Download - PDT UNIT14 Students)
SF027 1
A group of A group of
phenomena phenomena
associated with associated with
magnetic field.magnetic field.
UNIT 14: MAGNETISMUNIT 14: MAGNETISM
SF027 2
Learning outcomes:
14.1 Magnetic field� Explain magnetic field
� Identify magnetic field sources
� Sketch the magnetic field lines.
� Determine the direction of magnetic field
14.2 Force on a moving charged particle in a uniform magnetic field
► Explain the effect on a charge moving in a uniform magnetic field
► Use to explain the factors affecting the magnitude of the force
θsinqvBF =
14.3 Force on a current-carrying conductor in a uniform magnetic field
� Explain the effect on a current-carrying conductor in a uniform magnetic field
� Determine the direction of the force using both Right Hand Rule and Fleming’s Left Hand Rule.
� Use to explain the factors affecting the magnitude of the force.
θsinIlBF =
SF027 3
� Definition – the region around a magnet where a magnetic force can be experienced.
� A stationary electric chargestationary electric charge is surrounded by an electric field only.
� When an electric charge movesan electric charge moves, it is surrounded by an electric field and a magnetic field.The motion of the electric charge produces the magnetic fieldmotion of the electric charge produces the magnetic field.
� Magnetic field has two poles; north (N)north (N) and south (S)south (S). The magnetic poles are always found in pairs.found in pairs.A single magnetic pole has never been found. Characteristics:
�� Like poles (NLike poles (N--N or SN or S--S) repelS) repel each other.
�� Opposite poles (NOpposite poles (N--S) attractS) attract each other.
14.1.1 Magnetic field lines
� Magnetic field lines are used to represent a magnetic field.to represent a magnetic field.
� By convention, magnetic field lines leave the north pole and enters the south pole of a magnet.
� Magnetic field lines can be represented by straight lines or curves. The tangent to a curved field line at a point indicates the direction of the magnetic field at that point as shown in figure 14 a.
14.1 Magnetic Field
Fig. 14 aFig. 14 a
direction of magnetic field at direction of magnetic field at
point P.point P.PP
SF027 4
• Magnetic field can be represented by crossescrosses or by dotted circlesdotted circles as shown in figures 14 b and 14 c.
• A uniform field is represented by parallel lines of force. 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 in figure 14 d.
• A non-uniform field is represented by non-parallel lines. The number of magnetic field lines varies at different unit cross-sections as shown in figure 14 e.
Fig. 14 b : magnetic field lines Fig. 14 b : magnetic field lines enterenter
the page perpendicularlythe page perpendicularly
XX XX XX XX
XX XX XX XX
XX XX XX XX
Fig. 14 c : magnetic field lines Fig. 14 c : magnetic field lines leaveleave the the
page perpendicularlypage perpendicularly
unit crossunit cross--sectional areasectional area
Fig. 14 dFig. 14 d
SF027 5
� The number of lines per unit crossnumber of lines per unit cross--sectional area is proportional to the sectional area is proportional to the
magnitude of the magnetic fieldmagnitude of the magnetic field. (N α B )
� Magnetic field lines do not intersectdo not intersect one another.
14.1.2 Magnetic field lines Pattern
� The pattern of the magnetic field lines can be determined by using two methods.
�� Using compass needlesUsing compass needles (shown in figure 14 f)
stronger field in stronger field in AA11
Fig. 14 eFig. 14 e
AA11 AA22
weaker field in weaker field in AA22
Fig. 14 f : plotting a magnetic field Fig. 14 f : plotting a magnetic field
line of a bar magnet.line of a bar magnet.
SF027 6
�� Using sprinkling iron filings on paper Using sprinkling iron filings on paper (shown in figure 14 g).
� Figure below shows the various pattern of magnetic field lines around the magnets.
Fig. 14 g : Thin iron filing indicate the magnetic field lines Fig. 14 g : Thin iron filing indicate the magnetic field lines
around a bar magnet.around a bar magnet.
a. Bar magneta. Bar magnet
SF027 7
b. Horseshoe or U magnetb. Horseshoe or U magnet
c. Two bar magnets (unlike pole) c. Two bar magnets (unlike pole) -- attractiveattractive
d. Two bar magnets (like poles) d. Two bar magnets (like poles) -- repulsiverepulsive
Neutral point (point where the Neutral point (point where the
resultant magnetic force is zero)resultant magnetic force is zero)
SF027 8
14.1.3 Direction of magnetic field.
Magnetization of a Soft Iron
� There are two methods to magnetized the soft iron.
�� Using the permanent magnet.Using the permanent magnet.
� One permanent magnet
� A permanent magnet is bring near to the soft iron and touching the surface of the soft iron by following the path in the figure 14 h.
� This method is called induced magnetization.
� The arrowsarrows in the soft iron represent the magnetization direction with the arrowhead being the north pole and arrow tail being the south pole. It is also known as domains domains ( the tiny magnetized region because of spin magnetic moment of the electron).
Fig. 14 hFig. 14 h
NN SS
SF027 9
� In an demagnetized piece of soft iron, these domains are arranged randomly but it is aligned in one direction when the soft iron becomes magnetized.
� The soft iron becomes a temporary magnet with its south pole facing the north pole of the permanent magnet and vise versa as shown in figure 14h.
� Two permanent magnets
� Bring and touch the first magnet to one end of the soft iron and another end with the second magnet as shown in figure 14i and 14j.
Fig. 14 iFig. 14 iNN SS
Fig. 14 jFig. 14 jNN NN SSSS
SF027 10
�� Using Electrical circuit.Using Electrical circuit.
� A soft iron is placed inside a solenoid (a long coil of wire consisting of many loops of wire) that is connected to the power supply as shown in figure 14k.
� When the switch S is closed, the current I flows in the solenoid and produces magnetic field.
� The directions of the fields associated with the solenoid can be found by
viewing the current flows in the solenoid from both endviewing the current flows in the solenoid from both end as shown if figure 14k or applying the right hand grip rule below.
NN SS
II II
Fig. 14 kFig. 14 k
SSNN
Switch, Switch, SS
Thumb Thumb –– north polenorth pole
Other fingers Other fingers –– direction of current in solenoid.direction of current in solenoid.
Important
Current Current --
anticlockwiseanticlockwiseCurrent Current -- clockwiseclockwise
SF027 11
a.a. NNSS
� Other examples:
� Note :
� If you drop drop a permanent magnet on the floor or strikestrike it with a hammer, you may jar the domains into randomnessdomains into randomness. The magnet can thus lose some or alllose some or all of its magnetism.
�� HeatingHeating a magnet too can cause a loss of magnetism.
II II
b.b.
II II
SS NN
SF027 12
� The permanent magnet also can be demagnetized by placing it inside a placing it inside a
solenoid that connected to an alternating sourcesolenoid that connected to an alternating source.
� Example 1 : (exercise)
Sketch the magnetic field lines pattern around the bar magnets as shown in figures below.
a. b.
SF027 13
� When a current flows in a conductor wire or coil, the magnetic field will be produced.
� The direction of magnetic fielddirection of magnetic field around the wire or coil can be determined by using the
right hand grip ruleright hand grip rule as shown in figure 14k.
Magnetic field of a long straight conductor (wire) carrying current
� The magnetic field lines pattern around a straight conductor carrying current is shown in figures 14l and 14m.
or
Fig. 14 lFig. 14 l
I
Br
Br
I
Br
Br
I
Current out of the page
SF027 14
or
I
Fig. 14 mFig. 14 m
Br
Br
ICurrent into the page
Br
Br
IXX
XX
SF027 15
I IXX
SS
NN
Magnetic field of a Circular Shaped Coil
� The magnetic field lines pattern around a circular shaped coil carrying current is shown in figures 14n.
or
R
I
I I Fig. 14 nFig. 14 n
SS
NN
SF027 16
Fig. 14 oFig. 14 o
I Ior
Magnetic field of a Solenoid
� The magnetic field lines pattern around a solenoid carrying current is shown in figures 14o.
SSNN
I
IXX XX XX XX
I
I
I
I
I
I
SSNN
SF027 17
� The equation of magnetic flux density around a solenoid carrying current I can be determined by using Ampere’s law.
� Consider a very long solenoid with closely packed coils, the field is nearly uniform and parallel to the solenoid axes within the entire cross section, as shown in figure 14p.
� To find the magnetic field inside the solenoid and at the centre, we choose and draw the rectangle closed path abcd as shown in figure 14p(clockwise) for applying Ampere’s law.
� By considering this path consists of four segment : ab, bc, cd and da, then Ampere’s law becomes
Fig. 14 pFig. 14 p
) ()( smallvery0outsideB =r
Br
d
ab
c
l
ldr
ldr
ldr
ldr
SF027 18
� Example 2 :
Two long straight wires are placed parallel to each other and carrying the same current, I.
Sketch the magnetic field lines pattern around both wires
a. when the currents are in the same direction.
b. when the currents are in opposite direction.
Solution:
a.
I I
I
I
I
Ioror
SF027 19
b.
oror
I
I
I
I
I IXX
SF027 20
� A stationarystationary electric charge in a magnetic field will not experience any forcenot experience any force. But if
the charge is movingcharge is moving with a velocity, v in a magnetic field, B then it will experience a
force. This force known as MAGNETIC FORCEMAGNETIC FORCE.
� In vector formvector form,
� The magnitudemagnitude of the magnetic force can be calculated by using the equation below
14.2 Magnetic Force on a Moving Charge
where force magnetic :Fdensityflux magnetic :Bcharge a of velocity :vcharge theof magnitude:q
Bvθrr and between angle :
(14.2 a)(14.2 a)
(14.2 b)(14.2 b)
SF027 21
� The direction of the magnetic force can be determined by using the Fleming’s hand
rule.
�� FlemingFleming’’s right hands right hand rule : - for negativenegative charge
�� FlemingFleming’’s left hands left hand rule : - for positivepositive charge
� Example 3 :
Determine the direction of the magnetic force, exerted on a charge in each problems
below.
a. b.
Fig. 14 rFig. 14 rFig. 14 qFig. 14 q
shown in figures 14q
and 14r
Thumb Thumb ––
First finger First finger ––
Second fingerSecond finger ––
Important
Fr
+
Br
vr
−
Br
vr
SF027 22
c. d. e.
Solution:
a. By using Fleming’s left hand rule,
c. By using Fleming’s right hand rule,
−
Br v
r
XX XX XX XX
XX XX XX XX
XX XX XX XX
vr
+
I
I
−vr
+
Br
vr
b. By using Fleming’s right hand rule,
−
Br
vr
−
Br v
r
XX XX XX XX
XX XX XX XX
XX XX XX XX
d. Using right hand grip rule to determine the
direction of magnetic field forms by the current
I on the charge position. Then apply the
Fleming’s right hand rule,
I
−vr
Br
XX XX XX XX
XX XX XX
SF027 23
e. Using right hand grip rule to determine the direction of magnetic field forms by the current I on the charge position. Then apply the Fleming’s left hand rule, thus
� Example 4 : (exercise)
Determine the sign of a charge in each problems below.
a. b.
Ans. : positive charge, positive charge
� Example 5 : (exercise)
Determine the direction of the magnetic force exerted on a positive charge in each figures below when a switch S is closed.
vr
+
I
Br
XXXX
XX
XX
XX
XX
XX XX
Br
vr
Fr B
r
vrF
r
SF027 24
a. b.
Ans. :a. into the page, b. out of page
� Example 6 :
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)
Solution: v=3x107 m s-1, B=1.5 T, q=1.60x10-19 C
a. Given θ = 90° then by applying the equation of magnetic force on a
moving charge, thus
Switch, SSwitch, S
+ vr
Switch, SSwitch, S
+vr
NxF 102.7 12−=∴
SF027 25
b. Given θ = 50° then by applying the equation of magnetic force on a
moving charge, thus
� Example 7 : (exercise)
An electron experiences the greatest force as it travels 2.9 x 106 m s-1 in a magnetic field when it is moving north. The force is upward and of magnitude 7.2 x 10-13 N. Find the magnitude and direction of the magnetic field. (Giancolli, pg.705, no.22)
(Given the charge of the electron is -1.60 x 10-19 C)
Ans : 1.6 T to the east.
� Example 8 : (exercise)
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 the charge of the electron is -1.60 x 10-19 C)
Ans : 1.2 T
θsinBqvF =
N10x55F 12 . −=
SF027 26
� When a current-carrying conductor is placed in a magnetic field B, thus a magnetic force will acts on that conductor.
� In vector formvector form,
� The magnitudemagnitude of the magnetic force exerts on the current-carrying conductor is given by
14.3 Magnetic Force on a current-carrying conductor
where force magnetic :Fdensityflux magnetic theof magnitude :B
current :Iconductor theoflength :L
BIθr and ofdirection between angle :
(14.3 a)(14.3 a)
(14.3 b)(14.3 b)
SF027 27
The direction of the magnetic force can be determined by using the FlemingFleming’’s left hand rules left hand rule
as shown in Figure 14 s
Fig. 14 sFig. 14 s
Thumb Thumb ––
First finger First finger ––
Second fingerSecond finger ––
Important
SF027 28
Note :
� From eq. (14.3 a),
� the magnetic force on the conductor has its maximum value when the conductor (and therefore the current) and the magnetic field are perpendicular (at right angles) to each other then θ=90° (shown in figure 14 t).
� the magnetic force on the conductor is zero when the conductor (and therefore the current) is parallel to the magnetic field then θ=0° (shown in figure 14 u).
�� One One teslatesla is defined
as the magnetic flux density of a field in which a force of 1 as the magnetic flux density of a field in which a force of 1 newtonnewton acts on a 1 acts on a 1
metremetre length of a conductor which carrying a current of 1 ampere and length of a conductor which carrying a current of 1 ampere and is is
perpendicular to the field.perpendicular to the field.
o90BILF sinmax =BILF =max
Br
o90=θI
Fig. 14 tFig. 14 t
o0BILF sin= 0F =
Br
o0=θI
Fig. 14 uFig. 14 u
SF027 29
� Example 15 :
Determine the direction of the magnetic force, exerted on a conductor carrying current, I
in each problems below.
a. b.
Solution:
For both problems, use Fleming’s left hand rule :
a.
Br I
XX XX XX XX
XX XX XX XX
XX XX XX XX Br I
XX XX XX XX
XX XX XX XX
XX XX XX XX
Br I
XX XX XX XX
XX XX XX XX
XX XX XX XX Br I
XX XX XX XX
XX XX XX XX
XX XX XX XX
b.
SF027 30
� Example 16 :
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 is 12 A, find the magnitude of magnetic field was supplied.
Solution: L=20x10-2 m, B=0.40 T, θ=90°
a. Given I = 12 A.
By applying the equation of magnetic force on a current-carrying
conductor, thus
b. Given I = 12 A and L = 30x10-2 m
By applying the equation of magnetic force on a current-carrying
conductor, thus
θBILF sin=
θBILF sin=
SF027 31d
2I
2I
1I
1I
X Y
Fig. 14 vFig. 14 v
c. Given I = 12 A, L = 20x10-2 m , θ=90° and F = 60x10-2 N
By applying the equation of magnetic force on a current-carrying
conductor, thus
� 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 14v.
θBILF sin=
θIL
FB
sin=
6.8 Forces between two current-carrying conductors
1Br
2Br
� The conductors are in vacuum and their separation is d.
� The magnitude of the magnetic flux density, B1at 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.
P
d2
IB 101
π
µ= Direction : into the Direction : into the
page/paperpage/paper
12Fr
21Fr
Q
THE END