Various definitions & features related to THEM

Bar Magnet

A system composed of two poles, equal in magnitude but opposite in polarity, placed at a smalldisplacement apart is known as a bar magnet. It is also known as a magnetic dipole.

N S

–mp+mp N S

(b)(a)

Pole strength (mp)

(Ia) The strength of a magnetic pole to attract magnetic materials towards it, is known as pole strength.(b) Greater the number of unit poles in a magnetic pole, greater will be its strength.

(c) pF Magnetic forcemB Magneticinduction

(d) Its unit is ampere-meter.

Illustration 1: The magnetic flux density of magnetic field is 1.5 Wb/m2. A proton enters this fieldwith a velocity of 2 107 m/s in a direction making an angle of 30 with the field. Theforce acting on the proton will be :(A) 2.4 10–12 Newton (B) 0.24 10–12 Newton(C) 24 10–12 Newton (D) 0.024 10–12 Newton

Solution : (A) 19 7 12F qvBsin 1.6 10 2 10 1.5 0.5 2.4 10- -= q = ´ ´ ´ ´ ´ = ´

Magnetic lines of force

(a) The imaginary lines which represent the direction of magnetic field, are known as magnetic lines offorce.

(b) The imaginary path traced by an isolated (imaginary) unit north pole is defined as a line of force.(c) Magnetic lines of force are closed curves. Outside the magnet their direction is from north pole to

south pole and inside the magnet these are from south to north pole.

N S

(d) They neither have origin nor end.(e) These lines do not intersect, because if they do so then it would mean two value of magnetic field at

a single point, which is not possible.(f) The tangent drawn at any point to the line of force indicates the direction of magnetic field at that

point.(g) At the poles of the magnet the magnetic field is stronger because the lines of force there are

crowded together and away from the poles the magnetic field is week. i.e. magnetic field intensitynumber of lines of force.

(h) The number of magnetic lines of force passing through unit normal area is defined as magneticinduction (B) whereas the number of lines of force passing through any area is known as magneticflux.

(i) The lines of force can emerge out of the north pole of magnet at any angle and these can merge intothe south pole at any angle.

Magnetic field

(a) The space around a magnet in which a torque acts on a magnetic needle is known as magnetic field.(b) The space around a magnet in which a net force acts on a magnetic test pole is known as magnetic

field.(c) The space around a magnet in which its effect is experienced is known as magnetic field.(d) There are four types of magnetic field :

(i) Uniform magnetic field: (a) The magnetic field, in which the intensity of magnetic field issame at all points, is known as uniform magnetic field.

(b) In such a magnetic field the magnetic lines of force are parallel and equidistant. e.g.the magnetic dines of force of earth's magnetic field.

(ii) Non-uniform magnetic field: (a) The magnetic field, in which the intensity of magneticfield at different points is different, is known as non-uniform magnetic field.

(b) It is represented by non-parallel lines of force.(iii) Varying magnetic field: (a) The magnetic field, which keeps on changing withrespect to time is known as a variable magnetic field.

(b) Example :– B = B0 sin t or B = B0 cos t(iv) Non-varying magnetic field: (a) The magnetic field which does not change with time isknown as a constant magnetic field.

(b) The direction of magnetic field is that in which a force acts on a unit test pole.(c) It can be produced by moving charges, current carrying loops, and variations in

electric currents.

Magnetic dipole

(a) The structures, which tend to align along the direction of magnetic field, are known as magneticdipoles.

(b) Bar magnet, current carrying solenoid, current carrying coil, current carrying coil, current loop,magnetic needle etc. are the examples of magnetic dipole.

(c) It is not possible to separate out the two poles of a magnet.(d) Magnetic dipole is not a system composed of two poles because the existence of monopoles is not

possible.A loop of single turn is also a magnetic dipole. One face of the loop behaves as north pole and theother face behaves as south pole. The face of the coil, in which current is anticlockwise, behaves asnorth pole and the face in which current is clockwise, behaves as south pole.

Magnetic flux

(a) The number of lines of force passing through a given area is defined as magnetic flux.(b) The magnetic flux passing through unit normal area is defined as magnetic induction (B).

(c) When the magnetic field is normal to the plane then 2 then

A Normal tosurface

Vibrative linesOf force

Bn̂

Surface

(d) When magnetic field makes an angle with the normal to the plane :(i) Magnetic flux linked with the plane = Area of the plane

(A) × Component of magnetic field normal to the plane (B cosi.e.

A

or

BBcos

Bcos

B

n̂

A(ii) direction of magnetic

field (A cosi.e.

Acos

A

BIn both cases q is the angle between ˆB and n

.

(iii) When B and A

are mutually parallel then o and

(iv) When B and A

are mutually perpendicular, o and

(v) When B and A

are antiparallel, then o and BA.(e) When the angle between B and the plane of coil is BA sin

If the number of turns in the coil is N then

O

B

B cos

B sinn̂

B

A180o

(i) When the plane of coil is parallel to B

then o and

(ii) When B

and the plane of coil are mutually perpendicular i.e.

(iii) When B

and the plane of coil are mutually antiparallel i.e. o, then

(f) Magnetic flux linked with a small surface element dA d B.dA BdA cos

where d A

= area of small element.n̂ = unit normal vector.

DA

n̂

B

(g) The flux linked with total area of the surface A

d B.dA B dA cos

(h) Positive magnetic flux: When the magnetic induction B

and the unit normal vector n̂ are in thesame direction then

B

A

(i) Negative magnetic flux: When the magnetic induction B

and unit normal vector are mutually inopposite directions then

– BA(j) The flux emanating out of a surface is positive and the flux entering the surface is negative.(k)(l) The net magnetic flux coming out of a closed surface is always zero, i.e.

A

B.dA 0 or .B 0

Magnetic flux density or magnetic induction

(a) The magnetic lines of force passing through unit normal area in a magnetic field is defined asmagnetic induction.

(b) 2B when A 1m. then BA

L

B

i

i

F

(c) Direction of magnetic induction: The direction in the magnetic field in which if a current carryingconductor is placed then no force acts on it, is known as the direction of magnetic induction.

(d) Magnetic induction is a vector quantity.(e) The magnetic induction due to a bar magnet

(i) In axial position 32KMBr

(ii) In equatorial position 3KMBr

Here 70K 10 Weber / A.m4

-m= =p

Magnetising field or intensity of magnetic field H

(a) The ratio of magnetic induction produced in vacum (Bo) and magnetic permeability of vacuum is

defined as magnetising field (H), i.e.o

o

BH

(b) The intensity of magnetic field due to a pole of strength mp at a distance r from it is p2

mH

r

(c) Due to a small magnet H = 23

M 1 3cosr

Illustration 2: 1000 turns per meter are wound over a Rowland ring of ferromagnetic material. Onpassing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced init. The magnetising force generated in the material will be :(A) 1.2 10–3 A/m (B) 2.6 10–3 A/m(C) 2.6 10–4 A/m (D) 2 103 A/m

Solution: (D) 3 3NiH ni 10 2 2 10 A /m2 r= = = ´ = ´p

Illustration 3: In the above problem, the value of intensity of magnetisation in A/m will be :(A) 7.96 106 (B) 7.96 10–6

(C) 3.98 103 (D) zero

Solution: (A) ( )0B H= m I +

3 67

0

B 10H 2 10 7.96 10 A /m4 10-

\ I = - = - ´ = ´m p ´

Illustration 4: In the above question 115 the relative permeability of the material will be :(A) 4.98 103 (B) 4.98 10–3

(C) 2.98 10–3 (D) 3.98 103

Solution: (D)6

3r 3

7.96 101 X 1 1 3.98 10H 2 10I ´m = + = + = + = ´

´

Magnetic moment (M)

(a) If a magnet of length l and magnetic moment M is bent in the form of a semicircular are then its new

magnetic moment will be M' =2M

N S

mp mp

l

N S

l

M = mpI

N S

l = r

2r

(b) The magnetic moment of an electron due to its orbital motion is 1 B whereas that due to its spin

motion it is B

2

.

i.e. Morbital =l l

eh ehl4 m 2m

and Mspin = s BB

s s

eh ehs s4 m 2m 2

Here B = Bohr magneton

(i) The value of Bohr magnetoneh

4 m(ii) –23 Amp-m2

(c) Other formulae of M:(i) M = ni 2

(ii)2 2 2eVr er er 2 f erM

2 2 2 T

(iii)e

2mJ(iv) M = n B

(d) Resultant magnetic moment :(i) When two bar magnets are lying mutually perpendicular to each other, then

2 21 2 pM M M 2m l

S

l

N

–mp

mp

S l N

M2

–mp mp

2

2

1

2

M

M

M

M1

(ii) When two coils, each of radius r and carrying current i, are lying concentrically with theirplanes at right angles to each other, then

2 2 21 2M M M 2i r if M1 = M2

O r

r

M1

M2

Illustration 5: A square loop OABCO of side carries a current i. It is placed as shown in figure. Findthe magnetic moment of the loop.

Solution: Magnetic moment of the loop can be written as,

M i BC CO

Here,

BC k

3CO cos60 i sin60 j i j

2 2

3M i ( k) i j2 2

or

2iM j 3 i2

Ans.

Illustration 6: Find the magnitude of magnetic moment of the current carryingloop ABCDEFA. Each side of the loop is 10 cm long and current inthe loop is i 2.0 A.

C

D

B

EA

F

i

Solution : By assuming two equal and opposite currents in BE, two current carrying loop (ABEFAand BCDEB) are formed. Their magnetic moments are equal in magnitude butperpendicular to each other. Hence

2 2netM M M 2 M

Where M ia (2.0)(0.1)(0.1) 0.02 A-m2

Mnet ( 2 ) (0.02) A – m2 0.028 A-m2

C

D

B

EA

F

Illustration 7: Two identical bar magnets each of length L and pole–strength m are placed at rightangles to each other with the north pole of one touching the south pole of the other.Evaluate the magnetic moment of the system.

Solution: As magnetic moment is a vector

MR 2 2 1/ 21 2 1 2(M M 2M M cos ) with 2

1 2

M sintanM M cos

and as here M1 = M2 =

mL and = 90°, so

MR = (M2 + M2 + 2MM cos 90)1/2 = ( 2)mL

And,Msin90tan 1,

M Mcos90

i.e., = tan–1(1) = 45°

S

N N

M1

M2

(A)

1MRM

2M

(B)

Intensity of magnetisation (I)

(a) The magnetic moment per unit volume of a material is defined as intensity of magnetisation (I). i.e. I

=MV

, when

V = 1m3 then I = M.(b) The unit of intensity of magnetisation is ampere/meter and its dimensions are M0L–1T0A1

(c) The pole strength per unit area of cross-section is defined as intensity of magnetisation. i.e.

pmI

A when A = 1m2 then A=1m2 then I = mp

(d) It is a vector quantity whose direction is along the magnetic field.(e) In para- and ferro-magnetic materials its direction is in the direction of H and in dia-magnetic

materials it is opposite to that of H.(f) I is produced in materials due to spin motion of electrons.(g) The value of I and its direction in a material depend on the nature of that material.(h) I - M curve

(i) For para magnetic materials (ii) For diamagnetic materials

(iii) For ferromagnetic materials

(i) I - H curve

H

IFerromagnetic

Diamagnetic

(j) Its value depends on temperature.(k) It is produced on account of induction in a material.(l) For low magnetising field I H. i.e. I H or I = Ht(m) is a dimensionless constant i.e. it carries no unit.(n) Other types of intensity of magnetisation:

(i) Mass intensity of magnetisation (Imass)-

mass mIntensity of magnetisation 1I or I

densityof magnetic material

(ii) Molar intensity of magnetisation-

Imolar = IMW = Mo. Wt.× mass intensity of magnetisation = MW × Im

(iii) Molecular intensity of magnetisation-

MWmolecular M

IMolar int ensityof magnetisationI IAvogadro No. N

Magnetic susceptibility or ( or K)

(a) The ratio intensity of magnetisation (I) in a material and the magnetising field (H) is defined asmagnetic susceptibility ().

i.e.IH

When H = 1 Oersted then = I.(b) The intensity of magnetisation induced in a material by unit magnetising field is defined as magnetic

susceptibility.(c) It has no unit and no dimensions.(d) It is a measure of ease with which a material can be magnetised by a magnetised by a magnetising

field (H).

(e) Magnetic susceptibility of various materials-(i) For diamagnetic materials – = low and negative(ii) For paramagnetic materials – = low but positive(iii) For ferromagnetic materials – = high and positive

(f) For paramagnetic substances it is inversely proportional to temperature i.e.IT

(g) For low magnetising field the value of is constant.(h) Different types of magnetic susceptibility-

(i) Volume susceptibility

VI XH

(ii) Mass or specific susceptibility m

vm

Volume susceptibilitydensity of meterial

(iii) Molar susceptibility MW

MW = m × MW

= Specific susceptibility × Mol. Wt.(iv) Molecular susceptibility m

m = Atomic wt. × specific susceptibility

= A mAIH

(i) –T curve

(j) –1T

:

T(a)

T–1(b)

For paramagnetic substances For paramagnetic substances

Illustration 8: The magnetic induction along the axis of an air solenoid is 0.03 Tesla. On placing an iron coreinside the solenoid the magnetic induction becomes 15 Tesla. The relative permeability ofiron core will be :(A) 300 (B) 500(C) 700 (D) 900

Solution: (B) mm 0 r

0

B 15B H, B H 500B 0.03

m= m = m \ m = = = =m

Absolute magnetic permeability (

(a) The ratio of magnetic induction (B) to magnetising field (H) is defined as magnetic permeability ((b) The extent to which magnetic permeability of that medium.(c) It is the characteristic property of a magnetic material because it represents the amplification of

magnetising field in that material.(d) Its value is always positive and is different for different materials.(e) For materials its value can be greater or less than 0.(f) Its value depends on H and T.(h) 0 [1 + ]

(i) 0 r

(j) (i) For feeromagnetic materials(ii) For paramagnetic materials(iii) For diamagnetic materials

(k) The unit of magnetic permeability is Weber per ammere-meter or Henry per meter and itsdimensions are M1L0T–2A–2.

Relative permeability ( r)

(a) The ratio of magnetic permeability of medium ( 0) is

defined as relative permeability ( r). i.e. r0

(b) rNumber of lines of force passing throughunit area in mediumNumber of lines of force passing throughunit area in vacuum

(c) rmagnetic flux density in mediumMagnetic flux density in vacuum

(d) The limit unto which a magnetic field penetrates matter, is known as relative permeability of thatmaterial.

(e) It has no unit and no dimensions.(f) r = 1 + (g) Relative permeability of various substances-

(i) For diamagnetic substances the value of r is slightly less than one i.e. r < 1.(ii) For paramagnetic substances the value of r is slightly greater than one i.e. r > 1.(iii) For ferromagnetic substance the value of r is much greater than one i.e. r >> 1.

MAGNETIC DIPOLE IN A UNIFORM MAGNETIC FIELD

(i) When a dipole is placed (or suspended) in a uniformmagnetic field, then no net force acts on it i.e. theresultant force acting on it is zero.

(ii) When a dipole is placed in a uniform magnetic fieldthen a torque acts on it which is given by

M B

Newton-meter(iii) The magnitude of torque is given by

MB sin 2 ml Bsin Where m is the pole strength and 2l is the length ofdipole.

(iv) The units of M are ampere-meter2 or Joule/Tesla and its dimensions are M0L2A1

(v) When a magnetic dipole is placed in a non-uniform magnetic field, then a torque as well as a forceboth act on it. The force acting on the dipole is given by

dBF M Newtondr

(vi) (a) The work done in rotating a dipole in a magnetic field from an angle 1 to angle 2

with the field direction is given by-W = MB (cos 1 – cos 2) Joule

(b) If initially the dipole is lying along the field direction, then 1 = 0– cos

(vii) Potential energy (U): (a) The potential energy of a dipole in a magnetic field is given by

U = –M.B Joule

(b) Special cases:WhenU = – MBWhen o, U = 0When o, U = – MB (–1) = MB

Illustration 9: A Magnet is suspended in the magnetic meridian with a untwisted wire. The upper endof the wire is rotated through 180 to deflect the magnet by 30 from magneticmeridian. Now this magnet is replaced by another magnet. Now the upper end of thewire is rotated through 270 to deflect the magnet by 30 from magnetic meridian.Compare the magnetic moments of magnets.

Solution: If be the twist of the wire, then C , where C being restoring couple per unit twistof wire. Here1 180 – 30 150 (150 /150) radian2 (270 – 30 ) 240 (240 /180) radianIf M be the magnetic moment and H, the horizontal component of earth’s field, then MH sin

C MH sinIf M1 and M2 be the magnetic moments of the two magnets respectively, then

C 1 M1H sin for first magnet

C 2 M2H sin for second magnet

1 1 1

2 2 2

150 /180M M 5orM M 240 /180 8

M1 : M2 5 : 8

Classification of materials

(i) The root cause of magnetism in matter is the motion of electric charges.(ii) The motion of electrons and protons in atoms is responsible for their magnetic properties.(iii) The variation in the number of fundamental charged particles and variation in their arrangement in

different materials are responsible for differences in their magnetic properties.(iv) On the basis of mutual interactions or behaviour of various materials in an external magnetic field,

the materials are divided in three main categories.(1) Diamagnetic substances (2) Paramagnetic substances(3) Ferromagnetic substances

Comparative study of these materials:

PropertyDiamagneticsubstances

Paramagneticsubstances

Ferromagneticsubstances

Cause ofmagnetism

Orbital motionsubstances

Spin motion of electrons Ferromagnetic substances

Explanation ofmagnetism

On the basis of orbitalmotion of electrons

On the basis of spin andorbital motion of electrons

On the basis of domainsformed.

Principle Electron principle Electron principle Domain principleBehaviour in a These are repelled in an These are feebly attracted in These are strongly attracted in

non-uniformmagnetic field

external magnetic fieldi.e. have a tendency tomove from high to lowfield region.

an external magnetic field i.e.have a tendency to movefrom low to high field region.

an external magnetic field. i.e.have an easy tendency tomove from low to high fieldregion.

State ofmagnetisation

These are weaklymagnetised in adirection opposite tothat of appliedmagnetic field

These get weekly magnetisedin the direction of appliedmagnetic field

These get strongly magnetisedin the direction of appliedmagnetic field.

When a rod ofthe material issuspendedbetween thepole pieces of amagnet.

The materials alignthemselves at rightangles to the directionof magnetic field.

(a)N S

(b)

N S

The materials alignthemselves along thedirection of magnetic field.

(b)

N S

(a)

N S

The materials easily alignthemselves in the direction ofmagnetic field.

Liquid or powerin a watch glasswhen placedbetween thepole pieces(a) when polesare far apart.(b) When polesare close toeach other.

(a) The liquid getsbulged in the middle

(b) The liquid getsdepressed in themiddle.

(a) The liquid gets depressedin the middle

(b) The liquid gets bulged inthe middle.

(a) The liquid is very muchdepressed in the middle.

(b) The liquid gets very muchbulged in the middle.

When thematerial in theform of liquid isfilled in the U-tube and placedbetween polepieces.

Liquid level in that limbgets depressed

N S

liquid

Liquid level in that limb risesup.

N S

liquid

Liquid level in that limb risesup very much.

N S

liquid

On placing thegaseousmaterialsbetween polepieces.

The gas expands atright angles to themagnetic field.

The gas expands in thedirection of magnetic field.

The gas rapidly expands in thedirection of magnetic field.

The value ofmagneticinduction B

B < B0 B > B0 Here B0 = magneticinduction in vacuum

B > > B0

Magneticsusceptibility

Low and negative 1

Low but positive 1 Positive and high 102.

Dependence of ontemperature

Does not depend ontemperature (except Biat low temperature)

Inversely proportional totemperature

1 CorT T

c µ c = . This is

called Curie law where C =Curie constant.

C C

1 CorT T T T

c µ c =- -

This is called Cure-Weiss law.TC = Curie temperature.

Dependence of on H

does not depend does not depend does not depend

Relativepermeability (r)

r < 1 r > 1 r > > 1 r 102

Intensity ofmagnetisation()

is a direction oppositeto that of H and itsvalue is very low.

is in the direction of H butvalue is low.

is in the direction of H andvalue is very high.

-H Curves

Magneticmoment (M)

The value of M is verylow ( 0 and is inopposite direction toH.)

The value of M is very lowand is in the direction of H.

The value of M is very high andis in the direction of H.

Transition ofmaterials (atCurietemperature)

These do not change On cooling, these getconverted to ferro- magneticmaterials at Curietemperature.

These get converted intoparamagnetic materials aboveCurie temperature.

–T Curve

The property ofmagnetism

Diamagnetism is foundin those materials inthe atoms of which thenumber electrons iseven.

Paramagnetism is found inthose materials in the atomsof which the majority ofelectron spins are in thesame direction.

Ferro-magnetism is found inthose materials which whenplaced in an external magneticfield are strongly magnetised.

Examples Cu, Ag, Au, Zn, Bi, Sb,NaCl, H2O air anddiamond etc.

Al, Mn, Pt, Na, CuCl2, O2 andcrown glass.

Fe, Co, Ni, Gd, Fe3O4 etc.

Nature of effect Distortion effect Orientation effect Hysteresis effect

Curie law and Curie temperature

(a) Curie law: (i) The magnetic susceptibility of paramagnetic substances in inversely proportional to its

absolute temperature i.e.1T

c µ

(ii)CT

c µ Where C = Curie constant, T = absolute temperature

(iii) On increasing temperature, the magnetic susceptibility of paramagnetic materials decreasesand vice versa.

(iv) The magnetic susceptibility of ferromagnetic substances does not change according to curielaw.

(b) Curie temperature (TC): (i) The temperature above which a ferromagnetic material behaves like aparamagnetic material is defined as curie temperature (TC).(ii) The minimum temperature at which a ferromagnetic substance is converted into

paramagnetic substance is defined as curie temperature.(iii) For various ferromagnetic materials its values are different. e.g. for Ni

NiCT 358 C= °

for FeFeCT 770 C= °

for COCOCT 1120 C= °

(iv) At this temperature the ferromagnetism of the substances suddenly vanishes.

Curie-Weiss law

–T Cure

Other important features and formulae

(A) Angle of declination and Geographical meridian:

qf

Angle of dip

Angle ofdeclinetion

Magnetic meridian

Geographicalmeridian

BH

BVB

(i) The horizontal component of earth's magnetic field is from S to N.(ii) If at any place the angle of dip is and magnetic latitude is then,

tan = 2 tan

The total intensity of earth's magnetic field2

0 1 3sinI = I + l

here 0 3MR

I =

Here M and R are the magnetic moment of bar magnet of earth and radius of earthrespectively.

(iii) At magnetic equator of earth = 0 and at poles = 90.

pole equator2I = I(iv) At the poles and equator of earth, the values of total intensity are 0.66 and 0.33 oersted

respectively.

(B) (i) In vacuum :– B0 = 0H(ii) In medium :– B = H(iii) Resultant magnetic field due to and H :–

(a) B = B1 + BH= 0 + 0H

B = 0 ( + H)B = 0H (1 + /H)

or B = 0 (1 + )H + = H(iv) (a) = 0 [1 + ]

(b) = 0r

(c)BH

m=

(v) (a) r0

mm =m

(b) r 1m = + c(vi) (a) r 1c = m -

(b)HIc =

(vii) The magnetic potential due to a small magnet at a point distant r is given by :–

2McosV

rq=

(viii) The mutual interaction force between two small magnets of moments M1 and M2

is given by

1 24

6M MF Kr

=

(C) Magnetic torque (a) = MB sin (b) = BiNA sin

(c) M Bt = ´

(D) Magnetic potential energy (UB)

(a) BU M B= ×

(b) ( )BU M B 1 cos= × - q

Assignment

1. The magnetism of atomic magnet is due to -(A) only spin motion of electrons(B) only orbital motion of electrons(C) both spin and orbital motion of electrons(D) the motion of protons.

Solution: (C) The magnetism of atomic magnet is due to both spin an orbital motions of electrons.

2. The earth's magnetic field inside an iron box as compared to that outside the box, is-(A) less (B) more (C) zero (D) same.

Solution: (A) The earths magnetic field inside an iron box is less than that outside the box.

3. The magnetic susceptibility of a paramagnetic materials varies with absolute temperature T as -

(A) T (B) Te (C) 1T (D) cons tant .Solution: (C) The magnetic susceptibility of a paramagnetic substance is inversely proportional to the

absolute temperature i.e.1T

4. There are two points A and B on the extended axis of a 2 cm long bar magnet. Their distances fromthe centre of the magnet are x and 2x respectively. The ratio of magnetic fields at points A and B willbe-(A) 8 : 1 approximately (B) 4 : 1 (approximately)(C) 4 : 1 (D) 8 : 1.

Solution: (A) The intensity of magnetic field on the axis

03

2mB4 r

3 31 2

2 1

B r 2x 8 :1B r x

5. The resultant magnetic moment of neon atom will be-

(A) infinity (B) zero (C) mB (D) B

2

.

Solution: (B) Neon atom is diamagnetic, hence its net magnetic moment is zero.

6. A magnetic material of volume 30 cm3 is placed in a magnetic field of intensity 5 oersted. Themagnetic moment produced due to it is 6 amp-m2. The value of magnetic induction will be-(A) 0.2517 Tesla (B) 0.025 Tesla(C) 0.0025 Tesla (D) 25 Tesla.

Solution: (A) B = m0 (I + H)

I = 56

M 6 2 15 amp /mV 30 10

H = 5 oersted = 35 amp /m

4 10m0 = 4 –7 Wb/amp-m

53

5B 4 10 7 2 104 10

= 0.2517 Tesla

7. The mass of an iron rod is 80 gm and its magnetic moment is 10 A–m2. If the density of iron is8gm/C.C. hen the value of intensity of magnetisation will be-(A) 106 A/m (B) 104 A/m (C) 102 A/m (D) 10 A/m

Solution: (A)M Md Magnetic moment densityI IV m mass

3

310 8 1080 10

= 106 A/m

8. The main difference between electric lines of force and magnetic lines of force is-(A) Electric lines of force are closed curves whereas magnetic lines of force are open curves.(B) Electric lines of force are open curves whereas magnetic lines of force are closed curves.(C) Magnetic lines of force cut each other whereas electric lines of force do not cut.(D) Electric lines of force cut each other whereas magnetic lines of force do not cut.

Solution: (B) The magnetic lines of force are in the form of closed curves whereas electric lines offorce are open curves.

9. The mass of a specimen of a ferromagnetic material is 0.6 kg. and its density is 7.8×103 kg/m3. If thearea of hysteresis loop of alternating magnetising field of frequency 50 Hz is 0.722 MKS units thenthe hysteresis loss per second will be-(A) 277.7×10–5 Joule (B) 277.7×10–6 Joule(C) 277.7×10–4 Joule (D) 277.7×10–4 Joule.

Solution: (A) WH = VAftm Aftd

or H 30.6W 0.722 50

7.8 10

= 277.7 × 10–5 Joule

10. The horizontal component of flux density of earth's magnetic field is 1.7×10–5 tesla. The value ofhorizontal component of intensity of earth's magnetic field will be?(A) 24.5 A/m (B) 13/5A/m (C) 0.135 A/m (D) 1.35 A/m.

Solution: (B)5 2

0

B 1.7 10 Wb /mH4 10 7 Wb / A m

= 13.5 A/m

11. A magnetising field of 2×103 amp/m produces a magnetic flux density of 8relative permeability of the rod will be-(A) 102 (B) 10o (C) 104 (D) 101

Solution: (C) r0 0

BH

or 4r 3 7

8 102 10 4 10

12. In a hydrogen atom the electron is revolving in a circular path of radius 5.1×10–11m with a frequency6.8×1015 Hz. The equivalent magnetic moment will be-(A) 8 × 10–24 A–m2 (B) 8 × 10–22 A–m2

(C) 8.9 × 10–20 A–m2 (D) 8 × 10–18 A–m2

Solution: (A) M = iA = ef 2

or M = 1.6×10–19×6.8×1015×3.14×(5.1×10–11)2 = 8.9 ×10–24 A–m2

13. A cube of side l is placed in a magnetic field of intensity B. The magnetic flux emanating out of it willbe-(A) zero (B) Bl2 (C) 2Bl2 (D) 6Bl2

Solution: (A) The net flux coming out of a closed surface is zero. Hence the flux coming out of the cubewill be zero.

14. A circular disc of area 3 2ˆ ˆ(4i 5 j) 10 m is placed in a uniform magnetic field of intensity

ˆ ˆ(0.2i 0.3 j) Tesla. The flux crossing the disc will be-(A) 23 Weber (B) 23×10–2 Weber(C) 23×10–3 Weber (D) 23×10–4 Weber

Solution: (C) B.A

ˆ ˆ ˆ ˆ(0.2i 0.3 j).[0.004i 0.005 j] = 0.0008 + 0.0015= 0.0023 Weber= 23×10–4 Weber

15. The total magnetic flux in a material, which produces a pole of strength mp when a magneticmaterial of cross-sectional area A is placed in a magnetic field of strength H, will be-(A) 0 (AH + mp) (B) 0 AH(C) 0 mp (D) 0 [mp AH + A]

Solution: (A) = BA ....(A)

0B (I H) ....(B)

From eqs. (A) and (B)

0(I H)A

pmI

A .....(C)

From eqs. (B) and (C)

p0 0 p

mA H [m AH]

A

16. The relation between and H for a specimen of iron is as follows-

–40.4 12 10 Henry /meterH

The value of H which produces flux density of 1 Tesla will be-(A) 250 A/m (B) 500 A/m (C) 750 A/m (D) 103 A/m

Solution: (B) B = H

or 040.4B 12 10 HH

or 1 = 0.4 + 12×10–4 H H = 500 A/m

17. The correct curve between X and1T

for paramagnetic materials is-

(A)

1/T

(B)

1/T

(C)

1/T

(D)

1/T

Solution: (A)1XT

18. The SI unit of magnetic flux is-(A) Weber (B) Maxwel (C) Tesla (D) Gauss

Solution: (A)2

2mBA Weber Weberm

19. The intensity of magnetic field due to an isolated pole of strength mp at a point distant r from it willbe-

(A) p2

mr

(B) mpr2 (C)2

p

rm

(D) pmr

Solution: (A) The magnetic intensity due to an isolated pole of strength mp at a distance r = p2

mr

20. A uniform magnetic field is directed from left towards right in the plane of paper. When a piece ofsoft iron is placed parallel to the field. The magnetic lines of force passing through it will be-

(A) (B)

(C) (D)

Solution: (C) The magnetic lines of force in a ferrmagnetic material are crowded together.

21. A bar magnet of magnetic moment M is cut into two equal parts. The magnetic moment of either ofthe parts will be-

(A)M2

(B) 2M (C) 2M (D) 2I H

Solution: (A)l MM ml M' m2 2

22. A loop of area 0.5 m2 is placed in a magnetic field of strength 2 Tesla in direction making an angle of60o with the field. The magnetic flux linked with the loop will be-

(A)1 Weber2

(B)3Weber2

(C) 2 Weber (D) 3 Weber

Solution: (A) o1 1BA cos , 2 cos602 2

23. The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet oflength 5 cm and pole strength of 200A-m on its axial line will be(A) 2.5×10–2 N (B) 2.5×10–3N(C) 5.0×10–2N (D) 5.0×10–3N.

Solution: (A) 03

2m'lF mB m4 x

7

310 2 200 0.05 100

8 10

= 2.5×10–2 N

24. The magnetic susceptibility of a paramagnetic substance is 3×10–4. It is placed in a magnetising fieldof 4×104 amp/m. The intensity of magnetisation will be-(A) 3 108 A/M (B) 12 108 A/M(C) 12 A/M (D) 24 A/M

Solution: (C) I = XH = 3×10–4×4×103 = 12 A/m

25. Volt-second is the unit of-(A) B (B) (C) I (D) x

Solution: (B)de or et volt secdt

26. The volume susceptibility of a magnetic material is 30×10–4. Its relative permeability will be-(A) 31 10–4 (B) 1.003 (C) 1.0003 (D) 29 10–4

Solution: (B) 4r

0

1 X 1 30 10 1.003

27. A current of 2 ampere is passed in a coil of radius 0.5 m and number of turns 20. The magneticmoment of the coil is-

(A) 0.314 A–m2 (B) 3.14 A–m2

(C) 314 A–m2 (D) 31.4 A–m2

Solution: (D) M = i 2N = 2×3.14×0.25×20 = 31.4 A–m2

28. The value of magnetic susceptibility for super-conductors is-(A) zero (B) infinity (C) +1 (D) –1

Solution: (D) For superconductorr = 1 + X = 0 X = –1

29. A current of 1 ampere is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment willbe-(A) 0.314 A-m2 (B) 3140 A-m2

(C) 100 A-m2 (D) 0 A-m2

Solution: (A) M = iA = R2 Ni= 3.14×0.01×10×1=0.314 Am2

30. The magnetic moment of a magnet of mass 75 gm is 9×10–7 A-m2. If the density of the material ofmagnet is 7.5×103 kg/m3 then intensity of magnetisation will be-(A) 0.9 A/m (B) 0.09 A/m (C) 9 A/m (D) 90 A/m

Solution: (B)–7 3

–3M Md 9 10 7.5 10I 0.09 A /mV m 75 10

For test

1. The area of hysteresis loop of a material is equivalent to 250 Joule. When 10 kg material ismagnetised by an alternating field of 50Hz then energy lost in one hour will be if the density ofmaterial is 7.5 gm/cm3.(A) 6×104 Joule (B) 6×104 Erg(C) 3×102 Joule (D) 3×102 Erg.

Solution: (A)mE nA Vt nA td

43

50 250 10 3600 6 10 Joule7.5 10

2. The corecivity of a bar magnet is 100 A/m. It is to be demagnetised by placing it inside a solenoid oflength 100 cm and number of turns 50. The current flowing the solenoid will be-(A) 4A (B) 2A (C) 1A (D) zero.

Solution: (B) H = niH 100i 2An 50

3. A current i is flowing in a conductor of length l. When it is bent in the form of a loop then itsmagnetic moment will be-

(A)2l i

4(B)

2l4

(C) 24l i

(D) 4l2i.

Solution: (A) M = iA = i r2 But 2r = l2 2

2l ilM i

4 4

4. A rod of ferromagnetic material with dimensions 10 cm×0.5 cm×0.2 cm is placed in a magnetic fieldof strength 0.5×104 amp/m as a result of which a magnetic moment of 5 amp-m2 is produced in therod. The value of magnetic induction will be-(A) 0.54 Tesla (B) 0.358 Tesla(C) 2.519 Tesla (D) 6.28 Tesla.

Solution: (D) 0 0MB (I H) HV

76

54 3.14 10 5000 6.28 Tesla10

.

5. The ratio of intensities of magnetic field in the axial and equatorial positions of a magnet will be-(A) 1 : 4 (B) 4 : 1 (C) 1 : 2 (D) 2 : 1

Solution: (D) 0

a

0c3

2MB 4 3 2 :1

MB4 r

M2

r

O r M

6. The resultant magnetic moment due to two current (i) carrying concentric coils of radius r, mutuallyperpendicular to each other will be-

(A) 22i r (B) 22ir (C) 2ir (D) 22 rSolution: (A) M' = 2M = 22 r i

7. The magnetic susceptibility of a paramagnetic material at –73oC is 0.0075 then its value at –173oCwill be-(A) 0.0045 (B) 0.0030 (C) 0.015 (D) 0.0075.

Solution: (C) For paramagnetic materials1XT

2 12

1 2

X T 0.0075 200or X 0.015X T 100

8. The area of cross-section of three magnets of same length are A, 2A and 6A respectively. The ratio oftheir magnetic moments will be-(A) 6 : 2 : 1 (B) 1 : 2 : 6 (C) 1 : 4 : 36 (D) 36 : 4 : 1

Solution: (B) M m Aµ µ (Area of cross-section)

1 2 3M : M : M 1: 2 : 6=

9. If the radius of a circular coil is doubled and the current flowing in it is halved then the new magneticmoment will be if its initial magnetic moment is 4 units-(A) 8 units (B) 4 units (C) 2 units (D) zero.

Solution: (A) 2M R Ni= p2

1 1 12

2 2 2 2

M R i 4 1or 2 M 8 unitsM R i M 4

æ ö æ ö= = ´ \ =ç ÷ ç ÷è ø è ø

10. Newton/Weber is the unit of-

(A) H

(B) M

(C) B

(D)

Solution: (A)F iF Bil il or H

A lf= = = =

f

11. The inner and the outer radii of a toroid are 9cm and 11cm respectively and the number of turns in itis 3140. A magnetic field of 2.5 Tesla is produced in it when a current of 0.5 ampere is passed in it.The permeability of core material is (in Henry/meter)-(A) 10–1 (B) 10–2 (C) 10–3 (D) 10–4

Solution: (C) 3B B 2.5 2 3.14 0.1 10 Henry /meterNiH 3140 0.52 r

-´ ´ ´m= = = =´

p

12. In the above problem the relative permeability of the core will be-(A) 684.5 (B) 864.7 (C) 369.4 (D) 796.2

Solution: (D)3

r 70

10 796.212.57 10

-

-

mm = = =m ´

13. A magnetising field of 5000 A/m produces a magnetic flux of 5×10–5 Weber in an iron rod. If the areaof cross-section of the rod is 0.5 cm2, then the permeability of the rod will be (in henry/m)-(A) 1 10–3 (B) 2 10–4 (C) 3 10–5 (D) 4 10–6

Solution: (B) 54

4 3

B HA

5 10 2 10 Henry /mAH 0.5 10 5 10

--

-

f= = m

f ´\ m= = = ´´ ´ ´

14. In the above problem the magnetic susceptibility of the rod will be-(A) 158.2 (B) 198.0 (C) 295.3 (D) 343.6

Solution: (A)r

04

70

1 X

2 10X 1 1 159.2 1 158.212.56 10

-

-

mm = = +m

m ´\ = - = - = - =m ´

15. In the above problem the intensity of magnetisation will be (in A/m)-(A) 7.9×102 (B) 7.9×102 (C) 7.9×102 (D) 7.9×105

Solution: (D) 6XH 158.2 5000 7.9 10 Amp /mI = = ´ = ´

16. At any place on earth, the horizontal component of earth's magnetic field is 3 times the verticalcomponent. The angle of dip at that place will be-(A) 60o (B) 45o (C) 90o (D) 30o

Solution: (D) V V

H V

B BtanB 3 B

q = =

1 tan30 303

= = ° \ q = °

17. The horizontal component of earth's magnetic field at any place is 0.36×10–4 Weber/m2. If the angleof dip at that place is 60o then the value of vertical component of earth's magnetic field will be-(inWb/m2)-(A) 0.12 10–4 (B) 0.24 10–4 (C) 0.40 10–4 (D) 0.62 10–4

Solution: (D) 4V HB B tan 0.36 10 tan60-= q = ´ °4 4 20.36 10 3 0.62 10 Wb /m- -= ´ ´ = ´

18. The value of angle of dip at a place on earth is 45o. If the horizontal component of earth's magneticfield is 5×10–5 Telsa then the total magnetic field of earth will be-

(A) 55 2 10 Tesla (B) 510 2 10 Tesla

(C) 515 2 10 Tesla (D) zero.

Solution: (A) 2 2 2 2 2V H H H HB B B B tan 45 B B 2= + = ° + =

5 25 2 10 Wb /m-= ´

19. The ratio of intensities of magnetic field, at distances x and 2x from the centre of a magnet of length2cm on its axis, will be-(A) 4 : 1 (B) 4 : 1 approx (C) 8 : 1 (D) 8 : 1 approx.

Solution: (D) 0 02 2 3 / 2 3

2M 2MB4 (l x ) 4 xm m= »p + p

3

1 2

2 1

B x 8 :1 approximatelyB x

æ ö\ = =ç ÷è ø

20. The length of a bar magnet is 10 cm and its pole strength is 10–3 Weber. It is placed in a magneticfield of induction 4 10–3 Tesla in a direction making an angle of 30o with the field direction. Thevalue of orque acting on the magnet will be-(A) 2 10–7 N-m (B) 2 10–5 N-m(C) 0.5 102 N-m (D) 0.5 N-m.

Solution: (A) MBsin m /Bsint = q = q3 3 710 0.1 4 10 0.5 2 10 N m- - -= ´ ´ p ´ ´ = p ´ -

21. A magnetic needle of magnetic moment 60 amp-m2 experiences a torque of 1.210–3 N-m directed ingeographical north. If the horizontal intensity of earth's magnetic field at that place is 40 2,then the angle of declination will be-(A) 30 (B) 45 (C) 60 (D) 90

Solution: (A) MBsint = q3

61.2 10 1sin 30

MB 60 40 10 2

-

-

t ´\ q = = = \ q = °´ ´

22. The ratio of total intensities of magnetic field at the equator and the poles will be-(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 1 : 4

Solution: (A) Total intensity of magnetic field remains constant

e

p

B1:1

B\ =

23. The magnetic flux density of magnetic field is 1.5 Wb/m2. A proton enters this field with a velocity of2×107 m/s in a direction making an angle of 30o with the field. The force acting on the proton will be-(A) 2.4 10–12 Newton (B) 0.24 10–12 Newton(C) 24 10–12 Newton (D) 0.24 10–12 Newton

Solution: (A) 19 7F qvBsin 1.6 10 2 10 1.5 0.5-= q = ´ ´ ´ ´ ´122.4 10-= ´

24. The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensityat another point Y on its equatorial axis. The ratio of distances of X and Y from the centre of themagnet will be-(A) (B)–3 (B) (B)–3 (C) 23 (D) 21/3

Solution: (D) 1/ 30 03 3

2M M dx 24 d x 4 d y dym m

= \ =p p

25. Two magnets A and B are equal in length, breadth and mass, but their magnetic moments aredifferent. If the time period of B in a vibration magnetometer is twice that of A, then the ratio ofmagnetic moments will be-

(A)12

(B) 2 (C) 4 (D) 12

Solution: (C) 21T 2 M

MB TI= p \ µ

2

A B

B A

M T 2 4M T 1

æ ö æ ö\ = = =ç ÷ç ÷ è øè ø

26. The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal partslengthwise then the time period of each part will be-(A) 4 sec. (B) 2 sec. (C) 0.5 sec. (D) 0.25 sec.

Solution: (A)2mlT 2 2 4sec

MB 12 m/BI= p = p =

´

27. The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the sameplace in a similar manner. The size of two magnets is the same but the magnetic moment of B is fourtimes that of A. The time period of B will be-

(A)T4

(B)T2

(C) 2T (D) 4 T

Solution: (B) A B AB

B A A

M T M Tor T TM T 4M 2= = ´ =

28. The value of current enclosed by a circular path of radius 0.30 cm is 9.42 ampere. The value ofmagnetic field along the path will be-(A) 500 amp/m (B) 1000 Amp/m(C) 5 104 Amp/m (D) Zero

Solution: (A) 3i 9.42H 500 amp /m

2 r 2 3.14 3 10-= = =p ´ ´ ´

29. A magnetic wire is bent at its midpoint at an angle of 60o. If the length of the wire is l and its magneticmoment is M then the magnetic moment of new shape of wire will be-

(A) 2 M (B) M (C)M2

(D)M2

Solution: (C)1 MM' m2 2

= × =

30. A current of 2 ampere is flowing in a coil of radius 50 cm and number of turns 20. The magneticmoment of the coil will be-(A) 3.14 amp-m2 (B) 31.4 amp-m2

(C) 314 amp-m2 (D) 0.314amp-m2

Solution: (B) 2 2M R Ni 3.14 0.25 20 2 3.14 amp /m= p = ´ ´ ´ =

1. 1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing acurrent of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising forcegenerated in the material will be-(A) 1.2 10–3 A/m (B) 2.6 10–3 A/m(C) 0.6 10–4 A/m (D) 2 10–3 A/m

Solution: (D) 3 3NiH ni 10 2 2 10 amp /m2 r= = = ´ = ´p

2. A magnet makes 10 oscillation per minute at a place where the horizontal component of earth'smagnetic filed (H) is 0.33 oersted. The time period of the magnet at a place where the value of H is0.62 oersted will be-(A) 4.38 S (B) 0.38 S (C) 2.38 S (D) 8.38 S

Solution: (A) 12 1

2

H1 0.33T T T 6 4.38 sH 0.62H

µ \ = = ´ =

3. The magnetic induction inside a solenoid is 6.5×10–4T. When it is filled with iron medium then theinduction becomes 1.4T. The relative permeability of iron will be-(A) 1578 (B) 2355 (C) 1836 (D) 2154.

Solution: (D) r 40

B 1.4 2154B 6.5 10-

m = = =´