NEET-PHYSICS-VOL-II MAGNETISM

MAGNETISM
SYNOPSIS Geometrical length (L) : The actual length of
magnet is called geometric length
uur
Magnet: A body which attracts Iron,Cobalt,
Nickel, like substances and which exhibits
( )
Magnetic length 2l The shortest distance
between two poles of a magnet along the axis is
directive property is called Magnet.
called magnetic length or effective length. As
Types of Magnet: the poles are not exactly at the ends the magnetic
i) Natural magnets: a) The magnet which is length is always lesser than geometric length of
found in nature is called a natural magnet
a magnet. Effective length depends only on the
Eg: magnetite. ( F e3 O4 ) . positions of the poles but not on the magnet
b) Generally they are weak magnets. Examples :
ii) Artificial magnets:The magnets which are 2l
artificially prepared are known as artificial
magnets. These are generally made of iron, steel S N
S N
and nickel. L 2R
PROPERTIES OF MAGNETS :
Magnetic length = 2l Magnetic length = 2R
1) Attractive property : The property of Geometrical length = L Geometrical
attracting pieces of iron, steel , cobalt , nickel
etc by a magnet is called attractive property. It length= pR
was found that when a magnet is dipped into Magnetic length is a vector quantity. its direction
iron filings the concentrations of iron fillings is is from south pole to north pole along its axis
maximum at ends and minimum at centre. The 5
places in a magnet where the attracting power Magnetic length = Geometrical length
is maximum are called poles. 6
2. Directive property : If a magnet is suspended Pole Strength (m) : The ability of a pole to
freely, its length becomes parallel to N-S attract or repel another pole of a magnet is called
direction. This is called directive property. The pole strength S.I Unit : ampere - meter. Pole
pole at the end pointing north is called north strength is a scalar It depends on the area of
pole while the other pointing south is called south cross section of the pole. Its dimensional formula
pole. is M 0 LT 0 A1
Ø Magnetic poles always exist in pairs If a magnet Inductive property: When a magnetic
is broken into number of pieces, each piece substance such as iron bar is kept very close to
becomes a magnet with two equal and opposite a magnet an opposite pole is induced at the
poles This implies that monopole does not exist. nearer end and a similar pole is induced at the
Ø. The two poles of a magnet are found to be equal farther end of the magnetic substance.This
in strength and opposite in nature. property is known as inductive property.
Ø Unlike poles attract each other and like poles A magnet attracts certain other magnetic
repel each other substance through the phenomenon of magnetic
Ø There can be magnets with no poles. induction. induction precedes attraction.
Eg: Solenoid and toroid has properties of Ø Repulsion is a sure test of magnetism.A pole of a
magnet but no poles. magnet attracts the opposite pole while repels
Magnetic axis and magnetic meridian similar pole.How ever a sure test of magnetism
The line joining the poles of a magnet is called is repulsion but not attraction.Because attraction
magnetic axis and the vertical plane passing can takes place between opposite poles or
through the axis of a freely suspended magnet is
between a pole and a piece of
called magnetic meridian
unmagnetized material due to induction.
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Magnetic Moment Ø When the magnet is cut into ‘x’ equal parts
Magnetic dipole and magnetic dipole parallel to its length and ‘y’ equal parts
moment (M) : A configuration of two perpendicular to its length, then
magnetic poles of opposite nature and equal N S
strength separated by a finite distance is
called as magnetic dipole. x parts
The product of pole strength (either pole) and
magnetic length of the magnet is called
magnetic dipole moment or simply magnetic y parts
moment. pole strength of each part = m/x
If ‘m’ be the pole strength of each pole and Length of each part = 2 l / y
' 2l ' be the magnetic length, then magnetic
moment M is given by M = m ´ 2l 2l m M
Magnetic moment of each part, M1 = y ´ x = xy
In vector form, M = 2 l m
Magnetic moment is a vector whose direction Variation of magnetic moment due to bending
is along the axis of the magnet from south to of magnets
north pole. The S.I. unit of magnetic moment is Ø When a bar magnet is bent, its pole strength
ampere-meter2 (A-m2) its dimensional formula remains same but magnetic length decreases.
[AL2 ] Therefore magnetic moment decreases.
Variation of magnetic moment due to Ø When a thin bar magnet of magnetic moment
cutting of magnets : M is bent in the form of -shape with the
Consider a bar magnet of length ' 2l ' , pole arms of equal length as shown in figure, then
strength ‘m’ and magnetic moment ‘M’ S N
Ø When the bar magnet is cut into ‘n’ equal
M/3 M/3
parts parallel to its length, then
N S N S
S N
M/3
Pole strength of each part = m/n Magnetic moment of
( Q area of cross section becomes (1/n) times each part = M / 3
of original magnet) Net magnetic moment of the combination,
Length of each part = 2 l (remains same)
( - j) + ( i ) + ( j ) = ( i )
M M M M
M1 =
m M
\ Magnetic moment of each part, M1 = 2l ´ n = n 3 3 3 3
M
Note: If it is cut ‘n’ times , parallel to its length then \ M1 =
magnetic moment of each part is 3
Ø When a thin magnetic needle of magnetic
m M
M 1 = 2l × = moment M is bent at the middle, so that the
n +1 n +1 two equal parts are perpendicular as shown
Ø When the magnet is cut into ‘n’ equal parts in figure, then
perpendicular to its length then N
N S 2 /
M/2

S
N S
M/2
Pole strength of each part =m ( Q area of cross M
section remains same) Magnetic moment of each part =
Length of each part = 2l / n 2
Net magnetic moment of the combination,
2l M
Magnetic moment of each part, M1 = ´ m = M1=
M
(-i ) + ( j )
M
\ M 1= 2´
M
=
M
n n 2 2 2 2

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 NEET-PHYSICS-VOL-II MAGNETISM
Ø When a thin bar magnet of magnetic moment Ø Effective length of the magnet increases. Hence
M is bent into an arc of a circle subtending an Magnetic moment increases
angle ' q ' radians at the centre of the circle, Ø New magnetic moment is given by
then its new magnetic moment is given by Mq
æq ö M1=
æq ö
2M sin çç ÷÷÷ 2sin çç ÷÷ ( q must be in radians)
çè 2 ø ( q must be in radians) çè 2 ø÷
M1 =
q Resultant Magnetic Moment due to
combination of Magnets :
y y
S N N2
M2
2l 2l
ie q = Þ R = θ/2 S2
R q R
θ Ø θ
ο M1
q S1 N1
from the figure, Effective length = 2y= 2R sin When two bar magnets of moments M1 and M2
2 are joined so that their like poles touch each other
æ æ öö
ççQ sin q = y Þ y = Rsin çç q ÷÷÷÷ and their axes are inclined at an angle ' θ ' , then
çè 2 R çè 2÷ø÷ø the resultant magnetic moment of the combination
\ New Magnetic Moment, ‘M1 ’ is given by
æ 2l ö q M1 = M12 + M 22 + 2 M1M 2 cos θ
M = m ´ 2 y = m ´ 2 çç ÷÷÷ sin
1
çè q ø 2 ( q = angle between the directions of magnetic
æq ö moments)
2 M sin ççç ÷÷÷ S2
è2ø (Q M = 2l ´m)
ÞM1=
q N2
θ
S Ø M1
(180 - θ) S
0

1 N1
p π
M2
Ø If q = radians, θ=
2 When two bar magnets of moments M1 and M2
2
are joined so that their unlike poles touch each
N
i.e., if the magnet is bent in the form of quadrant other and their axes are inclined at an angle ‘ θ ’,
of a circle, then then the resultant magnetic moment
p M1= M12+M 22 + 2M1 M2 cos(180 0 -q )
2M sin
4 = 2 2M [ Q angle between directions of magnetic moments is
M1 =
æp ö p (1800 - q )]
çç ÷÷
çè 2 ÷ø
\ M1 = M12 + M 22 − 2M1 M2 cos θ
Ø If q = p radians, i.e., if the magnet is bent in M1 M2
the form of a semi circle, then
p Ø
2M sin S1 N1N2 S2
2M
M =
1 2 = θ = π When two bar magnets of moments M1 and M2
p p S N (M1 >M 2 ) are placed coaxially with like poles
Ø If q = 2p radians, i.e., if the magnet is bent in in contact then resultant magnetic moment,
the form of a circle, then M1 =M1–M2
2M sin p ( Q angle between directions of magnetic moments,
M1 = =0
2p q =1800 )
Ø When a magnet in the form of an arc of a M M 1 2

circle making an angle ' q 'at the centre having Ø

magnetic moment 'M' is straightened, then S NS N 1 1 2 2

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When two bar magnets of moments M1 and M2 an equilateral triangle with unlike poles at each
(M1 >M2 ) are placed coaxially with unlike poles corner, resultant magnetic moment is given by
are in contact then resultant magnetic moment,
M 1 = (2M ) +M 2 +2(2M )(M )cos1200 = 3M
2
M1 =M1+M2
( Q angle between directions of magnetic moments, Ø When four bar magnets of moments M, 2M, 3M
q = 00 ) & 4M are arranged to form a square with unlike
S 2
N poles at each corner, then resultant magnetic
2

M moment is given by
Ø
2

M 3M 1 2M
S N S
S1 N1 N
When two bar magnets of magnetic moments 4M
=
2M 2M
M1 and M2 are placed one over the other with N
like poles on the same side, then resultant S N
M
S

= M1 +M2 (Q q = 0 0 )
M1 = (2M ) +( 2M) +2( 2M)(2 M )cos900 = 2 2M
magnetic moment, M1 2 2

M2
N2 S2
Ø When half of the length of a thin bar magnet of
Ø M1
magnetic moment M is bent into a semi circle as
S1 N1 shown in figure, then
When two bar magnets of magnetic moments M1 S N
and M2 are placed one over the other with unlike S N
poles on the same side, then resultant magnetic M2

moment, M1 = M1 : M2 . (Q q = 180 )
0
M1
resultant magnetic moment, M1 = M1 +M2
N1 N1 S1
S1 æM ö
S2 N2 N2 2 ççç ÷÷÷ æ 2 + p ö÷
è 2 ø M M M = M çç
çè 2p ø÷÷
Ø
S1 S2
N2 S2 N1 = + = +
p 2 p 2
When two bar magnets of magnetic moments Ø In the above case if the two parts are arranged
M1 and M2 are placed at right angles to each perpendicular to each other, then resultant
other then resultant magnetic moment, magnetic moment is
M1 = M12 + M22 (Q q = 90 ) .
0
æM ö æ Mö
2 2

M = M +M
1 2 2 = ççç ÷÷÷ + ççç ÷÷÷
S N
M
1 2 è p ø è 2ø
S N S
M M N S
Ø M
M

M1
(4 + p2 )
N S N
M
S M N
S N S
=
2p M2
M
N
When identical magnets each of magnetic
moment M are arranged to form a closed polygon NS
like a triangle (or) square with unlike poles at Magnetic field :
each corner, then resultant magnetic moment, M1 Ø Around a pole there exist a region called
= 0. magnetic field in which the influence of the pole
Ø In the above point , if one of the magnets is is felt.
reversed pole to pole then resultant magnetic
Ø The space around the magnet is said to be
moment, M 1 = 2 M associated with a field known as magnetic field,
S N 120
0 if another magnet is brought into the space, it is
0
acted upon by a force due to this energy.
3M 2M M M 2M 60 M
Ø
= + Ø Magnetic induction is the measure of magnetic
N S field both in magnitude and direction.
M
S M N Magnetic Field Lines :
When three bar magnets of equal length but
moments M, 2M and 3M are arranged to form Ø The imaginary path in which a free unit north
pole would tend to move in a magnetic field is
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 NEET-PHYSICS-VOL-II MAGNETISM
known as a magnetic line of force (or) simply (ix) When a soft iron ring is placed in magnetic field,
magnetic “field line”. then most of lines of force pass through the ring
and no lines of force pass through the space
inside the ring as shown in figure. The
phenomenon is known as magnetic screening or
N S
shielding.
Magnetic line of force with magnetic needle
Characteristics of lines of force :
(i) Magnetic lines of force are closed curves. B=0
Outside the magnet, their direction is from north
to south pole, while inside the magnet they are
from south to north pole. Hence they have neither
origin nor end. (x) If the magnetic lines of force are straight and
(ii) Tangent, at any point to the line of force gives parallel, and equally spaced the magnetic field
the direction of magnetic field at that point. is said to be uniform.
(iii) Two lines of force never intersect each other. If Magnetic Induction (or)
the two lines of force intersect, at the intersecting Induction Field Strength (B)
point the field should have two directions, which Magnetic induction field strength at a point
is not possible. in the magnetic field is defined as the force
(iv) The lines of force tend to contract longitudinally experienced by unit north pole placed at that
or length wise . Due to this property the two point. It is denoted by ‘B’.
unlike poles attract each other. If a pole of strength ‘m’ placed at a point in a
magnetic field experiences a force ‘F’, the
N S magnetic induction (B) at that point is given by
F
Magnetic lines of force between two unlike poles. B= i.e., F = mB
m
(v) The lines of force tend to repel each other Ø B is a vector quantity directed away from N-
laterally. Due to this property the two similar pole or towards S-pole.
poles repel each other.
B
N B S
Unit North Unit North
Pole Pole
N S S.I. Unit of B :
N J V -s wb
(or ) ( or ) 2 ( or ) 2 (or )tesla (T )
A-m A-m
2
m m
Magnetic lines of force between two like poles CGS Unit of B : gauss (G) 1G = 10–4 T
(vi) If in any point, in the combined field due to two Dimensions of B :
magnets, there are no lines of force, it follows
F [ MLT -2 ]
that the resultant field at that point is zero. Such B= = = [ MT -2 A-1 ]
points are called null or neutral points. m [ AL]
(vii)Lines of force in a field represent the strength of When placed in an external magnetic field, all
the field at a point in the field. Lines of force N-poles experience a force (F = mB) in the
are crowded themselves in regions where the direction of the field and all S-poles experience
field is strong and they spread themselves apart the same force in the direction opposite to the
at places where the field is weak. field.
(viii) Lines of force have a tendency to pass through
magnetic substances. They show maximum N F = mB
tendency to pass through ferro magnetic B
materials. F = mB S

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Magnetic induction at a point due to an section normal to the cross section is called
isolated magnetic pole : magnetic flux density.
Consider a magnetic pole of strength ‘m’ kept at B= φ B / A
the point ‘O’. Consider a point ‘P’ at a distance SI unit is weber metre-2 or tesla or NA–1 m–1 .
‘r’ from ‘O’. To find the magnetic induction at Its C.G.S. unit is gauss
the point ‘P’, imagine a unit north pole at P. 1 gauss = 10-4 tesla
O Ø Its dimensional formula is [M1 L0 T-2A-1 ]
m P Ø It is also known as magnetic induction and
r magnetic field.
µ m ×1 Ø The relation between B and H is B0 = µ 0H in
Force on unit north pole at P = o N vacuum and B = µ H in a material medium Where
4π r 2
µ is the absolute permeability of the medium.
Force on unit north pole at ‘P’ gives the
magnetic induction at that point. Ø The force experienced by a pole of strength ‘m’
∴ Magnetic induction at P is ampere meter in a field of induction B is F = m
B
µo m
B= newton/amp–metre (or) tesla (T) Couple acting on the bar magnet (or)
4 π r2 Torque on a Magnetic Dipole
Types of Magnetic Field Ø When a bar magnet of moment M and length 2l
Ø Uniform magnetic field: The magnetic field, is placed in a uniform field of induction B, then
in which the magnetic induction field strength is each pole experiences a force mB in opposite
same both in magnitude and direction at all directions.
points, is known as uniform magnetic field. B
Ø In such a magnetic field the magnetic lines of
force are equidistant and parallel straight lines. N mB
Ex: Horizontal component of earth’s magnetic 2l θ
field in a limited region.
Ø Non uniform magnetic field: The magnetic mB S
field, in which the magnetic induction or field
strength differs either in magnitude, in direction As a result the bar magnet experiences a couple
or both is known as non uniform magnetic field. and moment of couple is developed.
Ø It is represented by non-parallel lines of force Ø Moment of couple acting on the bar magnet is
Ex: The magnetic field near the pole of any C = Force x perpendicular distance between two
magnet forces.
Magnetic flux (φ ) : It is equal to the total C = ( m)( 2l ) B sinθ ( or) C = M B sinθ
number of magnetic lines of force passing normal Where θ is the angle between magnetic moment
through a given area. Its S.I. unit is weber and and magnetic field.
C.G.S. unit is maxwell In vector notation C = M × B
1 weber = 108 maxwell
rr Ø When the bar magnet is either along or opposite
φ = B.A = BAcosθ to the direction of magnetic field then moment
Where ‘θ ’ is the angle made by magnetic field of couple=0.
r Ø When the bar magnet is perpendicular to the
( )
B with the area ( n̂ ) direction of applied magnetic field, then the
r moment of couple is maximum. i.e. Cmax = MB
A = Anˆ A = area of the coil
It is a scalar. Dimensional formula is Ø In a uniform magnetic field a bar magnet
experiences only a couple but no net force.
M L2T −2 I −1  . Therefore it undergoes only rotatory motion.
Magnetic Flux Density (B): The number of Ø In a non-uniform magnetic field a bar magnet
magnetic flux lines passing per unit area of cross experiences a couple and also a net force. So it

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 NEET-PHYSICS-VOL-II MAGNETISM
undergoes both rotational and translational motion B1 sin450 2
Ø Two magnets of magnetic moments M1 and M2 = M B2 sin (75 - 30 ) ; \ 0 = 0
0
=
are joined in the form of a (+) and this B2 sin30 1
arrangement is pivoted so that it is free to rotate Ø A pivoted magnetic needle of length 2l and pole
in a horizontal plane under the influence of strength 'm' is at rest in magnetic meridian. It is
earth's horizontal magnetic field. If 'q ' is the held in equilibrium at an angle 'q ' with BH by
angle made by the magnetic meridian with M1 pulling its north pole towards east by a string.
in equilibrium position, then Then tension in the string is
t1 = t 2 ; i.e., M1BH sinq from the figure ,
y
M2 cos q = Þ y = l cos q In equlibrium
= M2 BH sin (90 - q) ; \ tan q = l
M1 BH
τ2 BH τ1 N
θ F
M2 90– θ
θ M1 l θ y
N2 N1
l
S

S1 S2 t tension = t BH ; i.e., Fl cos q = MBH sin q

Ø Two magnets of moments M1 and M2 are joined (or) Fl cos q = 2lm BH sin q
as shown in figure and the arrangement is pivoted
so that it is free to rotate in a horizontal plane \ F = 2mBH tan q
under the influence of magnetic field B. Then b) In the above case, if the magnetic needle is
net torque acting on the system is given by held in equilibrium at an angle 'q ' to a uniform
t = t1 + t 2 ; = M 1B s i n q1 + M 2B sin q2 magnetic induction field BH by applying a force
F at a distance 'r' from the pivot along a direction
= B (M1 sin q1 + M 2 sin q 2 ) perpendicular to the field, then
B
BH

M1 N
θ1
N1 θ
S2 F
θ2 r

S
N2 S1
M2
Two uniform magnetic fields of strengths B1 and
MBH tan q
B2 acting at an angle 75 0 with each other in Frcos q = MBH sin q ; \ F=
horizontal plane are applied on a magnetic r
needle of moment M, which is free to move in ( 2l m) BH tan q
the horizontal plane. If the needle gets aligned =
r
B c) In the above case, if the force is applied at
at an angle 300 with B1 , then the ratio 1 is
B2 one end which is always perpendicular to length
In equilibrium position, of the magnetic needle, then
B2 BH

M
N
θ
0
75 l F
t tension = t BH
0
30
B1
S l
t1 = t 2 ; i.e., MB1 sin 300

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i.e., Fl = (2l m) BH sinq ; \ F = 2mBH sinq moment M2 is rotated through 300 . Then find
the ratio of M1 & M2 .
d) In the above case, if the force applied is Sol: C (a - q ) = MBsin q
always perpendicular to length of the magnetic
needle but at a distance 'r' from the pivot, then For first magnet, C (180 − 45) = M 1 B sin45 0 ----(1)
BH For second magnet, C (180 − 30) = M 2 B sin30 ----(2)
0

N Diving equation (1) by equation (2)

θ
135 M 1 M 9
r = × 2 ⇒ 1 =
Frsin900 = MBH sin q
F 150 M 2 M 2 10 2
l
S WE - 3 : A magnetic dipole is under the influence
of two magnetic fields. The angle between the
two field directions is 600 and one of the fields
MBH sin q 2lmBH sin q has a magnitudeof 1.2 × 10–2 T. If the dipole
\F= ; =
r r comes to stable equilibrium at an angle of 150
Ø A magnet of moment 'M' is suspended in the with this field, what is the magnitude of the
magnetic meridian with an untwisted wire. The other field?
upper end of the wire is rotated through an angle Sol. Here B1 = 1.2 × 10–2 T Inclination of dipole
' a ' to deflect the magnet by an angle ' q ' from with B1 is θ1 = 15 0 Therefore, inclination of
magnetic meridian. Then deflecting couple acting dipole with B2 is θ2 = 600 − 15 0 = 450 As the
on the magnet = MBH sinq dipole is in equilibrium, therefore the torque on
the dipole due to the two fields are equal and
α opposite. If M is magnetic dipole moment of the
dipole, then

θ1 =150
N
θ

45 0
B1

=
S N

θ
2
B2

Restoring couple developed in suspension wire

= C(a -q ) where C is couple per unit twist of 600
B1 sin q1 1.2 ´10-2 ´sin150
B2 = =
suspension wire. \ In equilibrium position, MB1 sin θ1 = MB2 sin θ2 or sin q2 sin45 0 ;
MBH sin q = C(a -q )
-2
1.2´10 ´0.2588 = 4.39´10-3 T
= ;
0.707
W.E-1 : When a bar magnet is placed at 90o t o W.E-4: A compass needle of magnetic moment
a uniform magnetic field, it is acted upon by 60A-m2 , pointing towards geographical north
a couple which is maximum. For the couple at a certain place where the horizontal
to be half of the maximum value, at what component of earth’s magnetic field is
angle should the magnet be inclined to the
magnetic field (B) ? 40µ wb/m2 experiences a torque of 1.2 × 10–3
Sol: We know that, τ = MB sin θ Nm. Find the declination at that place.
If θ = 90° then τ max = MB ..... (1) Sol. If θ is the declination of the place, then the
τmax torque acting on the needle is τ = M BH sin θ
= MBsin θ ..... (2)
2
From equations (1) and (2) τ 1.2 × 10−3 1
⇒ sin θ = = −
= \ q = 3 00
M BH 60 × 40 × 10 6 2
2 = sin1 θ or sin θ = 12 or θ = 30°
Work done in rotating a magnetic dipole in a
W.E-2 : A bar magnet of magnetic moment M1 is magnetic field
suspended by a wire in a magnetic field. The
upper end of the wire is rotated through 1800 , Ø The work done in deflecting a magnet from
then the magnet rotated through 450 . Under angular position θ1 to an angular position θ 2
similar conditions another magnet of magnetic
with the field is change in PE given as

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W = MB ( cos θ1 − cos θ 2 ) W.E - 7 : A bar magnet has a magnetic moment

2.5 J T–1 and is placed in a magnetic field of
Ø The work done in deflecting a bar magnet through 0.2 T. Calculate the work done in turning the
an angle θ from its state of equilibrium position in magnet from parallel to antiparallel position
a uniform magnetic field is given by relative to field direction.
W = MB(1 − cos θ )  here θ1 = 00 ,θ 2 = θ  Sol. Work done in changing the orientation of a dipole
of moment M in a field B from positionq1to q2 is
When it is released, this workdone converts into
rotational KE given by W = MB (cos q1 - cos q2 )
q1 = 00 and q2 = 180 0
1
MB(1-Cos θ )= Iω 2 Here,
2
Ø When a bar magnet is held at an angle θ with So, W = 2 MB = 2´ 2.5 ´0.2 = 1 J
the magnetic field, the potential energy W.E- 8 : A bar magnet with poles 25 cm apart and
possessed by the magnet is U = –MB cos θ pole-strength 14.4 A-m rests with its centre on
Ø When the bar magnet is parallel to the applied a frictionless pivot. It is held in equilibrium at
field, then θ = 00 and potential energy is (-MB).It 600 to a uniform magnetic field of induction
is said to be stable equilibrium. 0.25 T by applying a force F at right angles to
Ø When the bar magnet is perpendicular to the its axis, 10cm from its pivot. Calculate F. What
applied field, then θ = 900 and potential energy will happen if the force is removed?
is zero Sol. The situation is shown in figure. In equilibrium
Ø When the bar magnet is anti-parallel to the the torque on M due to B is balanced by torque
uur ur r ur
applied field, then θ = 1800 and potential energy due to F, i.e., i.e., M × B = r × F
is maximum i.e. U = +MB.It is said to be unstable mB
equilibrium. B
W.E-5: A magnet is suspended at an angle 600 in an 600
M
external magnetic field of 5 × 10 –4 T. What is the
work done by the magnetic field in bringing it in 0 F
its direction ? [The magnetic moment = 20 A-
m2 ] mB
Sol. Work done by the magnetic field,
( m × 2l ) B sinθ
W=MB( cosθ1 − cos θ2 ) Here θ1 = 600 and q2 = 00 MB sin θ = Fr sin900 or F=
r
\ W = 20 ´5 ´10 -4 é cos60 0 - cos0 0 ù ( as M = m x 2l) ; So substituting the given data,
ëê ûú

= 10 −2 1  −3
( )
14.4 × 25 × 10−2 × 0.25 ( 3/2 ) = 7.8 N
 2 − 1 = − 5 × 10 J . F=
10 × 10 −2
 
ur uur ur
W.E-6 : A magnetic needle lying parallel to a If the force F is removed, the torque M ´ B
magnetic field requires W units of work to will become unbalanced and under its action the
turn it through 60 0 . What is the torque magnet will execute oscillatory motion about the
needed to maintain the needle in this direction of B on its pivot O which will not be
positon? simple harmonic as sinθ ≠ θ
Sol. In case of a dipole in a magnetic field,
Field of a Bar Magnet
W = MB (cos q1 - cos q2 ) and C = MB sin q
Axial line: The magnetic induction at a point
Here, q1 = 0 0
and q 2 = 6 0 0  µ0 
2Md
on the axial line is Ba = 4π (d 2 − l 2 )2
 
q  
So, W = MB (1- cos q ) = 2MB sin
2

2 For a short bar magnet i.e. l <<<< d

q q  µ  2M
and , C = MB sin q = 2 MB sin cos then Ba =  4π0 
2 2   d3
æqö
So, W = cot çççè 2 ø÷÷÷ , i.e C = W cot30 = 3W
C 0 Ø The direction of magnetic induction on the axial
line is along the direction of magnetic moment.
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MAGNETISM NEET-PHYSICS-VOL-II

Equatorial line: The magnetic induction at a each other

uur ur
point on the equatorial line at a distance d from And as U = -M· B, the interaction energy of the
µ M uuur uuur
the centre is Be = 4π (d 2 + l 2 )3 / 2
0
system (i.e., P.E. of M 2 in the field of M1 or
uuur uuur
For a short bar magnet i.e. l < < d P.E. of M1 in the field of M 2 ) will be
 µ0  M uuur uur uuur uur m 2 M1 M2
then Be =   U = -M2 . B1 = -M1 .B 2 = - 0
 4π  d 3 4p r
3

Ø The direction of magnetic induction on the uuur uur

[as M 2 is parallel to B1 , i.e., q = 0 0 ].........(3)
equatorial line is in the direction opposite to uuur
magnetic moment. Now as F = - (dU/dr) so force on M1 due to
At any point in the plane of axial and uuur uuur uuur
equatorial lines: M 2 or force on M 2 due to M1 will be
 µ  M (3cos 2 θ + 1) d é m0 2M 1M 2 ù
F1 = F2 = - ê- ú
B=  0  dr êë 4p r3 úû
 4π  d3
θ = 00 for axial line ; θ = 900 for equatorial line m0 6M1M 2
=- ––––––(4)
Ø For a short bar magnet , at two equidistant 4p r 4
points , one on the axial and the other on From equation (4) it is clear that interaction
equitorial line Ba = 2 Be Ø force between the magnets varies as (1/r4 ).
When magnets are perpendicular to the line
Force between two magnets :When one
joining their centres
magnet is placed in the field of another magnet If the similar poles of two magnets face each
it usually experiences a couple or force o r uuur
both and has potential energy. Depending on the other as shown in Fig. (A), the field due to M1
orientation of the magnets relative to each other, uuur
at the position of M 2 , i.e., at O2 will be
the following situations are discussed .
Ø When magnets are along the line joining their N N N S
centres M 1
O1 F1 F2 O2 M2 M1 O1 F1 F2 O2 M2

If the opposite poles of two magnets face each S S S N

uur
other as shown in Fig.(A), the field due to M 1
uur B 2
r
B1 B2
r B1
at the position of M 2 , i.e., at O2 , will be :
Repulsion Attraction
ur m 2M (A) (B)
B1 = 0 3 1 with q = 00 [ as O2 lies
4p r uur m M
uuur B1 = 0 31 with f = 90 0 [as O2 lies on the
on the axis of M1 ] 4p r
uuur uuur uur uuur
So couple on M 2 due to M1 , i.e., B1 is equatorial line of M1 ]
uur uuur uur uuur uur uuur uur
C2 = M 2 ´ B1 = 0 [ as M 2 is Now as B1 is antiparallel to M 2 and B 2 to
uur uuur
parallel to B1 , i.e., q = 0 ]........(1) M1 , i.e., q = 1800 , so
uur uuur uur uur uuur uur
M1 M2 C2 = M 2 ´ B1 = 0 and C1 = M1´ B2 = 0
F1 F2 M1 M2
S O1 N S O2 N S O1 N
F1 F2
N O2 S
........... (5)
B2 B1 B2 B1
i.e., the magnets will not exert any couple on
each other
r r uur ur
And as U =-M.B , the interaction energy of
uuur uuur
Attraction Repulsion
(B)
(A)
uur uuur uur the system (i.e.,P.E. of M 2 in the field of M1
uuur uuur
Similarly, C1 = M1´ B2 = 0 or P.E. of M1 in the field of M 2 ) will be
uuur uur
[as M1 is parallel to B 2 i.e., q = 0 ] .......(2) uuur uur uuuruur m M M
U = -M 2 .B1 = -M1 .B2 = 0 1 3 2 ........... (6)
i.e., the magnets will not exert any couple on 4p r

180 NARAYANA MEDICAL ACADEMY

 NEET-PHYSICS-VOL-II MAGNETISM
uuur uur
[as M 2 is antiparallel to B1 , i.e., q = 1800 ] induction field due to the magnet is exactly same as
uuur that of earth’s horizontal component. These points
Now as F = -(dU/dr), so force on M1 due to
uuur uuur uuur are called null points. If the average distance of N1
M 2 or on M 2 due to M1 will be and N2 from the centre of the magnet is ‘d’ then
m 3M1M 2
Bmagnet = BH (horizontal component of earth’s
d æm M M ö
F1 = F2 = - çç 0 1 3 2 ÷÷÷ = 0 magnetic field)
dr çè 4p r ø 4p 4 .............. (7)
r
From equation (7) it is clear that interaction µ M
∴ 0 = B H → 4.16
force varies as (1/r4 ). ( )
4π d2 + l 2 3 / 2
Superposition of Magnetic fields
µ0 M
Neutral points and their location : For short magnet = BH → 4.17
In the combined field due to bar magnet and 4π d3
horizontal component of earth’s magnetic field North pole of the magnet pointing
(B H): towards geographic south:
Earth’s magnetic field is present every where
and its horizontal component extends from When a magnet is placed in the magnetic
south to north. When a magnet is placed any meridian with its north pole facing geographic
where, its field gets superimposed over the south, the field lines of the resultant magnetic
earth’s field, giving rise to resultant magnetic field are shown in the figure.
field. In this resultant magnetic field, there are
certain points where the resultant magnetic
induction field becomes zero. At these points, N1
N
the horizontal component of earth’s magnetic S
W E
field exactly balances the field due to the
magnet. These points are called null points O
N
or neutral points. S
“The points in the magnetic field where the N2
resultant magnetic induction field becomes
zero are called null points”.
North pole of the magnet pointing towards Magnetic lines of force when north pole
geographical north : When a magnet is placed of the magnet pointing towards
in the magnetic meridian, with its north pole geographic south Results:
facing geographic north, the combined magnetic
field lines due to earth and the bar magnet are Ø The directions of the two fields ( horizontal
as shown in the figure. component of earth’s magnetic field and the field
due to the magnet) are exactly opposite to each
other, on the axial line.
Ø As we deviate from the axial line, the two fields
N1 N2 differ in direction.
Ø The directions of the two fields at all points on
the equatorial line is the same.
Ø Along the axial line, the magnetic field due to
Magnetic lines of force when northpole
magnet decreases in magnitude on moving away
of the magnet pointing towards from the centre of the magnet. There will be
geographic north Results: points N1 and N2 situated at equal distances from
Ø Along the axial line, on both sides, the two fields the centre of the magnet where the fields are
have same direction. The magnitude of resultant exactly balanced by the earth’s horizontal
magnetic field is the sum of the magnitudes of component field. These points are called null
two fields. points.
Ø As we deviate from axial line, the two fields
m0 2Md
differ in direction. B= = BH
Ø 4p ( d 2 - l 2 )
2
On the equatorial line, the direction of the two
At null points,
fields are exactly opposite to each other.
Ø At N1 and N2 on the equatorial line, the magnetic (where BH is earth’s horizontal magnetic induction

181
NARAYANA MEDICAL ACADEMY
MAGNETISM NEET-PHYSICS-VOL-II

field) distance from the pole of the magnet where the

m0 2M
For short magnet, = BH µ m
BH = 0 2
4p d 3 neutral point is formed, then
4π d
If the horizontal component of earth’s magnetic Ø If the north pole is on the table, then the
field BH at the given place is known, the neutral point is formed towards geographic south
magnetic moment (M) of the magnet can be side of the pole.
determined by locating the neutral points. iv) If the south pole is on the table, then the
Magnet placed perpendicular to the neutral point is formed towards geographic north
magnetic meridian : When a bar magnet is side of the pole.
placed with its axial line perpendicular to the Note: A short bar magnet is kept along magnetic
magnetic meridian with its north pole facing east meridian with its north pole pointing north. A
of earth, the resultant magnetic field is shown in neutral point is formed at point 'P' at distance
the figure Along a line making an angle of 'd' from the centre of the magnet then
tan- 1 ( 2 ) with east - west line, there are two Ba
BH
points (N 1 and N2 ) where the resultant magnetic BH
N
BH BH BH
induction field is zero. Thus N1 (on the N-W n2 n1
line) and N2 ( on the S-E line) are the null points. d
S
d
2
d P 2d
At the null point, Be Be1 Be B1e 1
m M
BH = 0 3 1 + 3cos 2 q Where Tanq = 2 Ø At a distance 'd' on equatorial line, net megnetic
4p d induction B net = 0
N
m0 M
ie Be = BH Þ . 3 = BH
4p d
N1

Ø At a distance d/2 from the centre of the magnet

W O E on equatorial line, the net magnetic induction is
given by
m0 M
Bnet = Be| -BH = - BH = 7BH
N2

4p æ d ö3
çç ÷÷
S

1 Þ B = 2 m0 M çè 2 ÷ø
\ cos q =
4p d
H 3
3 Ø At a distance '2d' on equatorial line, the net
If a very long magnet is placed vertically with magnetic induction is given by
its one pole on a horizontal wooden table (or)
m M
when an isolated magnetic pole is kept in the Bnet = BH - B||e = B H - 0
earth’s magnetic field, then 4p (2d)3
S
BH 7BH
= BH - =
8 8
N Ø At a distance 'd' on axial line of the bar magnet,
the net magnetic induction is given by
BH BH Bnet = Ba + BH = 2Be + BH = 2BH + BH = 3BH.
BH Ø If the axis of the bar magent is rotated through
n
B
900 clockwise at the same position then the net
N S
BH magnetic magnetic induction at the same point
n 'P' is
5 BH (Q Ba = 2Be = 2BH )
B

Ø A single neutral point will be formed in the Bnet = Ba2 + BH

2 =
combined field on the horizontal table.
Ø If ‘m’ is the polestrength and ‘d’ is the

182 NARAYANA MEDICAL ACADEMY

 NEET-PHYSICS-VOL-II MAGNETISM

BH
æ B ö
BH ççQ Be = a = H ÷÷
5 B
Bnet = B2e + B2H =
N Bnet

1 1
2 çè 2 2 ÷ø
S N
P Ba Ø If axis of the magnet is rotated through 1800 ,then
S
magnetic induction at the point 'P' is
Bnet = Ba + BH = BH + BH = 2BH
Ø If the axis of the magnet is rotated through 1800 at
the same position, then net magnetic induction at Neutral points in the combined field due
the same point 'P' is Bnet = Be + BH = 2BH to isolated magnetic poles :
Note: A short bar magnet is kept along magnetic Ø When two like magnetic poles of pole strengths
meridian with its south pole pointing north. A m1 and m2 (m1 < m2 ) are separated by a distance
neutral point is formed at a point 'P' at a ‘d’, then neutral point is formed in between the
distance 'd' from the centre of the magnet poles and on the line joining them. Let ‘x’ be
then the distance of neutral point from weaker pole
Ø at a distance 'd' on axial line of the bar magnet of strength m1 .
net magnetic induction,
BH
N X (d−X) N
d n 1P
Ba
m1 B2 n B1 m2
S
BH
At neutral point, B1 = B2
m m m
Be
m2
Þ 0 . 21 = 0 .
BH

N 4p x 4 p ( d - x)2

m 0 2M on solving, we get x =
d
Þ . = BH
Bnet = 0 i.e., Ba = BH 4p d 3 m2
+1
d m1
Ø At a distance on axial line of bar magnet,
2 Ø When two unlike magnetic poles of strengths
netmagnetic induction is given by m1 and m2 (m1 < m2 ) are separated by a distance
m 0 2M ‘d’, then neutral point is formed outside and on
Bnet = Ba1 -BH = . - BH = 7BH
4p æ d ö3 the line passing through the poles. It always
çç ÷ ÷ lies closer to weaker pole.
èç 2 ÷ø B1 B2 N d S
Ø At a distance '2d' on axial line of the bar magnet, n X m1 m2
net magnetic induction is given by
At neutral point, B1 = B2
m 2M m m m
Bnet = BH - B||a = B H - 0 Þ 0 . 21 = 0 .
m2
4 p (2d)3 4p x 4p ( d + x )
2

BH 7B H
= BH - = d
8 8 on solving,we get x =
m2
Ø At a distance 'd' on equatorial line of the bar -1
magnet, net magnetic induction is m1
B Neutral points in the combined field due
Bnet = Be +BH = a + BH
2 to short bar magnets :
B 3 Two short bar magnets of magnetic moments
= H + BH = B H M1 and M2 (M1 < M2 ) are placed at a distance
2 2
‘d’ between their centres with their magnetic
Ø If axis of the magnet is rotated through 90 0
axes oriented as shown in the figure, Then two
clockwise at the same position, then net magnetic
neutral points are formed (i) in between and (ii)
induction at the same point 'P' is given by
outside and on the line passing through centres
of the magnets. In either case, null point is always
183
NARAYANA MEDICAL ACADEMY
MAGNETISM NEET-PHYSICS-VOL-II

closer to magnet of weaker moment. between the poles

B1 B2 M1 B1 B2 M2 b) For unlike poles the neutral point is situated
n2 x S1 N1 x n1
N2 S2 on line joining the poles. But not in between
d them.
c) In either case null point is always closer to
B1 S1 B1
N2
the weaker pole.
Ø If two short bar magnets of magentic moments
x x
n2 n1
M1 and M 2 ( M 1 < M 2 ) are placed along the
N1
B2
M1
B2 S2
M2
same line with like poles facing each other and
‘d’ is the distance between their centres, the
Case i) : If the neutral point is formed in between the distance of null point from M1 is
magnets, then B1 = B2
m 2M m 2M 2 x=
d
Þ 0 . 31 = 0 .  M2 
1/3
4p x 4p (d - x)3
  ±1
d
 M1 
on solving,we get x = a) + for null point formed between the magnets.
æ M 2 ÷ö
1
çç ÷
3
+1 b) − for null point formed outside the magnets.
çè M1 ÷÷ø c) When unlike poles face each other, null point
Case ii) : If the neutral point is formed outside the Time period of Suspended Magnet in
combination, then the Uniform Magnetic Field
m0 2M1 m0 2M2 Ø Principle : When a bar magnet is suspended
Þ . = .
4p x 3 4p (d + x )3 freely in a uniform magnetic field and displaced
from its equilibrium, it starts executing angular
d
on solving , we get x = SHM.
æ M 2 ö÷ Ø
1
Time period of oscillation and frequency of
çç
3
÷ -1
çè M1 ø÷÷ magnet is
I 1 1 M BH
Note:No null points are obtained when unlike poles T = 2π and n = =
MB H T 2π I
of the magnets are placed closer to each other
Ø When two or more magnetic fields are where M magnetic moment, BH Horizontal
superimposed in the same region, according to component of earth magnetic induction and I
the resultant magnetic field the space in the (l 2 + b2 )
region gets modified. moment of inertia, I = m
12
Ø The magnetic field of induction at any point is ml 2
the resultant of all the fields superimposed at for a thin bar magnet I =
12
that point.
where m is mass, l is length and b is breadth
Null Point (or) Neutral Point : The point at
of the magnet.
which the resultant magnetic field is zero is
called null point. Ø For small percentage changes in moment of
∆T 1 ∆I
Ø If two poles of pole strengths m1 and m2 inertia × 100 = × 100
T 2 I
( m1 < m2 ) are separated by a distance d, then Ø As I increases , T increases
the distance of the neutral point from the first Ø For small percentage changes in magnetic
pole m1 is ∆T − 1 ∆M
moment ×100 = ×100
T 2 M
d  + forlikepoles  Ø As M increases , T decreases
x=  
m2
±1  − forunlikepoles  Comparision of magnetic moments :
m1
Ø If two magnets of moment M1 and M2 of same
a) For like poles the neutral point is situated in dimensions and same mass are oscillating in the

184 NARAYANA MEDICAL ACADEMY

 NEET-PHYSICS-VOL-II MAGNETISM
same field separately, then T22 M1 + M2 æ ö
= ççQ T a 1 ÷÷
T1 M2 T12 M1 - M 2 èç M ø÷
= (Bar magnets of equal size)
T2 M1 M1 T22 +T12
Þ = 2 2
æç 1 ö÷ M2 T 2 -T1
ççQ T µ ÷
è M ÷ø b) If n1 and n2 are the corresponding
Ø A magnet is oscillating in a magnetic field B
M1 n 12 + n 22
and its time period is Tsec. If another identical frequencies, then =
magnet is placed over that magnet with similar M2 n11 - n 22
poles together, then time period remain Ø When same bar magnet used in the vibration
unchanged. magnetometer at two different places 1 and 2, then
(Q I| = 2I and M| = 2M, æ ö
T22 ççQT a 1 ÷÷
÷
BH1
= çç
T| = 2p
I|
=2p
2I
=2 p
I
=T ) BH2 T12 è BH ÷ø
M| B 2MB MB
Ø A magnet is oscillating in a magnetic field B Ø When two bar magnets of moments M1 and M2
and its time period is T sec. If another identical are placed one over the other such that (i) like
magnet is placed over that magnet with unlike poles together (ii) unlike poles together and (iii)
poles together, then time period becomes their axes are perpendicular to each other. When
infinite. i.e., it does not oscillate. vibrated in the same magnetic field, the ratio of
their time periods respectively is
é | ù
ê M = M - M = 0;T = 2 p I
=µú
ê 0 ´B ú T = 2p
I
Þ Ta
1
êë úû
MB M
Ø The time period of a thin bar magnet is T. It is
cut into 'n' equal parts by cutting it normal to its 1 1 1
\ T1 : T2 : T3 =
( )
: :
length. The time period of each piece when M1 + M2 M1 - M 2 M 12 + M 22
oscillating in the same magnetic field will be
æ
çç æ m öæ l ö
2
÷÷ö
Ø If T0 is the time period of oscillation of the
çç ÷÷ çç ÷÷ ÷÷
çç ç ÷ç ÷
ççQ I 1 = è n øè n ø = I & M 1 = M ÷÷
experimental magnet oscillating in BH . An
n ÷÷÷
T
T| = çç 12 n 3 external field B is applied due to a bar magnet
n çç ÷÷
çè ÷ø in addition to BH at the point where the first
magnet is oscillating. Then its new time period
I1 T
\ T 1 = 2p = is T.
M 1B n
Ø T0 Br uur ur uuur
Then T = B where Br = B + BH
The time period of a thin bar magnet is T. It is
cut into 'n' equal parts by cutting it along its
uuur
H
length. The time period of each piece remains ur
a) If B and BH are along the same direction,
unchanged, when oscillating in the same field.
æ m ÷ö 2 Br = B + B H Þ T < T0
ç ÷l ur uuur
M | ççè n ÷ø I b) If B and BH are in opposite directions,
(Q M = & I =
|
=
n 12 n Br = B - BH Þ T0 < T
|
I I/n
Þ T = 2p = 2p = T) c) If B and BH are in opposite directions, and
|
|
MB M
n
B
also if B = BH then Br = B : BH = 0
Ø a) Two magnets of magnetic moments M1 and Þ T =µ (i.e.,
uuur
it does not oscillate)
M2 (M1 >M2 ) are placed one over the other. ur
d) If B and BH are perpendicular to each other,,
If T1 is the time period when like poles touch
each other and T2 is the time peiod when unlike Br = B2 + BH2 Þ T < T0
poles touch each other, then m0 2M
Here B = (if the point is on the axial line)
4p d 3
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MAGNETISM NEET-PHYSICS-VOL-II

m0 M W.E - 11 : A short magnet oscillates in a vibration

B= (if the point is on the equatorial line) magnetometer with a time period of 0.1s
4p d 3
e) If a straight wire carries current vertically up where the horizontal component of earth’s
or down placed on the east or west or north or magnetic field is 24µT . An upward current of
m0 i 18A is established in the vertical wire placed
south side, then B = (From ampere’s law 20 cm east of the magnet. Find the new time
2p r period ?
in electro magnetism)
Ø If n1 and n2 are frequencies of oscillation of the T2 B1
Sol. T = B Where B1 = BH =24 x 10–6 T
bar magnet in uniform magnetic field when B 1 2

supports BH and when B opposes BH µi

and B2 = BH : B = BH : 0
n1 B + BH 2π r
then n = B - B (let B>BH ) −7
4π × 10 × 18
2 H −6
= 24 × 10 : = 6 × 10−6 T
2π × 0.2
B n +n 2 2
Þ = 1 2
T2 24 × 10−6
BH n - n 2
1
2
2
∴
0.1
=
6 × 10−6
=2 ⇒ T2 = 0.2s

1 W.E-12: A magnet is suspended so as to swing

Ø For a bar magnet , T µ B (or) n µ BH horizontally makes 50 vibrations/min at a
H
place where dip is 300 , and 40 vibrations / min
If B1 and B2 be the earth’s magnetic induction at where dip is 450 . Compare the earth’s total
two different places having angles of dip
fields at the two places.
q1 and q2 then Sol. nα BH
T1 BH 2 B2C o s q 2
= = n1 BCos θ1 50 B1 Cos 300
T2 B H1 B1C o s q1 ⇒ = 1
ie = ×
n2 BCos
2 θ2 40 B2 Cos 450
n1 BH 1 BCos q1
or n = B = BCos q
1 25 B1 3 B 25
⇒ = × (or) B = 8 6
1

2 H2 2 2 16 B2 2 2
W.E-9: Two bar magnets placed together in a WE - 13: When a short bar magnet is kept in tan
vibration magnetometer take 3 seconds for 1 A position on a deflection magnetometer, the
vibration. If one magnet is reversed, the magnetic needle oscillates with a frequency’f’
combination takes 4 seconds for 1 vibration. and the deflection produced is 450 . If the bar
Find the ratio of their magnetic moments. magnet is removed find the frequency of
Sol. Given that, T1 = 3s and T2 = 4s osciullation of that needle ?
M1 T22 + T12 4 2 + 3 2 16 + 9 25 M1
= 2 = 2 = = or = 3.57 ⇒
n1
=
B1
M2 T2 − T1 4 − 3
2 2
16 − 9 7 M2 Sol. nα B n2 B2
WE - 10 : A bar magnet makes 40 oscillations per
Where B1 = B 2 + BH2 = ( BH tan450 ) + BH2
2
minute in a vibration magnetometer. An
identical magnet is demagnetised completely
and is placed over the magnet in the = 2 BH & B2 = BH
magnetometer. Calculate the time taken for n 2BH n f
40 oscillations by this combination. Ingore ∴ 1 = = 21 / 4 ⇒ n2 = 11/ 4 = 1 / 4
n2 BH 2 2
induced magnetism.
Sol. In the first case, frequency of oscillation, W.E-14: Two bar magnets of the same length and
breadth but having magnetic moments M and
1 MB
n= 2M are joined with like poles together and
2p I
suspended by a string. The time of oscillation
In the second case, frequency of oscillation,
of this assembly in a magnetic field of strength
1 MB n1 1 T1
n =
1
Þ = Þ = 2 B is 3 sec. What will be the period of
2p 2I n 2 T oscillation, if the polarity of one of the
(or) T 1 = 2T (or) 40T1 = 2 × 40T magnets is changed and the combination is
(or) t = 2t = 2 minute = 1.414 minute
1 again made to oscillate in the same field ?

186 NARAYANA MEDICAL ACADEMY

 NEET-PHYSICS-VOL-II MAGNETISM
Sol. As magnetic moment is a vector, so when magnets The value of H is independent of medium .
are joined with like poles together Intensity of magnetising field is a vector in the
M1 = M + 2 M = 3 M, so direction of magnetic field and has unit
Wb / m 2 V ×s A
T = 2π
( I1 + I2 ) = =
Ω× s× m m
Dimensions AL−1
........... (1) H /m
3MB
2) Intensity of magnetisation I : When a
When the polarity of one of the magnets is magnetic material is magnetised by placing it in
reversed, M2 = M ~2M = M; a magnetising field, the induced dipole-moment
( I1 + I2 ) per unit volume in the specimen is called
so T ' = 2π ............ (2) intensity of magnetisation.
MB uur
Dividing Eq. (2) by (1), r M uur r
i.e I = but as M = mLn and V = SL
T'
T
= 3, i. e., T ' = ( 3 ) T = 3 3 sec r mr
V

Magnetic Materials I = n i.e , intensity of magnetisation is

S
Ø Curie and Faraday discovered that all the numerically equal to the induced pole-strength
materials in the universe are magnetic to some per unit area of cross -section.It is a vector
extent. These magnetic substances are quantity having direction of magnetising field
categorized mainly into two groups. or opposite to it as shown in figure. Its unit is
Ø Weak magnetic materials come under (A/m) and dimensions  AL−1 
diamagnetic and paramagnetic materials. Strong
magnetic materials are Ferro-magnetic materials. Magnetic Susceptibility ( χm) : The ratio of
Ø According to the modern electron theory of magnitude of intensity of magnetisation to that
magnetism, the magnetic response of any material of magnetising field strength is called magnetic
is due to the circulating electrons in the atoms. I
Each circulating charge constitutes a magnetic susceptibility χm =
H
moment in a direction perpendicular to the plane It is a scalar with no units and dimensions. It
of circulation. physically represents the ease with which a
Ø In magnetic material all these magnetic moments magnetic material can be magnetised.i.e large
due to the orbital and spin motion of all the value of χ m implies that the material is more
electrons in the atoms of the material, vectorially susceptible to the field and hence can be easily
add up to a resultant magnetic moment. The magnetised.
magnitude and direction of this resultant
magnetic moment is responsible for the magnetic Magnetic permeability ( µ ) : When a
behaviour of the material. magnetic material is placed in a magnetising
Ø Magnetic material are studied interms of the field, the ratio of magnitude of total field inside
following physical parameters the material to that of intensity of magnetising
uur field is called magnetic permeability; i.e.,
Intensity of Magnetising field( H ) :
B
Any magnetic field in which a magnetic material µ = , i.e., B = µ H
is placed for its magnetization is called H
It measures the degree to which a magnetic
magnetising field. material can be penetrated by the magnetising
In a magnetising field the ratio of magnetising
uur field or ability of the material to allow magnetic
field Bo to the permeability of free space is lines of force. It is a scalar having unit Hm −1
called intensity of magnetising field
uur (
and dimensions MLT −2 A−2 . )
uur B uur uur
In air . H =
o
or Bo = µo H Relative permeability ( µr ) :
µ o It is the ratio of magnitudes of total field inside
B the material to that of magnetising field or it is
In a medium H = the ratio of permeability of a medium to that of
µ free space.
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B µH µ W.E-19 : An iron bar of length 10 cm and diameter

µr = = = 2 cm is placed in a magnetic field of intensity
B0 µ0 H µ 0
1000Am–1 with its length parallel to the
It has no units and dimensions direction of the field. Determine the magnetic
Relation between relative permeability moment produced in the bar if permeability
and susceptibility : of its material is 6.3 × 10–4 TmA –1 .
æ I ö÷ Sol. we know that, µ = µ0 (1 + χ)
We know B = m0 ( H + I ) or, = m0 çççè1 + ø÷÷
B
H H µ 6.3 × 10−4
⇒χ= −1 = − 1 = 500.6
B I µ0 4π × 10− 7
or, m = m0 (1 + c ) [as = m and =c]
H H Intensity of magnetisation,
m I = χ H = 500.6 × 1000 = 5 × 105 Am–1
(or) m = 1 + c \ mr = 1 + c \ magnetic moment, M = I x V = I ´ pr l
2
0

= 5 × 105 × 3.14 × (10−2 ) × (10 × 10− 2 ) =17.70 A-m2

2
This is the desired result.
W.E-15 : A magnetising field of 1600Am- 1
produces a magnetic flux of 2.4 × 10–5 weber Electron Theory of Magnetism
2
in a bar of iron of cross section 0.2 cm . Ø i) Molecular theory of magnetism was first given
Calculate permeability and susceptibility of by Weber and was later developed by Ewing.
the bar.
−5
Ø ii) Electron theory of magnetism was proposed
Sol: Magnetic induction, B = Aφ = 2.4 × 10 −4 = 1.2 Wb/m2 by Langevin.
0.2 × 10
B 1.2 Ø iii) The main reason for the magnetic property
i) Permeability, µ = = = 7.5 × 10−4 TA−1 m
H 1600 of a magnet is spin motion of electron. Most of
ii) As m = m0 (1 + c ) then the magnetic moment is produced due to electron
m 7.5 × 10−4 spin. The contribution of the orbital revolution
Susceptibility, c = m - 1 = − 1 = 596.1 is very small.
0 4π × 10−7
W.E-16 : The permeability of substance is A) Explanation of diamagnetism:
6.28×10 –4 wb/A-m. Find its relative
permebility and suscepibility ? Ø i) Since diamagnetic substance have paired
electrons, magnetic moments cancel each other
µ 6.28 × 10−4
Sol. µγ = µ = 4π × 10−7 = 500 and there is no net magnetic moment.
0

µγ = 1 + χ ∴ χ = µγ − 1 = 500 − 1 = 499
Ø ii) When a diamagnetic substance is placed in
an external magnetic field each electron
W.E-17 : The magnetic moment of a magnet of experiences radial force F = Bev either inwards
mass 75 gm is 9×10–7 A-m2 . If the density of
the material of magnet is 7.5×103 kg m–3 , then or outwards. Due to this the angular velocity,
find intensity of magnetisation is current, and magnetic moment of one electron
mass ( m )
increases and of the other decreases. This results
M
Sol. I = Where volume, V = density ( ρ ) in a non-zero magnetic moment in the substances
V
in a direction opposite to the field.
M×ρ 9 × 10− 7 × 7.5 × 103
= = = 0.09A / m Ø iii) Since the orbital motion of electrons in atoms
m 75 × 10−3
is an universal phenomenon, diamagnetism is
WE- 18 : A magnetic field strength (H) 3×103Am–
1 produces a magnetic field of induction (B) of present in all materials. Hence diamagnetism is
12πT in an iron rod. Find the relative a universal property.
permeability of iron ? Properties of Dia-magnetic substances
B 12π
Sol. µ = = = 4π × 10− 3 Ø The substances which when placed in a external
H 3 × 103
magnetic field acquire feeble magnetism
µ 4π × 10 −3
∴ µr = = = 104 opposite to the direction of the magnetising field
µ0 4π × 10− 7
are known as dia-magnetic substances.
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 NEET-PHYSICS-VOL-II MAGNETISM

Ex: Bismuth (Bi), Zinc (Zn), Copper (Cu), Silver (Ag),

I
Gold (Au). Salt (Nacl), Water (H2 O), Mercury
(Hg), Hydrogen (H2 O) etc. H
O
Ø When a bar of dia-magnetic substance is
suspended freely between two magnetic poles
−I
[see figure] , then the axis of the bar becomes
perpendicular to magnetic field. Ø The magnetic susceptibility c (I / H) is small
and negative(Because I is small and opposite in
direction to H). This is independent of
N N S S temperature as shown in figure.

x
Ø When a dia-magnetic material is placed inside
a magnetic field, the magnetic field lines become
O T
less dense in the material.
Ø If one limb of a narrow U-tube containing a Ø The relative permeability is less than unity
dia-magnetic liquid is placed between the poles because m r = (1 + c ) and c is negative.
of an electromagnet, then on switching the field,
the liquid shows a depression. This is shown in Ø The origin of diamagnetism is the induced
figure. dipole moment due to change in orbital motion of
electrons in atoms by the applied field. Dia-
magnetism is shown only by those substances
which do not have any permanent magnetic
moment.
N S B) Explanation of Paramagnetism:
Ø i) Paramagnetic materials have a permanent
magnetic moment in them. The moments arise
from both orbital motion of electrons and the
Ø When a dia-magnetic substance is placed in a spinning of electrons in certain axis.
non-uniform field, then it tends to move towards Ø ii) In atoms whose inner shells are not completely
the weaker part from the stronger part of the field filled, there is a net moment in them since more
as shown in figure. number of electrons spin in the same direction.
This permanent magnet behaves like a tiny bar
magnet called atomic magnet.
Ø iii) In absence of external magnetic field atomic
N S N S magnets are randomly oriented due to the thermal
agitation and the net magnetic moment of the
a) Magnet closely b) pole pieces substance is zero.
spaced moved apart Ø iv) When it is placed in an external magnetic
Ø Dia magnetic substances acquire feeble field the atomic magnets align in the direction of
magnetism in a direction oppoite to magnetising the field and thermal agitation oppose them to
field. The intensity of magnetisation I is very do so.
small, negative and is directly proportional to Ø v) At low fields the total magnetic moment would
magnetising field H as shown in figure. be directly proportional to the magnetic field B

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and inversely proportional to temperature T.

Properties of Paramagnetic substances
(i) The substances which when placed in a magnetic N S N S
field, acquire feeble magnetism in the direction fig(a) fig(b)
of magnetising field are known as paramagnetic
are near to each other. In figure (b), the
substances.
distance between the poles is
Ex: Aluminium (Al), Platinum (Pt), Manganese (Mn),
increased. i.e., the field is stronger near the
Copper chloride (CuCl2 ), Oxygen (O 2 ), solutions
poles.
of salts of iron etc. are examples of paramagnetic
(vi) The intensity of magnetisation I is very small
substances.
and compared to one. It follows from the relation
(ii) When a bar of paramagnetic substance is placed
in a magnetic field, it tries to concentrate the mr = 1 + cm . The variation of mr or c with H
lines of force into it as shown in figure is shown in figure. As is clear from the figure,
the variation is non-linear. The large value of

N s n S mr is due to the fact that the field B inside the

material is much stronger than the magnetising
field due to ‘pulling in’ of a large number of
This shows that the magnetic induction B in it is lines of force by the material.
numerically slightly greater than the applied field
H. So the permeability m is greater than one
Y

because m = ( B / H) .
(iii) When the bar of paramagnetic material is µr or x

suspended freely between two magnetic poles,

O H X
its axis becomes parallel to magnetic field.
Move over, the poles produced at the ends of Curie’s law : Curie law states that far away
the bar are opposite to nearer magnetic poles. from saturation, the suceptibility c (I/H) of
(iv) If a paramagnetic solution is poured in a U-tube paramagnetic substance is inversely proportional
and if one limb is placed between the poles of to absolute temperature, i.e.,
an electromagnet in such a way that liquid level
1 C
is parallel to field, then on switching the field, cµ or c=
T T
the liquid rises. This is shown in figure.
where C is constant and is called as Curie
constant.
When a ferromagnetic material is heated, it
becomes paramagnetic at a certain temperature.
N S
This temperature is called as Curie temperature
and is denoted by TC. After this temperature, the
susceptibility varies with temperature as
C|
c=
(v) In a non-uniform magnetic field, the paramagnetic ( T - Tc )
substances are attracted towards the stronger
parts of the magnetic field from the weaker parts where C | is another constant. For iron, Tc = 1043
of the field. The situation is shown in figure. In K. = 7700 C
figure (a), the field is stronger in the middle as
the poles
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Properties of Dia, Para and Ferror Magnetic materials

DIA PARA FERRO

1. They are feebly repelled 1. They are feebly attracted 1. They are strongly attracted
by a magnet. by a magnet a magnet
2. The net magnet moment 2. The net magnetic moment 2. The net magnetic moment
due to all the electrons in atoms due to all electrons in atoms is very strong.
the atom is zero is not zero.

3. When subjected to the 3. Magnetised feebly in the 3. Magnetized strongly in the

magnetising field they are direction of magnetising direction of magnetising
feebly magnetised in field. field.
opposite direction to the
magnetising field

4. When suspended inside the 4. They align with their 4. They align with their
magnetic field, they align length along the direction length along the direction
their length perpendicular of magnetic field. of magnetic field.
to the magnetic field.

5. Magnetic lines of force 5. Few lines pass through the 5. Almost all lines prefer to
prefer to move out of the specimen. move through the
specimen. specimen.

6. They move from stronger 6. They move from weaker to 6. They move from weaker to
part of the magnetic field stronger part of the stronger part of the
to the weaker part of the magnetic field magnetic field.
magnetic field

7. µr < 1 7. µr > 1 7. µr >>> 1

8. Intensity of magnetization 8. I is small and positive 8. I is high and positive

(I) is small and negative.

9. χm is small and negative 9. χ m is small and positive 9. χ m is highly positive

10. χm is independent of 10. χ m is dependent on 10. χ m is dependent on

temperature. temperature. temperature
11.Doesn’t obey Curie law. 11.Obey Curie law 11. Obey Curie law and at
Curie temperature they are
turned to paramagnetic
materials.
12.Substances following 12.Substances following 12.Substances following
Diamagnetism are paramagnetism are ferromagnetism are Iron,
Bismuth, Copper, lead, Aluminum, Platinum, Cobalt, Nickel and alloys
silicon, water, glass etc. Manganese, Chromium, like alnico
Calcium, Oxygen,
Nitrogen (at STP)

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Hysteresis: When a bar of ferromagnetic material is Elements of Earth’s Magnetism

magnetized by a varying magnetic field H and (Terristrial Magnetism ): There are three
the intensity of magnetization I induced is elements of earth’s magnetism
measured. The graph of I versus H is as shown (i) Angle of declination (ii) Angle of dip
is figure. (iii) Horizontal component of earth’s field.
Ø Earth’s Magnetic Field: The earth’s magnetic
I
B A field Be in the magnetic meridian may be
C O H
G
resolved into a horizontal component BH and
D F
vertical component BV at any place.
Ø When magnetising field is increased from O the Geographical
Geographical
L meridian
intensity of magnetisation I increases and north

becomes maximum i.e at point (A). This Magnetic P α

O
north BH θ
maximum value is called the saturation value. S
N
Ø When H is reduced, I reduces but is not zero Magnetic
meridian BV
M BP
when H = 0. The remainder value OB of R
magnetisation when H = 0 is called the residual
magnetism or retentivity. OB is retentivity. θ =dip (or) inclination α = declination
Ø When magnetic field H is reversed, I reduces Ø Horizontal component of earth’s magnetic field
and becomes zero i.e., for H = OC, I = 0. This BH = Be cos θ ....... (1)
value of H is called the coercivity. Ø Vertical component of earth’s magnetic field
Ø When field H is further increased in reverse
direction, the intensity of magnetisation attains BV = Be sin θ ....... ( 2)
saturation value in reverse direction (i.e., point
D). When H is decreased to zero and changed Be = (B 2
H + BV2 )
direction in steps, we get the part DFGA Ø Dividing equation (2) by equation (1), we have
Properties of soft iron and steel: For soft
BV Be sin θ
iron, the susceptibility, permeability and = = tan θ
retentivity are greater while coercivity and BH Be cos θ
hysteresis loss per cycle are smaller than those Geographical Meridian: A vertical plane
of steel. passing through the axis of rotation of the earth
I I is called the geographic meridian.
Hysterisis
curve Magnetic Meridian: A vertical plane passing
Hysterisis H H through the axis of a freely suspended magnet is
curve Soft Magnetic Hard Magnetic
Material Material
called the magnetic meridian.
Ø Permanent magnets are made of steel and cobalt Angle of Declination (α ) : The acute angle
while electromagnets are made of soft iron. between the magnetic meridian and the
Ø Diamagnetism is universal. It is present in all geographical meridian is called the ‘angle of
materials. But it is weak and hard to detect declination’ at any place.
if substance is para or ferromagnetic
Ø The value of declination at equator is 170
Shielding from magnetic fields: For
shielding a certain region of space from magnetic Ø Declination varies from place to place
field, we surround the region by soft iron rings. Ø The lines joining the places of equal declination
Magnetic field lines will be drawn into the rings are called isogonal lines.
and the space enclosed will be free of magnetic Ø The lines joining the places of zero declination
field. are called agonic lines.

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 NEET-PHYSICS-VOL-II MAGNETISM
W.E - 20 Considering the earth as a short magnet Thus, one can get the true dip d without locating
with its centre coinciding with the centre of the magnetic meridian.
earth, show that the angle of dipφ is related More about angle of dip (δ ) :
to magnetic latitude λ through the relation (i) At a place on poles, earth’s magnetic field is
tan φ = 2tan λ perpendicular to the surface of earth, i.e.,
Sol. Considering the situation for dipole, at position d = 90 0
( r, θ) we have \ Bv = Bsin90 0 = B
Bθ
Br
Further, BH = B cos 900 = 0
So, except at poles, the earth has a horizontal
S r component of magnetic induction field.
λ

θ
(ii) At a place on equator, earth’s magnetic field
N

is parallel to the surface of earth, i.e., d = 0 0

µ 2Mcos θ µ 0 Msin θ \ BH = Bcos00 = B

Br = 0 and Bθ = Further BV = BH sin 00 = 0
4π 4π r
3 3
r
So, except at equator, the earth has a vertical
BV Br
and as tan φ = B = − B , so in the light of Eq. (1)
component of magnetic induction field.
H θ
(iii) In a vertical plane at an angle q to magnetic
tan φ = −2cot θ ; But From figure θ = 900 + λ meridian
B'H = BH cos q and BV = BV
'

So, tan φ = −2cot ( 90 + λ ) ; i.e., tan φ = 2tan λ

0

So, the angle of dip d ¢ in a vertical plane making

Apparent Dip: If the dip circle is not kept in an angle q to magnetic meridian is given by
the magnetic meridian, the needle will not show
B|V BV
the correct direction of earth’s magnetic field.
tan d ' = =
The angle made by the needle with the horizontal BH BH cos q
|

is called the apparent dip for this plane. If the

tan d æç BV ö
dip circle is at an angle q to the meridian, the or tan d ' = çQ = tan d÷÷÷
effective horizontal component in this place is cos q çè BH ÷ø

B¢H = BH cos q . The vertical component is still (a) For a vertical plane other than magnetic
Bv . If d1 is the apparent dip and d is the true meridian, q > 00 andcos q < 1, i.e., d1 > d
dip, we have ( angle of dip increases )
(b) For a plane perpendicular to magnetic
B BV
tan d1 = v = meridian, q = 900
¢
BH BH cos q
tan d
\ tan d 1 = = ¥ or d1 = 900
tan d ç æ ö
çQ tan d = V ÷÷÷ ..... (1)
B cos90
or tan d1 =
cos q çè BH ø÷ This shows that in a plane perpendicular to
magenetic meridian, the dip needle will become
Now suppose, the dip circle is rotated through
vertical.
an angle of 900 from this position. It will now
make an angle ( 90°-q ) with the meridian. The C.U.Q
effective horizontal component in this plane is MAGNETIC MOMENT AND
B¢¢H = BH sin q . if d 2 be the apparent dip, we RESULTANT MAGNETIC MOMENT
shall have 1. The dimensional formula for magnetic
tan d
B BV moment is
tan d2 = V'' = or tan d2 =
BH sin q sin q
BH
.........(2) 1) M0 L2 T0 A1 2) M0 L1 T0 A2
3) M0 L2 T0 A2 4) M0 L0 T1 A1
From (1) and (2) cot d1 + cot d2 = cot d
2 2 2

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2. If two bar magnets of different magnetic MAGNETIC FIELD

lengths have equal moments, then the pole 9. S.I. unit of Magnetic flux is
strength is 1) ampere-meter 2) amp.m2
1) equal for both the magnets 3) weber 4) weber/m2
2) less for shorter magnet 10. The Source of magnetic field is
3) more for longer magnet 1) isolated Magnetic pole
4) more for shorter magnet 2) static electric charge
3. A bar magnet of moment M is bent into arc, 3) current loop 4) moving light source
its moment 11. The earth’s magnetic field
1) decreases 2) increases 1) varies in direction but not in magnitude
3) does not change 4) may change 2) varies in magnitude but not in direction
4. A bar magnet is cut into two equal halves by 3) varies in both magnitude and direction
a plane parallel to the magnetic axis of the 4) is centred exactly about the centre of the earth
following physical quantities the one which 12. The electric and magnetic field lines differ in
remains unchanged is that
1) pole strength 2) magnetic moment 1) electric lines of force are closed curves while
3) intensity of magnetisation 4) moment of inertia magnetic field lines are not
5. Two magnets of magnetic moments of 2) magnetic field lines are closed while electric
M 1 , M 2 are placed one over the other with like lines are not
poles touching, the resultant magnetic 3) electric lines of force can give direction of
moment is the electric field while magnetic lines can not
1) M1 + M2 2) M1 - M2 4) magnetic lines can give direction
3) M 12 + M 22 4) M 12 − M 22 of magnetic field while electric lines can not.
6. A bar Magnet consists of 13. The incorrect statement regarding the lines
1) two poles of different nature and different of force of the magnetic field B is
strength 1) Magnetic intensity is a measure of lines of
2) equal poles in magnitude force passing through unit area held normal to it
3) equal and opposite magnetic poles 2) Magnetic lines of force form a closed curve
4) opposite poles 3) Inside a magnet, its magnetic lines of force
7. A small hole is made at the centre of the move from north pole of a magnet towards its
magnet then its magnetic moment south pole
1) decreases 2) increases 4) Magnetic lines of force never cut each other
3) remains same 14. Two bar magnets are placed on a piece of cork
4) depends on the nature of the magnetic material which floats on water. The magnets are so
placed that their axis are mutually
8. A magnetised wire of magnetic length ‘2l ’,
perpendicular. Then the cork
pole strength ‘m’ and magnetic moment ‘M’
1) rotates 2) moves a side
is bent at angle is 'θ ' radian at the centre of 3) oscillates 4) neither rotates nor oscillates
the circle, then
15. When a bar magnet of magnetic moment M
1) Its pole strength remains same
2) Its length decreases and becomes is placed in a magnetic field of induction field
strength B , each pole experiences a force of
 θ 
 4l sin  2   F then the distance between the South and
  
North pole of the magnet measured inside it
 θ 
  is
MB F FB
3) Its new magnetic moment becomes 1) MBF 2) 3) 4)
F MB M
 θ  16. Lines of force due to earth’s horizontal
 2 M sin  2 
   magnetic field are
 θ  4) All the above are correct 1) parallel and straight 2) elliptical
  3) concentric circles 4) curved lines
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 NEET-PHYSICS-VOL-II MAGNETISM
17. Magnetic lines of force are 25. A magnet is kept fixed with its length parallel
1) continuous 2) discontinuous to the magnetic meridian. An identical magnet
3) some times continuous and some times is parallel to this such that its center lies on
discontinuous 4) nothing can be said perpendicular bisector of both. If the second
magnet is free to move, it will have
18. In case of a bar magnet, lines of magnetic
1) translatory motion only
induction 2) rotational motion only
1) start from the north pole and end at the south 3) both translatory and rotational motion
pole. 4) vibrational motion only
2) run continuously through the bar magnet and 26. There is no. couple acting when two bar
outside. magnets are placed co-axially separated by a
3) emerge in circular paths from the middle of distance because
the bar 1) there are no forces on the poles.
4) are produced only at the north pole like rays 2) the forces are parallel and their lines of action
of light from a bulb do not coincide
19. The total number of magnetic lines of force 3) the forces are perpendicular to each other
originating or terminating on a pole of strength 4) the forces act along the same line
27. Find the wrong statement among the following.
‘m’ is Two unlike isolated magnetic poles are at
µm m some distance apart in air.
1) 0 2) µ 3) m 2 4) µ0 m 1) the resultant induction at a point beween the
4π 0

COUPLE ACTING ON THE BAR MAGNET poles is B1 + B2 on the line joining them
20. A magnetic needle is kept in a non uniform 2) The resultant induction is B1 : B2 at any point
magnetic field. It experiences out side the poles on the line joining them
1) a force and a torque 3) No neutral point is formed on the line joining
2) a force but not a torque them if the pole strengths are equal.
3) torque but not a force 4) A neutral point is formed in between the poles
and nearer to weak pole on the line joining them.
4) neither a torque nor a force 28. A magnetic field is produced and directed
21. A magnetic field is produced and directed along y-axis. A magnet is placed along y-axis.
along y-axis. A magnet is placed along x-axis The direction of torque on the magnet is
.The direction of the torque on the magnet is 1) in the x-y plane 2) along y-axis
1) in the x-y plane 2) along z-axis 3) along z-axis 4) Torque will be zero
3) along y-axis 4) torque will be zero FIELD OF A BAR MAGNET
22. A bar magnet of moment M is in a magnetic 29. The magnetic intensities at points lying at the
field of induction B . Then the couple is same distance from the magnetic pole are
1) same both in magnitude and direction
1) M x B 2) B x M 2) same in magnitude and different in direction
3) M . B 4) B . M 3) different in magnitude but same in direction
23. If a bar magnet of moment is suspended in a 4) different both in magnitude and direction
SUPERPOSITION OF MAGNETIC FIELDS
uniform magnetic field B it is given an angular
30. When N-pole of the given bar magnet is placed
deflection, w.r.t equilibrium position. Then the on a table pointing geographic north, the null
restoring torque on the magnet is points are formed due to the superposition of
1) MB sin θ 2) M B cos θ the magnetic field of the bar magnet and the
3) MB tan θ 4) MB sin θ
2 earth’s magnetic field. The two null points are
24. The effect due to uniform magnetic field on a located
freely suspended magnetic needle is as follows 1) on the axial line at equidistant on either sides
2) on the equitorial line at equidistant on either
1) both torque and net force are present sides
2) torque is present but no net force 3) on the axial line only on one side of the magnet
3) both torque and net force are absent 4) on the equitorial line only on one side of the
4) net force is present but no torque magnet
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31. When S-pole of the given bar magnet is placed 39. The time period of a freely suspended
on a table pointing geographical N-pole magnetic needle does not depend upon
1) two null points are located on the axial line 1) length of the magnet 2) pole strength
at equidistant on either sides 3) horizontal component of earth’s magnetic field
2) two null points are located on the equitorial 4) length of the suspension fibre
line at equidistant on either sides 40. A magnetic needle suspended by a silk thread
3) two null points are located on the axial line is oscillating in the earth’s magnetic field. If
only on one side of the magnet the temperature of the needle is increased by
4) two null points are located on the equitorial 5000 C, then
line only on one side of the magnet 1) the time period decreases
32. A very long magnet is held vertically with its 2) the time period increases
south pole on a table. A single neutral point is 3) the time period remain unchanged
located on the table to the 4) the needle stops vibrating
1) East of the magnet 2) North of the magnet TYPES OF MAGNETIC MATERIALS
3) West of the magnet 4) South of the magnet 41. The following instrument i.e. used to measure
33. The null points are on the axial line of a bar magnetic field
Magnet when it is placed such that its south 1) Thermometer 2) Pyrometer
pole points 3) Hygrometer 4) Fluxmeter
1) South 2) East 3) North 4) West 42. A watch glass containing some powdered
34. The null point on the equatorial line of a bar substance is placed between the pole pieces
magnet when the north pole of the magnet is of a magnet. Deep concavity is observed at
pointing the centre. The substance in the watch glass
1) North 2) South 3) East 4) West is (assume poles are far)
35. When the N - pole of a bar magnet points 1) iron 2) chromium 3) carbon 4) wood
towards the south and S- pole towards the 43. Permanent magnets are made from
north, the null points are on the 1) diamagnetic substances
1) magnetic axis 2) magnetic centre 2) paramagnetic substance
3) perpendicular division of magnetic axis 3) ferromagnetic substances 4) wood
4) N and S pole 44. Out of dia, para and ferromagnetism, the
TIME PERIOD OF SUSPENDED MAGNET IN universal property of all substances is
THE UNIFORM MAGNETIC FIELD 1) diamagnetism 2) paramagnetism
36. The restoring couple for a magnet oscillating 3) ferromagnetism 4) antiferromagnetism
in the uniform magnetic field is provided by 45. The following one is a diamagnetic
1) horizontal component of earth’s magnetic field 1) Liquid oxygen 2) Air
2) gravity 3) Water 4) Copper sulphate
3) torsion in the suspended thread 46. The following one is para-magnetic
4) magnetic field of magnet 1) Bismuth 2) Antimony
37. Vibration of suspended magnet works on the 3) Water 4) Chromium
principle of 47. Ferromagnetic ore properties are due to
1) torque acting on the bar magnet and rotational 1) filled inner sub-shells
inertia 2) vacant inner sub-shells
2) force acting on the bar magnet and rotational 3) partially filled inner sub-shells
inertia 4) all the sub-shells equally filled
3) both the force and torque acting on the bar 48. The major contribution of magnetism in
magnet substances is due to
4) neither force nor torque 1) orbital motion of electrons
38. The factors on which the period of oscillation 2) spin motion of electrons
of a bar magnet in uniform magnetic field 3) equally due to orbital and spin motions of
depend electrons
1) nature of suspension fibre 4) hidden magnets
2) length of the suspension fibre 49. If the magnetic moment of the atoms of a
3) vertical component of earth’s magnetic substances is zero, the substance is called
induction 1) diamagnetic 2) ferromagnetic
4) moment of inertia of the magnet 3) paramagnetic 4) antiferromagnetic

196 NARAYANA MEDICAL ACADEMY

 NEET-PHYSICS-VOL-II MAGNETISM
50. A uniform magnetic field exists in certain space 55. Curie temperature is the temperature above
in the plane of the paper and initially it is which
directed from left to right. When a rod of soft 1) a paramagnetic material becomes ferro
iron is placed parallel to the field-direction, the magnetic
magnetic lines of force passing within the rod 2) a ferromagnetic material becomes
will be represented by figure paramagnetic
3) a paramagnetic material becomes diamagnetic
4) a ferromagnetic material becomes diamagnetic
1) 2) 56. For a paramagnetic material, the dependence
of the magnetic susceptibility χ on the
absolute temperature T is given by
1) χαT 2) χα constant× T
3) 4)
1
3) χα 4) χ = constant
51. A rod of a paramagnetic substance is placed T
in a non-uniform magnetic field. Which of the 57. The area enclosed by a hysteresis loop is a
following figure shows its alignment in the field measure of
? 1) retentivity 2) susceptibility
3) permeability 4) energy loss per cycle
58. A material produces a magnetic field which
N S
1) N S
2) helps the applied magnetic field, then it is
1) diamagnetic 2) paramagnetic
3) electro magnetic 4) all the above
59. A material produces a magnetic field which
N S N S
3) 4) oppose the applied magnetic field, then it is
1) diamagnetic 2) para magnetic
52. The relative permeability of silicon is 0.99837 3) electro magnetic 4) ferro magnetic
and that of palladium is 1.00692, choose the 60. The permeability of a material is 0.9. The
correct options of the following material is
1) silicon is paramagnetic and palladium is 1) diamagnetic 2) para magnetic
ferromagnetic 3) ferro magnetic 4) non-magnetic
2) silicon is ferromagnetic and palladium is 61. The susceptibility of a diamagnetic substance
paramagnetic is
3) silicon is diamagnetic and palladium is 1) ∞ 2) zero
paramagnetic 3) small but negative 4) small but positive
4) Both are paramagnetic 62. Liquids and gases never exhibit
53. The relative permeability is represented by
1) diamagnetic properties
µr and susceptibility is denoted by χ for a 2) para magnetic properties
magnetic substance then for a paramagnetic
3) ferro magnetic properties
substance.
4) electro magnetic properties
1) µr < 1, χ < 0 2) µr < 1, χ > 0 63. Alnico is used for making permanent magnets
3) µr > 1, χ < 0 4) µr > 1, χ > 0 because it has
54. Two like poles of strengths m1 and m 2 aree 1) High coercivity and high retentivity
2) high coercivity and low retentivity
at far distance apart. The energy required to
3) low coercivity and low retentivity
bring them r0 distance apart is 4) low coercivity and high retentivity
µ 0 m 1m 2 µ 0 m 1m 2 64. A mariners compass is used
1) 4π r 2) 8π r 1) to compare magnetic moments
0 0
2) for determination of H
µ 0 m1m 2 µ 0 m 1m 2
3) 16π r 4) 2π r 3) for determination of direction
0 0 4) for determination of dip at a place
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65. The hysteresis cycle for the material of a 75. Relative permitivity and permeability of a
permanent magnet is material are ε r and µr . respectively. Which
1) Short and wide 2) tall and narrow
of the following values of these quantities are
3) tall and wide 4) short and narrow
allowed for a diamagnetic material
66. The relation between µr and χ is
1) ε r = 1.5, µ r = 0.5 2) ε r = 0.5, µr = 0.5
1) µr = 1 + χ 2) χ = µ r + 1
3) ε r = 1.5, µ r = 1.5 4) ε r = 0.5, µ r = 1.5
3) χ = µ0 µr 4) χ = µ r / µ0 76. Susceptibility is large and positive for
67. The curie weiss law is obeyed by iron 1) para magnetic 2) diamagnetic
1) at all temperatures 3) ferro magnetic 4) electromagnetic
2) above the curie temperature 77. For soft iron, in comparison with steel
3) below the curie temperature 1) hysteresis loss is more
4) at the curie temperature 2) hysteresis loss is same
68. Which of the following quantities: 3) hysteresis loss is less
(I) magnetic declination (II) dip is used to 4) hysteresis loss is negligible
determine the strength of earths magnetic
field at a point on the earths surface 78. χ 1 and χ 2 are susceptibilities of diamagnetic
1) Both I & II 2) Neither I nor II substance at temperatures T1K and T2 K
3) I Only 4) II Only respectively, then
69. Domain formation is the necessary feature of
1) ferro magnetism 2) paramagnetism 1) χ1T1 = χ2T2 2) χ 1 = χ 2
3) diamagnetism 4) electro magnetism 3) χ1 T1 = χ 2 T2 4) χ1T2 = χ 2T1
70. The magnetic force required to demagnetise
the material is 79. Ferromagnetic materials have their properties
1) retentivity 2) coercivity due to
3) energy loss 4) hysterisis 1) vacant inner subshells
71. Substances in which the magnetic moment of 2) partially filled inner subshells
a single atom is zero 3) filled inner subshells
1) dia magnetic 2) ferro magnetic 4) completely filled outer shells
3) para magnetic 4) electro magnetic 80. When a diamagnetic liquid is poured into a U-
72. Property possessed by ferro magnetic tube and one arm of the U-tube is placed
substance only is between the two poles of strong magnet with
1) attracting magnetic substance 2) hysterisis the meniscus along the lines of the field, then
3) directional property the level of the liquid in the arm where
4) susceptibility independent of temperature magnetic field is applied will
73. Needles N1, N2 , N3 are made of a 1) fall 2) rise 3) oscillate 4) remain unchanged
ferromagnetic, paramagnetic and a 81. At Curie temperature, in ferromagnetic
diamagnetic substance respectively. A magnet materials
when brought close to them will 1) attract all 1) the atomic dipoles get aligned
three of them 2) the atomic dipoles lose alignment
3) the atomic dipoles lose alignment
2) attract N1 and N 2 strongly but repel N3 4) magnetism is zero
weakly 82. A sensitive magnetic instrument can be
3) attract N1 strongly, N 2 weakly and repel N3 shielded very effectively from outside
weakly magnetic fields by placing it inside a box of
4) attract N1 strongly, but repel N2, N3 1) wood 2) plastic
3) metal of high conductivity
weakly 4) soft iron of high permeability
74. The substance used for preparing electro
83. The value of susceptibility for super conductor
magnets is
is
1) soft iron 2) steel 3) nickel 4) copper
1) 0 2) ∞ 3) +1 4) −1
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 NEET-PHYSICS-VOL-II MAGNETISM
84. In a permanent magnet at room temperature 93. The core of electromagnet is made of soft iron,
1) magnetic moment of each molecules is because
zero a) the susceptibility of soft iron is very high
2) the individual molecules have non-zero b) coercivity of soft iron is very low
magnetic moments which are all perfectly 1) only a is correct
aligned 2) only b is correct
3) domains are partially aligned 3) both a and b are correct
4) domains are all perfectly aligned 4) both a and b are wrong
TERRESTRIAL MAGNETISM 94. The angles of dip at the poles and the equator
respectively are
85. The angle of dip at a place on the earth's
surface gives 1) 300, 6 0 0 2) 900, 0 0
1) direction of earth’s magnetic field 3) 300, 9 0 0 4) 00 , 00
2) horizontal component of earth's magnetic field 95. Select the correct answer.
3) vertical component of earth's magnetic field a) When ‘n’ identical magnets are arranged
4) location of geographic poles in the form of closed polygon with unlike poles
86. A point near the equator has nearer, the resultant magnetic moment is
1) BV >> BH 2) BH >> BV zero.
b) If one magnet is removed from the polygon,
3) BV = BH 4) BV = BH = 0 the resultant magnetic moment becomes ‘M’.
87. If I is the intensity of earth's magnetic field, c) If one magnet is reversed in the polygon,
H its horizontal component and V the vertical the resultant magnetic moment of
component, then these are related as combination becomes 2M
1) I = V + H 2) I = H 2 + V 2 1) a, b and c are correct
2) a and b are correct but c is wrong
3) I = H 2 − V 2 4) I 2 = V 2 − H 2 3) only a is correct
88. A line joining places of zero declination is 4) a, b and c are wrong
called 96. Arrange the following in the descen-
1) agonic 2) isoclinic ding order of their resultant magnetic
3) isodynamic 4) isogonal moments consider two magnets of same
89. A line joining places of equal declination is moment
called a) They are kept one upon the other with like
1) aclinic 2) isoclinic poles in contact
3) isodynamic 4) isogonal b) They are kept one upon the other with
90. The needle of a dip circle when place at a unlike poles in contact
geomagnetic pole stays along c) They are arranged in perpendicular
1) south north direction only directions
2) east west direction only d) They are inclined 600 with like poles in
3) vertical direction contact
4) horizontal direction 1) a, c, d, b 2) a, b, c, d
91. The value of angle of dip is zero at the 3) a, d, c, b 4) d, b, c, a
magnetic equator because on it 97. Among the following statements:
1) V and H are equal A) A magnet of moment M is bent into a
2) the value of V and H are zero semicircle, then its magnetic moment
3) the value of V is zero decreases
4) the value of H is zero B) Magnetic moment is directed parallel to
92. Earth's magnetic field always has a horizontal axial line from south pole to north pole
component except at 1) A is true & B is false
1) equator 2) magnetic pole 2) A is false & B is true
3) a latitude of 600 4) an inclination of 600 3) A and B are true
4) A and B are false
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98. When a bar magnet is suspended freely in a 103. When a bar magnet is suspended in an uniform
uniform magnetic field, identify the correct magnetic field, then the torque acting on it
statements will be
a) The magnet experiences only couple and List-I List-II
undergoes only rotatory motion a) maximum e) θ = 450 with the field
b) The direction of torque is along the
b) half of the f) θ = 60 0 with the
suspension wire
c) The magnitude of torque is maximum when maximum value field
the magnet is normal to the field direction c) 3 / 2 times g) θ = 30 with the
0

1) only a and c are correct the maximum field

2) only a and b are correct d) 1/ 2 times h) θ = 900 with the
3) only b and c are correct
the maximum field
4) a, b, c are correct
99. Among the following statements: 1) a-h, b-g, c-f, d-e 2) a-e, b-f, c-g, d-h
(A) The resultant induction at a point on the 3) a-f, b-e, c-g, d-h 4) a-h, b-g, c-f, d-e
axial line of a bar magnet is parallel to 104. Match the following
magnetic moment. LIST - 1 LIST- 2
(B) The resultant induction at a point on the a) Magnetic moment d) Am 2

equatorial line is antiparallel to magnetic b) Pole strength e) Am

moment c) Relative f) weber
1) A is true & B is false 2) A is false & B is true Wb
3) A and B are true 4) A and B are false permeability g)
100. (i) Soft iron conducts electricity Am
(ii) Soft iron is magnetic material H
(iii) Soft iron is used for permanent magnets h)
m
(iv) Soft iron is used as electro magnet 1) a → e b → d c → g
Out of the statements given above
1) (i) and (ii) are correct 2) a → g b → e c → d
2) (i) ,(ii) and (iii) are correct 3) a → d b → e, f c → g , h
3) (ii) and (iv) are correct 4) a → f b → e c → d
4) (i), (ii) and (iv) are correct
101. Match the following: ASSERTION & REASON
Physical quantity Unit 1) Both A and R are true and R is the correct
a) Magnetic moment e) Amp-m explanation of A.
b) Magnetic flux f) Amp/m 2) Both A and R are true and R is not correct
density explanation of A.
c) Intensity of g) N-m3 /wb 3) A is true, But R is false
magnetic field 4) A is false, But R is true
d) Pole strength h) Gauss 105. Assertion (A) : The net magnetic flux coming
1) a-e, b-f, c-g, d-h 2) a-g, b-h, c-f, d-e out of a closed surface is always zero.
3) a-g, b-f, c-h, d-e 4) a-e, b-f, c-h, d-g Reason (R) : Unlike poles of equal strength
102. Some physical quantities are given in the list exist together
I the related units are given in the list II. 106. Assertion (A): A magnet remains stable, If it
Match the correct pairs in the lists aligns itself with the field
List-I List-II Reason (R): The P.E. of a bar magnet is
a) Magnetic field e) Wb m-1 minimum, if it is parallel to magnetic field.
intensity 107. Assertion (A) : To protect any instrument
b) Magnetic flux f) Wb m-2 from external magnetic field, it is put inside
c) Magnetic potential g) Wb an iron box
d) Magnetic induction h) Am-1 Reason (R) : Iron is a ferro magnetic
1) a-e, b-f, c-g, d-h 2) a-h, b-g, c-e, d-f substance
3) a-h, b-e, c-g, d-f 4) a-f, b-g, c-e, d-h

200 NARAYANA MEDICAL ACADEMY

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108. Assertion(A): χ − T graph for a diamagnetic 120. Assertion: A magnet suspended freely in an
material is a straight line parallel to T − axis uniform magnetic field experiences no net
Reason (R): This is because susceptibility of force, but a torque that tends to align the
a diamagnetic material is not affected by magnet along the field when it is deflected
temperature from equilibrium position
109. Assertion(A): If one arm of a U-tube Reason: Net force mB − mB = 0 , but the
containing a dia magnetic solution is placed in forces on north and south poles being equal,
between the poles of a strong magnet with unlike and parallel make up a couple that
the level in line with the field, the level of the tends to align the magnet, along the field.
solution falls, 121. Assertion: Basic difference between an
Reason(R): Diamagnetic substances are electric line and magnetic line of force is that
repelled by the magnetic field former is discontinuous and the latter is
110. Assertion(A): The ferro magnetic substances continuous or endless.
do not obey curie's law
Reason(R) : At curie point ferro magnetic Reason: No electric lines of forces exit inside
charged conductor but magnetic lines do exist
substances start behaving as a para magnetic inside magnet.
substances
111. Assertion(A): Earth's magnetic field inside a 122. Assertion: The earth’s magnetic field is due
closed iron box is less as compared to the out to iron present in its core.
side Reason: At a high temperature magnet losses
Reason(R) : The magnetic permeability of its magnetic property or magnetism.
iron is low 123. Assertion: The properties of paramagnetic
112. Assertion: Magnetic moment of an atom is and ferromagnetic substances are not
due to both, the orbital motion and spin motion affected by heating.
of every electron. Reason: As temperature rises, the alignment
Reason: A charged particle at rest produces of molecular magnets gradually decreases.
a magnetic field.
113. Assertion: Electromagnetis are made of soft 124. Assertion: A soft iron core is used in a moving
iron. coil galvanometer to increase the strength of
Reason: Coercivity of soft iron is small. magnetic field.
114. Assertion: Time period of vibrations of a pair Reason: From soft iron more number of the
of magnets in sum position is always smaller magnetic lines of force passes.
than in difference position.
Reason: T = 2π I / MBH , where symbols C.U.Q - KEY
have their standard meaning 1) 1 2) 4 3) 1 4) 3 5) 1 6) 3
115. Assertion: Magnetism is relativistic 7) 1 8) 4 9) 3 10) 3 11) 3 12) 2
Reason: When we move along with the 13) 3 14) 4 15) 2 16) 4 17) 1 18) 2
charge, so that there is no motion relative to 19) 4 20) 1 21) 2 22) 1 23) 1 24) 2
us, we find no magnetic field associated with 25) 3 26) 4 27) 4 28) 4 29) 2 30) 2
the charge 31) 1 32) 2 33) 3 34) 1 35) 1 36) 1
116. Assertion: Steel is attracted by a magnet 37) 1 38) 4 39) 4 40) 2 41) 4 42) 1
Reason: Steel is not a magnetic substance 43) 3 44) 1 45) 3 46) 4 47) 3 48) 2
117. Assertion: When radius of a circular wire 49) 1 50) 2 51) 1 52) 3 53) 4 54) 1
carrying current is doubled, its magnetic 55) 2 56) 3 57) 4 58) 2 59) 1 60) 1
moment becomes four times 61) 3 62) 3 63) 1 64) 3 65) 3 66) 1
Reason: Magnetic moment is directly 67) 2 68) 1 69)1 70) 2 71) 1 72) 2
proportional to area of the loop 73) 3 74) 1 75) 1 76) 3 77) 3 78) 2
118. Assertion: It is not necessary that every 79) 2 80) 1 81) 3 82) 4 83) 4 84) 3
magnet has one north pole and one south pole. 85) 1 86) 2 87) 2 88) 1 89) 4 90) 3
Reason: It is a basic fact that magnetic poles 91) 3 92) 2 93) 3 94) 2 95) 1 96) 3
occur in pairs 97) 3 98) 4 99) 3 100) 4 101) 2 102) 2
119. Assertion: Relative magnetic permeability 103) 1 104) 3 105) 1 106) 1 107) 1 108) 1
has no units and no dimensions 109) 1 110) 2 111) 3 112) 3 113) 1 114) 1
Reason: µr = µ / µ0 , where the symbols have 115) 1 116) 3 117) 1 118) 4 119) 1 120) 1
their standard meaning. 121) 1 122) 4 123) 4 124) 1

201
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9. A bar magnet of magnetic moment M is bent

LEVEL - I ( C.W ) in ‘ ’ shape such that all the parts are of
equal lengths. Then new magnetic moment
MAGNETIC MOMENT AND is
RESULTANT MAGNETIC MOMENT
1) M/3 2) 2M 3) 3M 4) 3 3 M
1. The geometric length of a bar magnet is
10. A thin bar magnet of length ' l ' and magnetic
24 cm. The length of the magnet is moment 'M' is bent at the mid point so that the
1) 24cm 2) 28.8cm 3) 20cm 4) 30cm two parts are at right angles. The new magnetic
2. The magnetic moment of a bar magnet is length and magnetic moment are respectively
3.6x10-3 A.m2 . Its pole strength is 120 milli l M M l
amp. m. Its magnetic length is 1) 2l , 2M 2) 2 , 2 3) 2l, 2 4) , 2M
2
1) 3cm 2) 0.3cm 3) 33.33cm 4) 3x10−2 cm 11. The resultant magnetic moment for the following
3. Two magnets have their lengths in the ratio arrangement is
2 : 3 and their pole strengths in the ratio 3 : 4. M
0
The ratio of their magnetic moment is 60
1) 2 :1 2) 4 :1 3) 1 : 2 4) 1 : 4 M
4. The length of a magnet is 16 cm. Its pole 60 M
0

strength is 250 milli. amp. m. When it is cut M

into four equal pieces parallel to its axis, the
magnetic length, pole strength and moments 1) M 2) 2M 3) 3M 4) 4M
12. Three magnets of same length but moments
of each piece are: (respectively) M,2M and 3M are arranged in the form of an
1) 4 cm; 62. 5 milli Am; 250 milli amp. cm2 equilateral triangle with opposite poles nearer,
2) 8 cm ; 500 milli Am; 400 milli amp. cm2 the resultant magnetic moment of the arrangement
3) 16 cm; 250 milli Am; 4000 milli amp. cm2 is
4) 16 cm; 62.5 milli Am; 0.01 A.m2 3
5. A bar magnet of magnetic moment M1 is 1) 6M 2) zero 3) 3M 4) M
2
axially cut into two equal parts. If these two 13. A bar magnet of moment M is cut into two
pieces are arranged perpendicular to each identical pieces along the length. One piece is
other, the resultant magnetic moment is M2 . bent in the form of a semi circle. If two pieces
M1 are perpendicular to each other, then resultant
Then the value of M is (2007M) magnetic moment is
2 2 2
 
2
 
2
 M  M 
1 1 M M 2)   +  
  + 
1)
1) 2) 1 3) 4) 2  π   2  π   2 
2 2 2
6. The resultant magnetic moment for the 2
M  M 
2
M M
following arrangement (non coplanar vectors) 3)     − 4) +
π   2  π 2
M′
MAGNETIC FIELD
14. A magnetic pole of pole strength 9.2 A m. is
600 M placed in a field of induction 50x10-6 tesla.
M The force experienced by the pole is
1) M 2) 2M 3) 3M 4) 4M 1) 46N 2) 46x10-4N 3) 4.6x10-4N 4) 460N
7. Two magnets of moments 4Am 2 and 3Am 2 15. The magnetic induction at distance of 0.1 m
are joined to form a cross (+), then the from a strong magnetic pole of strength
magnetic moment of the combination is 1200 Am is
1) 12x10-3 T 2) 12x10-4 T
1) 4Am 2 2) 1Am 2 3) 7Am 2 4) 5Am 2 -3
3) 1.2x10 T 4) 24x10-3 T
8. A magnet of magnetic moment M and length
2l is bent at its mid-point such that the angle 16. If area vector A = 3i + 2 j + 5k m flux
2

of bending is 600 . The new magnetic moment density vector B = 5i + 10 j + 6k ( web / m2 ) . The
is. magnetic flux linked with the coil is
M M 1) 31Wb 2) 9000Wb
1) M 2) 3) 2M 4) 3) 65Wb 4) 100Wb
2 2
202 NARAYANA MEDICAL ACADEMY
 NEET-PHYSICS-VOL-II MAGNETISM
17. P and Q are two unlike magnetic poles. 25. A bar magnet is at right angles to a uniform
Induction due to ‘P’ at the location of ‘Q’ is magnetic field. The couple acting on the
B, and induction due to ‘Q’ at the location of magnet is to be one fourth by rotating it from
P is B/2. The ratio of pole strengths of P and the position. The angle of rotation is
Q is 1) Sin-1(0.25) 2) 900 -Sin-1(0.25)
-1
1) 1 : 1 2) 1 : 2 3) 2 : 1 4) 1 : 2 3) Cos (0.25) 4) 900 - Cos-1(0.25)
18. Two north poles each of pole strength m and ∧ ∧
a south pole of pole strength m are placed at 26. A bar magnet of moment M = i + j is placed
the three corners of an equilateral triangle ur ^ ^ ^
in a magnetic field induction B = 3i + 4j + 4k .
of side a. The intensity of magnetic induction
field strength at the centre of the triangle is The torque acting on the magnet is
∧
µ0 m µ 0 6m µ 0 9m µ0 m 1) 4 ∧i -4 j + k∧ 2) ∧i + k∧
1) 2) 2 3) 2 4)
4π a 4π a 4π a 4π 2a
2 2
∧ ∧ ∧ ∧ ∧
19. The pole strength of a horse shoe magnet is 3) i - j 4) i + j + k
90 Am and distance between the poles is 6 27. A bar magnet of magnetic moment 1.5 J/T is
cm. The magnetic induction at mid point of aligned with the direction of a uniform
the line joining the poles is,
magnetic field of 0.22 T. The work done in
1) 10 −2 T 2) Zero 3) 2 × 10 −2 T 4) 10 −4 T turning the magnet so as to align its magnetic
20. The force acting on each pole of a magnet moment opposite to the field and the torque
when placed in a uniform magnetic field of 7 acting on it in this position are respectively.
A/m is 4.2x10-4 N. If the distance between the 1) 0.33J, 0.33N-m 2) 0.66J, 06.66N-m
poles is 10 cm, the moment of the magnet is 3) 0.33J, 0 4) 0.66J, 0
15 π 28. The work done in turning a magnet of
1) π 2) 2
15 Am magnetic moment M by an angle of 900 from
3) 7.5 x 10-12 Am
2
4) 6x10-6 Am
2 the meridian is n times the corresponding
21. An iron specimen has relative permeability work done to turn it through an angle of 600 ,
of 600 when placed in uniform magnetic field where n is given by
of intensity 110 amp /m. Then the magnetic 1 1
flux density inside is....... tesla. 1) 2) 2 3) 4) 1
2 4
1) 18.29 x 10-3 2) 8.29 x 10-2
3) 66 x 10 3
4) 7.536 x 10-4 29. A bar magnet of moment 4Am 2 is placed in a
COUPLE ACTING ON THE BAR MAGNET nonuniform magnetic field. If the field
strength at poles are 0.2 T and 0.22 T then
22. A magnetic needle of pole strength 'm' is pivoted
at its centre. Its N-pole is pulled eastward by a
the maximum couple acting on it is
string. Then the horizontal force required to 1) 0.04Nm 2) 0.84Nm3) 0.4 Nm 4) 0.44Nm
produce a deflection of θ from magnetic 30. A magnet of length 10 cm and pole strength
meridian (B H horizontal componet of earths 4x10-4 Am is placed in a magnetic field of
magnetic field) induction 2x10-5 weber m-2, such that the axis
1) mBcos θ 2) mBsin θ 3) 2 mBtan θ 4) mBcot θ of the magnet makes an angle 300 with the
23. Two identical bar magnets are joined to form a lines of induction. The moment of the couple
cross. If this combination is suspended freely in acting on the magnet is
a uniform field the angles made by the magnets 1) 4x10-10 Nm 2) 8x10-10 Nm
with field direction are respectively
1) 60°, 30° 2) 37°, 53° 3) 45°, 45° 4) 20°, 70° 3) 4x10-6 Nm 4) 3 x10-11 Nm
24. A bar magnet of length 16 cm has a pole 31. A bar magnet of magnetic moment 2 Am 2 is
strength of 500 milli amp.m. The angle at free to rotate about a vertical axis passing
which it should be placed to the direction of through its center. The magnet is released
external magnetic field of induction 2.5 gauss from rest from east - west position. Then the
so that it may experience a torque of 3 x10- KE of the magnet as it takes N-S position is
( BH = 25µT )
5
Nm is
1) π 2) π / 2 3) π / 3 4) π / 6
1) 25 µ J 2) 50µ J 3) 100µ J 4) 12.5µ J

NARAYANA MEDICAL ACADEMY

203
MAGNETISM NEET-PHYSICS-VOL-II
32. A bar magnet of length 10cm and pole 40. A short magnetic needle is pivoted in a
strength 2 Am makes an angle 600 with a uniform magnetic field of induction 1T. Now,
uniform magnetic field of induction 50T. The simultaneoulsy another magnetic field of
couple acting on it is induction 3T is applied at right angles to
1) 5 3Nm 2) 3Nm the first field; the needle deflects through an
3) 10 3Nm 4) 20 3Nm angle θ where its value is (EAM 2010)
FIELD OF A BAR MAGNET 1) 30 0 2) 45 0 3) 90 0 4) 60 0
33. The magnetic induction field strength due to 41. Two magnetic poles of pole strengths 324 milli
a short bar magnet at a distance 0.20 m on amp.m. and 400 milli amp m are kept at a
the equatorial line is 20x10 -6 tesla. The distance of 10 cm in air. The null point will be
magnetic moment of the bar magnet is at a distance of ...... cm, on the line joining
1) 3.2Am2 2) 6.4Am2 3) 1.6Am2 4) 16Am2 the two poles, from the weak pole if they are
34. The magnetic induction field strength at a like poles.
distance 0.3 m on the axial line of a short bar 1) 4.73 2) 5 3) 6.2 4) 5.27
magnet of moment 3.6 Am2 is TIME PERIOD OF SUSPENDED
1) 4.5 × 10-4 T 2) 9 × 10-4 T MAGNET IN THE UNIFORM
3) 9 × 10 T-5
4) 2.6 × 10-5 T
35. A magnet of length 10 cm and magnetic MAGNETIC FIELD
moment 1Am 2 is placed along the side of an 42. With a standard rectangular bar magnet, the
equilateral triangle of the side AB of length time period in the uniform magnetic field is 4
10 cm. The magnetic induction at third vertex sec. The bar magnet is cut parallel to its length
C is into 4 equal pieces. The time period in the
1) 10 -9 T 2) 10 -7 T 3) 10-5 T 4) 10 -4T uniform magnetic field when the piece is used
36. The length of a magnet of moment 5Am2 is (in sec) (bar magnet breadth is small)
14 cm. The magnetic induction at a point, 1) 16 2) 8 3) 4 4) 2
equidistant from both the poles is 3.2x10-5 Wb/ 43. A bar magnet of moment of inertia
m2 . The distance of the point from either pole 1×10-2 kgm 2 vibrates in a magnetic field of
is induction 0.36 × 10 -4 tesla. The time period of
1) 25 cm 2) 10 cm 3) 15 cm 4) 5 cm vibration is 10 s. Then the magnetic moment
37. A pole of pole strength 80 Am is placed at a of the bar magnet is (Am2 )
point at a distance 20cm on the equatorial line 1) 120 2) 111 3) 140 4) 160
from the centre of a short magnet of magnetic 44. Two bar magnets are placed together
moment 20Am2 . The force experienced by it suspended in the uniform magnetic field
is vibrates with a time period 3 second. If one
1) 8 x 10-2 N 2) 2 x 10-2 N magnet is reversed, the combination takes 4s
-2
3) 16 x 10 N 4) 64 x 10-2 N for one vibration. The ratio of their magnetic
38. A short bar magnet produces magnetic fields moments is
of equal induction at two points one on the 1) 3 : 1 2) 5 : 18 3) 18 : 5 4) 25 : 7
axial line and the other on the equatorial line. 45. A bar magnet of length ‘l’ breadth ‘b’ mass
The ratio of their distances is ‘m’ suspended horizontally in the earths
1) 2:1 2) 21/2 :1 3) 21/3 :1 4) 21/4 :1 magnetic field, oscillates with period T. If ‘l’,
SUPERPOSITION OF MAGNETIC FIELDS m, b are doubled with pole strength remaining
39. Two short bar magnets with magnetic the same, the new period will be
moments 8Am2 and 27Am2 are placed 35cm 1) 8T 2) 4T 3) T/2 4) 2T
apart along their common axial line with their 46. The time period of a suspended magnet is T0 .
like poles facing each other. The neutral point Its magnet is replaced by another magnet
is whose moment of inertia is 3 times and
1) midway between them magnetic moment is 1/3 of that of the initial
2) 21 cm from weaker magnet magnet. The time period now will be
3) 14 cm from weaker magnet
4) 27 cm from weaker magnet To T
1) 3To 2) To 3) 4) o
3 3
204 NARAYANA MEDICAL ACADEMY
 NEET-PHYSICS-VOL-II MAGNETISM
47. A magnetic needle is kept in a uniform LEVEL - I (C. W ) KEY
magnetic field of induction 0.5 x 10-4 tesla. It 1) 3 2) 1 3) 3 4) 4 5) 4 6) 2
makes 30 oscillations per minute. If it is kept 7) 4 8) 2 9) 1 10)2 11) 2 12) 3
in a field of induction 2 x 10-4 tesla. Then its 13) 2 14) 3 15)1 16) 3 17) 3 18) 2
frequency is 19) 3 20) 1 21) 2 22) 3 23) 3 24) 3
1) 1 oscillation/s 2) 60 oscillations/s 25) 2 26) 1 27) 4 28) 2 29) 2 30) 1
3) 15 oscillations/min 4) 15 oscillations/s 31) 2 32) 1 33) 3 34) 4 35) 4 36) 1
48. A magnet is suspended horizontally in the 37) 2 38) 3 39) 3 40) 4 41) 1 42) 3
earth’s field. The period of oscillation in the 43) 2 44) 4 45) 4 46) 1 47) 1 48) 4
place is T. If a piece of wood of the same 49) 1 50) 3 51) 2 52) 4 53) 1 54) 2
moment of inertia as the magnet is attached
to it, the new period of oscillation would be LEVEL-I (C. W ) HINTS
5
1)
T
2) T/2 3) T/3 4) 2T 1. 2l = [ geometriclength]
2 6
49. A magnet freely suspended makes 30 M
vibrations per minute at one place and 20 2. 2l =
m
vibrations per minute at another place. If the
value of BH at first place is 0.27 tesla. The M 1 m1 ( 2l )1
= ×
value of BH at other place is 3. M 2 m2 ( 2l )2
1) 0.12 T 2) 2.1 T 3) 5.4 T 4) 0.61 T 4. Magnetic length remains same
TYPES OF MAGNETIC MATERIALS m
Pole strength m =
1

50. A magnet has a dimensions of 4

25 cm x 10 cm x 5 cm and pole strength of M
Magnetic moment = M =
1
200 milli ampm The intensity of magnetisation
4
due to it is
M1 M
1) 6.25A/m 2) 62.5A/m 3) 40A/m 4) 4A/m
5. =
51. The mass of iron rod is 110g, its magnetic M2 M / 2
moment is 20 Am2 . The density of iron is 8g/
( )
2
cm3 . The intensity of magnetization is nearly 6. M res = 3M +M2
1) 2x105 Am-1 2) 2.26x106 Am-1
3) 1.6x106 Am-1 4) 1.4 x 106 Am-1 7. M = M 12 + M 22
52. Relative permeability of iron is 5500, then its
θ
magnetic susceptibility will be: 8. M 1 = M sin
2
1) 5500 × 107 2) 5500 × 10 −7
2l
M 1 = m× ( 2 l ) , ( 2l ) =
1 1
3) 5501 4) 5499 9.
53. A specimen of iron is uniformly magnetised 3
θ θ
by a magnetising field of 500Am −1 . If the 10. M = M sin , 2l = ( 2l ) sin
1 1

magnetic induction in the specimen is 0.2 2 2

Wbm −2 . The susceptibility nearly is 11. M R = M 12 + M 22 + 2 M 1M 2 cosθ
1) 317.5 2) 418.5 3) 217.5 4) 175
12. M R = M 12 + M 22 + 2 M 1M 2 cosθ
54. The magnetic susceptibility of a rod is 499.
The absolute permeability of vacuum is 2M  M
13. M 1 =  , M 2 =
4π × 10−7 H / m . The absolute permeability of π 2  2
the material of the rod is
M 1 = M 12 + M 22
1) π ×10 −4 H / m 2) 2π × 10−4 H / m
14. F = mB
3) 3π × 10−4 H / m 4) 4π × 10−4 H / m
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205