conductors
which have negligible of cross sectional areas
and 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.
6519.5 Forces between two parallel current-
carrying conductors.
Example 19.5.1
Two very long parallel wires are placed 2.0 cm
apart in air. Both wires carry a current of 8.0 A
and 10 A respectively. Find the magnitude of the
magnetic force in newton, on each metre length of
wire.
6619.5 Forces between two parallel current-
carrying conductors.
Exercise
1. Two long parallel wires are 5.0 cm apart. They
each exerts a force of attraction per unit length
on the other of 6 x 10 -7 Nm-1 . The current in
one wire is 400 mA. (i) Calculate the current in
the second wire. (ii) In which direction is the
current in the second wire, relative to the first
?
I2 0.375 A (same direction)
67CHAPTER 19 MAGNETIC FIELD
19.1 Magnetic field 19.2 Magnetic field
produced by current-carrying conductor 19.3
Force on a moving charged particle in a uniform
magnetic field 19.4 Force on a
current-carrying conductor in a uniform
magnetic field 19.5 Forces between two
parallel current-carrying
conductors 19.6 Torque on a coil 19.7 Motion
of charged particle in magnetic field and
electric field
6819.6 Torque on a coil (1 hour)
- Learning Outcomes
- At the end of this lesson, the students should be
able to - Use torque, where N number
of turns. - Explain the working principles of a moving coil
galvanometer.
6919.6 Torque on a coil.
- Consider a rectangular loop with length a and
width b is pivoted so that it can rotate about a
vertical axis (shown in figure) which is at right
angle to a uniform magnetic field of flux density
B.
Axis of rotation
Top view
7019.6 Torque on a coil.
- When a steady current I passes round the coil, a
magnetic force acts on each side of the coil. - No magnetic forces act on sides 1 and 3 because
these wires are parallel to the field.
Axis of rotation
Top view
7119.6 Torque on a coil.
- The two magnetic forces on sides 2 and 4 each of
length a are equal and opposite and have the
value F where, - F2 F4 BIL BIa
Side view
- The forces exerted a torque that tends to rotate
the coil clockwise.
7219.6 Torque on a coil.
- The magnitude of this torque for each side is
abA (area of the coil) ? angle between B
and the normal to plane of the coil
7319.6 Torque on a coil.
Example 19.6.1
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.
7419.6 Torque on a coil.
Solution 19.6.1
(a)
(b)
(c)
7519.6 Torque on a coil.
Exercise
1. A rectangular loop of wire has an area of 0.30
m2 . The plane of the loop makes an angle of 30o
with a 0.75 T magnetic field. What is the torque
on the loop if the current is 7.0 A ?
Solution
7619.6 Torque on a coil.
Exercise
2. 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.
Solution
7719.6 Torque on a coil.
A moving coil galvanometer
Structure of a moving-coil galvanometer
Structure of a moving-coil galvanometer
7819.6 Torque on a coil.
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.
7919.6 Torque on a coil.
A moving coil galvanometer
- 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.
8019.6 Torque on a coil.
A moving coil galvanometer
- 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 ( t NABI )
applied by the magnet. - This causes the pointer ( attached to the coil)
to move in relation to the scale.
8119.6 Torque on a coil.
A moving coil galvanometer
- 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.
8219.6 Torque on a coil.
- 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 ?.
8319.6 Torque on a coil.
- This working principles of a moving coil
- galvanometer also used in voltmeter
- (multiplier), ammeter (shunt) , ohmmeter
- and multimeter.
8419.6 Torque on a coil.
- Exercise
- The moving coil of a galvanometer has 100 turns
and an area of 1.5 x 10-4 m2 . It is suspended
by a wire with a torsional constant of 2.6 x
10-8 Nm rad-1 . The coil is placed in a radial
magnetic field of 0.1 T. Calculate the current
flowing in the coil if a deflection of 1.2 rad is
observed. - ( I 2.08 x 10-5 A )
85CHAPTER 19 MAGNETIC FIELD
19.1 Magnetic field 19.2 Magnetic field
produced by current-carrying conductor 19.3
Force on a moving charged particle in a uniform
magnetic field 19.4 Force on a
current-carrying conductor in a uniform
magnetic field 19.5 Forces between two
parallel current-carrying
conductors 19.6 Torque on a coil 19.7 Motion
of charged particle in magnetic field and
electric field
8619.7 Motion of charged particle in magnetic
field and electric field (1 hour)
- Learning Outcomes
- At the end of this lesson, the students should be
able to - Explain the motion of a charged particle in both
magnetic field and electric field. - Derive and use velocity, in a
velocity selector.
8719.7 Motion of charged particle in magnetic
field and electric field
- Consider a charged particle q moves with a
velocity v in combined electric and magnetic
fields (the electric and magnetic fields are
perpendicular), the particle experiences no
resultant force ( a 0).
- The particle will continue to move in the same
direction with the same velocity.
- For this to happen, the electric force downward
must balance the magnetic force upwards (refer
diagram),
8819.7 Motion of charged particle in magnetic
field and electric field
89Velocity Selector
19.7 Motion of charged particle in magnetic
field and electric field
- A velocity selector uses this property of crossed
electric and magnetic fields to select a single
velocity of particle only particles traveling at
this velocity will be undeflected.
9019.7 Motion of charged particle in magnetic
field and electric field
Velocity selector
9119.7 Motion of charged particle in magnetic
field and electric field
Example 19.7.1
What is the velocity of protons (1 e) injected
through a velocity selector if E 3 x 105 V/m
and B 0.25 T ?
Solution
9219.7 Motion of charged particle in magnetic
field and electric field
Exercise
1. A velocity selector is to be constructed to
select ions (positive) moving to the right at 6.0
kms-1 . The electric field is 300 Vm-1
upwards. What should be the magnitude and
direction of the magnetic field?
Solution
9319.7 Motion of charged particle in magnetic
field and electric field
- When the magnetic field only is applied, the
particle moves in an arc of a circle of radius r
under the action of the centrally-directed
magnetic,
9419.7 Motion of charged particle in magnetic
field and electric field
Only B is applied
Both E and B are applied
9519.7 Motion of charged particle in magnetic
field and electric field
9619.7 Motion of charged particle in magnetic
field and electric field
A mass spectrometer
- A mass spectrometer is a device used for
separating atoms or molecules according to their
mass. - The atoms or molecules are ionized and then
accelerated through an electric field, giving
them a speed which depends on their mass (their
kinetic energies are all the same). - Then they enter a region of uniform magnetic
field, which bends them in a circular path.
9719.7 Motion of charged particle in magnetic
field and electric field
A mass spectrometer
- The radius of the path depends on the momentum of
the particle if the kinetic energies are the
same and the masses are different, the momentum
will be different as well. - A detector can be placed to detect particles
whose path has a particular radius, thereby
selecting a particular mass.
9819.7 Motion of charged particle in magnetic
field and electric field
A mass spectrometer
99(No Transcript)