C H APT E R

MAGNETISM

"""" ~.,., • • •• ~ ~ " > ~. .- >, .

5.1 INTRODUCTION1. What do you mean by the terms magnet and

magnetism ? What are natural magnets ? What is theorigin of the word magnetism ? o

Magnets and magnetism. A magnet is a material thathas both attractive and directive properties. It attracts smallpieces of iron, nickel, cobalt, etc. This property ofattraction is called magnetism. When suspended freely,a thin long piece of magnet comes to rest nearly in thegeographical north-south direction.

As early as the 6th century B.C., the Greeks hadsome knowledge of natural magnets. Thales of Miletusknew that pieces of a naturally occurring iron ore,lodestone or magnetite or black iron oxide Fe304 had theproperty of attracting small pieces of iron. The wordmagnetism originates from the place - Magnesia - aprovince in.the upper part of Greece - where this orewas found. Later on the Chinese discovered that thinlong pieces of lodestone, if suspended horizontally andfreely with a string, would naturally orient themselvesroughly in the geographical north-south direction. Infact, the word lodestone means a leading stone whichexpresses this directional property, valuable totravellers. By about A.D. 1000, the Chinese were usingthe magnetic compass for navigation. It was WilliamGilbert who, in his book 'De Magnete' of 1600, firstsuggested that the earth itself was a huge magnet,causing the alignment of compass needles.

In the early part of the nineteenth century, Oersteddiscovered that moving charges or currents are thesources of magnetic fields. However, the science ofmagnetism was known long before the nineteenthcentury.

5.2 ARTIFICIAL MAGNETS2. What are artificial magnets ? What are their

common shapes ?Artificial magnets. Generally, the natural magnets

are not strong enough magnetically and have

s -:;.._-

/sBar magnet Magnetic needle

S N

€t=! ======--_@Horse shoe

magnetBall ended magnet

Fig. 5.1(5.1)

Different forms of artificial magnets.

N

5.2

inconvenient shapes. The pieces of iron and othermagnetic materials can be made to acquire the properties ofnatural magnets. Such magnets are called artificialmagnets. The main advantage of these magnets is thatthey can be made much more stronger than the naturalmagnets and also of any convenient shape and size.They are generally available in the following forms:

1. Bar magnet. It is a bar of circular or rectangularcross-section.

2. Magnetic needle. It is a thin magnetised steelneedle having pointed ends and is pivoted at its centreso that it is free to rotate in a horizontal plane.

3. Horse shoe magnet. It has the shape of ahorse-shoe and thus it has been named so.

4. Ball-ended magnet. It is a thin bar of circularcross-section ending in two spherical balls.

5.3 BASIC PROPERTIES OF MAGNETS3. State the important properties of magnets.Basic properties of magnets :

1. Attractive property. A magnet attracts smallpieces of iron, cobalt, nickel, etc. When a magnet isbrought near a heap of iron filings, the ends of themagnet show the greatest attraction. These ends, wherethe magnetic attraction is the maximum, are calledpoles of the magnet. Thus every magnet has two poles.

Fig. 5.2 Poles of a bar magnet.

2. Directive property. When a magnet is suspendedor pivoted freely, it aligns itself in the geographicalnorth-south direction. The pole of the magnet whichpoints towards the geographical north is called thenorth-seeking or north (N) pole. The other pole whichpoints towards the geographical south is called thesouth-seeking or south (S) pole of the magnet.

Fig. 5.3 A magnet points north-south whenfreely suspended.

PHYSICS-XII

3. Like poles repel and unlike poles attract. If theN-pole of a magnet is brought near the N-pole of asuspended magnet, the poles are found to repel eachother. Two S-poles also repel each other. In contrast, N-and S- poles always attract each other. This action canbe described by the law of magnetic poles which statesthat like magnetic poles repel, and unlike magnetic polesattract each other.

( ~

Repulsio~

Fig. 5.4 Like poles repel and unlike poles attract.

4. Magnetic poles always exist in pairs. If we try toisolate the two poles of a magnet from each other bybreaking the magnet in the middle, each broken part isfound to be a magnet with N- and S-poles at its ends. Ifwe break these parts further, each part again is foundto be a magnet. So unlike electric charges, magneticmonopoles do not exist. Every magnet exists as a dipole.

IN 5 IIN 571 N 51IN 5 f/fC}7IN 5 fiN 5 I

Fig. 5.5 Poles always exist in pairs.

5. Magnetic induction. A magnet induces mag-netism in a magnetic substance placed near it. Thisphenomenon is called magnetic induction. When N-poleof a powerful magnet is placed close to a soft iron bar,the closer end of the bar becomes S-pole and the fartherend N-pole. As a result, the magnet attracts the ironbar. Thus induction precedes attraction.

For Your Knowledge

~ Repulsion is the surer test of magnetism. A magnetcan attract another magnet. Also it can attractmagnetic substances like iron, nickel, cobalt, etc.However, a magnet can repel another magnet only.So repulsion is the surer test of magnetism.

5A SOME IMPORTANT DEFINITIONSCONNECTED WITH MAGNETISM

4. Define the terms magnetic field, uniform magneticfield, magnetic poles, magnetic axis, magnetic equatorand magnetic length with reference to a bar magnet.

MAGNETISM

Some important definitions connected withmagnetism:

1. Magnetic field. The space around a magnet withinwhich its influence am beexperiencedis calledits magneticfield.

2. Uniform magnetic field. A magnetic field in aregion is said to be uniform if it has same magnitude anddirection at all points of that region. At a given place, themagnetic field of the earth can be considered uniform.The field due to a bar magnet is not uniform.

A uniform magnetic field acting in the plane of paperis represented by equidistant parallel lines [Fig. S.6(a)].A uniform magnetic field acting perpendicular to thepaper and directed outwards is represented by dots[Fig. S.6(b)]. A uniform magnetic field acting perpen-dicular to the plane of paper and directed inwards isrepresented by crosses [Fig. S.6(c)].

~ • • • • X X X X

~ •• • • • • X X X XB

X X X X~ • • • •~ • • • • X X X X

(a) (b) (c)

Fig. 5.6 Representations of a uniform magnetic field.

3. Magnetic poles. These are the regions of apparentlyconcentrated magnetic strength in a magnet where the magneticattraction is maximum. The poles of a magnet lie some-what inside the magnet and not at its geometrical ends.

: Magnetic: Iaxis

I 0- - - - - - - - - - -:- - j --------0 IN ' 5

: Magnetic: equator

Fig. 5.7 A bar magnet.

4. Magnetic axis. The line passing through the poles ofa magnet is called the magnetic axis of the magnet.

5. Magnetic equator. The line passing through thecentre of the magnet and at right angles to the magnetic axisis called the magnetic equator of the magnet.

6. Magnetic length. The distance between the twopoles of a magnet is called the magnetic length of the magnet.It is slightly less than the geometrical length of the magnet.

It is found that Magnetic length =0.84Geometrical length

I-- Magnetic length ---+!

oN 50

I<--- Geometrical length ---I

Fig. 5.8 Magneticand geometricallengths of a magnet.

5.3

5.5 COULOMB'S LAW OF MAGNETIC FORCE5. State Coulomb's law of magnetic force. Hence define a

unit magnetic pole.Coulomb's law of magnetic force. This law states that

the force of attraction or repulsion between two magneticpoles is directly proportional to the product of their polestrengths and inversely proportional to the square of thedistance between them.

If q n and q are the pole strengths of the twoI 1 nI2

magnetic poles which are distance r apart, then theforce between them is given by

qnI1qm2F cc ?-

or

where k is a proportionality constant which dependson the nature of the medium as well as on the systemof units chosen. For SI units and for poles in vacuum,

!-L qm1qnI2F = -.Jl.47t ?-

where!-Lois the permeability of free space and is equalto 47t x 10 -7 henry/metre. We can define unit magneticpole from Coulomb's law :

If q = q = 1 unit; r = 1 m, thennil nI2

F = !-Lo=10-7 N47t

Hence a unit magnetic pole may be defined as thatpole which when placed in vacuum at a distance of onemetre from an identical pole repels it with a force of 10-7

newton.

5.6 MAGNETIC DIPOLE AND MAGNETICDIPOLE MOMENT

6. What is a magnetic dipole? Give some examples.Magnetic dipole. In electricity, the fundamental or

simplest structure that can exist is a point charge. Heretwo equal and opposite charges separated by a smalldistance constitute an electric dipole, which is described

~by an electric dipole moment p . In magnetism, isolated

magnetic poles do not exist. Here the simple structurethat can exist is the magnetic dipole which is described

~by a magnetic dipole moment m.

An arrangement of two equal and opposite magneticpoles separated by a small distance is called amagnetic dipole.

5.4

Every bar magnet is a magnetic dipole. A currentcarrying loop behaves as a magnetic dipole. Even anatom acts as a magnetic dipole due to the circulatorymotion of the electrons around its nucleus.

7. Define magnetic dipole moment. Is it a scalar orvector quantity? Give its SI unit.

Magnetic dipole moment. The magnetic dipolemoment of a magnetic dipole is defined as the productof its pole strength and magnetic length. It is a vectorquantity, directed from S-pole to N-pole.

-4

where qm is the pole strength and 2 I is the magneticlength of the .dipole measured in the direction 5 - toN -pole.

We shall see later on that the 51 unit of magneticdipole moment is ampere metre2 (Am 2) or joule pertesla (JT - 1).

For Your Knowledge

~ Basic d p n between elect 'icity and m gnetism.In electricity a point charge is the simplest source ofelectricity and can be used as a test object also. Bymeasuring the force on a test charge at various placesin a given electric field, we can map out the entirefield. In magnetism, on the other hand, isolatedmagnetic poles do not exist. A magnetic dipole is thesimplest source of a magnetic field. It can be used as atest object for mapping a magnetic field. The mappingis done by measuring the torque experienced by a testmagnet or a magnetic dipole at various points in themagnetic field.In short, a small magnet is described by a vector iitwhile an electric charge by the scalar charge q on it. Amagnet experiences a torque in a magnetic field whilean electric charge experiences a force in an electricfield.

~ The pole strength qm is also called magnetic charge.Thus we assign magnetic charge + qm to the northpole and - qm to the south pole.

~ The direction of magnetic dipole moment iit is fromS-pole to N-pole. This is analogous to the direction ofthe electric dipole moment p of an electric dipolefrom negative charge to positive charge.

~ When a magnet of pole strength qrn is cut into twoequal parts.(i) along its axis (longitudinally), the pole strength ofeach half becomes q rn / 2(il) perpendicular to its axis (transversely), the polestrength of each half still remains qrn •

PHYSICS-XII

Examples based on'::,'/~:"~'~C9~loWp~~s1-~W.~~ri~~QJR~let.: -~;v:'~1. .' .... "Moment of a ';iagnet <__ 't:'~';o: ,if]

Formulae Used1. Magnetic dipole moment, m = qrn X 21

110 q,"J «:2. Coulomb's law, F = _ . 241t r2

Units UsedPole strength is in Am, force in newton, distancein metre.

Constant Used110 = 41t x 10-7 TmA -1.

Example 1. Two magnetic poles, one of which is four timesstronger than the other, exert a force of 5 g f on each otherwhen placed at a distance of 10 em. Find the strength of eachpole.

Solution. Let the pole strengths of the two dipolesbe qm and 4qm'

Here F = 5 g f = 5 x 10-3 kg f = 5 x 10-3 x 9.8 N,r=lO ern =0.1 m

Using Coulomb's law of magnetism,

F = 110 qml qm241t . r2

5 x 10- 3 x 9.8 10-7

x qm X 4qm(0.1)2

q 2 = 5 x 9.8 x (0.1)2 x 104 =25 x 49III 4or

or qlll = 5 x 7 = 35 Am4qm = 4 x 35 = 140 Am.and

Example 2. Two similar magnetic poles, having polestrengths in the ratio 1 : 2 are placed 1 m apart. Find thepoint where a unit pole experiences no net force due to thetwo poles.

Solution. Let the pole strengths of the two magneticpoles be qm and 2 qm' Suppose the required point islocated at distance x from the first pole. Then at thispoint,

Force on unit pole due to first pole=Force on unit pole due to second pole

i-lo qm x 1 _ i-lo 2 qlll X 1or 41t --.;z- - 41t' (1- xl

2x2 = (I-xl1x=~=0.414m.

1+ ,,2

or or .fix = I-x

or

MAGNETISM

Example 3. Calculate the force acting between twomagnets of length 15 em each and pole strength 80 Am eachwhen the separation between their north poles is 10 em andthat between south poles is 40 em.

Solution. The situation is shown in Fig. 5.9. Hereqllll = q11l2 =80 Am

I+-- 15em--1--10 em+-- 15em--.j

I!•..•-------40 em--------+i·1

Fig. 5.9

Force of repulsion between poles N1 and N2 is

F = 110. qml qlll2 = 10-7 x80x80 =0.064 N1 4lt r2 (0.10)2

Force of repulsion between poles 51 and 52 is

F = 10-7 x80x80 =0.004 N2 (0.40)2

Force of attraction between N1 and 52 is

F = 10-7 x80x80 =0.010 N3 (0.25)2

Force of attraction between N2 and 51 is

F4= F3=0.010 NResultant force between the two magnets is

F=F1+F2-F3-F4

= 0.064 + 0.004 - 0.010 - 0.010= 0.048 N (repulsive).

Example 4. A magnetic dipole of length 10 em has polestrength of20 Am. Find the magnetic moment of the dipole.

Solution. Here 21 =10 em =0.10 m, qm =20 Am

:. Magnetic moment,m= qm X 21 =20 x 0.10 Am2 =2.0 Am 2.

Example 5. A bar magnet of magnetic moment 5.0 Anlhas poles 20 em apart. Calculate the pole strength.

[CBSE D92C]

Solution. Here m = 5.0 Am 2, 21 = 20 em = 0.20 m

As m= qm x 21Pole strength,

«; = ~ = ~~~ = 25 Am.

Example 6. A steel wire of length I has a magnetic momentm. It is bent into a semicircular arc. What is the newmagnetic moment ?

Solution. Pole strength, qm =!!!.I

5.5

When the wire is bent into a semicircular arc, theseparation between the poles changes from I to 2r,where r is the radius of the semicircular arc. Thus

I1= ttr or r =-1t

. m 212mNew magnetic moment = q x 2 r = - x - = - .111 I 1t rr

j2)roblems ForPractice

1. Two magnetic south poles are located 4.0 cm apart.If the poles of each magnet have a strength of8.0Am and are 20.0 ern apart, find the force exertedby one south pole on the other. (Ans. 4.0 x 1O-3N)

2. Two equal and unlike poles placed 5 cm apart in airattract each other with a force of 14.4x 10-4 N. Howfar from each other should they be placed so thatthe force of attraction will be 1.6x10-lN?

(Ans. 0.15 m)3. Two magnetic poles, one of which is 10 times as

strong as the other, exert on each other a force equalto 9.604 mN, when placed 10 cm apart in air. Findthe strength of the two poles.

(Ans. 9.8 Am, 98 Am)4. Two like magnetic poles of strengths 5 Am and

20 Am are situated 1.0 m apart. At what point onthe line joining the two poles, will the magneticfield be zero?(Ans. 0.33 m from 5 Am pole towards 20 Am pole,

1 m from 5 Am pole away f~om20 Am pole)5. Two bar magnets of length 0.1 m and pole strength

75 Am each, are placed on the same line. Thedistance between their centres is 0.2 m. What is theresultant force due to one on the other when (i) thenorth pole of one faces the south pole of the otherand (ii) the north pole of one faces the north pole ofthe other? [Ans. (i) 3.4x10-2N (attraction),

(ii) 3.4x10-2 N (repulsion)]6. A magnetic dipole of length 15 ern has a dipole

moment of 1.5 Am 2. What is the pole strength?(Ans. 10 Am)

7. A magnetised steel wire 31.4 ern long has a polestrength of 0.2 Am. It is bent in the form of asemicircle. Calculate the magnetic moment of thesteel wire. (Ans. 0.04 Am2)

8. Two thin bar magnets of pole strengths 25 Am and48 Am respectively and lengths 0.20 m and 0.25 mrespectively are placed at right angles to each otherwith the N-pole of first touching the S-pole of thesecond. Find the magnetic moment of the system.

(Ans. 13 Am 2)

5.6

HINTS

I! 0 qml qm 2 10-7 x 8 x 8 -31. F = - . --2- = 22 N = 4.0 x 10 N.

4n, (4 x 10 )

_1.10 qn, q~ . 12. F----2-I.e., Focz4n, r2.1i_'2

"--2"F2 '1

~

1 14.4 x 10--4 -2'2 = E

2· 'i = --4 x5xl0 =0.15 m.1.6 x 10

F = 1.10 q,"J qn'24n· ?

10-7 xq xlO q.. 9.604 x 10-3 = m m

(0.1)2

or q; = 96.04or qm = 9.8 Amand 10 q", = 98 Am.

4. Let the magnetic field be zero at distance x from thepole of strength 5 Am.

3. As

I+-- x -- .••·1•...' --- 1- x ---+1'1

•• •o

Fig. 5.10

Magnetic field at 0 due to 5 Am pole= Magnetic field at 0 due to 20 Am pole

1.10 5 1.10 204n . x2 = 4n . (1- x)2

or 4x2=(I-x)2or 2x = ± (1 - x)

or x = -1 m, + 0.33 m.5. Proceed as in Example 2 on page 5.4.

m 1.5Am2

6. q = - = = 10 Am.m 21 0.15m

7. Proceed as in Example 6 on page 5.5.8. Here n; = 25 x 0.20 = 5.0 Am 2

~ = 48 x 0.25 = 120 Am 2

Resultant magnetic moment of the two magnetsplaced perpendicular to each other is

m = ~n;2 + ~2 = ~52 + 122 =13 Am2.

5.7 MAGNETIC FIELD LINES8. What are magnetic lines of force ? Give their

important properties.Magnetic lines of force. Michael Faraday, the cele-

brated physicist of London (1791-1867) introduced the

PHYSICS-XII

concept of the magnetic lines of force to represent amagnetic field visually. Magnetic lines of force do notreally exist but they are quite useful in describingmany different magnetic phenomena.

A magnetic line of force may be defined as thecurve the tangent to which at any point gives thedirection of the magnetic field at that point. It may alsobe defined as the path along which a unit north polewould tend to move if free to do so.

Properties of lines of force :1. Magnetic lines of force are closed curves which

start in air from the N-pole and end at the S-poleand then return to the N-pole through theinterior of the magnet.

2. The lines of force never cross each other. If theydo so, that would mean there are two directionsof the magnetic field at the point of intersection,which is impossible.

3. They start from and end on the surface of themagnet normally.

4. The lines of force have a tendency to contractlengthwise and expand sidewise. This explainsattraction between unlike poles and repulsionbetween like poles.

5. The relative closeness of the lines of force givesa measure of the strength of the magnetic fieldwhich is maximum at the poles.

9. Describe a method for plotting the magnetic field ofa bar magnet.

Plotting magnetic field of a bar magnet. Themagnetic field around a magnet can be traced withthe help of a magnetic compass needle. Iteonsists of asmall and light magneticneedle pivoted at the centreof a small circular brass caseprovided with a glass top, asshown in Fig. 5.11. Thenorth pole of the magneticneedle is generally painted Fig. 5.11 Compass needle.black or red.

Fig. 5.12 To plot magnetic field of a bar magnetwith a compass needle.

Imagine a unit north-pole placed at point P. ThenTherefore, the strength of the magnetic field B at from Coulomb's law of magnetic forces, the force exerted

point P is by the N-pole of the magnet on unit north-pole is

BaxI·aJ = Force experienced bya J..l qFN = ~ . -1!!.2' along NPunit north - pole at point P 41t x

- F - F - J..loqm [_1 1_] Similarly, the force exerted by the S-pole of the- N S - 41t (r _1)2 (r + 1)2 magnet on unit north-pole is_ J..loqm 4rl-4;-. (r2 _12)2

MAGNETISM

To plot the magnetic field of a magnet, the magnetNS is placed on a white paper. The compass needle isplaced near its N-poJe, as shown in Fig 5.12. Thepositions of the two ends S and N of the needle aremarked by pencil dots on the paper. The needle is nowdisplaced to the new position so that its south poleexactly comes over the mark previously made againstthe north pole and again the new position of the northpole is marked. The process is repeated till the southpole is reached. The various dots are joined together bya smooth curve which gives a line of force.

Similarly, other lines of force are drawn. A completepattern of the magnetic field around a bar magnet isshown in Fig. 5.24 on page 5.18.

5.8 MAGNETIC FIELD OF A BAR MAGNETAT AN AXIAL POINT

10. Derive an expression for the magnetic fieldinieneitv at a point on the axis of a bar magnet. What isthe direction of the field ?

Magnetic field of a bar magnet at an axial point(end-on position). Let NS be a bar magnet of length 21and of pole strength qm. Suppose the magnetic field isto be determined at a point P which lies on the axis ofthe magnet at a distance r from its centre, as shown inFig. 5.13.

~I'-------r+l------~-If.s uuu+O-u--N .•..}--u----- ~ p

1--1 '" 1 '1' r-1 ~I' r -I

Fig. 5.13 Magnetic field of a bar magnet at an axial point.

Imagine a unit north pole placed at point P. Thenfrom Coulomb's law ofmagnetic forces, the force exertedby the N-pole of strength qm on unit north pole will be

F =Ilo .J«: alongNPN 41t . (r _1)2 '

Similarly, the force exerted by S-pole on unit northpole is

F=llo ~s 41t . (r + 1)2 '

~along PS

5.7

But qm .21 = m, is the magnetic dipole moment, so

B . = Ilo 2mraxial 4n . (? _12)2

For a short bar magnet, I < < r, therefore, we have110 2m ~

Baxial = 4n . "?" ' along NP ...(1)

Clearly, the magnetic field at any axial point of magneticdipole is in the same direction as that of its magnetic dipolemoment i.e., from S-pole to N-pole, so we can write

~B. = J..lo 2 maxial 41t· r3

5.9 MAGNETIC FIELD OF A BAR MAGNETAT AN EQUATORIAL POINT

11. Derive an expression for the magnetic fieldintensity at a point on the equatorial line of a bar magnet.What is the direction of this field?

Magnetic field of a bar magnet at an equatorialpoint (broadside-on position). Consider a bar magnetNS of length 21 and of pole strength qm. Suppose themagnetic field is to be determined at a point P lying onthe equatorial line of the magnet NS at a distance rfrom its centre, as shown in Fig. 5.14.

Q

Rt+--~

,'T,,x .,,,,,,

.: e

,-t-\ ~

\~'"',, ", ...."

\--''=?,,,,,

r

5 N

Fig.5.14 Magnetic field of a bar magnet atan equatorial point.

F - J..lo qm along PSs - 41t· x2 ' r

5.8

As the magnitudes of FN and Fs are equal, so theirvertical components get cancelled while the horizontalcomponents add up along PR.

Hence the magnetic field at the equatorial point PisBequa = Net force on a unit N-pole placed at point P

= FN cas 8 + Fs cas 8

=2 FN cas 8 r. FN = Fsl

=2 ~o ~ I. 41t . x2 . x

or B = ~o mequa 41t' (1+ 12)3/2

where m = qm .21, is the magnetic dipole moment.Again for a short magnet, I < < r, so we have

~o mBeq = -. 3"' along PR ...(2)

ua 41t r

Clearly, the magnetic field at any equatorial point of amagnetic dipole is in the direction opposite to that of itsmagnetic dipole moment i.e., from N-pole to 5-pole. Sowe can write

~~ _ ~o mBequa - - 4nr3

On comparing equations (1) and (2), we note thatthe magnetic field at a point at a certain distance on the axialline of a short magnet is twice of that at the same distance onits equatorial line.

Formulae UsedMagnetic field of a bar magnet of length 21 anddipole moment m at a distance r from its centre,

1. B . I = ~ o. /mr2 2 (on the axial line)axia 41t (r - I )

2. B - ~o mequa - 4lt . (r2 + 12)3/2

For a short magnet, I< < r, so_ ~o 2m

3. Baxial - 4lt .7 (on the axial line)

4. B - ~ 0 m (on the equatorial line)equa - 41t . r3

(on the equatorial line)

Units UsedMagnetic field B is in tesla, distances r and I inmetre and magnetic moment in JT-1 or Am2.

Example 7. What is the magnitude of the equatorial andaxial fields due to a bar magnet of length 5 cm at a distance of50 em from the midpoint? The magnetic moment of the bar ormagnet is 0.40 A~. [NCERT]

PHYSICS-XII

Solution. Here m =0.40 Am2, r = 50 em =0.50 m,21 = 5.0 cm

Clearly, the magnet is a short magnet (I « r).

(i) B = ~. m = 10-7

x 0.4 = 3.2 x 10-7 T.equa 4 1t r3 (0.5)3

(ii) B. I = ~ . 2 3m= 6.4 x 10-7 T.

axia 4 1t r

Example 8. A bar magnet of length 10 cm has a polestrength of 10 Am. Calculate the magnetic field at a distanceof 0.2 m from its centre at a point on its (i) axial line and(ii) equatorial line.

Solution. Here 21 = 10 cm or 1=5 cm =0.05 m,qm =10 Am, r=0.2 m

Magnetic moment,m= qm X 21 =10 x 0.1 =1 Am2

(i) Magnetic field on axial line is

B. = ~ 0 2 mr 10-7 x 2 x 1x 0.2axial 41t . (1_12)2 (0.22 _0.052)2

10-7 x 0.4 T 5= 2.84 x 10- T.(0.0375)2

(ii) Magnetic field on equatorial line is

B = ~o m 10-7 xl Tequa 41t' (1+ 12)3/2 (0.22+0.052)3/2

10-7 10-7-----::-= T = T(0.0425)3/2 8.76 x 10-3

= 1.14 x 10-5 T.

Example 9. Two small magnets are placed horizontally,perpendicular to the magnetic meridian. Their north polesare at 30 em east and 20 em west from a compass needle. Ifthe compass needle remains undeflected, compare the magneticmoments of the magnets.

I_S N_~20 ~~ £ --30-c;; - ~'__N s__1Fig. 5.15

Solution. The compass needle at C lies on the axialline of the two magnets. As it remains un deflected, thefields of the two magnets at C must be equal and opposite.

Ilo 2~ _ ~o 2111z-'-3---'-3-41t r

141t r2

or

MAGNETISM

Example 10. Two short magnets P and Q are placed oneover another with their magnetic axes mutually perpen-dicular to each other. It is found that the resultant field at apoint on the prolongation of the magnetic axis of P is inclinedat 30° with this axis. Compare the magnetic moments of thetwo magnets.

Solution. Let A be any point on the prolongation of-7 -7

the axis of magnet P. Let ~ and ~ be the fields of themagnets P and Q respectively at the point A. Let-7 -7m1 and rrIz be the magnetic moments of the two magnets.

I+---r--l

Q rr 52I

III

P :~-------- -~'-+_-.ml 1- -A",--.-L,;...;."..---.,'S1

N1

Fig 5.16

As point A lies on the axial line of P, therefore,_ 110 2 I~

Bl--'-3-4rc r

The point A lies on the broad-side-on position of Q,therefore,

~=110 111.;,

4rc . r 3

But the resultant field ~ is inclined at 30° with ~,so

Hence

~ =tan300=~~ 131 _ rrIz

13-2~ or

Example 11. Two identical magnetic dipoles of magneticmoments 1.0 Ant each are placed at a separation of2 m withtheir axes perpendicular to each other. What is the resultantmagnetic field at ~ point mid-way between the dipoles?

Solution. The situation is shown in Fig. 5.17.

Fig. 5.17

5.9

The magnetic fields of the two magnets at the mid-point Pare

B =110 2m= 1O-7x/X1=2 xlO-7T,1 4rc' r3 1

in horizontal direction

~ = 110 n; = 10-7 T, in vertical direction4rc r

BR = ~ B/ + ~2 =.J5 x 10-7 T

If the resultant field BR makes angle 8 with ~, then

R 10-7tan 8 = ~ = =0.5

~ 2 x 10-7

8 = 26.57°

~rOblems For Practice

1. A bar magnet is 0.10m long and its pole strength is12 Am. Find the magnitude of the magnetic field ata point on its axis at a distance of 20 ern from it.

(Ans. 3.4 x 10-5 1')

2. Calculate the magnetic field due to a bar magnet2 cm long and having a pole strength of 100Am at apoint 10 cm from each pole. (Ans. 2 x 10-4 T)

3. A bar magnet has a length of 8 cm. The magneticfield at a point at a distance of 3 em from the centrein the broad-side on position is found to be 4 x 10-6 T.Calculate the pole strength of the magnet.

(Ans. 6xl0-5 Am)4. The magnetic moment of a current-loop is

2.1x 10- 25 Am 2. Find the magnetic field on the axisof the loop at a distance of 1.0 A from the loop.

(Ans. 4.2 x10-2 T)5. If the earth's magnetic field has a magnitude of

3.4 x 10-5 T at the magnetic equator of the earth,what would be its value at the magnetic poles of theearth? (Ans. 6.8 x 10-5 T)

6. The intensities of magnetic field at two points onthe axis of a bar magnet at distances 0.1m and 0.2mfrom its middle point are in the ratio 12.5 : 1.Calculate the distance between the poles of themagnet. (Ans. 0.1 m)

7. Two short magnets a and b of magnetic moments0.108Am2 and 0.192Am2 are placed alongmutually perpendicular straight lines meeting at apoint P. Find the magnitude and direction ofmagnetic field at point P, if it lies at distances 30 ernand 60 cm respectively from the centres of the twomagnets.

(Ans. 8.24 x 10-7 T, at 14°with the axis of a)

5.10

8. Two short magnets of magnetic moments a and b ofmagnetic moments 32 Am 2 and 27 Am 2 are placedon a table, as shown in Fig. 5.18. Find the magni-tude and direction of the magnetic field produced

5

40 em----------~PII

:30cm

a

N

Fig. 5.18

by these magnets at point P situated at the equa-torial lines of both the magnets at distances 40 cmand 30 cm respectively from the centres of the twomagnets. (Ans. 1.12 x 10-4T, at an angle

of 26.57° with the equatorial line of a)

HINTS

110 2mr 110 2xqmx2/xr1. Baxia1 = 41t . (,1 _ /2)2 = 41t' (r2 _12)2

10-7 x2x12xO.10xO.20(0.22 _ 0.052)2

= 3.4 x 10-5 T.

B _110 m_llo qmx212. equa - 41t . ,3 - 41t . -,3-

107 x 100 x 0.02 -4----;;-3-- = 2 x10 T.

(0.10)

3. Use B 110 Inequa = 41t . (,2 _ 12 )3/2

110 qm x21= 41t . (,2 _12)3/2 .

4. The current loop is short magnetic dipole. So

B. =110 2m=1O-7x2x2.1xlO-25axial 41t . r3 (1.0 x 10-10)3

= 4.2 x10-2T.5. At the equator,

110 m -5B = Bequa = - . 3" = 3.4 x 10 T41t r

At the poles,

B r = B 'aJ = ~ 2m = 2 BaXI 41t',3

= 6.8 x 10-5 T.

7. As shown in Fig. 5.19, the point P lies on the axialline of both the magnets a and b.

PHYSICS-XII

Bb B/-30 em

Is N ~-IP B.a III

IIII 60 emII

1N

b

5

Fig. 5.19

B =110 2n; = 1O-7x2xO.108a 41t 1. 3 (0.30)3

= 8 x 10-7 T (along the axis of a)

R _ 110 2~ _ 10-7 x2 xO.192"'b - • 3 - 341t r2 (0.60)

= 2 x 1O-7T (along the axis of b)The resultant field at Pis

B = ~ B/ + 1),2= 10-7 ~82 + 22 = 8.24 x 10-7 T

If the field B makes angle ewith the direction of Ba 'then

tan e = 1), = ~ = 0.25 orBa 8

8. As shown in Fig. 5.20, the point P lies on the equa-torial line of both the magnets a and b.

a 5B. B

40 em .--1(':----------~:P, BI b

:30 emIII

N

51Fig. 5.20 b

B - ~ n; _ 10-7 x32 - 0 5 10-4T.. a- '3- 3-'x

41t 1. (0.40) (antiparallel to "1 )

1),= 110 .1= 10-7

x;7 = lO-4T41t r2 (0.30) (antiparallel to Inz)

The two fields are perpendicular to each other. Sothe resultant field at point Pis

B = ~ B; + ~ = 10-4 ~(0.5)2 + 12 =1.12 x 10-4 T

If the field B makes an angle e with the direction ofBv, then

B 0.5 x 10-4tan e = -1!... = 4 = 0.5 or e = 26.57°

1), 10

MAGNETISM

5.10 TORQUE ON A MAGNETIC DIPOLEIN A MAGNETIC FIELD

12. Derive an expression for the torque on a magneticdipole placed in a uniform magnetic field. Hence definemagnetic dipole moment.

Torque on a magnetic dipole in a uniform magneticfield. Consider a bar magnet NS of length 21 placed in a

--7

uniform magnetic field B. Let qm be the pole strength

of its each pole. Let the magnetic axis of the bar magnet--7

make an angle 8 with the field B, as shown in

Fig. S.21(a).--7

Force on N-pole = qm B; along B--7

Force on S-pole = qm B, opposite to B

•• It!

q",B

B••

q"'B •..• -~------~~------------------~..~ ~ ~ ~T =m x B

(b)(a)

Fig. 5.21 (a) Torque on a bar magnet in a magnetic field.

(b) Relation between the directions of t>, ;;, B.

The forces on the two poles are equal and opposite.They form a couple. Moment of couple or torque isgiven by

T = Force x perpendicular distance

= qm B x 21 sin 8 = (q", x 21) Bsin 8

or T=mBsin8

where m = qm X 2/, is the magnetic dipole moment ofthe bar magnet. In vector notation,

--7 --7 --7

T=mxB--7

The direction of the torque 1 is given by the right

hand screw rule as indicated in Fig. S.21(b). The effect--7

of the torque 1 is to make the magnet align itself--7

parallel to the field B . That is why a freely suspended

magnet aligns itself in the north-south directionbecause the earth has its own magnetic field whichexerts a torque on the magnet tending it to align alongthe field.

5.11

Special Cases

1. When the magnet lies along the direction of themagnetic field,

8=0°, sin 8=0, 1=0,Thus the torque is minimum.

2. When the magnet lies perpendicular to thedirection of the field,

8 = 90°, sin 8 = 1, 1 = mBThus the torque is maximum

1max = 11IB

Definition of magnetic dipole moment. If in Eq. (1),B=l, 8 =90°, then

1=m

Hence the magnetic dipole moment may be defined asthe torque acting on a magnetic dipole placed perpendicularto a uniform magnetic field of unit strength.

SI unit of magnetic moment. As

1m=-----Bsin 8

SI . f 1Nmunits 0 m = ---IT .1

= NmT-1 or JT-1 or Am2.

5.11 POTENTIAL ENERGY OF A MAGNETICDIPOLE IN A MAGNETIC FIELD

...(1)

13. Derive an expression for the potential energy of adipole placed in a uniform magnetic field at an angle ewith it. When will the magnetic dipole be in the positionsof stable and unstable equilibrium?

Potential energy of a magnetic dipole. As shown inFig. S.21(a), when a magnetic dipole is placed in a

--7

uniform magnetic field B at angle 8 with it, it expe-

riences a torque1= mBsin 8

This torque tends to align the dipole in the--7

direction of B ....(2)

If the dipole is rotated against the action of thistorque, work has to be done. This work is stored aspotential energy of the dipole.

The work done in turning the dipole through asmall angle d8 is

dW = 1d8 = mB sin 8 d8If the dipole is rotated from an initial position 8 = 81

to the final position 8 = 82, then the total work donewill be

5.12

92W = f dW = f mB sin 9 d9 = mB [- cos 9] ~

~= - mB(cos 92 - cos 91)

This work done is stored as the potential energy Uof the dipole.

:. U=-mB(cos92-cos91)

The potential energy of the dipole is zero when~ ~m .1 B. So potential energy of the dipole in any orien-tation 9 can be obtained by putting 91 =90° and 92 = 9in the above equation.

U = - mB(cos 9 - cos 90°) ~~ ~

or U = - mB cos 9 = - m . B

Special Cases1. When 9 =Ob, U =- mBcosO° =- mB

Thus the potential energy of a dipole is~ ~

minimum when m is parallel to B. In this state,the magnetic dipole is in stable equilibrium.

2. When 9 =90°, U = - mB cos 90° =0.3. When 9 = 180°, U = - mB cos 180° = + mB.

Thus the potential energy of a dipole is maximum~ ~

when m is aniiparallel to B . In this state, the magneticdipole is in unstable equilibrium.

5.12 CURRENT LOOP AS A MAGNETIC DIPOLE

14. Show that a current carrying loop behaves as amagnetic dipole. Hence write an expression for itsmagnetic dipole moment.

Current loop as a magnetic dipole. We know thatthe magnetic field produced at a large distance r fromthe centre of a circular loop (of radius a) along its axis isgiven by

or

where I is the current in the loop and A = na2 is itsarea. On the other hand, the electric field of an electricdipole at an axial point lying far away from it is givenby

E=_l_. 2: ...(2)4 n EO r

where p is the electric dipole moment of the electricdipole.

PHYSICS-XII

On comparing equations (1) and (2), we note that

both Band E have same distance dependence (:3 ).Moreover, they have same direction at any far awaypoint, not just on the axis. This suggests that a circularcurrent loop behaves as a magnetic dipole of magneticmoment,

m= IAIn vector notation,

~ ~ Am = I A = IAn

This result is valid for planar current loop of anyshape. Thus the magnetic dipole moment of any currentloop is equal to the product of the current and its loop area.Its direction is defined to be normal to the plane ofthe loop in the sense given by right hand thumb rule.

Right hand thumb rule. If we curl the fingers of theright hand in the direction of current in the loop, then theextended thumb gives the direction of the magnetic momentassociated with the loop.

A

11

N-Face Area A

S-Face

Fig. 5.22 Current loop as a magnetic dipole.

...(1)

It follows from the above rule that the upper face ofthe current loop shown in Fig. 5.22. has N-polarity andthe lower face has S-po)arity. Thus a current loop behaveslike a magnetic dipole.

If a current carrying coil consists of N turns, then

m= NIAThe factor NI is called amperes turns of current loop.

So,

Magnetic dipole moment of current loop= Ampere turns x loop area

Clearly, dimensions of magnetic moment

= [A] [L2] = [AL2]

51 unit of magnetic dipole moment is Am2. It isdefined as the magnetic moment associated with one turnloop of area one square metre when a current of one ampereflows through it.

MAGNETISM

Table 5.1 Analogy between electricand magnetic dipoles

~;. .~.1

Free space constant

Dipole moment-+m

Axial field--_.-

Equatorialfield

Torque in external field-+ -+ -+ -+pxE mxB

---r-----------+-+

-m.B-+-+

-p.EP.E. in external field

5.13 MAGNETIC DIPOLE MOMENT OF AREVOLVING ELECTRON

15. Derive an expression for the magnetic dipolemoment of an electron revolving around a nucleus.Define Bohr magneton and find its value.

Magnetic dipole moment of a revolving electron.According to Bohr model of hydrogen-like atoms,negatively charged electron revolves around the posi-tively charged nucleus. This uniform circular motionof the electron is equivalent to a current loop whichpossesses a magnetic dipole moment = fA. As shown inFig. 5.23, consider an electron revolving anti clockwisearound a nucleus in an orbit of radius r with speed vand time period T.

1

e+ -II

-e

Fig. 5.23 Orbital magnetic moment of a revolving electron.

Equivalent current,

f_ Charge _ e _ e _ ev- -------

Time T 21tr / v 21trArea of the current loop, A = n,z

5.13

Therefore, the orbital magnetic moment (magneticmoment due to orbital motion) of the electron is

ev 2III = fA = - . 1tT

2nrevr

1l1=-2

...(1)or

As the negatively charged electron is revolvingantic1ockwise, the associated current flows clockwise.According to right hand thumb rule, the direction ofthe magnetic dipole moment of the revolving electronwill be perpendicular to the plane of its orbit and in thedownward direction, as shown in Fig. 5.23

Also, the angular momentum of the electron due toits orbital motion is

1= mevr ...(2)

The direction of r is normal to the plane of theelectron orbit and in the upward direction, as shown inFig. 5.23.

Dividing equation (1) by (2), we get!:1. = evr / 2 = _e_I mevr 2me

The above ratio is a constant called gyromagneticratio. Its value is 8.8 x 1010 C kg-I. So

eIII =-1

2me

Vectorially,-+ e-tIl --- I

1- 2me

The negative sign shows that the direction of r is-t

opposite to that of Il r According to Bohr's quantisationcondition, the angular momentum of an electron in anypermissible orbit is integral multiple of h / 2n, where his Planck's constant, i.e.,

I=nh h 1232n' w eren=, " .....

1l1=n[~)4n me

This equation gives orbital magnetic moment of anelectron revolving in n th orbit.

Bohr magneton. It is defined as the magnetic momentassociated with an electron due to its orbital motion in thefirst orbit of hydrogen atom. It is the minimum value of IIIwhich can be obtained by putting n = 1in the aboveequation. Thus Bohr magneton is given by

ehIl B = (Ill )rnin = -4--

n me

5.14

Putting the values of various constants, we get1.6 x 10-19 C x 6.63 x 10-34 JS

IlB = 4 x 3.14 x 9.11 x 10-31 kg

= 9.27 x 10-24 Am 2.->

Besides the orbital angular momentum I , an->

electron has spin angular momentum 5 due to itsspinning motion. The magnetic moment possessed by anelectron due to its spinning motion is called intrinsic mag-netic moment or spin magnetic moment. It is given by

-> e->11 =-- 5

5 me

The total magnetic moment of the electron is thevector sum of these two momenta. It is given by

->->-> e->->11 =11/+115=--(1 +25)

2me

Examples based on

•

-> -> ->1. Torque, 't = mB sin 8 or r = m x B

2. Work done in turning the dipole or P.E. of a dipole,W = U = - mB (cas 82 - cos ~)

3. If initially the dipole is perpendicular to the field,U =- mBcas 8

(1') -> ->When m is parallel to B, 8 = 0°, U = - mB

Potential energy of the dipole is minimum. Itis in a state of stable equilibrium.

(ii) When iit is perpendicular to B, e = 90°, U = O.

(iii) When 1:;; is antiparalleJto B, e = 180°, U = + mB

Potential energy of the dipole is maximum. Itis in a state of unstable equilibrium.

4. Magnetic moment of a current loop, m = NIA5. Orbital magnetic moment of an electron in nth orbit,

11 / = ~r = 2:e I= n ( 4::J6. Bohr magneton is the magnetic moment of an

electron in first (n = 1)orbit.ell

11 B = (11/) . = -- .nun 4nme

Units Used

Torque 1: is in Nm, magnetic moment m in JT -1 orArn2, field B in tesla, potential energy U in joule.

PHYSICS-XII

Example 12. A magnetised needle of magnetic moment4.8 x 10-2 J T -1 is placedat 30° with the direction of uniformmagnetic field of magnitude 3 x 10-2 T. What is the torqueacting on the needle ? [CBSE D OIC]

Solution. Here m = 4.8 x 10-2 JT -1, e = 30°,B=3xlO-2 T

:. Torque, r = mB sin 8= 4.8 x 10-2 x 3 x 10-2 x sin 30°= 7.2 x 10-4 J.

Example 13. A short bar magnet placed with its axis at 30°to a uniform magnetic field of 0.2 T experiences a torque of0.06 Nm. (i) Calculate the magnetic moment of the magnet.(ii) Find out what orientation of the magnet corresponds toits stable equilibrium in the magnetic field. [CBSE OD 02]

Solution. (i) Here B=0.2 T, 8 =30°, T =0.06 Nm

Magnetic moment,r 0.06m=---=----

B sin 8 0.2 sin 30°0.06 = 0.6 Am 2.

0.2 x 0.5

(ii) The P.E. of a magnetic dipole in a uniformmagnetic field is

U = - mBcos 8In stable equilibrium, the P.E. is minimum. So

cos Oe I or 8=0°Hence the bar magnet will be in stable equilibrium

->when its magnetic moment m is parallel to the

->magnetic field B.

Example 14. In an iron bar (5 cm x 1 cmx 1 em) themagnetic moment of an atom is 1.8 x 10-23 Ant-. (i) Whatwill be magnetic moment of the bar in the state of magneticsaturation ? (ii) What torque will have to be applied to keepthe bar perpendicular to an external magnetic field of 15,000gauss? Density of iron = 7.8 g cm-3, its atomic mass = 56.

Solution. (i) Mass of iron bar = volume x density= 5 cm3 x 7.8 gcm-3 =39 g

Number of atoms in 56 g of iron = 6.02 x 1023

:. Number of atoms in 39 g of iron

6.02 x 1023

x 39 = 4.19 x 102356

Magnetic moment of each atom= 1.8 x 10-23 Am2

Magnetic moment of the iron bar in the state ofmagnetic saturation is

m = 1.8 x 10-23 x 4.19 x 1023 = 7.54 Am 2.

(ii) Here 8 =90°, B=15,o00 G =15000 x 10--4TRequired torque, 't = mB sin 8

= 7.54 x 15000 x 10-4 x sin 90° = 11.3 Nm.

MAGNETISM

Example 15. A planar loop of irregular shape encloses anarea of7.5 x 10--4 nl- and carries a current of 12 A. The senseof flow of current appears to be clockwise to an observer.What is the magnitude and direction of the magneticmoment vector associated with the current loop?

[NCERT]

Solution. Here A = 7.5 x 10-4 m 2, I = 12 A

Magnetic moment associated with the loop ism = IA = 12 x 7.5 x 10--4

= 9.0 x 10-3 JT-1

Applying right hand rule, the direction of magneticmoment is along the normal to the plane of the loopaway from the observer.

Example 16. A current of 5 A isflowing through a 10turncircular coil of radius 7 em. The coil lies in the x-y plane.What is the magnitude and direction of the magnetic dipolemoment associated with it ?

If this coil were to be placed in a uniform externalmagnetic field directed along the x-axis, in which planewould the coil lie, when in equilibrium? (Take 1t= 22 /7)

[CBSE Sample Paper 03]

Solution. Magnetic dipole moment,m = NIA = NI x 1tr2

22 (7)2= 10 x 5 x --;;-x 100

= 0.77 Am2

The direction of magnetic dipole moment isperpendicular to the plane of the coil. Hence it is alongz-axis.

Torque on the current loop of magnetic momentmis

t = mBsin a

where a is angle between rr; and B. For stable~

equilibrium torque is zero, so a =0°. For this B shouldbe perpendicular to the plane of the coil. Hence the coilwill lie in y-z plane in the condition of stableequilibrium.

Example 17. A bar magnet with poles 25 cm apart and ofpole strength 14.4 Am rests with its centre on a frictionlesspivot. It is held in equilibrium at 60° to a uniform magneticfield of induction 0.25 T by applying a force F, at rightangles to its axis, 12 em from its pivot. Calculate F. Whatwill happen if the force F is removed ? [lIT]

Solution. Here m = qm X 21 = 14.4 x 0.25 = 3.6 Am 2

9=60°, B=0.25T, r=12cm=0.12mTorque, t = Fr = mB sin 9

5.15

F = mB sin 9 = 3.6 x 0.25 x sin 60°r 0.12

= 3.6 x 0.25 x 0.866 = 6.5 N0.12

When the force F is removed, the magnet alignsitself in the direction of field B.

Example 18. An electron in an atom revolves around thenucleus in an orbit of radius 0.5 A. Calculate the equivalentmagnetic moment if the frequency of revolution of theelectron is 1010 MHz. [lIT 88, CBSE D 98]

Solution. The electron revolving around thenucleus in a circular orbit is equivalent to a currentloop. Its magnetic moment is m = IA = ev x 1t,z

Here e = 1.6 x 10-19 C v = 1010 MHz = 1016 Hz,r =0.5 A =0.5 x 1O-lOm

:. m=1.6x 1O-19x1016x3.14x(0.5xlO-1O)2

= 1.2S6x 10-23 Am2.

Example 19. An electron moves around the nucleus in ahydrogen atom of radius0.51 A with a velocity oj 2 x 106m/s.Calculate the following:

(I) the equivalent current due to orbital motion of electron(ii) the magnetic field produced at the centre of the nucleus

(iii) the magnetic moment associated with the electron.[CBSEOD 08]

Solution. Here r =0.51x 1O-lOm, v =2 x 105ms-1

(i) I = ~ = ~ = 1.6 x 10-19 x 2 x 10

5 = 10-4 AT 21tr 21t x 0.51 x 10-10

41tx 10-7 x 10-4------.;;- = 1.23 T.2 x 0.51 x 10-10

(iii) m = IA = ~ x if = evr21tr 2

1.6 x 10-19 x 2 x 105 x 0.51 x 10-10

2= 8.16 x 10-25 Am 2.

Example 20. Two magnets of magnetic moments m and.J3 m are joined to form a cross (+ ~ The combination issuspended freely in a uniform magnetic field. In equilibriumposition, the magnet of magnetic moment m makes an angle9 with the field. Find 9.

Solution. When the magnet of moment in makesangle 9 with the field B, the other magnet of moment.J3 m will make angle (90° - 9) with the field B. In theequilibrium position,

Torque experienced by first magnet= Torque experienced by second magnet

or mB sin 9 = .J3 mB sin (90° - 9)

5.16

or sin 8 =.J3 cas 8

sin 8 =.J3 or tan 8 = .J3cas 8

or

8 = 60°.

Example 21 . A bar magnet having a magnetic moment of1.0 x 104 J T -1 is free to rotate in a horizontal plane. Ahorizontal magnetic field of 4 x 10-5 T exists in space. Findthe work done in rotating the magnet slowly from a directionparallel to the field to a direction 60° from the field.

Solution. Here m = 1.0 x 104 J T -I, B= 4 x 10-5 T,

81 =0°, 82 =60°Work done,

W = - mB(cos 82 - cas 81)

= -1.0 x 104 x 4 x 10-5 (cas 60°- cas 0°)

= 1.0 x 104 x 4 x 10-5 x .!= 0.2 J.2

Example 22. A current of 7.0 A is flowing in a planecircular coil of radius 1.0 cm having 100 turns. The coil isplaced in a uniform magnetic field of 0.2 Wbm-2

. If the coilisfree to rotate, what orientations would correspond to its (i)stable equilibrium and (ii) unstable equilibrium? Calculatethe potential energy of the coil in these cases. [CBSE 00 92)

Solution. Here N = 100, A = 7.0 A,r=1.0cm=1.0x10-2m, B=O.2 Wbm-2

Magnetic moment associated with the coil is

m = NI A = Nl x 1<?

= 100 x 7.0 x 22 x (1.0 x 10- 2)2 =0.22 Am27

-t(i) The stable equilibrium corresponds to m parallel-t

to B. The potential energy is then minimum.

Umin = - mBcos 0° = -0.22 x 0.2 xl = - 0.044 J.-t

(ii) The unstable equilibrium corresponds to m anti--t

parallel to B. The potential energy is then maximum.

Umax = - mB cas 180° = - 0.22 x 0.2 x (-1)

= + 0.044 J.Example 23.A short bar magnet placed with its axis at30°experiences a torque of 0.016 Nm in an external field of800 G.(a) What is the magnetic moment of the magnet? (b) What isthe work done by an external force in moving it from its moststable to most unstable position? (c) What is the work doneby the force due to the external magnetic field in the processmentioned in part (b) ? (d) TIle bar mTet is replaced by asolenoid of cross-sectional area2 x 10 1~ and 1000 turns,but the same magnetic moment. Determine the currentflowing through the solenoid. [NCERT)

PHYSICS-XII

Solution. (a) Here 8 = 30°, B= 800 G = 800 X 10-4T,t =0.016 Nm

Magnetic moment,1:m=---

Bsin 8

0.016 = 0.40 Am 2.800 x 10-4 x sin 30°

(b) For most stable position, 8 =0° and for mostunstable position 8 = 180°. So the required work doneby the external force,

W = - mB(cos 180° - cos 0°) =2mB= 2 x 0.40 x 800 x 10-4 = 0.064 J.

(c) Here the displacement and the torque due to themagnetic field are in opposition. So the work done by themagnetic field due to the external magnetic field is

W8 = -0.064 J.(d) Here A = 2 x 10-4 m2, N = 1000Magnetic moment of solenoid,

nls = m=0.40 Am2

But ms= NlAm 0.40

•. Current, 1=_5 = = 2ANA 1000 x 2 x 10-4

~rOblems For Practice

1. A short bar magnet of magnetic moment 0.9 JT -1 isplaced with its axis at 30° to a uniform magneticfield. It experiences a torque of 0.063J. (i) Calculatethe magnitude of the magnetic field. (ii) In whichorientation will the bar magnet be in stableequilibrium in the magnetic field ? [CBSE F 121

...• -t(Ans. 0.14 T, m II B for stable equilibrium)

2. A circular coil of 300 turns and diameter 14 emcarries a current of 15 A. What is the magnitude ofmagnetic moment associated with the loop?

[Haryana OIl (Ans. 69.3J T -1 )

3. Calculate the magnitude of the torque required tohold a bar magnet of magnetic moment 200 Am2along a direction making an angle of 30° with thedirection of a uniform magnetic field of 0.36 G.

(Ans. 3.6 x10-3 Nm)4. Calculate the torque acting on a magnet of length

20 em and of pole strength 2 x 10-5 Am, placed inearth's magnetic field of flux density 2 x10-5 T,when (i) magnet is parallel to the field (ii)magnet isperpendicular to the field.

[Ans. (i) Zero (ii) 0.8x10-10 Nm]

MAGNETISM

5. The magnetic dipole moment of the earth is6.4 x 1021Am 2. If we consider it to be due to acurrent loop wound round the magnetic equator ofthe earth, then what should be the magnitude of thecurrent? Assume the earth to be sphere of radius6400 km. ( .6.. s 5 x 107 A)

6. A straight solenoid of length 50 ern has 1000 turnsand a mean cross-sectional area of 2 x 10-4m 2. It isplaced with its axis at 30°, with a uniform magneticfield of 0.32 T. Find the torque acting on thesolenoid when a current of 2 A is passed through it.

( \ns. 0.064 Nm)7 A current of 3 A flows through a plane circular coil

of radius 4 cm and having 20 number of turns. Thecoil has been placed in a uniform magnetic field of0.5 T. Find (i) dipole moment of the coil(ii) potential energy of the dipole.

(Ans 0.3 Am 2, - 0.15 J)8. A bar magnet placed in a uniform magnetic field of

strength 0.3 T with its axis at 30° to the field, expe-riences a torque of 0.06 Nm. What is the magneticmoment of the bar magnet? [ISeE 98]

(Ans. 0.4 Am 2)9. A short bar magnet placed with its axis at 30° with a

uniform external magnetic field of 0.16 T expe-riences a torque of magnitude 0.032 J.(a) Estimate the magnetic moment of the magnet.(b) If the bar were free to rotate, which orienta-

tions would correspond to its(i) stable, and (ii) unstable equilibrium?What is its potential energy in the field forcases (i) and (ii) ?[Ans. (a) 0.40 JT -I, (b) (i) - 0.064 J (ii) + 0.064 JJ

10. Calculate the work done in rotating a magnet ofmagnetic moment 3.0 J T -1 through an angle of 60°from its position along a magnetic field of strength0.34 x 10-4T. (Ans. 5.1 x 10-5 J)

11. A bar magnet of magnetic moment 2.5 Am 2 is freeto rotate about a vertical axis through its centre. Themagnet is released from rest from the east-westdirection. Find the kinetic energy of the magnet as italigns itself in the north-south direction. Thehorizontal component of earth's magnetic field is0.3 G. ( !1S. 75 JlJ)

HINTS

1. Here m=0.9JrI, 9=30°, ,=0.063J

(i) B= -'- = 0.063 = 0.14 Tm sin 9 0.9 x sin 30°

(ii) When 9 = 0°, U = -mB cosO° = -mB

5.17

The P.E. of the magnet is minimum. Hence the barmagnet will be in stable equilibrium when n; is

->parallel to B .

2 Here N = 300, I = 15 A, r = 7 em = 7 x 10-2 m:. m = NIA = NI x 1tr2 = 300 x 15 x 3.14 x (7 x 10-2)2

= 69.3 JT -1.

5. Use 111= LA = I x 1tr2.

6. r > I11Bsin 9 = NIA Bsin 9= 1000 x 2 x 2 x 10-4 x0.32 xsin 30°

= 0.064 Nm.

7. (i) m = NIA = NI x 1tr2 = 20 x 3 x 3.14 x (0.04)2= 0.3 Am2

(ii) In equilibrium position,

U = - mB = - 0.3 x 0.5 = - 0.15 J., Q06 28. 111= --- = = 0.4 Am .

Bsin 9 0.3 sin 30°

9. (a) Here 9 = 30°, B= 0.16 T, r = 0.032 J

Magnetic moment,

111=_'_ = 0.032 =0.40 JT-1.Bsin e 0.16 x sin 30°

(b) Potential energy of the dipole in a magnetic field->B is given by

-> ->U = - 111. B = - mB sin 9

(i) The bar will be in stable equilibrium when its-> ->

magnetic moment m is parallel to B (9 = 0°).Its potential energy is then minimum and isgiven by

Umin = - I11Bcos 0° = - mB

= - 0.40 x 0.16 = - 0.064 J.(ii) The bar will be in unstable equilibrium when

n; is antiparallel to B (9 = 180°). Its potentialenergy is then maximum and is given byUmax = - mB cos 180° = + mB = + 0.064 J

10 W=-I11B(cos92-cos~)

= - 3.0 x 0.34 x 10- 4 (cos 60° - cos 0°)

= 5.1 x10-s J.11. Here ~ = 90°, 92 = 0°, 111= 2.5 Am 2,

B= 0.3 G = 0.3 x 10-4 T

Kinetic energy = Loss in P.E.

= Ui - Uf = - mBcos 90° + mBcos 0°

= 2.5 x 0.3 x 10- 4 J = 75 JlJ.

5.18

5.14 BAR MAGNET AS AN EQUIVALENTSOLENOID

16. State some similarities between a current carryingsolenoid and a bar magnet.

Similarities between a current carrying solenoidand a bar magnet. When a current is passed through asolenoid, it behaves like a bar magnet. Some observa-tions of similar behaviour are as follows:

1. A current carrying solenoid suspended freelyalways comes to rest in north-south direction.

2. Two current-carrying solenoids exhibit mutualattraction and repulsion when brought closerto one another. This shows that their endfaces act as N-and S-poles like that of a barmagnet.

3. Figure 5.24 shows the lines of force of a barmagnet while Fig. 5.25 shows the lines of forceof a finite solenoid. The two patterns have astriking resemblance.

Fig. 5.24 Field lines of a bar magnet.

If we move a small" compass needle in theneighbourhood of the bar magnet and thecurrent carrying finite solenoid, we shall findthat deflections of the needle are similar in thetwo cases. This again supports the similaritybetween the two fields.

+

Fig. 5.25 Field lines of a current carrying finite solenoid.

PHYSICS-XII

4. The magnetic fields of both the bar magnet andcurrent carrying solenoid at any far away axialpoint are given by the same expression:

B " = 110 2maxial 41t . r3

Thus a bar magnet and a solenoid producesimilar magnetic fields.

17. Explain how is a current carrying solenoidequivalent to a bar magnet.

A solenoid as an equivalent bar magnet. A solenoidcan be regarded as a combination of circular loopsplaced side by side, as shown in Fig. 5.26(a). Each turnof the solenoid can be regarded as a small magneticdipole of dipole moment fA. Then the solenoidbecomes an arrangement of small magnetic dipolesplaced in line with each other, as shown in Fig. 5.26(b).The number of such dipoles is equal to the number ofturns in the solenoid. The north pole of one touches thesouth of the adjacent one. The opposite polesneutralise each other except at the ends. Thus, acurrent carrying solenoid can be replaced by just asingle south pole and a single north pole, separated bya distance equal to the length of the solenoid. Hence acurrent carrying solenoid is equal to a bar magnet asshown in Fig. 5.26(c).

14---------L --------~

(a)

E:BE:BE:BE:BE:BEJs ns n5 ns ns "5 n

(b)

Is. .NI (c)14-- - - - - - - L - - - - - - --~

Fig. 5.26 A solenoid as an equivalent bar magnet.

A bar magnet and a finite solenoid produce similarmagnetic field patterns, as shown in Fig. 5.24 andFig. 5.25 respectively. It may be noted that themagnetic field inside the solenoid is in directionopposite to that we expect on the basis of the abovepole model (N ~ 5).

18. Explain how is a bar magnet equivalent to acurrent carrying solenoid.

A bar magnet as an equivalent solenoid. We canexplain this by Ampere's hypothesis according towhich all magnetic effects are produced by current-loops.The electrons in an atom keep on revolving around itsnucleus and hence set up electric currents. These atomic

MAGNETISM

currents are equivalent to small circular current-loops.In a magnet, these current-loops are arranged parallelto each other and have currents in the same sense.

Figure 5.27 shows the atomic current loops in across-section of a cylindrical bar magnet. At any pointinside the magnet, the currents from the adjacent loopscancel each other and hence the net current is zero. Butthere is a net current on the surface. Due to this surfacecurrent, the bar magnet is equivalent to a closely-wound, current carrying solenoid. Hence a bar magnetproduces a magnetic field similar to the solenoid.

Fig. S.27 A bar magnet as an equivalent solenoid.

It may be noted here that at the ends of the magnet,the current loops behave differently from those insidethe magnet. As a result, the magnetic poles are locatedslightly inside the bar magnet. That is why themagnetic length of a bar magnet is slightly less than itsgeometrical length.

5.15 GAUSS'S LAW IN MAGNETISM19. State Gauss's law in magnetism. What are its

important consequences ?Gauss's law in magnetism. Gauss's law in elec-

trostatics states that the surface integral of the-4

electrostatic field E over a closed surface 5 is equal to

1/ EO times the total charge q enclosed by the surface 5,i.e.,

f E.is =_5...s EO

Suppose that the closed surface 5 encloses anelectric dipole which consists of two equal and oppo-site charges. Then the total charge enclosed by 5 is zeroso that the surface integral of the electrostatic field of adipole over the closed surface is also zero, i.e.,

f ~ipo)e' dS = 05

Now a magnetic field is produced only by amagnetic dipole because isolated magnetic poles donot exist, so the above equation for a magnetic field canbe written as

f B.dS =05

5.19

This is Gauss's law in magnetism which states thatthe surface integral of a magnetic field over a closed surface isalways zero. But the surface integral of a magnetic fieldover a surface gives magnetic flux through that surface.So Gauss's law in magnetism can also be stated asfollows:

The net magnetic flux through a closed surface is zero.Consequences of Gauss's law:1. Gauss's law indicates that there are no sources or

sinks of magnetic field inside a closed surface. Sothere is no point at which the field lines start orthere is no point at which the field linesterminate. In other words, there are no freemagnetic charges. Hence isolated magnetic poles(also called monopoles) do not exist.

2. The magnetic poles always exist as unlike pairsof equal strengths.

3. If a number of magnetic lines of force enter aclosed surface, then an equal number of lines offorce must leave that surface.

Closedsurface

Fig. 5.28 Magnetic field lines never terminate.

\

For Your Knowledge

• Gauss's law of magnetism formally expresses the factthat magnetic monopoles do not exist. Hence the mostelementary magnetic element is a magnetic dipole ora current loop. All magnetic phenomena can beexplained in terms of an arrangement of magneticdipoles and/or current loops.

> Basic difference between electric and magnetic linesof force. An important consequence of the fact thatmagnetic monopoles do not exist is that magneticlines of force are continuous and form closed loops.They do not start or end at a point. In contrast, theelectric lines of force start from a positive charge andend on a negative charge or they fade out at infinity incase of isolated charges.

5.16 MAGNETIC FIELD OF THE EARTH20. Give some experimental evidences which support

the existence of earth's magnetic field.

5.20

Magnetic field of the earth. Earth is a powerfulnatural magnet. Its magnetic field is present every-where near the earth's surface. TIUsfield can be approxi-mated to the field of a magnetic dipole of dipolemoment 8.0 x 1022Am2 assumed to be located at thecentre of the earth. The axis of the dipole makes anangle of about 20° with the axis of rotation of the earth.The magnetic north pole Nm of the earth lies somewherenear the geographic south pole S while the magnetic southpole Sm lies somewhere near gthe geographic north poleNg' The magnitude of the magnetic field on the earth'ssurface is typically about lO-4T which is equal to1 gauss (G). A gauss is also often called an oersted. Thusthe earth's magnetic field is of the order of 1 oersted.

The branch of physics that deals with the study ofearth's magnetism is called terrestrial magnetismor geomagnetism.

Experimental evidences in support of earth'smagnetism:

1. A freely suspended magnetic needle comes to restroughly in north-south direction. TIUs suggeststhat the earth behaves as a huge magnet with itssouth pole lying somewhere near the geographicnorth pole and its north pole lying somewherenear the geographic south pole.

2. An iron bar buried in the earth becomes weak magnetafter some time. The magnetism is induced byearth's magnetic field.

3. Existence of neutral points near a bar magnet indi-cates the presence of earth's magnetic field. Atthese points, the magnetic field of the magnet iscancelled by the earth's magnetic field.

5.17 ORIGIN OF EARTH'S MAGNETIC FIELD21. Give a brief account of different theories regarding

the source of earth's magnetism.Origin of earth's magnetic field. The magnetic field

of the earth is approximately like that of a giant barmagnet embedded deep inside the earth. Many theorieshave been proposed about the cause of earth'smagnetism from time to time. Some of these arementioned below :

1. In 1600, William Gilbert in his book 'De Magnete'first suggested that the earth behaves as a bar magnetand its magnetism is due to the presence of magneticmaterial at its centre, which could be a permanentmagnet. However, the core of the earth is so hot that apermanent magnet cannot exist there.

2. Prof Blackett suggested that the earth'smagnetism is due to the rotation of the earth about itsown axis. Every substance is made of charged particles

PHYSICS-XII

such as protons and electrons. As these particles rotatealong with the earth, they cause circulating currentswhich, in turn, magnetise the earth.

3. Cosmic rays cause the ionisation of gases in theearth's atmosphere. As the earth rotates, strong electriccurrents are set up due to the movement of the chargedions. These currents may be the source of earth'smagnetism.

4. According to Sir E. Bullard (U.K.) and w.M.Elaster (U.S.A.), there are large deposits of ferro-magnetic materials like iron, nickel, etc. in the core ofthe earth. The core of the earth is very hot and molten.The circulating ions in the highly conducting liquid regionof the earth's coreform current loops and hence produce amagnetic field. At present, this hypothesis seems mostprobable because our moon, which has no molten core,has no magnetic field. Venus, which has a slower rateof rotation, has a weaker magnetic field while Jupiter,with a faster rate of rotation has a stronger magneticfield.

The changes in the earth's magnetic field are socomplicated and irregular that the exact cause ofearth's magnetism is yet to be known.

5.18 SOME DEFINITIONS IN CONNECTIONWITH EARTH'S MAGNETISM

22. Define the terms geographic axis, magnetic axis,magnetic equator, magnetic meridian and geographicmeridian in connection with geomagnetism.

S( me definitions in connection with earth'smagnetism, Fig. 5.29 shows the magnetic lines of forcearound the earth.

Magneticequator

Geographic G hinorth Olel . eograp ~c.: ::~tt."on

/ Earth's~magnetic

.Yt:::::---\-_ south pole

Earth'smagnetic --

north pole / / / / /7Geographicsouth pole

Fig. 5.29 Magnetic field of the earth.

1. Geographic axis. The straight line passing throughthe geographical north and south poles of the earth is calledits geographic axis. It is the axis of rotation of the earth.

MAGNETISM

2. Magnetic axis. The straight line passing through themagnetic north and south poles of the earth is called itsmagnetic axis.

The magnetic axis of the earth makes an angle ofnearly 20° with the geographic axis. At present, themagnetic south pole Sm is located at a point inNorthern Canada at a latitude of 70.5°N and alongitude of 96°W. The magnetic north pole NIII islocated diametrically opposite to Sm i.e., at a latitude of70.5°5 and a longitude of84°E. The magnetic poles arenearly 2000 km away from the geographic poles. Themagnetic equator intersects the geographic equator atlongitudes of 6°W and 174°E.

3. Magnetic equator. It is the great circle on the earthperpendicular to the magnetic axis.

4. Magnetic meridian. The vertical plane passingthrough the magnetic axis of a freely suspended smallmagnet is called magnetic meridian. 171eearth's magneticfield acts in the direction of the magnetic meridian.

5. Geographic meridian. The vertical plane passingthrough the geographic north and south poles is calledgeographic meridian.

5.19 ELEMENTS OF EARTH'SMAGNETIC FIELD

23. What are the elements of earth's magnetic field?Explain their meanings. Show these elements in a labelleddiagram and deduce various relations between them.

Elements of earth's magnetic field. The earth'smagnetic field at a place can be completely described by threeparameters which are called elements of earth's magneticfield. They are declination, dip and horizontal componentof earth's magnetic field.

1. Magnetic declination. 171e angle between thegeographical meridian and the magnetic meridian at a placeis called the magnetic declination (a) at that place.

Magnetic declination arises because the magneticaxis of the earth does not coincide with its geographic axis.

To determine magnetic declination at a place, setup a compass needle that is free to rotate in ahorizontal plane about a vertical axis, as shown inFig. 5.30. The angle a that this needle makes with thegeographic north-south (N g - Sg) direction is themagnetic declination. By knowing declination, we candetermine the vertical plane in which the earth'smagnetic field lies. In India, the value of a is small. It is0°41' E for Delhi and 0°58' W for Mumbai. This meansthat the N-pole of a compass needle almost points inthe direction of geographic north.

5.21

DeclinationNg

Fig. 5.30 Determination of declination at a place.

2. Angle of dip or magnetic inclination. The angle~

made by the earth's total magnetic field B with the horizontaldirection in the magnetic meridian is called angle of dip (8) atany place.

Vertical-+

B Sill

Fig. 5.31 Determination of dip at a place.

The angle of dip is different at different places onthe surface of the earth. Consider a dip needle, which isjust another compass needle but pivoted horizontallyso that it is free to rotate in a vertical plane coincidingwith the magnetic meridian. It orients itself so that itsN-pole finally points exactly in the direction of the

~earth's total magnetic field B. The angle between the

horizontal and the final direction of the dip needlegives the angle of dip at the given location.

At the magnetic equator, the dip needle restshorizontally so that the angle of dip is zero at the magneticequator. The dip needle rests vertically at the magneticpoles so that the angle of dip is90° at the magnetic poles. Atall other places, the dip angle lies between 0° and 90°.

3. Horizontal component of earth's magneticfield. It is the component of the earth's total magnetic

~field B in the horizontal direction in the magnetic

5.22

meridian. If 8 is the angle of dip at any place, then the~

horizontal component of earth's field B at that place isgiven by

BH= Bcos 8

At the magnetic equator, 8 = 0°, BH = B cos 0° = BAt the magnetic poles, 8 = 90°, BH= B cas 90° = 0

Thus the value of BHis different at different places onthe surface of the earth.

Geographicmeridian

Magneticmeridian

Fig. 5.32 Elementsof earth's magnetic field.

Relations between elements of earth's magneticfield. Fig. 5.32 shows the three elements of earth'smagnetic field. If 8 is the angle of dip at any place, thenthe horizontal and vertical components of earth's

~magnetic field B at that place will be

andBH·= Bcos 8Bv = Bsin 8

Bv = Bsin 8BH Bcos 8

Bv = tan 8~

or

AlsoBH

2 + Bv2 = W( cos2 8+ sin2 8)= W

or B=~BH2+Bv2 ...(3)

Equations (1), (2) and (3) are the different relationsbetween the elements of earth's magnetic field. Byknowing the three elements, we can determine themagnitude and direction of the earth's magnetic fieldat any place.

PHYSICS-XII

For Your Knowledge

~ Magnetic maps. These are the detailed charts whichindicate on the world map the lines passing through allsuch places where one of the three magnetic elements hasthe same value. Three types of lines are drawn on suchmaps. These are

1. Isogonic lines. The lines joining the places of equaldeclination are called isogonic lines. The line ofzero declination is called agonic line.

2. Isoclinical lines. The lines joining. the places ofequal dip or inclination are called isoclinical lines.The line of zero dip is called aclinic line ormagnetic equator. The points of 90° dip are calledmagnetic poles. The magnetic equator crossesthe geographic equator twice once in Atlanticand then in Pacific ocean.

3. Isodynamic lines. The lines joining the placeshaving the same value of the horizontal componentof earth's magnetic field are called isodynamic lines.The horizontal component is zero at poles andmaximum at the magnetic equator.

5.20 GLOBAL VARIATIONS IN THE EARTH'SMAGNETIC FIELD· .

...(1)

24. Describe the variations of earth's magnetic fieldfrom place to place.

Global variations in the earth's magnetic field.Earth's magnetic field changes both in magnitude anddirection from place to place. Some of the noticeableglobal variations are as follows:

1. The magnitude of the magnetic field on earth'ssurface is small, nearly 4 x 10-5 T.

2. Still smaller is the background field of our owngalaxy, the Milky Way, being about 2 pT i.e.,2 x 10-12 T.

3. If we assume that the earth's field is due todipole of 8.0 x 1022 Am2 located at its centre,then the earth's magnetic field will be less than1 u'T (l0-6T) at a distance of 5 times the radius ofthe earth i.e., at about 32,000 km. Upto thisdistance, the magnetic field is entirely governedby the earth.

4. At distances greater than 32,000 km, the patternof the earth's magnetic field gets severelydistorted by the solar wind.

5. Solar wind causes ionisation of atmospherenear the magnetic poles of the earth. This inturn causes beautiful displays of colours highup in the sky and is known as aurora.

...(2)

MAGNETISM

25. What is solar wind? How does it affect earth'smagnetic field?

Solar wind. The solar wind is a stream of hot chargedions, composed of equal numbers of protons and electronscontinuously flowing radially outward from the SUIl with aspeed of approximately 400 kmls. A long magneto tailstretches out for several thousand earth diameters in adirection away from the sun.

At distances greater than 32,000 km, the dipolefield pattern of the earth's magnetic field gets severelydistorted by the solar wind, as shown in Fig. 5.33.

Solarwind

Fig. 5.33 Distortion of earth's magnetic field bythe solar wind.

26. What are aurora borealis and aurora australis ?Can this effect be seen anywhere in India ?

Aurora borealis and aurora australis. This is aspectacular display of light seen in the night sky athigh altitudes, occurring most frequently near theearth's magnetic poles. The displays of aurora appearas gaint curtains high up in the atmosphere. Theaurora is caused when the charged particles of thesolar wind get attracted by the magnetic poles of theearth and there they ionise the atmospheric atoms ormolecules. The aurora in the northern hemisphere iscalled aurora borealis or northern lights and theaurora in southern hemisphere is called auroraaustralis or southern lights.

5.21 TEMPORAL VARIATIONS IN THEEARTH'S MAGNETIC FIELD'

27. Describe the variations of earth's magnetic fieldthat have occurred with the passage of time.

Temporal variations in the earth's magnetic field.The earth's magnetic field changes both in magnitudeand direction as time passes. These changes are of twotypes:

(i) Short term changes. The position of the magneticpoles of the earth keeps shifting slowly with thepassage of time. In a period of 240 years, from 1580 to1820, the magnetic declination at London has changedby 35°. The magnetic south pole in the northern Arcticregion of Canada has been found to shift in thenorth-west direction at the rate of 10 km per year inrecent times.

5.23

(ii) Long term changes. The changes in earth'smagnetic field over long term or geological time scalesare interesting. The studies of basalt reveal that earth'smagnetic field reverses its direction every millionyears or so. This means that once in a million years orso, the currents in earth's core cool down, come to ahalt and then pick up speed in the opposite direction.

Basalt which contains iron, is emitted duringvolcanic activity on the ocean floor. As it cools, itsolidifies and provides a picture of earth's magneticfield. Its age can be determined by other means.

5.22 NEUTRAL POINT28. Define neutral point. How will you find the

magnetic moment of a bar magnet by locating its neutralpoints, when the magnet is placed with its north poletowards (i) north pole of the earth and (ii) south pole ofthe earth?

Neutral point. It is the point where the magnetic fielddue to a magnet is equal and opposite to the horizontalcomponent of earth's magnetic field. The resultant magneticfield at the neutral point is zero. If a compass needle isplaced at such a point, it can stay in any position.

(i) Magnet placed in the magnetic meridian with itsnorth pole pointing north. Fig. 5.34 shows themagnetic lines of force of a bar magnet placed in themagnetic meridian with its north-pole pointingtowards the geographic north of the earth. The fieldsdue to the magnet and the earth are in same directionsat points on the axial line and are in opposite directionsat points on the equatorial line. So the resultant field isstronger at axial points and weaker at equatorialpoints. The two neutral points P and Q lie on theequatorial line.

Let

South

Fig. 5.34 Field lines of a bar magnet with itsN-pole towards north.

5.24

Then megnetic field strength at each neutralpoint is

B = !-Io mequa 41t . (r2 + [2l/2

For a short magnet, [ «r, therefore,

_!-Io mBequa - 41t .?

At the neutral point, the field of the magnet isbalanced by the horizontal component BH of the earth'smagnetic field so that

BrI = !-I0 .!!.:41t'r3

Knowing rand BrI, the value of the magnetic dipolemoment m can be determined.

(ii) Magnet placed in the magnetic meridian withits south-pole pointing north. Fig. 5.35 shows themagnetic lines of force of a bar magnet placed in themagnetic meridian with its south-pole pointingtowards the geographic north of the earth. Here thefields due to the magnet and the earth are in the samedirection at points on the equatorial line and are inopposite directions at points on the axial line of themagnet. So the resultant field is weaker at axial pointsand is stronger at equatorial points. In this case the twoneutral points P and Q lie on the axial line near theends of the magnet.

NW+EEarth's Sfield

South

Fig. 5.35 Field lines of a bar magnet withS-pole towards north.

Suppose r be the distance of each neutral pointfrom the centre of the magnet. Let 21 be the length ofthe magnet. Then magnitude of the magnetic field ateither of the neutral points will be

B.=!-Io 2mraxial 41t . (,1 _ z2)2

PHYSICS-XII

For a short magnet, 1 «« r, thereforeB. =!-Io 2maxial 41t . r 3

Again, at the neutral point, the field of the magnetis balanced by the horizontal component BrI of theearth's magnetic field, so we have

B =!-Io 2mH 41t' r 3

Knowing the values of r and BrI, the magneticdipole moment m of the magnet can be determined.

Fo nu e Used

1. Declination (a) = Angle between geographicmeridian and magnetic meridian.

2. Relations between elements of earth's magneticfield are

BH = Beas 8 and

~= tan 8 andBH

3. For a magnet placed with its N-pole pointingnorth, neutral points lie at its equatorial line.

Bv = Bsin 8

B= ~ B~ + l\i

B = !-I0 mH 471:' (r2 + [2)3/2

-!-Io m- 471: . r3 [for i.l hort magnet]

4. For a magnet placed with its N-pole pointingsouth, neutral points lie on its axialline.

B = !-I0 2mrH 471:' (r2 _[2)2

Ilo 2m- - - [for a short magnet]- 471: . r3

Urn s Used

Magnetic fields B, BH and Bv are in tesla,distances r and [ in metre, magnetic moment m inJT-1 or Am 2, angle a and 8 are in degrees.

Co so sed

Ilo = 471: x 1O-7TmA -1.

Example 24 The declination at a place is 15°west of north.In which direction should a ship be steered so that it reaches aplace due east?

Solution. As the ship is to reach a place due easti.e., along OP (Fig. 5.36), so it should be steered at angleof 15° + 90° = 105° with the direction of the compassneedle.

MAGNETISM

MN GN

_-ME

PGW--------.-~~----_.---GE

GS MS

Fig. 5.36

Example 25. A ship is to reach a place 10° south of west. Inwhich direction should it be steered if the declination at theplace is 18° west of north?

Solution. As the ship is to reach a place 10° south ofwest i.e., along OP (Fig. 5.37), so it should be steeredwest of magnetic north at angle of 90 -18 + 10 = 82 0.

MN GN

PMW---

_-ME

GW-------c~~----------GE

GS MS

Fig. 5.37

Example 26. In the magnetic meridian of a certain place,the horizontal component of the earth's magnetic field is 0.26 Gand the dip angle is 60°. What is the magnetic field of theearth in this location ? [NCERT]

Solution. Here BH =0.26 G, 8=60°

As ~=Bcas8

B= ~ = 0.26 G = 0.26 G =0.52 G.cos 8 cos 60° 0.5

Example 27. A compass needle whose magnetic moment is60 An? pointing geographical north at a certain place wherethe horizontal component of earth's magnetic field is40 IlWb / ni experiences a torque of1.2 x 10-3 Nm. What isthe declination of the place? [Roorkee 82]

Solution. In stable equilibrium, a compass needlepoints along magnetic north and experiences no torque.

5.25

When it is turned through declination a, it points alonggeographic north and experiences torque,

't = mBsin a. r 1.2 x 10-3 1

sm a = - = -------,-mB 60 x 40 x 10-6 2

a = 30°.or

Example 28. The horizontal and vertical components ofearth's field at a place are 0.22 gauss and 0.38 gauss respec-tively. Calculate the angle of dip and resultant intensity ofearth's field. [Haryana 93; Himachal 94]

Solution. Here BH = 0.22 G, By = 0.38 G

Now tan 8= ~ = 0.38 =1.7272BH 0.22

:. Angle of dip, 8 = 59°56'Resultant magnetic field of the earth is

B= ~ BH2 + By2 = ~0.22 2 + 0.382 = 0.427 G.

Example 29. If the horizontal component of earth's magneticfield at a place where the angle of dip is 60° is 0.4 X 1O-4T,calculate the vertical component and the resultant magneticfield at that place. [CBSE 00 97C ; Haryana 01]

Solution. Here 8 = 60°, BH = 0.4 x 10-4 T

By = BH tan 8 = 0.4 x 10-4 tan 60°

= 0.4 x 10-4 x .J3 = 0.69 x 10-4 T.

Resultant magnetic field,B 04 10-4B=_H_= . x =0.8x 10-4 T.

cos 8 cas 60°

Example 30. If 81 and 82 be the angles of dip observed intwo vertical planes at right angles to each other and 8 is thetrue angle of dip, prove that cot2 81 + cot2 82 = cot2 8.

[Punjab 96]

Solution. Let ~ and By be the horizontal and~

vertical components of earth's magnetic field B . Since 8is the true angle of dip, therefore

tan 8= ByBH

Bor cot 8 = -1i. ...(1)By

As shown in Fig. 5.38, suppose planes 1 and 2 are twomutually perpendicular planes and respectively makeangles 8 and 90° - 8 with the magnetic meridian. Thevertical components of earth's magnetic field remainsame in the two planes but the effective horizontalcomponents in the two planes will be

Br = BH cos 8 and ~ = BH sin 8

5.26

Plane 2Magneticmeridian

"--'----- ..•• '----- Plane 1

Fig. 5.38

The angles of dip 01 and 02 in the two planes aregiven by

or

tano=Bv= Bv11\ BHcos8

BH cas 8cot °1 = --'-'--Bv--

tan s =~ = Bv2 ~ ~sin8

B sin 8cot 02 =----'-'H __Bv

or

From equations (2) and (3), we have2 2

cot2 ° + cot2 ° = BH (cos2 8 + sin2 8) = BH1 2 Hi Hi

or cot2 01 + cot2 02 = cot2 0. [Using equation (1)]

Example 31. A dip circle shows an apparent dip of 60°at aplace where the true dip is 45°. If the dip circle is rotatedthrough 90°, what apparent dip will it show?

Solution. Here 01 =60° and 0=45°.

As cot2 s = cot2 01 + cot2 02

cot2 45°= cot2 60° + cot2 02

cot2 02 = (1)2 -(.1r = ~

cot 02 = 0.81602 = 51°.

Example 32. True value of dip at a place is 45°. The plane orof the dip circle is turned through 60° from the magneticmeridian. Find the apparent value of dip.

Solution. Here 0 = 45°, 8 = 60°, 0' =?

tan 0'= Bv = BH tan °~ ~ cas 8

= tan ° = tan 45° =2cas 8 cas 60°

or

or

:. Apparent dip,

PHYSICS-XII

Example 33. A bar magnet of length 10 em is placed in themagnetic meridian with its north pole pointing towards thegeographical north. A neutral point is obtained at a distanceof 12 em from the centre of the magnet. Find the magneticmoment of the magnet, if ~ =0.34 G.

Solution. Here, 21 = 10 em, 1=5 em = 5 x 1O-2mr=12 em =12 x 10-2 m, ~ =0.34 G =0.34 x 10-4 T

In this case, the neutral points lie on the equatorialline of the magnet so that at any neutral point,

or ~o m = B411:. (r2 + 12)3/2 H~qua = ~

...(2)

:. Magnetic moment

m= ~. 411:.(1+ 12)3/2~o

= 0.34 x 10-4 x _1_ [52 + 122]3/2(10-4)3/210-7

= 0.747 rr'.Example 34. The magnetic moment of a short bar magnetis 1.6 A~. It is placed in the magnetic meridian with northpole pointing south. The neutral point is obtained at distanceof 20 em from the centre of the magnet. Find the horizontalcomponent of earth's magnetic field. If the magnet bereversed, i.e., north pole pointing north, find the position ofneutral point.

Solution. Here m=1.6 Am2, r=20 em =0.20 m

...(3)

When N-pole of the magnet points south,neutral points lie on the axial line of the magnet.

Hence at the neutral point,B - - ~o 2maxial - ~ - 411:· r3

~ = 10-7 x 2 x 1.6 = 4 x 10-5 T(0.20l

the

or

When the magnet is reversed, its north pole pointsnorth. The neutral points will lie on the equatorial lineof the magnet. Hence

_ _~o mBequa - ~ - -. 1411: r:

r3 = ~o. ~ = 10-7

x 1.6 =4.0 x 10 -3 m3411: BH 4 x 10-5

r = (4.0 x 10-3)1/3

= 1.6 x 10- 1 m = 16 em.

Example 35. A magnet placed in the magnetic meridianwith its north pole pointing north of the earth produces aneutral point at a distance of 0.15 m from either pole. It isthen broken into two equal pieces and one such piece isplaced in a similar position. Find the position of the neutralpoint.

MAGNETISM

Solution. Here the neutral points lie on the equatorialline of the magnet at distance x from each of the twopoles.

_ _ 110 mBH - Bequa - 411:. x3

When the magnet is broken into two parts, its polestrength remains unchanged.

Original magnetic moment,

m= qlll x 21Magnetic moment of each part,

21 mm = «; X 2 =2

B =110 m/2H 411:' x,3

110 m _ 110 m411:. x3 - 411:. 2x,3

2x,3= x3

Hence

or, x

x = 21/3

0.15 f h Ix' = -- = 0.119 m, rom eac po e.1.26

Example 36. The magnetic field at a point on the magneticequator is3.1 x 10-5T. Taking the radius of the earth equal to6400 km, find the magnetic moment of the assumed dipole at~~~~~~ ~

Solution. Any point on themagnetic equator lies in thebroad side on position of theassumed magnetic dipole.Hence

or

or

or

B - 110 !!!...-equa - 411:. R3

411: 3 Fig. 5.39m= -. Bequa R

~lO

= 107 x3.1x 10-5 x (6400 x 103)3

= 8.1 x 1022 Am 2.

GS

Example 37. The earth's magnetic field at the equator isapproximately 0.4 G. Estimate the earth's dipole moment.Radius of the earth = 6400 km. [NCERT]

Solution. Here ~ = B = 110~equa 411:r

0.4 x 10-4 = 10-7 x m(6.4 x 106)3

m=1.04x 1023 Am2.

or

or

Example 38. A short bar magnet is placed in a horizontalplane with its axis in the magnetic meridian. Null points arefound on its equatorial line (i.e., its normal bisector)at 12.5em

5.27

from the centre of the magnet. The earth's magnetic field atthe place is 0.38 G and the angle of dip is zero.

(i) What is the total magnetic field at points on the axisof the magnet located at the same distance 02.5 em)as the null-points from the centre?

(ii) Locate the null points when the magnet is turnedaround by 180°.Assume that the length of the magnet is negligible ascompared to the distance of the nul/-points from thecentre of the magnet.

Solution. (a) At the neutral point on the equatorialline of a short magnet, we have

= 110 .!!!. = BBequa 411:. r3 H

Magnetic field of the magnet on its axial line at thesame distance will be

110 2mBaxial = -. 3 =2 BH =2 x 0.38 =0.76 G

411: rAt any point on the axial line, BH and Baxial are in

the same direction. So total magnetic field,

B = Baxial + BH = 0.76 + 0.38 = 1.14 G.

(b) When the magnet is turned through 180°, theneutral points lie on the axial line.

B. =110 2m=Baxial 411:. x3 H

But ~ = 110 m 110 m _110 2m411:. r3 . . 411:'r3 - 411:. ~

or x3 = 2r3

or x = (2 )1/3 r = 1.26 x 12.5 em = 15.75 cm.

cproblems For Practice

1. A ship is sailing due west according to Mariner'scompass. If the declination of the place is 15° east ofnorth, what is true direction of the ship?

(Ans. 75° west of north)2. A ship is sailing due east according to Mariner's

compass. If the declination of the place is 18°east ofnorth, what is the true direction of the ship?

(Ans. 18° south of east)3. The horizontal component of earth's magnetic field

is 0.2 G and total magnetic field is 0.4 G. Find angleof dip. [Haryana 96] (Ans.600)

4. Calculate earth's magnetic field at a place, wherethe angle of dip is 60° and vertical component ofearth's field is 0.40 G. (Ans. 0.462 G)

5. A magnetic needle free to rotate in a vertical planeparallel to the magnetic meridian has its north tipdown 60° with the horizontal. The horizontal

5.28

component of the earth's magnetic field at the placeknown to be 0.4 G. Determine the magnitude of theearth's magnetic field at the place. [CBSE F 111

(Ans. 0.8 G)6. The vertical and horizontal components of earth's

magnetic field at a place are 0.2 G and 0.3464 Grespectively. Calculate the angle of dip and earth'smagnetic field at that place. (Ans. 30°, 0.4 G)

7. A vertical wire in which current is flowing pro-duces a neutral point with the earth's horizontalfield at a distance of 5 em from the wire in air. Whatis current, if BH= 0.18 x 10-4 T ? (Ans. 4.5 A)

8. A compass needle whose magnetic moment is 60 Am2pointing geographical north at a certain place, wherethe horizontal component of earth's magnetic field is40IlT, experiences a torque of 1.2 x 10-3 Nm. Whatis the declination at that place? (Ans. 30°)

9. A magnetic needle free to rotate about the verticaldirection (compass) points 3.5° west of the geo-graphic north. Another magnetic needle free torotate in a vertical plane parallel to the magneticmeridian has its north tip pointing down at 18°withthe horizontal. The magnitude of the horizontalcomponent of the earth's magnetic field at the placeis known to be 0.40 G. What is the direction andmagnitude of the earth's magnetic field at the place?

(Ans. 0.42 G, directed at an angle of 18° with horizontalin the magnetic meridian towards the ground)

10. The true dip at a place is 30°. What is the apparentdip when the dip circle is turned 60° out of themagnetic meridian? (Ans. 49°71

11. The values of the apparent angles of dip in two planesat right angles to each other are 30° and 45°.Calculate the true value of the angle of dip at theplace. (Ans. 26.6°)

12. A dip circle lying initially in the magnetic meridianis rotated through an angle e in the horizontalplane. Show that the tangent of the angle of dip isincreased in the ratio sec e: 1

13. A short bar magnet of magnetic moment 0.5 J T-1 isplaced with its magnetic axis in the magnetic meri-dian, with its north pole pointing geographical north.A neutral point is obtained at a distance of 0.1 m fromthe centre of the magnet. Find the horizontal comfo-nent of the earth's magnetic field. (Ans. 10- T)

14. A bar magnet 30 cm long is placed in the magneticmeridian with its north pole pointing geographicalsouth. The neutral point is found at a distance of30 cm from its centre. Calculate the pole strength ofthe magnet. Given BH = 0.34 G. (Ans. 8.61 Am)

15. A neutral point is found on the axis of a bar magnetat a distance of 10 cm from its one end. If the lengthof the magnet be 10 cm and BH = 0.3 G, find themagnetic moment of the magnet. (Ans. 0.012 Am2)

PHYSICS-XII

16. A magnet placed in the north pointing northposition, balances the earth's magnetic field at apoint, which is 27 cm from either pole. If it is brokeninto three pieces and one such piece is similarlyplaced, find the position of the neutral point.

(Ans. 18.73 x 10-2 m, from either pole)

HINTS

4... B=~=~= 0.40 = 0.462 G.sin 0 sin 60° 0.866

5. Here, 0 = 60°, BH = 0.4 G, B= ?B= BH =~= 0.4 =0.8 G.

cos 0 cos 60° 0.511 I7. As-o- = BH2nr

1- 2r.:r BH _ 2r.:x5xlO-2 xO.18xl0-4 _

.. ---- 7 -4.5A.110 4r.:x 10-

8. 1" = 111 BHsin a.. 1.2 x 10-3 = 60 x40 x 10-6 xsin aor sin a = -1 . . a = 30° .

9. Here 3.5° is the magnetic declination and 18° is theangle of dip.As BH=Bcoso

B= BHsee 0 = 0.40 sec 18°= OAO x 1.0514 = 0.42 G.

11. cot2 0 = cot2 ~ + cot2 O2= cot2 30° + cot2 45° = 3 + 1= 4

cot 0 = 2 :. 8 = 26.6° .

12. The true angle of dip 0 is given by tan 0 = BvBH

When the dip circle is rotated through angle S, theapparent angle of dip 0' is given by

tan 0':. -- = sec O:1.

tan 0tan 0/ = Bv = tan o. see e

BHcos 0~o m13. Use BH=-. 3"'4r.: r

14. Here the neutral points lie on the axial line.110 2 mr

.. Baxia1 = BH or 41t . (? _12)2 = BH

41t B (r2 _12)2or m = _ . -,-,H,-,-_~_~o 2r

1 0.34 x 10-4 (0.302 - 0.152)2 2= --7 . = 2.582 Am

10- 2 x 0.30m 2582Pole strength, qm = - = -- = 8.61 Am.21 0.30

~o 2m .15. Use BH =-. -3 .Here r= 10+ 5= 15cm.41t r

16. Proceed as in Example 35 on page 5.26.

MAGNETISM

5.23 SOME IMPORTANT TERMS USED TODESCRIBE MAGNETIC PROPERTIESOF MATERIALS

29. Define the terms magnetising field, magneticinduction, intensiiu of magnetisation, magnetising fieldintensity, magnetic permeability, relative permeabilityand magnetic susceptibilitu. Write the various relationsamong these quantities.

1. Magnetising field. When a magnetic material isplaced in a magnetic field, a magnetism is induced init. The magnetic field that exists in vacuum and inducesmagnetism is called magnetising field. For example,consider a toroidal solenoid carrying current I andplaced in vacuum. If the solenoid has n turns per unitlength, then the magnetic field set up in the solenoid isgiven by

Bo=llonIThis field is called the magnetising field caused by

the so called free current in the solenoid.2. Magnetic induction. As shown in Fig. 5.40,

suppose the toroidal solenoid is wound round a ring of-+

magnetic material. Under the influence of field fb, the

magnetic moments of the atomic current loops of themagnetic material tend to align themselves with oragainst the magnetising field %. This gives rise to a netcurrent on the surface of the material and is calledmagnetisation surface current 1M as shown in Fig. 5.41.

Spinningcurrents

Fig. 5.40 Magnetic field in amagnetic material.

Fig. 5.41 Magnetisingsurface current.

-+This current induces magnetic field BM inside the

material which is given by

BM = 110 n 1MThe total magnetic field inside a magnetic material is the

sum of the external magnetising field and the additionalmagnetic field produced due to magnetisation of the material

-+and is calledmagnetic induction B. The magnetic inductionmay also be defined as the total number of magnetic lines offorce crossing per unit area normally through a material. orThus the 51 unit of magnetic induction is tesla (T) orweber metre=? (Wbm-2) which is equivalent toNm-1A-1 or JA-1m-2.

5.29

3. Magnetising field intensity. The ability ofmagnetising field to magnetise a material medium is

-+expressed by a vector H , called magnetising field inten-sity or magnetic intensity. Its magnitude may be defined asthe number of ampere-turns (nI ) flowing round the unitlength of the solenoid required to produce the givenmagnetising field. Thus

H = nI

%=llonI=lloH or H=%110

The dimensions of magnetic intensity are [L-1A]. Its51 unit is ampere metre-1 (Am -1) which is equivalent toNm-2T-1 or Jm-1Wb-1.

4. Intensity of magnetisation. When a magneticmaterial is placed in a magnetising field, it gets magne-tised. The magnetic moment developed per unit volume of amaterial when placed in a magnetising field is called intensityof magnetisaiion or simply magnetisation. Thus

-+-+ mM=-

VIf 1M is the surface magnetisation current set up in a

solenoid of cross-sectional area A and having n turnsper unit length, then magnetic moment developed perunit length of the solenoid is nIM A. Therefore,magnetic moment developed per unit volume or the

-+magnetisation M is given by

m nI AM---~M~-nI-V- A - M

Hence BM = 110 n 1M = 110 MAgain, consider a bar of magnetic material having

cross-sectional area a and length 21. Its volume isV = a x 21

Suppose the bar develops pole strength qrn whenplaced in a magnetising field, then its magnetic moment,

m=qrnx21

M = m = qrnx 21 =!1JJLVax 2l a

Hence intensity of magnetisation may also be defined as thepolestrength deoelopedper unit cross-sectionalareaoj a materia/.

As the total magnetic field or the magnetic induction-+B inside a magnetic material is the resultant of the

-+ -+magnetising field % and the field BM produced due tothe magnetisation of the material, therefore,

B= % + BM =lloH + l-loM

B=llo(H+M)Clearly, both Hand Mhave the same units, namely

Am-1.

5.30

5. Magnetic Permeability. Permeability is themeasure of the extent to which a material can be pene-trated or permeated by a magnetic field. The magneticpermeability of a material may be defined as the ratio of itsmagnetic induction B to the magnetic intensity H.

B1-1=-H

Clearly, 51 unit of 1-1tesla

ampere metre-1

= tesla metre ampere-lor TmA-1

Dimensions of 1-1= [MLr2A-2].6. Relative permeability. Permeability of various

magnetic substances can be compared with one anotherin terms of relative permeability I-Ir. It is defined as theratio of the permeability of the medium to the permeability offree space. Thus,

I-I=..e...r 1-10

For vacuum 1-1r = I, for air it is 1.0000004 and foriron, the value of 1-1r may exceed 1000.

7. Magnetic susceptibility. Magnetic susceptibilitymeasures the ability of a substance to take up magne-tisation when placed in a magnetic field. It is defined asthe ratio of the intensity of magnetisation M to themagnetisingfield intensity H.1t is denoted by Xm. Thus,

MX/II = H

As magnetic susceptibility is the ratio of twoquantities having the same units (Am -1), so it has nounits.

8. Relation between magnetic permeability andmagnetic susceptibility. If a linear magnetic material,subjected to the action of a magnetising field intensityH, develops magnetisation M and magnetic inductionB; then

ButB= l-Io(H + M)B=I-I H

1-1H =l-Io(H + M)

1-1=1-10(1+ ~) oror

or

5.24 CLASSIFICATION OFMAGNETIC MATERIALS

30. How are materials classified on the basis of theirbehaviour in a magnetic field ? Give examples of eachtype.

PHYSICS-XII

Classification of magnetic materials. On the basis oftheir behaviour in external magnetic fields, Faradayclassified the various substances into three categories:

1. Diamagnetic substances. Diamagnetic substancesare those which develop feeble magnetisation in the oppositedirection of the magnetising field. Such substances arefeeblyrepelled by magnets and tend to move from stronger toweaker parts of a magnetic field.

Examples. Bismuth, copper, lead, zinc, tin, gold,silicon, nitrogen (at 5TP), water, sodium chloride, etc.

2. Paramagnetic substances. Paramagnetic sub-stances are those which develop feeble magnetisation in thedirection of the magnetisingfield. Such substances arefeeblyattracted by magnets and tend to move from weaker tostronger parts of a magnetic field.

Examples. Manganese, aluminium, chromium,platinum, sodium, copper chloride, oxygen (at 5TP), etc.

3. Ferromagnetic substances. Ferromagnetic sub-stances are those which develop strong magnetisation in thedirection of the magnetising field. They are strongly attractedby magnets and tend to move from weaker to stronger partsof a magnetic field.

Examples. Iron, cobalt, nickel, gadolinium andalloys like alnico.

5.25 ORIGIN OF DIAMAGNETISM31. Explain the origin of diamagnetism. Why are the

diamagnetic substances repelled by magnets ?Origin of diamagnetism. In atoms of some

materials like Bi, Cu, Pb, the magnetic moments due todifferent electrons cancel out. In such atoms, electronsoccur in pairs with one of them revolving clockwise

111

v

111

(a) (b)

t 11l-/!.mB v+/!.v

(d)(c)

Fig. 5.42 An electron orbiting in an atomproduces a moment.

MAGNETISM

and other anticlockwise around the nucleus. Netmagnetic moment of an atom is zero, as shown inFig. 5.42(a) and (b).

~When such an atom is placed in a magnetic field B,

the speed of revolution of one electron increases andthat of other decreases. The magnetic moment of the

~ ~former electron increases to In + t'l m and that of the

~ ~latter electron decreases to m - t'l In. So each electron

~pair gains a net magnetic moment 2 t'l In which is

~proportional to the field B but points in its oppositedirection as shown in Figs. 5.42(c) and (d). A sufficientmagnetic moment is induced in the diamagnetic

~sample in the opposite direction of B. This sample

~moves from stronger to the weaker parts of the field B,i.e., a diamagnetic substance is repelled by a magnet.The behaviour of diamagnetic materials is

orindependent of temperature.

5.26 ORIGIN OF PARAMAGNETISM32. Explain the origin of paramagnetism. State

Curie's law of magnetism.Origin of paramagnetism. According to Langevin,

the atoms or molecules of a paramagnetic materialpossess a permanent magnetic moment either due to thepresence of some unpaired electron or due to thenon-cancellation of the spins of two electrons becauseof some special reason. In the absence of an externalmagnetic field, the atomic dipoles are randomlyoriented due to their ceaseless random motion, asshown in Fig. 5.43(a). There is no net magnetisation.

Bo=O Bo ~

-e- -e- -e--e--e- -e- -e--e--e- -e- -e--e-

(b)(a)

Fig. 5.43 (a) Randomly distributed atomic dipoles in a para-magnetic material in the absence of magnetic field.(b) Alignment of dipoles in the presence of magnetic field.

~When a strong enough field Ba is applied and the

~temperature is low enough, the field Ba tends to alignthe atomic dipoles in its own direction, producing a

~weak magnetic moment in the direction of Bo' Thematerial tends to move from a weak field region to astrong field region. This is paramagnetism.

5.31

At very high magnetic fields or at very low tempe-ratures, the magnetisation approaches its maximumvalue when all the atomic dipole moments get aligned.This is called the saturation magnetization value Ms.

Curie's law. From experiments, it is found that theintensity of magneiisaiion (M) of a paramagneticmaterial is

(i) directly propertional to the magneiising fieldintensity H, because the latter tends to align theatomic dipole moments.

(ii) inversely proportional to the absolute temperature T,because the latter tends to oppose the alignmentof the atomic dipole moments.

Therefore at low H / T values, we haveHMoc-T

M=C. HT

or

M C CH T or Xm =T

Here C is curie constant and Xm is the susceptibilityof the material. The above relation is called Curie's law.This law states that far away from saturation, the suscepti-bility of a paramagnetic material is inversely proportional tothe absolute temperature.

Figure 5.44 shows the variation of intensity ofmagnetisation M as a function of H / T. Beyond thesaturation value Ms' Curie law is not valid.

region

Curie law region

HIT~

Fig. 5.44 Magnetisation M as a function of HIT.

5.27 ORIGIN OF FERROMAGNETISMDOMAIN THEORY

33. Describe ferromagnetism on the basis of domaintheory. How does Curie's law get modified for ferro-magnetic substances ?

Origin of ferromagnetism. Weiss explainedferromagnetism on the basis of his domain theory. Inmaterials like Fe, Ni, Co, the individual atoms areassociated with large magnetic moments. The magneticmoments of neighbouring atoms interact with eachother and align themselves spontaneously in acommon direction over macroscopic regions called

5.32

domains. Each domain has a typical size of about 1 mmand contains about 1011 atoms. So each domainpossesses a strong magnetic moment. In the absence ofany external magnetic field, these domains are randomlydistributed so that the net magnetic moment is zero.

Domains

Fig. 45 Randomly oriented domains in a ferromagnetic substance.

When a ferromagnetic material is placed in amagnetic field, all the domains align themselves alongthe direction of the field leading to the strong magne-tisation of the material along the direction of the field.That is why the ferromagnetic substances are stronglyattracted by magnets. The alignment of domains mayoccur in either of the following two ways :

1. By displacement of the boundaries of domains.When the external field Bo is weak, the domains alignedin the direction of Ba grow in size while those oppositelydirected decrease in size, as shown in Fig. 5.46(b).

WeakBo~ Strong Bo~

(a) Unmagnetised (b) Magnetisation by (c) Magnetisation bysample growing of domains rotation of domains

Fig. 5.46 Magnetisation of a ferromagnetic sample.

2. By rotation of domains. When the external fieldBa is strong, the domains rotate till their magneticmoments get aligned in the direction of Ba as shown inFig. 5.46(c).

Modified Curie's law for ferromagnetic substances.When a ferromagnetic sample is heated, its magne-tisation decreases due to the increase in the randomi-sation of its domains. At a sufficiently high tempe-rature, the domain structure disintegrates and theferromagnetic substance becomes paramagnetic. Thetemperature at which a ferromagnetic substance becomesparamagnetic is calledCurie temperature or Curie point Te.

PHYSICS-XII

Above the curie point i.e., in the paramagneticphase, the susceptibility varies with temperature as

C'Xm = T _ I (T > Te)

e

where C' is a constant. This is modified Curie's law for aferromagnetic material above the Curie temperature. Itis also known as Curie-Weiss law. This law states thatthe susceptibilih) of a ferromagnetic substance above itsCurie temperature is inversely proportional to the excess oftemperature above the Curie temperature.

Table 5.2 Curie Temperatures of someFerromagnetic Materials

Material Tc (K)Cobalt 1394Iron 1043Fep3 893Nickel 631Gadolinium 317

5.28 PROPERTIES OF DIAMAGNETICSUBSTANCES

34. Describe some of the important properties ofdiamagnetic substances.

Properties of diamagnetic substances :

1. When placed in an external magnetic field, adiamagnetic substance develops feeble magnisation inthe opposite direction of the applied field.

2. When a rod of a diamagnetic material is placed ina magnetic field, poles are induced on it in a directionopposite to that of the inducing field. So the lines offorce prefer to pass through the surrounding air thanto pass through the material itself i.e., the lines offorce get expelled or repelled, as shown in Fig. 5.47.Consequently, the magnetic induction B inside thematerial becomes less than the magnetising field,Ba = J.1 0 H. The reduction is very small, about 1 partin 105.

Fig. 5.47 Reduction of lines of force in a diamagnetic rod.

MAGNETISM

3. When placed in a non-uniform magnetic field, adiamagnetic substance moves from stronger to the weakerparts of the field.

When a watch glass containing a diamagneticliquid is placed over two closely lying (3 - 4 mm apart)pole pieces of a magnet, the liquid is found to movetowards the poles causing a depression in the middle[Fig 5.4~(a)]. This indicates that the field is stronger inthe middle than that near the poles. Now if the polesare moved apart sufficiently, the magnetic field at themiddle becomes weaker than that near the poles.Consequently, the liquid accumulates in the middleand thins out near the poles [Fig. S.48(b)].r Diamagnetic liquid \

ntf~~Fig. 5.48 Effectof non-uniform magnetic field on a diamagneticliquid when (a) poles are quite close to each other, (b) poles aresufficiently apart.

4. When a rod of a diamagnetic material is sus-pended freely in a uniform magnetic field, it alignsitself perpendicular to the magnetising field (Fig. 5.49).

Fig. 5.49 A freely suspended diamagnetic rodin a uniform field.

5. As a diamagnetic substance dev ....lops a weakmagnetisation in the opposite direc .ion of themagnetising field, the susceptibility (Xnz = M/ H) ofdiamagnetic materials is small and negative. Forbismuth, Xm = - 0.00015.

6. The relative permeability Il r (= 1 + X", ) is positivebut less than 1 for a diamagnetic material.

7. The susceptibilityof diamagnetic sub- X'"stances is independentof the magnetising field 0 Tand the temperature, asshown in Fig. 5.50. Xdi3~-----

8. The magnetisationof a diamagnetic sub-stance lasts so long as Fig. 5.50 Xm-T graph for dia-the magnetising field is magnetic material.applied.

5.33

35. Briefly describe diamagnetism in superconductingmetals.

Diamagnetism in superconducting metals. When ametal is cooled to a temperature below its criticaltemperature in a magnetic field, it attains bothsuperconductivity and perfect diamagnetism. Themagnetic lines of force get completely expelled from itand it repels a magnet. For this material, X = -1 andIl r = O. This phenomenon of diamagnetism in super-conductors is calledMeissner effect. This effect forms thebasis for running magnetically levitated superfast trains.

5.29 PROPERTIES OF PARAMAGNETICSUBSTANCES

36. Describe some of the important properties oj.paramagnetic substances.

Properties of paramagnetic substances:1. When placed in an external magnetic field, a

paramagnetic substance develops feeble magne-tisation in the direction of the applied field.

2. When a rod of paramagnetic material is placedin a magnetic field, the lines of force prefer topass through it than through the surroundingair i.e., the lines of force get slightly moreconcentrated inside the material, as shown inFig. 5.51. The magnetic induction B becomesslightly greater than the magnetising field,Ba = lloH. The increase is very small, about1 part in 105.

Fig. 5.51 Slightly cocentrated lines of forcein a paramagnetic rod.

3. When placed in a non-uniform magnetic field, aparamagnetic substance moves from weaker to thestronger parts of the field.When a watch glass containing a paramagneticliquid is placed over two closely lying polepieces of a magnet, the liquid accumulates andelevates in the middle and thins out near thepoles [Fig. 5.52(a)]. This is because the field inthe centre is the strongest. When the poles aremoved apart, the field at the poles becomesstronger than that at the centre and the liquidmoves towards the poles [Fig. 5.52(b)].

5.34

Fig. 5.52 Effect of magnetic field on a paramagneticliquid when (a) poles are quite dose toeach other, (b) poles are farther apart.

4. When a rod of paramagnetic material is sus-pended freely in a uniform magnetic field, it alignsitself parallel to the magnetising field (Fig. 5.53).

Fig. 5.53 A freely suspended paramagnetic rod ina uniform magnetic field.

5. A paramagnetic material develops small magne-tisation in the direction of the magnetising field,so its susceptibility has small but positive value.For aluminium, X =1.8 x 10-6.

6. The relative permeability (ur = 1 + Xm) for aparamagnetic material has a value slightly greaterthan 1.

7. The magnetic susceptibility of a paramagneticmaterial varies inversely as the absolute tempe-rature, i.e.

1X ex:-

m T

or CXm =y'

where C is a constant called the Curie constantand this equation is known as Curie's law.

T~

Fig. 5.54 Xm -T graph for a paramagnetic material.

8. For a given temperature, the intensity of magne-tisation is proportional to the magnetising field,

PHYSICS-XII

so the susceptibility and permeability do not~

show any variation with the field Ea .

9. As soon as the magnetising field is removed, aparamagnetic substance loses its magnetism.

5.30 PROPERTIES OF FERROMAGNETICSUBSTANCES

37. Describe some of the important properties offerromagnetic substances.

Properties of ferromagnetic substances. Ferro-magnetic substances exhibit properties similar to thoseof paramagnetic substances but in a highly dominantmanner. These are as follows:

1. When placed in an external magnetic field, aferromagnetic material develops strong magne-tisation in the direction of the applied field.

2. When a ferromagnetic substance is placed in amagnetic field, the lines of force concentrategreatly into the material so that the magneticinduction B becomes much more than themagnetising field Ea.

Fig. 5.55 Highly concentrated lines of force in a ferromagnetic rod.

3. When a ferromagnetic substance is placed innon-uniform magnetic field, it moves fromweaker to the stronger parts of the field.

4. When a rod of a ferromagnetic material is sus-pended freely in a uniform magnetic field, itquickly aligns itself parallel to the magnetic field.

5. The intensity of magnetisation M is propor-tional to the magnetising field intensity H for itssmaller values. For moderate values of H, Mincreases rapidly and then finally attains constantvalue for large H. This indicates the attainmentof the saturation stage of magnetisation.

6. The susceptibility of a ferromagnetic materialhas a large positive value. This is because

MXIII = H

and M» H for a ferromagnetic material. It is ofthe order of several thousands.

7. The relative permeability (f..l r = 1 + XIII) of aferromagnetic material has a large positivevalue. It is of the order of several thousands. Foriron, f..lr = 1000.

MAGNETISM

8. The susceptibility of ferromagnetic materialdecreases with temperature in accordance withCurie-Weiss law:

C'XI1l =T-'[ (T>Tc)

c9. At a certain temperature called the Curie point,

the susceptibility suddenly falls and the ferro-magnetic substance becomes paramagnetic.

10. The magnetisation developed depends not onlyon the value of magnetising field but also on thepast magnetic and mechanical history of thematerial.

11. A ferromagnetic substance retains magnetismeven after the magnetising field is removed.

5.35

For Your Knowledge

~ In the presence of an external magnetic field, magneticmoments are induced in all materials. Hencediamagnetism is universal. But paramagnetism andferromagnetism are much stronger thandiamagnetism, so it is difficult to detect diamag-netism in para- and ferro-magnetic substances.

~ Magnetic materials are broadly classified asdiamagnetic, paramagnetic and ferromagnetic.However, there exist some other types of magneticmaterials with mysterious properties. These includeferrimagnetic, anti-ferromagnetic, spin glass, etc.

~ A very small variation in the value of XIII may lead to analtogether different magnetic behaviour: diamagneticvs. paramagnetic. For diamagnetic materials, XIII z: -10-5

where XIII = + 10-5 for paramagnetic materials.

Table 5.3 Comparative study of the properties of dic-, para- and ferromagnetic substances

.....1. Effect oj

ma nets2. In external

magnetic field

3. In a non-uniformmagnetic field

4. In a uniformmagnetic field

5. Susceptibilityvalue (XIII)

6. Relativepermeabilityvalue

7. Permeabilityvalue

8. Effect ojtemperature

I. .•. •. Paramagnetic substances

A freely suspended ferro-magnetic rod aligns itself

arallel to the field.

Tend to move slowly from Tend to move slowly fromstronger to weaker parts of weaker to stronger parts ofthe field. the field.

They are feebly repelled by They are feebly attracted by They are strongly attractedma ets. ma nets. b ma nets.

Ferromagnetic substances

A freely suspended diamag- A freely suspendednetic rod aligns itself paramagnetic rod aligns

er endicular to the field. itself arallel to the field.

Susceptibility is small and Susceptibility is small andnegative. - 1:::;Xm < 0 positive. 0 < Xm < s , where e

is a small number

Slightly less than 1O:::;l1r<l

Susceptibility is indepen-dent of temperature.

Magnetisation lasts aslong as the magnetisingfield is a lied.

M changes linearly with H.

Slightly greater than 1l<l1r<l+s

Susceptibility variesinversely as temferature :

Xm <x'y'

As soon as the magnetisingfield is removed, magneti-sation is lost.

M changes linearly with Hand attains saturation atlow temperature and inver stron fields.

Acquire strong magnetisa-tion in the direction of thema etisin field.

Tend to move quickly fromweaker to stronger parts ofthe field.

Susceptibility is very largeand positive. Xm > 1000

11.12. Physical state oj

the material

Of the order of thousandsIlr > 1000

11 »110

Susceptibility decreaseswith temperature in acomplex manner.

1Xm <X. T _ '[ (T> Tc)c

9. Removal ojmagnetising field

Magnetisation is retainedeven after the magnetisingfield is removed.

10. Variation oj Mwith H

M changes with H non-linearly and ultimatelyattains saturation.

Solid, liquid or gas. Normally solids only.Solid, liquid or gas.

Bvector shows h steresis.

Fe,Ni,Co,Gd,Fep3,Alnico.13. Examples Bi,Cu,Pb,Si, Nz(atSTP),Hp,NaCI

AI,Na,Ca,Oz(atSTP),CuClz

5.36

5.31 HYSTERESIS38. Explain the phenomenon of hysteresis in magnetic

materials. What is the significance of the area ofhysteresis loop ? State the practical importance ofhysteresis loops.

Hysteresis. When a ferromagnetic sample is placedin a magnetising field, the sample gets magnetised byinduction. As the magnetising field intensity H varies,the magnetic induction B does not vary linearly withH, i.e., the permeability Il( = BI H) is not constant butvaries with H. In fact, it also depends on the pasthistory of the sample.

Figure 5.56 shows the variation of magneticinduction Bwith magnetising field intensity H. Point 0represents the initial unmagnetised state of aferromagnetic sample. As the magnetising fieldintensity H increases, the magnetic induction B firstgradually increases and then attains a constant value.In other words, the magnetic induction B saturates at acertain value + Hmax'

B

Initialbuild up

-Hmax-----r----+-~--r-----------------~H+ Hmax,,,,,,D'

Saturation

DB = RetentivityDC = Coercivity

Fig. 5.56 Hysteresis loop for a ferromagnetic sample.

Now if the magnetising field intensity H isgradually decreased to zero, B decreases but along anew path AB. It is found that the magnetic induction Bdoes not become zero even when the magnetising fieldH is zero, i.e., the sample is not demagnetised evenwhen the magnetising field has been removed. Themagnetic induction (= 0 B) left behind in the sample after themagnetising field has been removed is called residualmagnetism or retentivity or remanence.

To reduce the magnetism to zero, the field H isgradually increased in the reverse direction, theinduction B decreases and becomes zero at a value ofH = OC. The value of reverse magnetising field intensity Hrequired for the residual magnetism of a sample to becomezero is called coercivity of the sample.

PHYSICS-XII

On further increasing H in the reverse direction to avalue - Hmax' we reach the saturation point D locatedsymmetrically to point A. Now if H is decreasedgradually, the point A is reached after going throughthe path DEFA.

The closed curve ABCDEFA which represents acycle of magnetisation of a ferromagnetic sample iscalled its hysteresis loop. Throughout the cycle, themagnetic field B lags behind the magnetising fieldintensity H, i.e., the value of B when H is decreasing isalways more than when H is increasing. The pheno-menon of the lagging of magnetic induction behind the magne-tisingfield is calledhysteresis. In fact, the word hysteresisoriginates from a Greek word meaning' delayed'.

Significance of the area of hysteresis loop. The

product BH = B(~J= .».,has the dimensions of11 Ilollr

energy per unit volume. Hence the area within the B-Hloop represents the energy dissipated per unit volume inthe material when it is carried through a cycle ofmagnetisation. The source is the source of emf used inmagnetising the material and the sink is the hystereticheat loss in the magnetic material.

Practical importance of hysteresis loops. A study ofhysteresis loop provides us information aboutretentivity, coercivity and hysteresis loss of a magneticmaterial. This helps in proper selection of materials fordesigning cores of transformers and electromagnetsand in making permanent magnets.

39. Distinguish between soft and hard ferromagneticmaterials. Draw their hysteresis loops. Give examples ofeach type.

Types of ferromagnetic materials. Ferromagneticmaterials can be divided into two categories:

1. Soft ferromagnetic materials or soft ferro-magnets. These are theferromagnetic materials in which themagnetisation disappears on the removal of the externalmagnetizing field. Such materials have narrow hyste-resis loop, as shown in Fig. 5.57(a). Consequently, they

B B

--+i-r-----...H --f--++---..H

(a) (b)

Fig. 5.57 Magnetic hysteresis loop for(a) soft, (b) hard ferromagnetic material.

MAGNETISM

have low retentivity, low coercivity, and low hyste-resis loss. But they have high relative magnetic per-meability. They are used as cores of solenoids and trans-formers. Examples. Soft iron, mu metal, stalloy, etc.

2. Hard ferromagnetic materials or hard ferro-magnets. These are theferromagnetic materials which retainmagnetisation even after the removal of the external magne-tisingfield. Such materials have wide hysteresis loop, asshown in Fig. 5.57(b). Consequently, they have highretentivity, high coercivity and large hysteresis loss.They are used for making permanent magnets.

Examples. Steel, alnico, lodestone, ticonal, etc.

Formulae Usedm1. Intensity of magnetisation, M = V

B 3. Il =~2. Il =-H r Il 0

M 5. C [Curie's law]4. Xm=fj Xm=T

6. B=llo (H + M) 7. Ilr = 1 + Xm·

Units UsedMagnetising field intensity His in Am -1, field Bintesla, magnetisation M in Am -l, permeability Il inTmA -1 or Hm -l, susceptibility Xm and relativepermeability Il r have no units.

Example 39. A magnet of magnetic moment 2.5 Arrtweighs 66 g. If the density of the material of the magnet is

7500 kgm-3, find the intensity of magnetisation.Solution. Volume,

66 x 10-3 kg 66 x 10-5 3----"-7= m7500 kgm -3 75

V = MassDensity

Magnetisation,

M= m 2.5V 66 x 10-5

--75

= 2.84 x 105 Am -t.

2.5 x 75 x 105

66

Example 40. Obtain the earth's magnetisation. Assumethat the earth's field can be approximated by a gaint barmagnet of magnetic moment 8.0 x 1022 Ant-. The earth'sradius is 6400 km. [NCERT)

Solution. Here magnetic moment,m = 8.0 x 1022 Am2

Radius of the earth, R = 6400 krn = 6.4 x 106 m

5.37

Magnetisation,m m 8.0 x 1022 x 3

M = V = i n? = 4 x 3.14 x (6.4 x 106)3

3= 72.9 Am-t.

Example 41. A domain inferromagnetic iron is in theformof a cube of side length l)lm. Estimate the number of ironatoms in the domain and the maximum possible dipolemoment and magnetisation of the domain. The molecularmass of iron is 55 g I mole and its densitu is 7.9 g I crrf.Assume that each iron atom has a dipole moment of9.27 x 10-24 AI;. [NCERT)

Solution. Each side of cubic domain,1= Lurn =10-6 m

Volume of the domain,V = 13 =(10-6 m)3 =10-18m3 =10-12 cm '

Mass of domain = Volume x density= 10-12 em:' x 7.9 gern-3

=7.9xl0-12gNumber of atoms in 55 g iron

= 1 mole = 6.023 x 1023

:. Number of atoms in 7.9 x 10-12 g iron

6.023 x 1023 x 7.9 x 10-12

55N = 8.65 x 1.otO atoms.

Dipole moment of each iron atom,m=9.27x 1O-24Arn2

The dipole moment of the domain will be maxi-mum when all its atomic dipoles get perfectly aligned.Its value will be

m = mN =9.27 x 10-24 x 8.65 x 1010max

= 8.0 x 10-13 Am 2.

The maximum possible magnetisation of the domain,

80 x 10-13 Am2M- mmax _. _--.;;--.".--_

- V 1O-18m3

= 8.0 x 105 Am-t.

Example 42. A magnetising field of 1500 AI m producesa magnetic flux of2.4 x 10-5 weber in a bar of iron of cross-section 0.5 crrf. Calculate permeability and susceptibility ofthe iron-bar used. [CBSE OD 08)

Solution. Here H = 1500 Am-I, <p == 2.4 x 10-5 Wb,A =0.5 x 10 -4m2

Magnetic induction,

_ <p _ 2.4 x 10-5 _ 0 48 Wb -2B--- - . mA 0.5 x 10--4

or

5.38

Permeability,

Il = ~ = 0.48 = 3.2 x 10-4 TmA-1H 1500

As Il = Il0 (1 + XIII )

., Susceptibility,

= ~ -1 = 3.2 x 10-4

-1XIII Ilo 4x3.14xlO-7

= 254.77 -1 = 253.77.

Example 43. Assume that each iron atom has a permanentmagnetic moment equal to 2 Bohr magnetons (1 Bohrmagneton =9.27 x 10-24 AI;). The number density ofatoms in iron is 8.52 x 1028 m-3. (i) Find the maximummagnetisation M in a long iron bar. (ii) Find the maximummagnetic induction B in the bar.

Solution. (i) Number of atoms per unit volume,n = 8.52 x 1028 m-3

Magnetic moment of each iron atom=2 IlB =2 x9.27x 10-24 Am2

As magnetisation M is the magnetic moment perunit volume, so the maximum value of magnetisation is

~ax = nx 21lB(when all the dipoles get aligned)

= 8.52 x 1028 x 2 x 9.27 x 10-24

= 1.58 x 106 Am -1.

(ii) Magnetic induction, B=llo (H + M)As no magnetising field is applied, so H = O. Hence

B= Ilo M = 411:X 10-7 x 1.58 x 106 = 1.985 T.

Example 44. A solenoid of 500 turns / m is carrying acurrent of3 A Its core is made of iron which has a relativepermeability of 5000. Determine the magnitudes of themagnetic intensity, magnetisation and the magnetic fieldinside the core. [NCERT]

Solution. Here n = 500 turns/m, I =3A Il r = 5000

Magnetic intensity,H = nI = 500 m -1 x 3 A = 1500 Am -1.

As Ilr=l+XI/l

X m = Il r - 1 = 5000 - 1 = 4999 =- 5000

Also, Il r = ~ = 5000 or Il = 5000 Il0Ilo

Magnetisation,M = Xm H =5000 x 1500

= 7.5 x 106 Am -1.

Magnetic field inside the core,

B= IlH = 5000 Il0H= 5000 x 411:x 10-7 x 1500 = 311:=- 9.4 T.

PHYSICS-XII

Example 45. The core of a toroid having 3000 turns hasinner and outer radii of 11 em and 12 cm respectively. Themagnetic field in the core for a current of 0.70 A is 2.5 T.What is the relative permeability of the core?

Solution. The magnetic field in the empty spaceenclosed by a toroid is given by

B = llonI

where n is the number of turns per unit length and I isthe current. If the space is filled by a core of perme-ability u, then

B = IlnI

Here B=2.5 T, I =0.70 AMean radius,

r = 11 + 12 em = 11.5 em = 11.5 x 10 -2 m2

3000 3000n----- 211:r - 2 x 3.14 x 11.5 x 10-2

HenceB 2.5 x 2 x 3.14 x 11.5 x 10-2

Il = -;;r = 3000 x 0.70

= 8.6 x 10-4 Tm A-1

Relative permeability,

Ilr = ~ = 8.6 x 10-4

= 684.4.Ilo 411:x 10-7

Example 46. An iron rod of volume 10-4m3 and relativepermeability 1000 is placed inside a long solenoid woundwith 5 turns per em. If a current of 0.5 A is passed throughthe solenoid, find the magnetic moment of the rod.

Solution. The relation between the magneticinduction B, magnetising field intensity H and themagnetisation M is given by

B= llo(H + M)B Il H

M=--H=--H [.: B=1l H]Ilo Ilo

= Il r H - H = (Il r -l)H

But for a long solenoid, we haveH= nI

where n is the number of turns per metre.

M = (Il r -1) nIHere Jl r =1000, I =0.5 A

. 5n = - turns / m = 500 turns / m

0.01M = (1000 -1) x 500 x 0.5 =2.5 x 105Am-1

Magnetic moment,m= Mx V =2.5 x 105 x 10-4 Am2 =25 Am2.

MAGNETISM

Example 47. The hysteresis loss for a specimen of ironweighing 12 kg is equivalent to 300 Im-3 cycle-I. Find theloss of energy per hour at 50 cycle s-l. Density of iron is7500 kg m-3.

Solution. Let Q be the energy dissipated per unitvolume per hysteresis cycle in the given sample. Thenthe total energy lost by the volume V of the sample intime t will be

W=QxVXyxt

where y is the number of hysteresis cycles per second.

Here Q =300 Jm-3 cycle-l, v = 50 cycle s-l,t =1 h =3600 s

Volume V = Mass =~ m3

, Density 7500

.'. Hysteresis loss,12W = 300 x ~- x 50 x 3600 J = 86400 J.

7500

Example 48. Thecoercivity of a certain permanent magnet is4.0 x 104 Am-I. This magnet is placed inside a solenoid 15 emlong and having 600 turns and a current is passed in thesolenoid to demangneiise it completely. Find the current.

Solution. The coercivity of 4 x 104 Am -1 of thepermanent magnet implies that a magnetic intensityH = 4 x 104 Am -1 is required to be applied in oppositedirection to demagnetise the magnet.

600 600Here n = ~~ = ? = 4000 turns / m15 ern 15 x 10- - m

As H=nI

C I = H = 4 x 104

= 10 A.'. urrent,n 4000

jOrOblems For Practice

1. A bar magnet made of steel has a magnetic momentof 2.5 Am2 and a mass of 6.6 g. If the density of steelis 7.9 x 103 kg m -3, find the intensity of magne-tisation of the magnet. (Ans. 3.0 x 106 Am-1)

2. The maximum value of permeability of u-metal(77% Ni, 16% Fe, 5% Cu, 2% Cr) is 0.126 TmA-1.Find the maximum relative permeability and -susceptibility. (Ans. fl r = 1.0 x 105, X=- 1.0 x 105)

3. Find the percent increase in the magnetic field Bwhen the space within a current-carrying toroid isfilled with aluminium. The susceptibility ofaluminium is 2.1 x 10-5. (Ans. 2.1 x 10-3)

4. The susceptibility of magnesium at 300 K is1.2 x 10-5. At what temperature will thesusceptibility increase to 1.8 x 10-5. (Ans. 200 K)

5.39

5. An iron rod of 0.2 em 2 cross-sectional area issubjected to a magnetising field of 1200 Am -1. Thesusceptibility of iron is 599. Find the permeabilityand the magnetic flux produced.

(Ans. 7.536 x 10-4Tm A-1, 1.81 x 10-5Wb)6. An iron rod 0.2 m long, 10 mm in diameter and of

permeability 1000 is placed inside a long solenoidwound with 300 turns per metre. If a current of0.5 ampere is passed through the solenoid, find themagnetic moment of the rod. (Ans. 0.2325 Am -1 )

7. An iron ring of mean circumferential length 30 emand cross-section 1em 2 is wound uniformly with300 turns of wire. When a current of 0.032 A flowsin the windings; the flux in the ring is 2 x 10-6 Wb.Find the flux density in the ring, magnetising fieldintensity and relative permeability of iron.

(Ans. 2 x 10-2 Wb m -2, 32 A turns m -1, 500 )

8. An iron ring having 500 turns of wire and a meandiameter of 12 em carries a current of 0.3 A. Therelative permeability of iron is 600. What is themagnetic flux density in the core ? What is themagnetisation field intensity? What part of the fluxdensity is due to the electronic loop currents in thecore? (Ans. 0.3 Wbm ", 397. 9 A turns m-I,

0.2995 Wbm -2 )

HINTS1. Proceed as in Example 39 on page 5.37.

2. Maximum relative permeability,

-~- 0.126 -10 105fl r - fl 0 - 41t X 10 7 - • x

Maximum susceptibility,X = fl r - 1 =- 1.0 x 105 •

3. In the absence of aluminium,Bo = floH

In the presence of aluminium

B = flH = fl 0 (1 + X) H

Increase in field = B - Bo = fl 0 X H

Percent increase = B - Bo x 100 = fl 0 X H x 100Bo floH

= X x 100 = 2.1 x 10-3•

.. 12 = Xl . 'Ii = 1.2 x 10-5

x:OO = 200 K.X2 1.8xl0

5. Here A = 0.2 em2 = 0.2 X 10-4 m 2, H = 1200 Am-1,Xm = 599

Permeability, fl =flo (1+ Xm)

= 41t x 10-7 x( 1+ 599) = 7.536 x10-4 TmA-1

5.40

Magnetic induction,B = ~H = 7.536 x 10-4 x 1200 = 0.904 T

Magnetic flux,<p = BA = 0.904 x 0.2 x 10-4 = 1.81 x 10-5 Wh.

6. Proceed as in Example 46 on page 5.38.7. Here I = 30cm = 0.30 m, A = lcm2 = 1O-4m2,

N = 300, I = 0.032 A, <p = 2 x 10-6 Wb,

n= N = 300 = 1000 m-II 0.30

Magnetic flux density,

B=.!=2x~-6 =2x10-2Whm-2

A 10

Magnetising field intensity,H= n1= 1000 x 0.032 = 32 A turns m-1

Permeability,

_ ~ _ 2 x 10-2 _ 6 25 104 T A-I~- - -. x m.H 32

11 6.25 x 10-4Relative permeability, ~l = - = 7 = 500.

r 110 41tx l0-

8. Here N = 500, D = 12cm = 0.12 rn, 1= 0.3 A,~r = 600

Magnetising field intensity,N N

H=n1=-.I=-.1I 1tD

500 x 0.3 -1= = 397.9 A turns m

1tx 0.12

Magnetic flux density, B = I1H = ~ r 110 H= 600x41t x 10-7 x397.9 = 0.3 Wbm-1

Flux density due to electronic current loop is~ oM =.B -11 oH = 0.3 - 41tx 10-7 x 397.9

= 0.3 - 0.0005 = 0.2995 Whm -2 •

5.32 PERMANENT MAGNETS ANDELECTROMAGNETS

40. Give a comparison of the magnetic properties ofsoft iron and steel.

Comparison of the magnetic properties of soft ironand steel. Fig. 5.58 shows the hysteresis loops for softiron and steel.

A study of these B - H loops reveals the followinginformation:

1. Permeability. For a given H, B is more for softiron than steel. So soft iron has a greater permeability(J..l = B/ H) than steel.

2. Susceptibility. As permeability of soft iron isgreater than steel, so soft iron has a greater suscep-tibility (Xm = u , -1) than steel.

PHYSICS-XII

B

---~L-H-:+-+----.H

Fig. 5.58 Hysteresis loops of soft iron and steel.

3. Retentioitq. The retentivity of soft iron (Ob') isgreater than the retentivity (Ob) of steel.

4. Coercivity. The coercivity of soft iron (Oc ) is lessthan the coercivity (Oc) of steel.

5. Hysteresis loss. As the area of the hysteresis loopof soft iron is much smaller than that of steel, so thehysteresis loss per unit volume per cycle is less for softiron than for steel.

We can summarise the above properties as follows:

1. Permeability2. Susceptibility3. Retentivity

4. Coercivity5. Hysteresis loss

1are greater for soft ironthan for steel

] are less for soft ironthan for steel

41. How will you select materials for making perma-nent magnets, electromagnets and cores of transformers ?

Selection of magnetic materials. The choice ofmagnetic materials for making permanent magnets,electromagnets and cores of transformers is decidedfrom the hysteresis loop of the material.

A. Permanent magnets. The material used formaking permanent magnets must have the followingcharacteristics:

1. High retentivity so that it produces a strongmagnetic field.

2. High coercivity so that its magnetisation is notdestroyed by stray magnetic fields, temperaturevariations or minor mechanical damage.

3. High permeability.Inspite of its slightly smaller retentivity than soft

iron, steel is favoured for making permanent magnets.Steel has much higher coercivity than soft iron. Themagnetisation of steel is not easily destroyed by strayfields. Once magnetised under a strong field, it retains

MAGNETISM

magnetisation for a long duration. Other suitablematerials for making permanent magnets are:

Cobalt steel 52% Fe, 36% Co, 7% W,3.5% Cr, 0.5% Mn, 0.7% C98% Fe, 0.86% C, 0.9% Mn55% Fe, 10% Al, 17% Ni,12% Co, 6% Cu42% Co, 26.5 Fe, 14% Ni, 8% AL6.5 Ti, 3% Cu

Carbon steelAlnico

Ticonal

B. Electromagnets. The material used for makingcores of electromagnets must have the followingcharacteristics:

1. High initial permeability so that magnetisation islarge even for a small magnetising field.

2. Low retentivity so that the magnetisation is lost asthe magnetising current is switched off.

So soft iron is more suitable than steel for cores ofelectromagnets.

C. Transformer cores. The material used for makingcores of transformers must have the following charac-teristics :

1. High initial permeability so that the magnetic fluxis large even for low magnetising fields.

2. Low hysteresis loss as the materials are subjected toalternating magnetising fields of high frequency.

3. Low resistivity to reduce losses due to eddycurrents.

Soft iron is preferred for making transformer coresand telephone diaphragms.

42. Mention three methods for making permanentmagnets.

Methods for making permanent magnets. A hardferromagnetic material like steel can be converted intoa permanent magnet by any of the following methods:

1. By holding the steel rod in north-south directionand hammering it repeatedly.

2. Hold a steel rod and stroke it with one end of abar magnet a number of times, always in thesame sense to make a permanent magnet.

3. The most efficient way of making a permanentmagnet is to place a steel rod in a solenoid andpass a strong current. The rod gets magnetiseddue to the magnetic field of the solenoid.

43. Briefly explain how an electromagnet is formed.State some uses of electromagnets.

Electromagnet. As shown in Fig. 5.59, take a softiron rod and wind a large number of turns of insulated

5.41

copper wire over it. When we pass a current through thesolenoid, a magnetic field is set up in the space withinthe solenoid. The high permeability of soft iron increasesthe field one thousand times. The end of the solenoid atwhich the current in the solenoid seems to flowanticlockwise acts as -pole and other one as S-pole.When the current in the solenoid is switched off, thesoft iron rod loses its magnetism almost completelydue to its low retentivity.

I

+ ~I- (.

Battery Key

Fig. 5.59 An electromagnet.

Uses of electromagnets:1. Electromagnets are used in electric bells, loud-

speakers and telephone diaphragms.2. Large electromagnets are used in cranes to lift

heavy machinery, and bulk quantities of ironand steel.

3. In hospitals, electromagnets are used to removeiron or steel bullets from the human body.

5.33 TANGENT GALVANOMETER·44. State tangent law of magnetism.Tangent law. This law states that if a freely

suspended small magnet is acted upon by two uniformmutually perpendicular magnetic fields Br and ~simultaneously, then the magnet comes to rest in such

B2 ---------------------

5

Fig. 5.60 Tangent law.

a position that the tangent of the angle e that themagnet makes with Br is equal to the ratio ~ / Br of thetwo fields. That is,

tan e = ~Bl

or ~ = Br tan e

5.42

45. Describe the principle, construction, theory andworking of a tangent galvanometer.

Tangent galvanometer. It is a device used tomeasure very small currents. It is a moving magnet typegalvanometer. Its working is based on tangent law.

Construction. It consists of a circular frame ofnon-magnetic material mounted on a horizontal turntable. Three coils having 2, 50 and 500 turns ofinsulated copper wire are wound over it. The ends ofthe coils are connected to three base terminals.

A compass box of non-magnetic material is fitted atthe centre of circular frame. It has a small magneticneedle pivoted at its centre with a long thin aluminiumpointer attached perpendicular to it. Both are free tomove over a horizontal circular scale graduated indegrees and divided into four quadrants of 0° - 90°each. A plane mirror is fixed at the base below thepointer. This removes error due to parallax whilereading the position of the pointer.

Shortmagneticneedle

Aluminiumpointer

Levellingscrews

Fig. 5.61 Tangent galvanometer.

Adjustments The tangent galvanometer is levelledwith the help of levelling screws and the coil is sorotated that its plane becomes parallel to the length ofthe magnetic needle. In this position, the plane of thecoil lies along the magnetic meridian. The compass boxis rotated, so that the pointer comes along 0° - 0° line.

Theory and working. The working of tangentgalvanometer is based on tangent law. When nocurrent is passed through the coil, the magnetic needleis influenced only by ~ of earth's magnetic field.When a current I is passed, there is a magnetic field Balong the axis of coil perpendicular to ~ , as shown inFig. 5.62. The magnetic needle is influenced by twoperpendicular magnetic fields and it comes to rest at anangle 8 with BHsuch that

PHYSICS-XII

B= ~ tan 8

This relation is known as Tangent law.

Coil inverticalplane, N-Tums,radius R

B BR (Resultant)

5

Fig. 5.62

Now magnetic field at centre of coil isB= 110 NI

2Rwhere N and R are number of turns and radius of thecoil.

11 N I_0__ = ~ tan 8

2R

I = 2 R BHtan e = K tan 81l0N

2R~where K = --- is a constant for the tangent galvano-1l0N

meter and is called its reduction factor.

5.34 OSCILLATIONS OF A FREELYSUSPENDED MAGNET

46. Show that the oscillations of a freely suspendedmagnet in a uniform magnetic field are simple harmonic.Hence deduce an expression for its time period.

Oscillations of a freely suspended magnet in amagnetic field. In the position of equilibrium, the

--+magnetic dipole lies along B. When it is slightlyrotated from this position and released, it begins tovibrate about the field direction under the restoringtorque,

t = -mB sin 8

The negative sign indicates that the direction oftoque t is such so as to decrease 8.

-->m

-->-----~~~~----~B

Fig. 5.63

MAGNETISM

For small angular displacement 9, sin 9 '" 9

r = -mB9

If Iis the moment of inertia of the magnet, then thedeflecting torque on the magnet is

d29't= la=l-dt2

In the equilibrium condition,

Deflecting torque = Restoring torque

d291-=-I11B9dt2

d29 =_ mB9=_0)29

dt2 I

i.e., angular acceleration d2~ ex; angular displacement 9.dt

or

Hence the oscillations of a freely suspended magneticdipole in a uniform magnetic field are simple harmonic. Thetime period of oscillation is given by

T=2n=2n0.0) f;;:;B

Example 49. In Fig. 5.64, a magnetic needle is freeto oscillate in a uniform magnetic field. The magnetic needlehas magnetic moment 6.7 A~ and moment of inertia1=7.5 x 10-6 km~. It performs 10 complete oscillations in6.70 s. What is the magnitude field? [NCERT]

- - - - - - - - - - - - - - - - - - ~

----~--->/.e-l'!.-----~ --+

------ ---------~B,-,-- - - - - - - - - - - - - - - - ~

5- - - - - - - - - - - - - - - - - - ~

Fig. 5.64

. 6.70 s 2Solution. Here T = -- = 0.67 s, m = 6.7 Am ,10

I =7.5 x 10-6 kg m2

As T =2n 0v-;;:;B

T2 = 4n2 _1_mB

The magnitude of the magnetic field is

B= 4n21 = 4x9.87x7.5xlO-6

mT2 6.7x (0.67)2

= 9.8 x 10-5 T.

or

5.43

5.35 VIBRATION MAGNETOMETER *47. Describe the principle, construction and working

of a vibration magnetometer. Mention its uses.Vibration magnetometer. It is an instrument used to

compare the magnetic moments of two magnets or to determinethe horizontal component of earth's magnetic field at a place.

Principle. When a magnet suspended freely in auniform magnetic field (like the one due to the earth),is displaced from its equilibrium position, it begins tovibrate simple harmonically about the direction of thefield. The period of vibration is given by

T=2n~ ImBH

where, m = magnetic moment of the magnet,BH = horizontal component of earth's

magnetic field,I= the moment of inertia of the magnet about

an axis of rotation through its centre of mass12 + b2

and I= Mass x --- 12Here Iis the length and b the breadth of the magnet.Construction. It consists of a short magnet enclosed

in a wooden box provided with glass windows. Thebox has a narrow tube fixed on its top at the middle.The magnet is suspended horizontally in a light brassstirrup by a silk thread which passes centrally downthe tube and is provided with a torsion-head at the topof the tube. The glass box protects the magnet from aircurrent and its top has two slits through whichvibrations of the magnet can be observed. A planemirror strip with a reference line on it is placedlengthwise at the base of the instrument just below theslits.

--- Torsion head

Glass tube

Silk thread

Fig. 5.65 Vibration magnetometer.

5.44

Adjustments

(i) Place a compass needle on the reference line.Rotate the box till the line becomes parallel tothe compass needJe. This sets the magneto-meter in the north-south direction.

(ii) To ensure that there is no twist in the thread,place a brass bar of the same size as that ofthe magnet in the stirrup and allow the stirrupto come to rest. Adjust the torsion head so thatthe brass bar is in the north-south direction.

(iii) Replace the brass bar by a small magnet withits N-pole pointing geographic north. Bring apowerful magnet near the box and remove it.The suspended magnet starts oscillating. Noteits period of vibration.

48. Explain the various uses of a vibration magneto-meter.

Uses of a vibration magnetometer:1. Measurement of magnetic moment of a magnet.

Set the vibration magnetometer in the north-southdirection. Place the bar magnet in its stirrup. Measurethe time period of the bar magnet.

As T=2n~m~

or

The moment of inertia Ican be determined from thegeometry of the magnet. Knowing Bw magneticmoment m can be determined.

2. Comparison of magnetic moments of twomagnets of same size and same mass. For the two barmagnets of same size and mass, the moments of inertiaare equal. With the help of vibration magnetometer,we measure the time periods of vibration Tl and T2 ofthe two magnets at a particular place. Then

Tl =2n~~~ and T2 =2n~~~

or

3. Comparison of magnetic moments of twounequal sizes and masses (sum and difference method).Sum position. Place the two magnets in the stirrup ofvibration magnetometer so that their north poles pointin the geographical north.

Moment of inertia of the combination = II + 12Magnetic moment of the combination = ~ + ~

PHYSICS-XII

Determine the period of vibration of the combi-nation. Let it be T1. Then

...(1)

I = II + 12

m= '11-~ ~

5 NN 5

(a) (b)

Fig. 5.66 (a) Sum position. (b) Difference position.

Difference position. Now place the two magnets inthe stirrup of the vibration magnetometer with theiropposite poles in the same direction.

Moment of inertia of the combination

= II + 12Magnetic moment of the combination

='11-111:2

Let period of vibration of the combination=T2

Then ...(2)

or

('11-~)~Dividing (1) by (2), we get

Tl_~'11-~T2 '11+~T12_ '11-~Ti - '11+ ~

By componendo and dividendo, we get

4. Comparison of horizontal components ofearth's magnetic field. With the help of a vibrationmagnetometer, we measure the periods of vibrationT and T' of the same magnet at the two given places.Let ~ and s;, be the horizontal components of earth'smagnetic field at these places.

Then T = 2 n ~ m~ and

.. ~ =~ or

T' =2n~m~