5 Magnetism and Matter

8/13/2019 5 Magnetism and Matter

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Question 5.1:

Answer the following questions regarding earths magnetism:

A vector needs three quantities for its specification. Name the three independent

quantities conventionally used to specify the earths magnetic field.

The angle of dip at a location in southern India is about 18.

Would you expect a greater or smaller dip angle in Britain?

If you made a map of magnetic field lines at Melbourne in Australia, would the lines

seem to go into the ground or come out of the ground?

In which direction would a compass free to move in the vertical plane point to, if located

right on the geomagnetic north or south pole?

The earths field, it is claimed, roughly approximates the field due to a dipole of magnetic

moment 8 1022 J T1 located at its centre. Check the order of magnitude of this numberin some way.

(f ) Geologists claim that besides the main magnetic N-S poles, there are several local

poles on the earths surface oriented in different directions. How is such a thing possible

at all?

Answer

The three independent quantities conventionally used for specifying earths magnetic

field are:

Magnetic declination,

Angle of dip, and

Horizontal component of earths magnetic field

(b)The angle of dip at a point depends on how far the point is located with respect to the

North Pole or the South Pole. The angle of dip would be greater in Britain (it is about

70) than in southern India because the location of Britain on the globe is closer to the

magnetic North Pole.

(c)It is hypothetically considered that a huge bar magnet is dipped inside earth with its

north pole near the geographic South Pole and its south pole near the geographic North

Pole.


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Magnetic field lines emanate

south pole. Hence, in a map

Melbourne, Australia would

(d)If a compass is located on

will be free to move in the h

magnetic poles. In such a cas

(e)Magnetic moment,M= 8

Radius of earth, r= 6.4 10

Magnetic field strength,

Where,

= Permeability of free sp

This quantity is of the order

(f)Yes, there are several loca

magnetised mineral deposit i

Question 5.2:

Answer the following questi

The earths magnetic

Does it also change with tim

The earths core is kn

source of the earths

The charged currents

thought to be respons

the source of energy)

The earth may have e

history of 4 to 5 billi

such distant past?

from a magnetic north pole and terminate at a

epicting earths magnetic field lines, the field l

seem to come out of the ground.

the geomagnetic North Pole or South Pole, the

rizontal plane while earths field is exactly vert

e, the compass can point in any direction.

1022

J T1

m

ce =

f magnitude of the observed field on earth.

l poles on earths surface oriented in different d

s an example of a local N-S pole.

ns:

field varies from point to point in space.

? If so, on what time scale does it change appr

own to contain iron. Yet geologists do not rega

agnetism. Why?

in the outer conducting regions of the earths c

ible for earths magnetism. What might be the

to sustain these currents?

ven reversed the direction of its field several ti

n years. How can geologists know about the ea

agnetic

nes at

the compass

ical to the

rections. A

ciably?

d this as a

re are

battery (i.e.,

es during its

ths field in


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Answer

Magnetic field strength,B =

Torque on the bar magnet, T

Angle between the bar magn

Torque is related to magneti

T=MB sin

Hence, the magnetic momen

Question 5.4:

A short bar magnet of magne

field of 0.15 T. If the bar is f

correspond to its (a) stable, a

the magnet in each case?

Answer

Moment of the bar magnet,

External magnetic field,B =

0.25 T

= 4.5 102

J

t and the external magnetic field,= 30

moment (M) as:

of the magnet is 0.36 J T1

.

tic moment m = 0.32 J T1

is placed in a unifor

ee to rotate in the plane of the field, which orie

d (b) unstable equilibrium? What is the potenti

= 0.32 J T1

0.15 T

magnetic

tation would

al energy of


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(a)The bar magnet is aligned

in stable equilibrium. Hence,

is 0.

Potential energy of the syste

(b)The bar magnet is oriente

equilibrium.

= 180

Potential energy = MB cos

Question 5.5:

A closely wound solenoid of

current of 3.0 A. Explain the

its associated magnetic mom

Answer

Number of turns in the solen

Area of cross-section,A = 2.

Current in the solenoid,I= 3

A current-carrying solenoid

along its axis, i.e., along its l

The magnetic moment associ

as:

M= n I A

along the magnetic field. This system is consid

the angle , between the bar magnet and the m

180 to the magnetic field. Hence, it is in unst

800 turns and area of cross section 2.5 104

sense in which the solenoid acts like a bar mag

ent?

id, n = 800

104

m2

.0 A

ehaves as a bar magnet because a magnetic fiel

ngth.

ated with the given current-carrying solenoid is

ered as being

gnetic field

ble

2carries a

et. What is

d develops

calculated


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= 800 3 2.5 104

= 0.6 J T1

Question 5.6:

If the solenoid in Exercise 5.

horizontal magnetic field of

solenoid when its axis makes

Answer

Magnetic field strength,B =

Magnetic moment,M= 0.6

The angle , between the axi

Therefore, the torque acting

Question 5.7:

A bar magnet of magnetic m

magnetic field of 0.22 T.

What is the amount of work

align its magnetic moment: (direction?

What is the torque on the ma

Answer

5 is free to turn about the vertical direction and

.25 T is applied, what is the magnitude of torq

an angle of 30 with the direction of applied fi

0.25 T

1

s of the solenoid and the direction of the applie

n the solenoid is given as:

ment 1.5 J T1

lies aligned with the direction o

equired by an external torque to turn the magne

i) normal to the field direction, (ii) opposite to t

gnet in cases (i) and (ii)?

uniform

e on the

ld?

field is 30.

a uniform

t so as to

e field


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(a)Magnetic moment,M= 1.5 J T1

Magnetic field strength,B = 0.22 T

(i)Initial angle between the axis and the magnetic field, 1 = 0

Final angle between the axis and the magnetic field, 2 = 90

The work required to make the magnetic moment normal to the direction of magnetic

field is given as:

(ii) Initial angle between the axis and the magnetic field, 1 = 0

Final angle between the axis and the magnetic field, 2 = 180

The work required to make the magnetic moment opposite to the direction of magnetic

field is given as:

(b)For case (i):

Torque,

For case (ii):

Torque,


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Question 5.8:

A closely wound solenoid of

a current of 4.0 A, is suspen

What is the magnetic mome

What is the force and torque

102

T is set up at an angle

Answer

Number of turns on the sole

Area of cross-section of the s

Current in the solenoid,I= 4

(a)The magnetic moment alo

M= nAI

= 2000 1.6 104

4

= 1.28 Am2

(b)Magnetic field,B = 7.5

Angle between the magnetic

Torque,

Since the magnetic field is u

solenoid is

2000 turns and area of cross-section 1.6 104

ed through its centre allowing it to turn in a hor

t associated with the solenoid?

on the solenoid if a uniform horizontal magneti

of 30 with the axis of the solenoid?

oid, n = 2000

olenoid,A = 1.6 104

m2

A

ng the axis of the solenoid is calculated as:

102

T

field and the axis of the solenoid, = 30

iform, the force on the solenoid is zero. The to

m2, carrying

izontal plane.

field of 7.5

que on the


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Question 5.9:

A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its

plane normal to an external field of magnitude 5.0 102

T. The coil is free to turn about

an axis in its plane perpendicular to the field direction. When the coil is turned slightly

and released, it oscillates about its stable equilibrium with a frequency of 2.0 s1

. What is

the moment of inertia of the coil about its axis of rotation?

Answer

Number of turns in the circular coil,N= 16

Radius of the coil, r= 10 cm = 0.1 m

Cross-section of the coil,A = r2

= (0.1)2

m2

Current in the coil,I= 0.75 A

Magnetic field strength,B = 5.0 102

T

Frequency of oscillations of the coil, v = 2.0 s1

Magnetic moment,M=NIA

= 16 0.75 (0.1)2

= 0.377 J T1

Where,

I= Moment of inertia of the coil


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Hence, the moment of inertia

Question 5.10:

A magnetic needle free to ro

its north tip pointing down atearths magnetic field at the

earths magnetic field at the

Answer

Horizontal component of ear

Angle made by the needle wi

Earths magnetic field streng

We can relateB andBHas:

Hence, the strength of earth

of the coil about its axis of rotation is

ate in a vertical plane parallel to the magnetic

22 with the horizontal. The horizontal compolace is known to be 0.35 G. Determine the mag

lace.

hs magnetic field,BH= 0.35 G

th the horizontal plane = Angle of dip =

th =B

magnetic field at the given location is 0.377 G.

eridian has

ent of thenitude of the

.


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Question 5.11:

At a certain location in Afric

north tip of the magnetic nee

points 60 above the horizonto be 0.16 G. Specify the dir

Answer

Angle of declination,= 12

Angle of dip,

Horizontal component of ear

Earths magnetic field at the

We can relateB andBHas:

Earths magnetic field lies in

making an angle of 60 (upw

Question 5.12:

A short bar magnet has a ma

magnitude of the magnetic fi

centre of the magnet on (a) t

magnet.

Answer

a, a compass points 12 west of the geographic

dle of a dip circle placed in the plane of magnet

al. The horizontal component of the earths fielction and magnitude of the earths field at the l

hs magnetic field,BH= 0.16 G

given location =B

the vertical plane, 12 West of the geographic

ard) with the horizontal direction. Its magnitud

netic moment of 0.48 J T1

. Give the direction

eld produced by the magnet at a distance of 10

e axis, (b) the equatorial lines (normal bisector

orth. The

ic meridian

d is measuredcation.

eridian,

is 0.32 G.

and

m from the

of the


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Magnetic moment of the bar

Distance, d= 10 cm = 0.1 m

The magnetic field at distanc

relation:

Where,


The magnetic field is along t

The magnetic field at a distamagnet is given as:

The magnetic field is along t

Question 5.13:

magnet,M= 0.48 J T1

e d, from the centre of the magnet on the axis is

ce =

e S N direction.

ce of 10 cm (i.e., d= 0.1 m) on the equatorial l

e N S direction.

given by the

ne of the


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A short bar magnet placed in

north-south direction. Null p

centre of the magnet. The ea

is zero. What is the total ma

distance as the nullpoint (i.

field due to a magnet is equa

magnetic field.)

Answer

Earths magnetic field at the

The magnetic field at a dista

Where,


M= Magnetic moment

The magnetic field at the sa

Total magnetic field,

Hence, the magnetic field is

Question 5.14:

a horizontal plane has its axis aligned along th

ints are found on the axis of the magnet at 14 c

ths magnetic field at the place is 0.36 G and th

netic field on the normal bisector of the magnet

., 14 cm) from the centre of the magnet? (At n

l and opposite to the horizontal component of ea

given place,H= 0.36 G

ce d, on the axis of the magnet is given as:

ce

e distance d, on the equatorial line of the magn

.54 G in the direction of earths magnetic field.

magnetic

m from the

e angle of dip

at the same

ll points,

rths

t is given as:


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If the bar magnet in exercise 5.13 is turned around by 180, where will the new null points

be located?

Answer

The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:

Where,

M = Magnetic moment

= Permeability of free space

H= Horizontal component of the magnetic field at d1

If the bar magnet is turned through 180, then the neutral point will lie on the equatorial

line.

Hence, the magnetic field at a distance d2, on the equatorial line of the magnet can be

written as:

Equating equations (1) and (2), we get:


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When the resultant field is in

The magnetic field at a dista

The resultant field is incline

Question 5.16:


Why does a paramagnetic sa

magnetising field) when cool

Why is diamagnetism, in co

clined at 45 with earths field,B =H

ced from the centre of the magnet on its axi

at 45 with earths field.

ns:

ple display greater magnetisation (for the sam

ed?

trast, almost independent of temperature?

s is given as:


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If a toroid uses bismuth for i

(slightly) less than when the

Is the permeability of a ferro

is it more for lower or higher

Magnetic field lines are alwapoint. (This fact is analogous

of a conductor at every point.

(f ) Would the maximum pos

order of magnitude as the ma

Answer

(a)Owing to therandom ther

disrupted at high temperatur

paramagnetic sample display

(b)The induced dipole mome

magnetising field. Hence, th

temperature) does not affect

(c)Bismuth is a diamagnetic

magnetic field slightly great

(d)The permeability of ferro

magnetic field. It is greater f

(e)The permeability of a ferr

than one. Hence, magnetic fi

materials at every point.

(f)The maximum possible m

order of magnitude as the ma

fields for saturation.

Question 5.17:


s core, will the field in the core be (slightly) gre

core is empty?

agnetic material independent of the magnetic

fields?

ys nearly normal to the surface of a ferromagneto the static electric field lines being normal to

.) Why?

sible magnetisation of a paramagnetic sample b

gnetization of a ferromagnet?

al motion of molecules, the alignments of dipo

s. On cooling, this disruption is reduced. Hence

s greater magnetisation when cooled.

nt in a diamagnetic substance is always opposit

internal motion of the atoms (which is related

he diamagnetism of a material.

ubstance. Hence, a toroid with a bismuth core

r than a toroid whose core is empty.

agnetic materials is not independent of the app

r a lower field and vice versa.

magnetic material is not less than one. It is alw

eld lines are always nearly normal to the surfac

gnetisation of a paramagnetic sample can be of

gnetisation of a ferromagnet. This requires hig

ns:

ater or

ield? If not,

at everythe surface

e of the same

les get

, a

e to the

o the

as a

lied

ays greater

of such

the same

magnetising


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Explain qualitatively on the basis of domain picture the irreversibility in the

magnetisation curve of a ferromagnet.

The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel

piece. If the material is to go through repeated cycles of magnetisation, which piece will

dissipate greater heat energy?

A system displaying a hysteresis loop such as a ferromagnet, is a device for storing

memory? Explain the meaning of this statement.

What kind of ferromagnetic material is used for coating magnetic tapes in a cassette

player, or for building memory stores in a modern computer?

A certain region of space is to be shielded from magnetic fields.

Suggest a method.

Answer

The hysteresis curve (B-Hcurve) of a ferromagnetic material is shown in the following

figure.

It can be observed from the given curve that magnetisation persists even when the

external field is removed. This reflects the irreversibility of a ferromagnet.

(b)The dissipated heat energy is directly proportional to the area of a hysteresis loop. A

carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat

energy.


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(c)The value of magnetisatio

magnetisation. These bits of

Hysteresis loops can be used

(d)Ceramic is used for coati

memory stores in modern co

(e)A certain region of space

soft iron rings. In such arran

Question 5.18:

A long straight horizontal ca

west to 10 north of east. Th

the geographic meridian. Th

angle of dip is zero. Locate t

(At neutral points, magneticthe horizontal component of

Answer

Current in the wire,I= 2.5

Angle of dip at the given loc

Earths magnetic field,H= 0

The horizontal component o

HH=Hcos

The magnetic field at the ne

relation:

Where,

is memory or record of hysteresis loop cycles

information correspond to the cycle of magnetis

for storing information.

g magnetic tapes in cassette players and for bui

puters.

an be shielded from magnetic fields if it is surr

ements, the magnetic lines are drawn out of the

le carries a current of 2.5 A in the direction 10

magnetic meridian of the place happens to be

earths magnetic field at the location is 0.33 G,

e line of neutral points (ignore the thickness of

field due to a current-carrying cable is equal anearths magnetic field.)

tion on earth, = 0

.33 G = 0.33 104

T

earths magnetic field is given as:

tral point at a distanceR from the cable is give

of

ation.

lding

unded by

region.

south of

0 west of

and the

the cable).

opposite to

by the


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Hence, a set of neutral points

distance of 1.51 cm.

Question 5.19:

A telephone cable at a place

1.0 A in the same direction eand the angle of dip is 35. T

magnetic fields at points 4.0

Answer

Number of horizontal wires i

Current in each wire,I= 1.0

Earths magnetic field at a lo

Angle of dip at the location,

Angle of declination, 0

For a point 4 cm below the

Distance, r= 4 cm = 0.04 m

The horizontal component o

Hh =Hcos B

Where,

ce =

parallel to and above the cable are located at a

as four long straight horizontal wires carrying

ast to west. The earths magnetic field at the plae magnetic declination is nearly zero. What ar

cm below the cable?

n the telephone cable, n = 4

A

cation,H= 0.39 G = 0.39 104

T

= 35

cable:

earths magnetic field can be written as:

normal

a current of

ce is 0.39 G,the resultant


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B = Magnetic field at 4 cm due to currentIin the four wires

= Permeability of free space = 4 10

7

Tm A

1

= 0.2 104

T = 0.2 G

Hh = 0.39 cos 35 0.2

= 0.39 0.819 0.2 0.12 G

The vertical component of earths magnetic field is given as:

Hv =Hsin

= 0.39 sin 35 = 0.22 G

The angle made by the field with its horizontal component is given as:

The resultant field at the point is given as:

For a point 4 cm above the cable:

Horizontal component of earths magnetic field:

Hh =Hcos +B

= 0.39 cos 35 + 0.2 = 0.52 G

Vertical component of earths magnetic field:

Hv =Hsin


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= 0.39 sin 35 = 0.22 G

Angle,

And resultant field:

Question 5.20:

A compass needle free to tur

of 30 turns and radius 12 cm.the magnetic meridian. Whe

east.

Determine the horizontal co

The current in the coil is rev

angle of 90 in the anticlock

needle. Take the magnetic de

Answer

Number of turns in the circul

Radius of the circular coil, r

Current in the coil,I= 0.35

Angle of dip, = 45

The magnetic field due to cu

Where,

= 22.9

in a horizontal plane is placed at the centre of

The coil is in a vertical plane making an anglethe current in the coil is 0.35 A, the needle poi

ponent of the earths magnetic field at the loca

rsed, and the coil is rotated about its vertical ax

ise sense looking from above. Predict the direc

clination at the places to be zero.

ar coil,N= 30

12 cm = 0.12 m

rentI, at a distance r, is given as:

circular coil

of 45 withnts west to

ion.

is by an

tion of the


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= 5.49 105 T

The compass needle points f

earths magnetic field is give

BH=Bsin

= 5.49 105

sin 45 = 3.88

When the current in the coil

angle of 90 , the needle will

point from East to West.

Question 5.21:

A magnetic dipole is under t

field directions is 60, and o

comes to stable equilibrium

other field?

Answer

Magnitude of one of the mag

Magnitude of the other magn

Angle between the two field

At stable equilibrium, the an

Angle between the dipole an

At rotational equilibrium, th

Torque due to fieldB1 = To

ce = 4 107

Tm A1

om West to East. Hence, the horizontal compo

n as:

105

T = 0.388 G

s reversed and the coil is rotated about its verti

reverse its original direction. In this case, the n

e influence of two magnetic fields. The angle b

e of the fields has a magnitude of 1.2 102

T.

t an angle of 15 with this field, what is the ma

netic fields,B1 = 1.2 102

T

etic field =B2

, = 60

le between the dipole and fieldB1, 1 = 15

fieldB2, 2 = 1 = 60 15 = 45

torques between both the fields must balance e

rque due to fieldB2

ent of

al axis by an

edle will

etween the

If the dipole

nitude of the

ach other.


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MB1 sin1 =MB2 sin2

Where,

M= Magnetic moment of th

Hence, the magnitude of the

Question 5.22:

A monoenergetic (18 keV) eto a horizontal magnetic fiel

down deflection of the beam

in this exercise are so chosen

magnetic field on the motion

TV set.]

Answer

Energy of an electron beam,

Charge on an electron, e = 1.

E= 18 103

1.6 1019

J

Magnetic field,B = 0.04 G

Mass of an electron, me = 9.1

Distance up to which the ele

We can write the kinetic ene

dipole

other magnetic field is 4.39 103

T.

ectron beam initially in the horizontal directionof 0.04 G normal to the initial direction. Estim

over a distance of 30 cm (me= 9.11 1019

C).

that the answer will give you an idea of the eff

of the electron beam from the electron gun to t

E= 18 keV = 18 103

eV

6 1019

C

1 1019

kg

tron beam travels, d = 30 cm = 0.3 m

gy of the electron beam as:

is subjectedate the up or

Note:Data

ct of earths

e screen in a


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The electron beam deflects a

The force due to the magneti

Let the up and down deflecti

Where,

= Angle of declination

Therefore, the up and down

Question 5.23:

A sample of paramagnetic sa

1.5 1023

J T1

. The sample

and cooled to a temperature

long a circular path of radius, r.

c field balances the centripetal force of the path.

n of the electron beam be

eflection of the beam is 3.9 mm.

lt contains 2.0 1024

atomic dipoles each of dip

is placed under a homogeneous magnetic field

f 4.2 K. The degree of magnetic saturation achi

ole moment

of 0.64 T,

eved is equal


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to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and

a temperature of 2.8 K? (Assume Curies law)

Answer

Number of atomic dipoles, n = 2.0 1024

Dipole moment of each atomic dipole,M= 1.5 1023

J T1

When the magnetic field,B1 = 0.64 T

The sample is cooled to a temperature, T1 = 4.2K

Total dipole moment of the atomic dipole,Mtot = n M

= 2 1024

1.5 1023

= 30 J T1

Magnetic saturation is achieved at 15%.

Hence, effective dipole moment,

When the magnetic field,B2 = 0.98 T

Temperature, T2 = 2.8K

Its total dipole moment =M2

According to Curies law, we have the ratio of two magnetic dipoles as:

Therefore, is the total dipole moment of the sample for a magnetic field of

0.98 T and a temperature of 2.8 K.


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Which of these relations is in accordance with the result expected classically? Outline the

derivation of the classical result.

Answer

The magnetic moment associated with the intrinsic spin angular momentum ( ) is given

as

The magnetic moment associated with the orbital angular momentum ( ) is given as

For current i and area of cross-sectionA, we have the relation:

Where,

e = Charge of the electron

r= Radius of the circular orbit

T= Time taken to complete one rotation around the circular orbit of radius r

Angular momentum, l=mvr

Where,

m = Mass of the electron

v = Velocity of the electron

Dividing equation (1) by equation (2), we get:


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Therefore, of the two relatio s, is in accordance with classical p ysics.

Documents

5 Magnetism and Matter