Chapter 14 Magnetism and Matter free study material by TEACHING CARE online tuition and coaching classes
The molecular theory of magnetism was given by Weber and modified later by Ewing. According to this theory.
Every molecule of a substance is a complete magnet in itself. However, in an magnetised substance the molecular magnets are randomly oriented to give zero net magnetic moment. On magnetising, the molecular magnets are realigned in a specific direction leading to a net magnetic moment.
(Unmagnetised) (Magnetised)
Note : @ On heating/hammering the magnetism of magnetic substance reduces.
Bar Magnet.
A bar magnet consist of two equal and opposite magnetic pole separated by a small distance. Poles are not exactly at the ends. The shortest distance between two poles is called effective length (L_{e}) and is less then its geometric length (L_{g}).
 Directive properties : When a magnet suspended freely it stays in the earth’s N–S direction (in magnetic meridian).
 Monopole concept : If a magnet is Broken into number of pieces, each piece becomes a This in turn implies that monopoles do not exist. (i.e., ultimate individual unit of magnetism in any magnet is called dipole).
 For two rods as shown, if both the rods attract in case (i) and doesn’t attract in case (ii) then, B is a magnetic and A is simple iron Repulsion is sure test of magnetism.
 Pole strength (m) : The strength of a magnetic pole to attract magnetic materials towards itself is known as pole
 It is a scalar
 Pole strength of N and S pole of a magnet is conventionally represented by +m and –m
 It’s SI unit is amp × m or N/Tesla and dimensions are [LA].
 Pole strength of the magnet depends on the nature of material of magnet and area of cross It doesn’t depends upon length.
(5) Magnetic moment or magnetic dipole moment
(M)
: It represents the strength of magnet.
r
Mathematically it is defined as the product of the strength of either pole and effective length. i.e. M = m(2l )
 It is a vector quantity directed from south to
 It’s I. unit amp×m^{2} or N–m/Tesla and dimensions [AL^{2}]
 Magnetic moment in various other
– m + m
L = 2l r
M
Current carrying coil
M Magnetic moment M = NiA N = number of turns, i i M i= current through the coil, A= Area of the coil 

Combination of bar magnet
M 2 S
N q M net a N S M 1
2 M M N
S 90^{o} M N 
Mnet =
tana =
2 M
S 
M ^{2} + M ^{2} + 2M M 1 2 1 2
M 2 sinq M1 + M 2 cosq
N
M S 60^{o} 3 M N M 
cosq
S 
S 
S 
N 
N 
M M 
M_{net} = 2M 
Revolving charge
(a) Orbital electron : In an atom electrons revolve around the nucleus in circular orbit and it is equivalent to the flow of current in the orbit. Thus the orbit of electrons is considered as tiny current loop with magnetic moment. M = enA = ewr 2 = 1 evr = e L = eh ; where, w = angular speed, n = frequency, v = linear speed and 2 2 2m 4pm L= Angular moment Iw. 
(b) For geometrical symmetrical charged rotating bodies : The magnetic moment given by M = QL = QIw ; where
2m 2m m = mass of rotating body, Q = charge on body, I = moment of inertia of rotating body about axis of rotation. w w w + + + + + + + + + R + + R + + + + + + + ++ + Q + + + + + + + Q + L + + Q + + + + + Disc Ring Rod 

I = mR 2 ,
2 
M = 1 QwR ^{2} I = MR ^{2} , M = 1 QwR ^{2} I = mL2 ,
4 2 12 
M = 1 Qw L^{2}
24 
Note : @Bohr magneton m _{B}
= eh =
4pm
9.27×10^{–24} A/m^{2} . It serves as natural unit of magnetic moment. Bohr
magneton can be defined as the orbital magnetic moment of an electron circulating in inner most orbit.
@ Magnetic moment of straight current carrying wire is zero.
@ Magnetic moment of toroid is zero.
@ If a magnetic wire of magnetic moment (M) is bent into any shape then it’s M decreases as it’s length (L) always decreases and pole strength remains constant.
L/3
L
– m +m
M = mL
L/2
L/2
L/3
L/3
M‘ = M /
M‘ = M / 3
M‘ = 2M / p
M¢ = 0
 Cutting of a bar magnet : Suppose we have a rectangular bar magnet having length, breadth and mass are L, b and w respectively if it is cut in n equal parts along the length as well as perpendicular to the length simultaneously as shown in the figure then
Length of each part
L‘ = L , breadth of each part
b‘ = b
, Mass of each part w‘ = w , pole strength of each
n
part m‘ =
, Magnetic moment of each part
M‘ = m‘ L‘ = ´ = M
n
If initially moment of inertia of bar magnet about the axes passing from centre and perpendicular to it’s length

æ 2 2 ö


is I = w ç ÷
then moment of inertia of each part
I‘ = I
ç 12 ÷ n2
Note : @For short bar magnet b = 0 so
L‘ = L ,w‘ = w , m‘ = m, M‘ = M and I‘ = I
n n n n3
@ Commonly asked question
Þ
L
S N A/2 Þ
S N A/2
L
(m, M)
m¢ = m/2, M¢ = M/2
(m, M)
m¢ = m, M¢ = M/2
Various Terms Related to Magnetism.
 Magnetic field and magnetic lines of force : Space around a magnetic pole or magnet or current carrying wire within which it’s effect can be experienced is defined magnetic field. Magnetic field can be represented with the help of a set of lines or curves called magnetic lines of
Isolated north pole Isolated south pole Magnetic dipole Inward magnetic field Outward magnetic field
(2) Magnetic flux (f) and flux density (B)
 The number of magnetic lines of force passing normally through a surface is defined as magnetic flux (f). It’s
S.I. unit is weber (wb) and CGS unit is Maxwell. Remeber 1 wb = 10^{8} maxwell.
 When a piece of a magnetic substance is placed in an external magnetic field the substance becomes The number of magnetic lines of induction inside a magnetised substance crossing unit area normal to
their direction is called magnetic induction or magnetic flux density (B). It is a vector quantity.
It’s SI unit is Tesla which is equal to
wb = N = J = volt ´ sec
m^{2} amp ´ m amp ´ m^{2} m ^{2}
and CGS unit is Gauss. Remember 1 Tesla = 10^{4} Gauss.
Note : @ Magnetic flux density can also be defined in terms of force
experienced by a unit north pole placed in that field i.e.
B = F .
m0
 Magnetic permeability : It is the degree or extent to which magnetic lines of force can enter a substance and is denoted by m. or
Characteristic of a medium which allows magnetic flux to pass through it is called it’s permeability. e.g.
permeability of soft iron is 1000 times greater than that of air.
Also m = m _{0} m_{r} ; where m_{0} = absolute permeability of air or free space = 4p ´ 10 ^{7} tesla ´ m / amp.
and m
= Relative permeability of the medium = B = flux density in material .

B_{0} flux density in vacuum
(4) Intensity of magnetising field (H)
field can magnetise a substance. Also H = B .
m
(magnetising field) : It is the degree or extent to which a magnetic
It’s SI unit is
A / m. =
N
m^{2} ´ Tesla
= N =
wb
J
m^{3} ´ Tesla
= J
m ´ wb
It’s CGS unit is Oersted. Also 1oersted = 80 A/m
 Intensity of magnetisation (I) : It is the degree to which a substance is magnetised when placed in a magnetic
It can also be defined as the pole strength per unit cross sectional area of the substance or the induced dipole moment per unit volume.
Hence I=
m = M . It is a vector quantity, it’s S.I. unit is Amp/m.
A V
 Magnetic susceptibility (c_{m}) : It is the property of the substance which shows how easily a substance can be It can also be defined as the ratio of intensity of magnetisation (I) in a substance to the magnetic
intensity (H) applied to the substance, i.e.
c = I . It is a scalar quantity with no units and dimensions.
m H
 Relation between permeability and susceptibility : Total magnetic flux density B in a material is the
sum of magnetic flux density in vacuum
B_{0} produced by magnetising force and magnetic flux density due to
magnetisation of material
Bm . i.e.
B = B0 + Bm
Þ B = m _{0} H + m _{0} I = m _{0} (H + I) = m _{0} H(1 + c _{m} ) . Also m_{r} = (1 + c _{m} )
Note : @ In CGS B = H + 4pI
Force and Field.
and
m = 1 + 4pc _{m}.
 Coulombs law in magnetism : The force between two magnetic poles of strength m_{1} and m_{2} lying at a
distance r is given by F = k. m1m2 . In S.I. units k = m0 = 10 ^{7} wb / Amp ´ m , In CGS units k = 1
r ^{2} 4p
(2) Magnetic field
 Magnetic field due to an imaginary magnetic pole (Pole strength m) : Is given by
B = F
also B = m 0 . m
m0
 Magnetic field due to a bar magnet : At a distance r from the cerntre of magnet
4p d ^{2}
 On axial position
r e B
Be g
B = m0
2Mr ;
Equatorial line
^{a} 4p (r ^{2} – l ^{2})^{2}
m 2M
q q + a


a r
If l<<r then
Ba = 0
2l Ba
4p r ^{3}
Axial line
M
 On equatorial position : B
= m _{0}
M ; If l <<r ; then B
= m_{0} M
e 4p (r 2 + l 2 )3 / 2 e 4p r 3
 General position : In general position for a short bar magnet
Bg =

Bar magnet in magnetic field : When a bar magnet is left free in an uniform magnetic field, if align it self in the directional
 Torque : t = MB sinq Þ t = M ´ B
 Work : W = MB(1 – cosq )
 Potential energy : U = MB cosq = –M . B ; (q = Angle made by the dipole with the field)
Note : @ For more details see comparative study of electric and magnetic dipole in electrostatics.
 Gauss’s law in magnetism : Net magnetic flux through any surface is always zero e. òB.ds = 0
Example: 1 The work done in turning a magnet of magnetic moment M by an angle of 90° form the meridian, is n times the corresponding work done to turn it through an angle of 60°. The value of n is given by [MP PET 2003]
(a) 2 (b) 1 (c) 1/2 (d) 1/4
Solution : (a) W = MB(1 – cosq ) Þ W0o ®90o = nx(W o o ) Þ MB(1 – cos 90 ) = n ´ MB(1 – cos 60 ) Þ n = 2
o o
Example: 2 The magnetic susceptibility of a material of a rod is 499, permeability of vacuum is
4p ´ 10^{–}^{7} H / m .
Permeability of the material of the rod in henry/metre is [EAMCET 2003]
(a)
p ´ 10 ^{–}^{4}
(b)
2p ´ 10 ^{–}^{4}
(c)
3p ´ 10 ^{–}^{4}
4p ´ 10 ^{–}^{4}
Solution : (b)
m_{r} = (1 + g _{m})Þ m_{r} = (1 + 499) = 500 Also m = m_{0}m_{r} = 4p ´ 10^{–}^{7} ´ 500 = 2p ´ 10^{–}^{4}
Example: 3 A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque required to maintain the needle in this position will be [MNR 1991; KCET 1994; MP PET 1996; AIEEE 2003]
(a)
W (b) – W (c) 3 W
2
 2W
Solution : (a) t = MB sinq
and W = MB(1 – cos q ) Þ W = MB(1 – cos 60^{o}) = MB . Hence t = MB sin 60^{o} =
2
= 3W
2
Example: 4 An iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment will be [CPMT 1984; MP Board 1986; NCERT 1975; MP PET/PMT 1988; EAMCET (Med.) 1995;
Manipal MEE 1995; RPMT 1996; BHU 1995; MP PMT 2002]
 M (b) 2M
p
(c)
M (d) Mp
p
Solution : (b) On bending a rod it’s pole strength remains unchanged where as it’s magnetic moment changes
New magnetic moment M‘ = m(2R) = mæ 2L ö = 2M S N Þ
ç ÷


è ø L
S L‘ = 2R N
Example: 5 A short bar magnet with its north pole facing north forms a neutral point at P in the horizontal plane. It the magnet is rotated by 90° in the horizontal plane, the net magnetic induction at P is : (Horizontal component of earth’s magnetic field = B_{H}) [EAMCET (Engg.) 2000]
(a) 0 (b)
Solution : (d) Initially
2 BH
(c)
5 BH
2
(d)
B_{H}
5 BH
B
Neutral point is obtained on equatorial line and at neutral point  BH = Be 
Where
BH =
Horizontal component of earth’s magnetic field and
Be = Magnetic field due to bar magnet on
it’s equatorial line
Finally
B_{a}
Point P comes on axial line of the magnet and at P, net magnetic field
B = =
= = 5BH
Example: 6 A bar magnet of magnetic moment 3.0 Amp ´ m is placed in a uniform magnetic induction field of
2 ´ 10^{–}^{5} T.
If each pole of the magnet experiences a force of 6 ´ 104 N the length of the magnet is [EAMCET (Med.) 2000]
Solution : (d)
(a) 0. 5 m (b) 0. 3 m (c) 0.2 m (d) 0.1 m
M = mL and F = mB ,Þ F = M ´ B Þ 6 ´ 10^{–}^{4} = 3 ´ 2 ´ 10^{–}^{5} Þ L = 0.1m
L L
Example: 7 Force between two identical bar magnets whose centres are r metre apart is 4.8 N when their axes are in the same line. If the separation is increases to 2r metre, the force between them is reduced to
[AIIMS 1995; Pb. CET 1997]
(a) 2.4 N (b) 1.2 N (c) 0.6 N (d) 0.3 N
1 F æ d ö4
4.8
æ 2r ö^{2}
Solution : (d) Force between two bar magnet F µ
4 Þ 1 = ç 2 ÷ Þ
= ç ÷
Þ F_{2} = 0.3N
Where d= separation
between magnets.
d F2 è d1 ø
F2 è r ø
Example: 8 Two identical magnetic dipoles of magnetic moments 1.0 Am^{2} each, placed at a separation of 2m with their axis perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is
[Roorkee 1995]
(a)
5 ´ 10^{–}^{7} T
(b)
´ 10^{–}^{7} T
(c)
T (d) None of these
2
Solution : (b)
B = m 0 . 2M = 10 ^{–}^{7} ´ 2 ´ 1 ; B
= m _{0}
M = 10 ^{–}^{7} ´ 1 ; B =
= ´ 10 ^{–}^{7} T
^{1} 4p
æ r ö ^{3}
ç 2 ÷
æ 2 ö ^{3}
ç 2 ÷
^{2} 4p æ r ö ^{3}
ç 2 ÷
æ 2 ö ^{3}
ç 2 ÷
net
è ø è ø è ø è ø
Example: 9 A magnet of magnetic moment 20 C.G.S. units is freely suspended in a uniform magnetic field of intensity 0.3
C.G.S. units. The amount of work done in deflecting it by an angle of 30° in C.G.S. units is [MP PET 1991]
(a) 6 (b)
3 (c)
3(2 – 3)
 3
Solution : (c)
W = MB(1 – cosq ) Þ W = 20 ´ 0.3(1 – cos 30^{o}) = 3(2 – 3)
Example: 10 The magnetic field at a point X on the axis of a small bar magnet is equal to the field at a point Y on the equator of the same magnet. The ratio of the distance of X and Y from the centre of the magnet is
[MP PMT 1990]
(a)
23
(b)
21 / 3
 2^{3}
(d)
21 / 3
Solution : (d) Suppose distances of points X and Y from magnet are x and y respectively then According to question
B = B
Þ m_{0} . 2M = m_{0} . M Þ x = 2^{1} ^{/} ^{3}
axial
equatorial
4p x^{3}
4p y^{3} y 1
Example: 11 A magnetising field of 2000 A/m produces a flux
6.28 ´ 104
weber in a rod. If the area of crosssection is
2 ´ 105 m2. Then the relative permeability of the substance is
(a) 0.75 ´ 10^{–}^{2}
(b) 1.25 ´ 10 ^{4}
f f
(c) 0.25 (d) 1.01
6.28 ´ 10^{–}^{4} _{4}
Solution : (b) By using
B = m_{0}m_{r} H
and B = A , Þ m_{r} =
Am_{0}H
= 2 ´ 10^{–}^{5} ´ 4p ´ 10^{–}^{7} ´ 2000 = 1.25 ´ 10
Example: 12 Due to a small magnet intensity at a distance x in the end on position is 9 Gauss. What will be the intensity at
a distance x on broad side on position
2
(a) 9 Gauss (b) 4 Gauss (c) 36 Gauss (d) 4.5 Gauss

Solution : (c) In C.G.S. Baxial = 9 = 2M
x
…..(i)
Bequaterial = M = 8M
æ x ö^{3} x3
ç 2 ÷
…..(ii)
è ø
From equation (i) and (ii) Bequaterial = 36 Gauss.
Example: 13 The magnetic moment produced in a substance of 1gm
then the intensity of magnetisation in A/m will be
is 6 ´ 107 ampere – metre2.
If its density is
5 gm / cm^{3},
(a)
8.3 ´ 10^{6}
(b) 3.0 (c)
1.2 ´ 10^{–}^{7}
(d)
3 ´ 10^{–}^{6}
Solution : (b)
I = M
V
= M , given mass = 1gm = 10^{–}^{3}
mass/density
kg , and density = 5gm / cm^{3}
= 5 ´ 10^{–}^{3} kg
(10^{–}^{2})^{3} m^{3}
= 5 ´ 10^{3}
kg / m^{3}
Hence I = 6 ´ 10^{–}^{7} ´ 5 ´ 10^{3} = 3
10 3
Example: 14 The distance between the poles of a horse shoe magnet is 0.1 m and its pole strength is 0.01 amp–m. The induction of magnetic field at a point midway between the poles will be
(a)
2 ´ 10^{–}^{5} T
(b)
4 ´ 10^{–}^{6} T
(c)
8 ´ 10^{–}^{7} T
 Zero
Solution : (c) Net magnetic field at mid point P, B = BN + BS
0.1 m
where BN
= magnetic field due to N pole
BS = magnetic field due to S pole
BN = BS = m0 m = 10^{–}^{7} ´ 0.01 = 4 ´ 10^{–}^{7} T
\ Bnet = 8 ´ 10^{–}^{7} T.
4p r ^{2}
æ 0.1 ö^{2}

ç ÷
è ø
Example: 15 A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetisation of
5.30 × 10^{3}Amp/m^{3}. What its magnetic dipole moment
(a)
1 ´ 10^{–}^{2} J / T
(b)
2.08 ´ 10^{–}^{2} J / T
(c)
3.08 ´ 10^{–}^{2} J / T
(d)
1.52 ´ 10^{–}^{2} J / T
Solution : (b) Relation for dipole moment is,
M = I ´ V , Volume of the cylinder
V = pr ^{2}l, Where r is the radius and l is the
length of the cylinder, then dipole moment,
M = I ´ pr ^{2}l = (5.30 ´ 10^{3}) ´ 22 ´ (0.5 ´ 10^{–}^{2})^{2} (5 ´ 10^{–}^{2})
7
= 2.08 ´ 10^{–}^{2} J / T
Example: 16 A bar magnet has a magnetic moment of
2.5 JT ^{–}^{1}
and is placed in a magnetic field of
0.2T. Work done in
turning the magnet from parallel to antiparallel position relative to field direction is
(a)
 J
 1J
 2J
 0 J
Solution : (b) Work done, W = –MB(cosq _{2} – cosq_{1}) = –MB(cos180^{o} – cos 0^{o} ) = –MB(1 1) = 2MB = 2 ´ 2.5 ´ 0.2 = 1J
Example: 17 A bar magnet with it’s poles 25 cm apart and of pole strength 24 amp×m rests with it’s centre on a frictionless pivot. A force F is applied on the magnet at a distance of 12 cm from the pivot so that it is held in equilibrium at an angle of 30° with respect to a magnetic field of induction 0.25 T. The value of force F is
(a) 5.62 N (b) 2.56 N (c) 6.52 N (d) 6.25 N
Solution : (d) In equilibrium
Magnetic torque = Deflecting torque Þ MB sin q = F.d
or F = mlB sin q
d
24 ´ 0.25 ´ 0.25 sin 30^{o} = 6.25 N
0.12
Example: 18 Two identical bar magnets with a length 10 cm and weight 50 gm – weight are arranged freely with their like poles facing in a arranged vertical glass tube. The upper magnet hangs in the air above the lower one so that the distance between the nearest pole of the magnet is 3mm. Pole strength of the poles of each magnet will be
 64 amp ´ m
 2 amp ´ m
 25 amp ´ m
 None of these
Solution : (a) The weight of upper magnet should be balanced by the repulsion between the two magnet
\ m . m2
4p r 2
= 50gm – wt
Þ 10^{–}^{7} ´
m2
(9 ´ 10^{–}^{6})
= 50 ´ 10^{–}^{3} ´ 9.8 Þ m = 6.64amp ´ m
Earth’s magnetic Field (Terrestrial Magnetism).
As per the most established theory it is due to the rotation of the earth where by the various charged ions present in the molten state in the core of the earth rotate and constitute a current.
 The magnetic field of earth is similar to one which would be obtained if a huge magnet is assumed to be buried deep inside the earth at it’s
 The axis of rotation of earth is called geographic axis and the points where it cuts the surface of earth are called geographical poles (N_{g}, Sg). The circle on the earth’s surface perpendicular to the geographical axis is called
 A vertical plane passing through the geographical axis is called geographical
 The axis of the huge magnet assumed to be lying inside the earth is called magnetic axis of the earth. The points where the magnetic axis cuts the surface of earth are called magnetic poles. The circle on the earth’s surface perpendicular to the magnetic axis is called magnetic
 Magnetic axis and Geographical axis don’t coincide but they makes an angle of 5° with each other.
 Magnetic equator divides the earth into two The hemisphere containing south polarity of earth’s magnetism is called northern hemisphere while the other, the southern hemisphere.
 The magnetic field of earth is not constant and changes irregularly from place to place on the surface of the earth and even at a given place in varies with time
 Direction of earth’s magnetic field is from S (geographical south) to N (Geographical north).
Elements of Earth’s Magnetic Field.
The magnitude and direction of the magnetic field of the earth at a place are completely given by certain. quantities known as magnetic elements.
 Magnetic Declination (q) : It is the angle between geographic and the magnetic meridian
Declination at a place is expressed at q ^{o} E
or q ^{o}W
depending upon whether the north pole of the compass
needle lies to the east or to the west of the geographical axis.
 Angle of inclination or Dip (f) : It is the angle between the direction of intensity of total magnetic field of earth and a horizontal line in the magnetic
 Horizontal component of earth’s magnetic field (B_{H}) : Earth’s magnetic field is horizontal only at the magnetic equator. At any other place, the total intensity can be resolved into horizontal component (B_{H}) and vertical component (B_{V}).
Also B_{H}=
B cos f
…… (i) and
BV = B sin f
……. (ii)
By squaring and adding equation (i) and (ii) B =
Dividing equation (ii) by equation (i)
tan f = BV
BH
Note : @ At equator q = 0 Þ
BH = B, BV
= 0 while at poles f
= 90 ^{o} Þ
BH = 0,
BV = B.
Magnetic Maps and Neutral Points.
 Magnetic maps (e. Declination, dip and horizontal component) over the earth vary in magnitude from place to place. It is found that many places have the same value of magnetic elements. The lines are drawn joining all place on the earth having same value of a magnetic elements. These lines forms magnetic map.
 Isogonic lines: These are the lines on the magnetic map joining the places of equal
 Agonic line: The line which passes through places having zero declination is called agonic
 Isoclinic lines : These are the lines joining the points of equal dip or
 Aclinic line : The line joining places of zero dip is called aclinic line (or magnetic equator)
 Isodynamic lines : The lines joining the points or places having the same value of horizontal component of earth’s magnetic field are called isodynamic
 Neutral points : At the neutral point, magnetic field due to the bar magnet is just equal and opposite to the horizontal component of earth’s magnetic
 Magnet is placed horizontally in a horizontal
N– pole of magnet is facing N– pole of earth N – pole of magnet is facing N– pole of earth
B_{H}



N BH
W
B_{H}
N_{1}
B

d

N B_{H} B_{H}
W B



E N1 N2 E S B B S
B_{H} B_{H}
d d N2
B
Two neutral points N_{1} and N_{2} are obtained on equatorial Two neutral points N_{1} and N_{2} are obtained on axial line
line of bar magnet as shown and at Neutral points
B = B Þ
of B or magnet and at neutral points
m 0 . 2M = B
B = BH
i.e.
H
 Magnet is placed vertically in a horizontal plane
4p d 3 H
N– pole of magnet is the horizontal plane S– pole of magnet is the horizontal plane
B_{H}
B_{N} B_{S}
q B_{H}
P
B_{H} B_{N}
B_{H} N
BN W
E
S
B_{H}

P q
B_{N}
B_{S} N
BH W

B_{S} BH E BS S
B_{H}
B_{S}
BN B_{N} = Magnetic field due to Npole B_{S} = Magnetic field due to Spole
M = Pole strength of each pole of the magnet
At neutral point P : B_{N} – B_{S} cosq = B_{H} (B_{S} < B_{N})
If suppose effect of Spole is neglected : As seen from top only one neutral point is obtained as shown and at
At neutral point P : B_{S} – B_{N} cosq = B_{H} (B_{S} < B_{N})
If suppose effect of Npole is neglected : As seen from top only one neutral point is obtained as shown and at
neutral point B_{N}
= B_{H} Þ
neutral point B_{S}
= B_{H} Þ

B_{H}
N BH
N P
B_{H} B_{H}
N W E
BS N


H H
B_{N} B_{N}
r BH S
S W E
BS r B BS
P BS H S
B_{N}
Example: 19 If the angles of dip at two places are 30° and 45° respectively, Then the ratio of horizontal components of earth’s magnetic field at the two places will be [MP PET 1989]
 3 :
 1:
 1:
1 : 2
Solution : (a) By using B
= B cos f Þ (B_{H} )_{1} = (cos f)_{1} = cos 30 =
^{H} (BH )2 (cos f)2 cos 45
Example: 20 At a place the earth’s horizontal component of magnetic field is
0.38 ´ 10^{–}^{4} weber / m^{2}.
If the angle of dip at
that place is 60°, then the vertical component of earth’s field at that place in weber/m^{2} will be approximately
[MP PMT 1985]
(a)
0.12 ´ 10 ^{–}^{4}
(b)
0.24 ´ 10 ^{–}^{4}
(c)
0.40 ´ 10 ^{–}^{4}
(d)
0.62 ´ 10 ^{–}^{4}
Solution : (d) By using
tan f = BV Þ tan 60 ^{o} = BV Þ B
= 0.38 ´ 10 ^{–}^{4} ´
= 0.62 ´ 10 ^{–}^{4}.
BH 0.38 ´ 10 ^{–}^{4} V
Example: 21 A dip circle is so set that it moves freely in the magnetic meridian. In this position, the angle of dip is 40°. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30° with the magnetic meridian. In this position, the needle will dip by the angle [Roorkee 1983]
 40° (b) 30° (c) More than 40° (d) Less than 40°
Solution : (c) By using
tan f ¢ =
tan f ; where f = 40 ^{o} , b = 30 ^{o}
cos b
As cos 30^{o} < 1
Þ 1 > 1
cos 30 ^{o}
Hence
tan f ¢ > 1Þ tan f ¢ > tan f = f ¢ > f
tan f
or f ¢ > 40 ^{o} .
Example: 22 Earth’s magnetic field may be supposed to be due to a small bar magnet located at the centre of the earth. If the magnetic field at a point on the magnetic equator is 0.3×10^{–4} T. Magnet moment of bar magnet is
(a) 7.8 ´ 10^{8} amp ´ m^{2} (b) 7.8 ´ 10^{22}amp ´ m^{2} (c) 6.4 ´ 10^{22}amp ´ m^{2}
(d) None of these
Solution : (b) When a magnet is freely suspended in earth’s magnetic field, it’s north pole points north, so the magnetic field of the earth may be suppose to be due to a magnetic dipole with it’s south pole towards north and as equatorial point is on the broad side on position of the dipole.
B = m 0 . M Þ 0.3 ´ 10 ^{–}^{4} = 10 ^{–}^{7} ´ M Þ M = 7.8 ´ 10 ^{22} A–m^{2}.
^{e} 4p r 3
(6.4 ´ ´10^{6} )^{3}
Example: 23 A short bar magnet is placed with its south pole towards geographical north. The neutral points are situated at
a distance of 20 cm from the centre of the magnet. If the magnet is
 9000 ab amp × cm^{2}
 900 ab – amp ´ cm^{2}
BH = 0.3 ´ 10 ^{–}^{4} wb / m^{2} then the magnetic moment of
 1200 ab – amp ´ cm ^{2}
 225ab – amp ´ cm ^{2}
Solution : (c) At neutral point magnetic field due to magnet = Horizontal component of earth’s magnetic field
Þ m0 . 2M = B
Þ 107 ´ 2 ´ M ´ 1 = 0.3 ´ 15^{4} Þ
M = 1.2amp ´ m^{2} = 1200ab – amp ´ cm^{2}.
4p r ^{3} ^{H}
(0.2)^{3}
Example: 24 Two magnets of equal mass are joined at right angles to each other as shown the magnet 1 has a magnetic moment 3 times that of magnet 2. This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium what angle will the magnet 1 subtend with the magnetic meridian
(a)
tan^{–}^{1}æ 1 ö

2
è ø
(b)
tan^{–}^{1}æ 1 ö

3
è ø
 tan^{–}^{1}(1)
 0°
Solution : (b) For equilibrium of the system torques on
M1 and
M 2 due to BH
must counter balance each other i.e.
M 1 ´ BH
= M _{2} ´ B_{H} . If q is the angle between
M1 and BH
then the angle between
M2 and BH
will be
(90 – q) ; so M1BH sinq = M2 BH sin(90 – q )
Þ tanq = M2 = M = 1 Þ q = tan^{–}^{1}æ 1 ö
M 3M 3
ç 3 ÷
_{1} è ø
Tangent Law and it’s Application.
When a small magnet is suspended in two uniform magnetic fields B and
BH which are at right angles to each
other, the magnet comes to rest at an angle q with respect to BH
such that
B = BH tanq . This is called tangent law.
Tangent galvanometer : It is an instrument which can detect/measure very small electric currents. It is also called as moving magnet galvanometer. It consists of three circular coils of insulated copper wire wound on a vertical circular frame made of nonmagnetic material as ebonite or wood. A small magnetic compass needle is pivoted at the centre of the vertical circular frame. This needle rotates freely in a horizontal plane inside a box made of nonmagnetic material. When the coil of the tangent galvanometer is kept in magnetic meridian and current passes through any of the coil then the needle at the centre gets deflected and comes to an equilibrium position under the action of two perpendicular field : one due to horizontal component of earth and the other due to field set up by the coil due to current (B).
In equilibrium
B = BH tanθ
where
B = m0 ni ; n = number of turns, r = radius of coil, i = the current to be
2r
measured, q = angle made by needle from the direction of B_{H} in equilibrium.
Hence
m_{0} Ni = B
tanq
Þ i = k tanq where k = 2rBH
is called reduction factor.
2r ^{H} m N
0
Note : @ Principle of moving coil galvanometer is uniform.
i µ tanq . Since
i µ tanq
so it’s scale is not
@When q = 45 ^{o} , reduction factor equals to current flows through coil.
@ Sensitivity of this galvanometer is maximum at q = 45^{o}.
@ This instrument is also called moving magnet type galvanometer.
Magnetic Instruments.
Magnetic instruments are used to find out the magnetic moment of a bar magnet, find out the horizontal component of earth’s magnetic field, compare the magnetic moments of two bar magnets.
(1) Deflection magnetometer
It’s working is based on the principle of tangent law. It consist of a small compass needle, pivoted at the centre of a circular box. The box is kept in a wooden frame having two meter scale fitted on it’s two arms. Reading of a scale at any point directly gives the distance of that point from the centre of compass needle.
Different position of deflection magnetometer : Deflection magnetometer can be used according to two following positions.
Tan A position Tan B position
Arms of magnetometer are placed along E–W direction such that magnetic needle is acted upon by only
Arms of magnetometer are placed along N–S direction such that magnetic needle align itself in the direction of
horizontal component of earth’s magnetic field (BH ) as shown
earth’s magnetic field (i.e.
BH ) as shown.
N

BH N W E

H
2l W E S

0 0 B




S N S 9

r B
r
It a bar magnet is placed on one arm with it’s length N S
parallel to arm, so magnetic needle comes under the influence of two mutual perpendicular magnetic field (i)
BH and (ii) Axial magnetic field of experimental bar
If a bar magnet is placed on one arm with it’s length
perpendicular to arm, so magnetic needle comes under
magnet.
the influence of two mutual perpendicular magnetic
In equilibrium B = B tanq Þ
fields (i) BH and (ii) equatorial magnetic field of
H
(M= Magnetic moment of experimental bar magnet)
experimental bar magnet.
In equilibrium B = BH and Þ
Note : @Deflection magnetometer also used to compare the magnetic moments either by deflection method or by null
M tanq
M æ d ö ^{3}
deflection method. Deflection method :
^{ } ^{1} = ^{1} , Null deflection method :
1 = ç 1 ÷
M 2 tanq _{2}
M 2 è d2 ø
where d_{1} and d_{2} are the position of two bar magnet placed simultaneously on each arm.
(2) Vibration magnetometer
Vibration magnetometer is used for comparison of magnetic moments and magnetic fields. This device works on the principle, that whenever a freely suspended magnet in a uniform magnetic field, is disturbed from it’s equilibrium position, it starts vibrating about the mean position.
Time period of oscillation of experimental bar magnet (magnetic moment M) in
earth’s magnetic field (BH ) is given by the formula. T = 2p
Where, I = moment of inertia of short bar magnet = wL2
12
(w = mass of bar
magnet)
(3) Use of vibration magnetometer
 Determination of magnetic moment of a magnet :
The experimental (given) magnet is put into vibration magnetometer and it’s time period T is determined. Now
T = 2p
Þ M =
4p ^{2} I
BH .T ^{2}
 Comparison of horizontal components of earth’s magnetic field at two
T = 2p
; since I and M the magnet are constant, so
T 2 µ 1 Þ (B_{H} )_{1}
T 2
= 2


BH (BH )2 2
 Comparison of magnetic moment of two magnets of same size and
T = 2p
; Here I and B_{H}
are constants. So
M µ 1
T 2
Þ M1 2





M 2 2
 Comparison of magnetic moments of two magnets of unequal sizes and masses (by sum and difference method) :
In this method both the magnets vibrate simultaneously in two following position.
Ts = 2p Is
Ms BH 
= 2p  I1 + I 2 
(M1 + M 2 )BH 
n d = 1
2p 
(M1 + M 2 ) BH 
(I1 + I 2 ) 
T M_{1}
T ^{2} + T ^{2}
n ^{2} + n ^{2}
From equation (i) and (ii) we get ^{s} =
Þ = d s = s d
T_{d} M 2
T 2 – T 2 n 2 –n 2
d s s d
 To find the ratio of magnetic field : Suppose it is required to find the ratio by magnet and B_{H} is the horizontal component of earth’s magnetic
B where B is the field created
BH
To determine
B a primary (main) magnet is made to first oscillate in earth’s magnetic field (B_{H}) alone and
BH
it’s time period of oscillation (T) is noted.
T = 2p
B_{H}
and frequency n = 1 M
2p
Now a secondary magnet placed near the primary magnet so primary magnet oscillate in a new field with is the resultant of B and B_{H} and now time period, is noted again.
There are two important possibilities for placing secondary magnet
B  æn ‘ ö ^{2}
= 1 – ç ÷ è n ø 
BH 
Example: 25 Two magnets are held together in a vibration magnetometer and are allowed to oscillate in the earth’s magnetic field. With like poles together 12 oscillations per minute are made but for unlike poles together only 4 oscillations per minute are executed. The ratio of their magnetic moments is [MP PMT 1996]
(a) 3 : 1 (b) 1 : 3 (c) 3 : 5 (d) 5 : 4
Solution : (d) By using
2 2


1 = d s ;
M_{2} T ^{2} – T ^{2}
where Ts
= 60
12
= 5sec and Td
= 60
4
= 15sec
\ M1
M2
= (15)^{2} + (5)^{2} = 5
(15)^{2} – (5)^{2} 4
d s
Example: 26 The magnetic needle of a tangent galvanometer is deflected at an angle 30° due to a magnet. The horizontal component of earth’s magnetic field 0.34×10^{–4} T is along the plane of the coil. The magnetic intensity is
[KCET 1999; AFMC 1999, 2000; BHU 2000; AIIMS 2000, 02]
(a)
1.96 ´ 10^{–}^{4} T
(b)
1.96 ´ 10^{–}^{5} T
(c)
1.96 ´ 10^{4} T
(d)
1.96 ´ 10^{5} T
Solution : (b)
B = BH tanq Þ B = 0.34 ´ 10^{–}^{4} tan 30^{o} = 1.96 ´ 10^{–}^{5} T
Example: 27 A magnet freely suspended in a vibration magnetometer makes 10 oscillations per minute at a place A and 40
oscillations per minute at a place B. If the horizontal component of earth’s magnetic field at A is 36 ´ 10^{–}^{6} T,
then its value at B is [EAMCET 2001]
(a)
36 ´ 10^{–}^{6} T
(b)
72 ´ 10^{–}^{6} T
(c)
144 ´ 10^{–}^{6} T
(d)
288 ´ 10^{–}^{6} T
Solution : (c) By using T = 2p
Þ T µ 1
Þ TA =
TB
Þ 60 / 10 =
60 / 20
Þ (BH )B
= 144 ´ 10 ^{–}^{6} T.
Example: 28 The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. By doing this the periodic time of the magnetometer will [MP PMT 2000]
 Increase by 19% (b) Increase by 11% (c) Decrease by 19% (d) Decrease by 21%
Solution : (b)
T = 2p
Þ T µ
1 Þ T1 =
T2
If M1 = 100 then M2 = (100 – 19) = 81 . So,
T1 = = 9 Þ T2 = 10 T1 = 11%T1
T2 10 9
Example: 29 A magnet makes 40 oscillations per minute at a place having magnetic field intensity
BH = 0.1 ´ 10 ^{–}^{5} . At
another place, it takes 2.5 sec to complete onevibration. The value of earth’s horizontal field at that place
[CPMT 1999; AIIMS 2000]
(a)
0.25 ´ 10^{–}^{6} T
0.36 ´ 10^{–}^{6} T
0.66 ´ 10^{–}^{8} T
1.2 ´ 10^{–}^{6} T
Solution : (b) By using T = 2p
Þ T1 =
T2
Þ 60 / 40 =
2,5
Þ (BH )2
= 0.36 ´ 10 ^{–}^{6} T .
Example: 30 When 2 amp. current is passed through a tangent galvanometer, it gives a deflection of 30°. For 60° deflection, The current must be [MP PET 2000]
 1 amp. (b) 2 amp. (c) 4 amp. (d) 6 amp.
Solution : (d) By using i µ tanq Þ i1
i2
= tanq_{1} tanq _{2}
Þ 2 = tan 30^{o}
i_{2} tan 60^{o}
= 1 Þ i_{2} = 6 amp.
3
Example: 31 In vibration magnetometer the time period of suspended bar magnet can be reduced by [CBSE PMT 1999]
(a) Moving it towards south pole (b) Moving it towards north pole
(c) Moving it toward equator (d) Any one them
Solution : (c) As we move towards equator
BH increases and it becomes maximum at equator. Hence T = 2p
, we
can say that according to the relation T decreases as BH
increases (i.e. as we move towards equator).
Example: 32 The time period of a freely suspended magnet is 2 sec. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be [MP PMT 1999]
(a) 4 sec (b) 2 sec (c) sec (d) 1 sec
Solution : (d)
T = 2p
; When a bar magnet is broken in n equal parts so magnetic moment of each part become
1 times
n
and moment of inertia becomes of each part becomes 1 times. Hence time period becomes 1 times i..e. T‘ = T
n^{3} n 4
In this question n = 2 so, T‘ = T = 2 = 1sec
2 2
Example: 33 A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where dip angle is 30^{o} and 15 oscillations per minute at a place where dip angle is 60^{o}. The ratio of total earth’s magnetic field at the two places is [MP PMT 1991; BHU 1997]
(a)
3 3 : 8
(b)
16 : 9
(c) 4 : 9 (d) 2 9
Solution : (b) By using T = 2p = 2p
Þ T µ
Þ T1 =
T2
Þ 60 / 20 =
60 / 15
Þ B1 = .
B2
Example: 34 If q_{1} and q_{2} are the deflections obtained by placing small magnet on the arm of a deflection magnetometer at the same distance from the compass box in tan A and tan B positions of the
magnetometer respectively then the value of
tanq_{1} tanq _{2}
will be approximately [MP PMT 1992]
(a) 1 (b) 2 (c)
1 (d)
2
Solution : (b) In tan A
position
m0 . 2M = BH tanq_{1}
……(i)
4p d^{3}
In tan B position
m0 . M = BH tanq _{2}
…..(ii)
4p d^{3}
Dividing equation (i) by equation (ii) tanq1 = 2 .
tanq _{2} 1
Example: 35 In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth’s magnetic field is 2 sec. When a magnet is brought near and parallel to it, the time period reduces to 1 sec. The ratio H/F of the horizontal component H and the field F due to magnet will be
[MP PMT 1990]
(a) 3 (b)
1 (c)
3
(d)
Solution : (b) Time period decreases i.e. field due to magnet (F) assist the horizontal component of earth’s magnetic field (see theory)
B æ T ö
F æ 2 ö^{2} H 1
Hence by using
B
= ç T‘ ÷ – 1 Þ H = ç 1 ÷
– 1 = 3 Þ = .
F 3
_{H} è ø è ø
Example: 36 A certain amount of current when flowing in a properly set tangent galvanometer, produces a
deflection of 45^{o}. If the current be reduced by a factor of 3, the deflection would
 Decrease by 30^{o} (b) Decreases by 15^{o} (c) Increase by 15^{o} (d) Increase by 30^{o}
Solution : (b) By using i µ tanq Þ i1
i2
= tanq_{1} Þ i1
tanq _{2}
= tan 45^{o} Þ
tanq _{2}
3 tanq _{2} = 1 Þ tanq _{2} =
Þ q _{2} = 30^{o}
So deflection will decrease by 45^{o} – 30^{o} = 15^{o} .
Example: 37 The angle of dip at a place is 60^{o}. A magnetic needle oscillates in a horizontal plane at this place with period T. The same needle will oscillate in a vertical plane coinciding with the magnetic meridian with a period
(a) T (b) 2T (c)
Solution : (d) When needle oscillates in horizontal plane
T (d) T
2
Then it’s time period is T = 2p……………………………………….. (i)
When needle oscillates in vertical plane i.e. It oscillates in total earth’s total magnetic field (B)
Hence T‘ = 2p
Dividing equation (ii) by (i)
T‘ = =
T
……(ii)
= = 1 Þ T‘ = T
Example: 38 A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 seconds. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 seconds. Then the angle of dip is
(a) 0^{o} (b) 30^{o} (c) 45^{o} (d) 90^{o}
Solution : (c) In vertical plane perpendicular to magnetic meridian.
T = 2p……………………………………. (i)
In horizontal plane T = 2p……………….. (ii)
Equation (i) and (ii) gives BV = BH
Hence by using tan f = BV
BH
Þ tan f = 1 Þ f = 45 ^{o}
Magnetic Materials.
 Types of magnetic material : On the basis of mutual interactions or behaviour of various materials in an external magnetic field, the materials are divided in three main
 Diamagnetic materials : Diamagnetism is the intrinsic property of every material and it is generated due to mutual interaction between the applied magnetic field and orbital motion of
 Paramagnetic materials : In these substances the inner orbits of atoms are The electron spins are uncoupled, consequently on applying a magnetic field the magnetic moment generated due to spin motion align in the direction of magnetic field and induces magnetic moment in its direction due to which the material gets feebly magnetised. In these materials the electron number is odd.
(a)
When no field is applied
(b)
On application of field (B)
(iii) Ferromagnetic materials : In some materials, the permanent atomic magnetic moments have strong tendency to align themselves even without any external field.
These materials are called ferromagnetic materials.
In every unmagnetised ferromagnetic material, the atoms form domains inside the material. The atoms in any domain have magnetic moments in the same direction giving a net large magnetic moment to the domain. Different domains, however, have different directions of magnetic moment and hence the materials remain unmagnetised. On applying an external magnetic field, these domains rotate and align in the direction of magnetic field.
Unmagnetised Magnetised
 Curie Law : The magnetic susceptibility of paramagnetic substances in inversely to its absolute
temperature i.e.
c µ 1 Þ
T
c µ C
T
where C = Curie constant, T = absolute temperature
On increasing temperature, the magnetic susceptibility of paramagnetic materials decreases and vice versa. The magnetic susceptibility of ferromagnetic substances does not change according to Curie law.
 Curie temperature (T_{c}) : The temperature above which a ferromagnetic material behaves like a paramagnetic material is defined as Curie temperature (T_{c}).
or
The minimum temperature at which a ferromagnetic substance is converted into paramagnetic substance is defined as Curie temperature.

For various ferromagnetic materials its values are different, e.g. for Ni, T
Ni
= 358 ^{o} C

for Fe, T
Fe
= 770^{o} C

for CO, T
CO
= 1120^{o} C
At this temperature the ferromagnetism of the substances suddenly vanishes.
 Curieweiss law : At temperatures above Curie temperature the magnetic susceptibility of ferromagnetic
materials is inversely proportional to (T – T_{c}) i.e.
c µ 1
T – Tc
Þ c = C
(T – Tc )
Here T_{c} = Curie temperature
cT curve is shown (for CurieWeiss Law)
(3) Comparative study of magnetic materials
Property  Diamagnetic substances  Paramagnetic substances  Ferromagnetic substances  
Cause of magnetism  Orbital motion of electrons  Spin motion of electrons  Formation of domains  
Explanation of magnetism  On the basis  of orbital  On the basis of spin and  On the basis of domains  
motion of electrons  orbital motion of electrons  formed  
Behaviour  In  a  non  These are repelled in an  These are feebly attracted  These are strongly attracted  
uniform magnetic field  external magnetic field i.e.  in an  external  magnetic  in an external magnetic  
have a tendency to move  field i.e., have a tendency  field i.e. they easily move  
from high to  low field  to move from low to high  from low to high field  
region.  field region  region  
State of magnetisation  These are  weekly  These  get  weekly  These get strongly  
magnetised in a direction  magnetised in the direction  magnetised in the direction  
opposite to that of applied  of applied magnetic field  of applied magnetic field  
magnetic field  
When the material in the  Liquid level in that limb  Liquid level in that limb  Liquid level in that limb  
form of liquid is filled in the
Utube and placed between
gets depressed rises up rises up very much
pole pieces. N S N S N S
On placing the gaseous materials between pole pieces
Liquid
The gas expands at right angles to the magnetic field.
Liquid
The gas expands in the direction of magnetic field.
Liquid
The gas rapidly expands in the direction of magnetic field
The value of magnetic induction B
B < B_{0} B > B_{0} B >> B_{0}
where B_{0} is the magnetic induction in vacuum
Magnetic susceptibility c Low and negative c » 1 Low but positive c » 1 Positive and high c » 10^{2}
Dependence of c on
Does not depend on
Inversely proportional to
c µ 1 or
temperature
temperature (except Bi at low temperature)
temperature
c µ 1
T
or T – Tc
c = C
This is called
c = C .
T
This is called
T – Tc
Curie Weiss law.
Dependence of c on H Does not depend
independent
Curie law, where C = Curie constant
Does not depend independent
T_{c} = Curie temperature
Does not depend independent
Relative permeability (m_{r})
m_{r} < 1 m_{r} > 1 m_{r} >> 1
m_{r} = 10^{2}
Intensity of magnetisation (I)
I–H curves
I is in a direction opposite to that of H and its value is very low
I is in the direction of H but value is low
+
I is in the direction of H
and value is very high.
I_{s}
H
– I H
Hs H
Magnetic moment (M) The value of M is very low
(» 0 and is in a direction opposite to H.)
The value of M is very low and is in the direction of H
The value of M is very high and is in the direction of H
Transition of materials (at Curie temperature)
These do not change. On cooling, these get
converted to ferromagnetic materials at Curie temperature
These get converted into
paramagnetic materials above Curie temperature
c c c
T T T_{C} T
 Hysteresis : For ferromagnetic materials, by removing external magnetic field e. H = 0. The magnetic moment of some domains remain aligned in the applied direction of previous magnetising field which results into a residual magnetism.
The lack of retracibility as shown in figure is called hysteresis and the curve is known as hysteresis loop.
 When magnetising field (H) is increased from O, the intensity of magnetisation I increases and becomes This maximum value is called the saturation value.
 When H is reduced, I reduces but is not zero when H = The remainder value OC of magnetisation when
H = 0 is called the residual magnetism or retentivity.
The property by virtue of which the magnetism (I) remains in a material even on the removal of magnetising field is called Retentivity or Residual magnetism.
 When magnetic field H is reversed, the magnetisation decreases and for a particular value of H, denoted by H_{c}, it becomes zero e., H_{c} = OD when I = 0. This value of H is called the corecivity.
 So, the process of demagnetising a material completely by applying magnetising field in a negative direction is defined Corecivity. Corecivity assesses the softness or hardness of a magnetic material. Corecivity signifies magnetic hardness or softness of substance :
Magnetic hard substance (steel) ® High corecvity Magnetic soft substance (soft iron) ® Low corecivity
 When field H is further increased in reverse direction, the intensity of magnetisation attains saturation value in reverse direction (e. point E)
 When H is decreased to zero and changed direction in steps, we get the part EFGB. Thus complete cycle of magnetisation and demagnetisation is represented by BCDEFGB.
Note : @ The energy loss (or hysteresis energy loss) in magnetising and demagnetising a specimen is proportional to the area of hysteresis loop.


 Comparison between soft iron and steel :
Example: 39 A ferromagnetic substance of volume 10^{–3} m^{3} is placed in an alternating field of 50 Hz. Area of hysteresis curve obtained is 0.1 M.K.S. unit. The heat produced due to energy loss per second in the substance will be
 5 J (b) 5 ´ 10^{–2} cal (c) 19 ´ 10^{–3} cal (d) No loss of energy
Solution : (c) By using heat loss = VAnt ; whre V = volume = 10^{–3} m^{3}; A = Area = 0.1m^{2}, n = frequency = 50 Hz and t = time = 1sec Heat loss = 10^{–3} ´ 0.1 ´ 50 ´ 1 = 5 ´ 10^{–3} J = 1.19 ´ 10^{–3} cal
Example: 40 A magnetising field of 1600 A–m^{–1} produces a magnetic flux of 2.4 ´ 10^{–5} Wb in an iron bar of crosssectional area 0.2 cm^{2}. The susceptibility of an iron bar is [BHU 2002]
(a) 298 (b) 596 (c) 1192 (d) 1788
Solution : (b) By using B = mH = m _{0} m_{r}
H and m_{r}
= (1 + c _{m}
) Þ m_{r}
= B =
m _{0} H
f
m _{0} HA
m_{r} =
2.4 ´ 10 ^{–}^{5}
(4p ´ 10 ^{–}^{7} ) ´ 1600 ´ (0.2 ´ 10 ^{–}^{4} )
= 596.8. Hence c _{m}
= 595.8 » 596
Example: 41 For iron it’s density is 7500 kg/m^{3} and mass 0.075 kg. If it’s magnetic moment is 8 ´ 10^{–7} Amp ´ m^{2}, it’s intensity of magnetisation is
(a) 8 Amp/m (b) 0.8 Amp/m (c) 0.08 Amp/m (d) 0.008 Amp/m
Solution : (c)
I = M
V
= Md m
= 8 ´ 10 ^{–}^{7} ´ 7500
0.075
= 0.08 Amp / m
Example: 42 The dipole moment of each molecule of a paramagnetic gas is 1.5 ´ 10^{–23} Amp ´ m^{2}. The temperature of gas is 27^{o}C and the number of molecules per unit volume in it is 2 ´ 10^{26} m^{–3}. The maximum possible intensity of magnetisation in the gas will be
(a) 3 ´ 10^{3} Amp/m (b) 4 ´ 10^{–3} Amp/m (c) 5 ´ 10^{5} Amp/m (d) 6 ´ 10^{–4} Amp/m
Solution : (a)
I = M
V
= m N V
= 1.5 ´ 10 ^{–}^{23} ´ 2 ´ 10^{26}
1
= 3 ´ 10^{3}
Amp / m
Example: 43 The coereivity of a small bar magnet is 4 ´ 10^{3} Amp/m. It is inserted inside a solenoid of 500 turns and length 1 m to demagnetise it. The amount of current to be passed through the solenoid will be
(a) 2.5 A (b) 5 A (c) 8 A (d) 10 A
Solution : (c) H = ni Þ i = H
n
= 4 ´ 10^{3}
500
= 8 A
Example: 44 The units for molar susceptibility
 m^{3} (b) kg–m^{–3} (c) kg^{–1} m^{3} (d) No units
Solution : (a) Molar susceptibility = Volume susceptibility
Density of material
 molecular weight = I / H ´ M =
r
I / H ´ M M / V
So it’s unit is m^{3}.
Example: 45 The ratio of the area of B–H curve and I–H curve of a substance in M.K.S. system is


 ^{2}
 1


0
 m_{0}
 1
m_{0}
Solution : (c) Area of BH loop = m _{0}
(Area of IH)