PDT UNIT14 Students)

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A group of A group of

phenomena phenomena

associated with associated with

magnetic field.magnetic field.

UNIT 14: MAGNETISMUNIT 14: MAGNETISM

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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 =

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� 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

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• 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

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� 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.

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�� 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

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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)

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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

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� 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

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�� 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

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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

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� 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.

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� 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

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or

I

Fig. 14 mFig. 14 m

Br

Br

ICurrent into the page

Br

Br

IXX

XX

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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

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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

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� 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

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� 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

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b.

oror

I

I

I

I

I IXX

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� 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)

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� 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

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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

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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


Determine the sign of a charge in each problems below.

a. b.

Ans. : positive charge, positive charge


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

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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−=∴

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b. Given θ = 50° then by applying the equation of magnetic force on a

moving charge, thus


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.


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 . −=

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� 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)

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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

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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

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� 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.

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� 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


conductor, thus

θBILF sin=

θBILF sin=

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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


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

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