L1_Ch-5_ Sol_ Set-1_ Magnetism and Matter.pmd

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SECTION - A

School / Board Exam. Type Questions

Very Short Answer Type Questions :

1. How can a paramagnetic and a diamagnetic rod be distinguished in a magnetic field?

Sol. When both rods are suspended in a magnetic field, diamagnetic rod stands perpendicular while paramagnetic

rod stands parallel to magnetic field.

2. What is curie temperature?

Sol. It is the temperature above which ferromagnetic material behaves like paramagnetic material.

3. How does the magnetic permeability differ for dia, para and ferromagnetic substances.

Sol. Magnetic permeability for diamagnetic material < 1

paramagnetic material > 1

ferromagnetic material >> 1

4. Earth’s core has ferromagnetic material like iron yet it is not considered as source of earth’s magnetism why?

Sol. These material in earth core is in molten state i.e., at temperature much above its curie temperature hence

it cannot retain magnetism.

5. How relative permeability of a magnetic material is related to its magnetic susceptibility?

Sol. (1 )r m

where r = relative permeability

m

= magnetic susceptibility.

6. What is angle of dip at equator?

Sol. Zero

7. What is SI unit of magnetic permeability and magnetic susceptibility?

Sol. SI unit of magnetic permeability is newton/ampere2 or tesla-meter/ampere or weber/ampere-meter and magnetic

susceptibility is unit less.

8. If a bar magnet is cut into two pieces transverse to its length, how will its pole strength and magnetic dipole

moment change?

Sol. On cutting a bar magnet transverse to its length, the pole strengths remains unchanged and magnetic dipole

moment becomes half of its original value.

Chapter 5

Magnetism and Matter

Level - I

Solutions (Set-1)

2 Magnetism and Matter Solutions of Assignment (Set-1) (Level-I)


9. Which magnetic field configuration always have a north and south pole?

Sol. Magnetic field configuration which is produced by a source whose magnetic dipole moment is not zero, always

have north and south pole.

10. If magnetic monopoles existed, how would Gauss’s law of magnetism be modified?

Sol. If monopole existed, then Gauss’s law in magnetism would have been similar to Gauss’s law in electrostatics

i.e., 0 m

B dS q��

� . Where qm

is magnetic charge enclosed by Gaussian surface.

Short Answer Type Questions :

11. What is the main use of hysteresis curve?

Sol. The main use of hysteresis curve lies in the selection of suitable material for different uses like for

electromagnet, permanent magnets. This selection is done on the basis of properties like retentivity, coercivity,

energy loss, which are revealed by the hysteresis curve.

12. When an iron bar magnet is heated above 1000ºC and then cooled in a magnetic field free space, will it retain

magnetism?

Sol. No, it will lose its magnetism because curie temperature for iron is 750ºC. Above theis temperature iron

becomes paramagnetic and lose its magnetism.

13. What do you mean by soft magnetic material? Also write their uses?

Sol. The ferromagnetic material which have low retentivity, low coercivity and low hysteresis loss are called as soft

magnetic material. e.g., soft iron, Mu metal and stalloy. These are used for making electromagnets, core of

transformer, motors etc.

14. If a compass is taken to magnetic south pole of the earth, what will be the direction of the needle?

Sol. At poles earth’s magnetic field is vertical. Compass needle is free to move in a horizontal plane, therefore

compass may rest in any arbitrary direction, not necessarily in north south direction.

15. How does the knowledge of declination at a place help in navigation?

Sol. For navigation, true north and south directions are very essential. If we know angle of declination at a place

we can find geographical meridian i.e., we can find geographical north and south directions.

16. If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when

the core is empty.

Sol. Bismuth is a diamagnetic material, therefore

< 0

0

1

0 0

1B

B

B < B0

Hence field (B) in core will be less than the field when core is empty (B0).

17. What is geomagnetic equator?

Sol. A plane perpendicular to the line joining the magnetic poles of earth and passing through the centre of the

earth intersects the earth’s spherical surface into a circle. This circle is called as geomagnetic equator.

18. What are the magnetic maps?

Sol. The values of declination, dip and horizontal component of earth’s field vary from place to place. These values

are depicted in charts, known as magnetic maps.

3Solutions of Assignment (Set-1) (Level-I) Magnetism and Matter


19. What is the maximum value of dip angle, where it is maximum and with what instrument the dip angle is

measured?

Sol. 90º, at magnetic poles, dip circle is the instrument used to measure dip angle.

20. Write down formula for magnetic field due to a short bar magnet at equatorial position. What is angle between

magnetic field and magnetic dipole moment of the bar magnet at equatorial position?

Sol.0

34

MB

r

0 = magnetic permeability of free space.

M = Magnetic dipole moment of the bar magnet.

r = Distance of the point from the bar magnet.

B = Magnetic field.

Angle between and M B

� �

at perpendicular bisector (equatorial line) is 180º.

21. Soft iron is preferred for making the core of a transformer, why?

Sol. For soft iron area enclosed by hysteresis loop is very small, therefore, energy dissipated per unit volume per

cycle of magnetisation is small, hence it is preferred for core of transformer.

22. State Curie law in magnetism.

Sol. According to curie law, intensity of magnetisation I of magnetic material is directly proportional to magnetic

induction B0 and inversely proportional to the absolute temperature T of the material.

I B0

...(i)

1IT

...(ii)

0B

IT

0

0( )

HI B H

T

∵

0

m

C

T

,

m = Magnetic susceptibility C = Curie constant

23. What do you mean by hard magnetic material?

Sol. Those magnetic material which have high retentivity, high coercivity and large hysteresis loss are called hard

magnetic material e.g., steel, alnico, alcomax and ticonal. These are used for making permanent magnets.

24. Why is diamagnetism almost independent of temperature?

Sol. The induced dipole moment in a diamagnetic substance is opposite to B and is independent to the internal

motion of the atom. The internal motion of atom is temperature dependent.

25. A short bar magnet has a magnetic moment of 0.64 J T–1. Find magnitude of magnetic field produced by magnet

at distance of 10 cm from its centre on the equatorial line of the magnet. Also write the direction of magnetic

field.

Sol. B = 0

34

M

r



= 7

3

0.6410

(0.1)

S N

B

10 cm=

7

3

0.6410

10

= 0.64 × 10–4 T

B = 6.4 × 10–5 T.

B is opposite to magnetic dipole moment.

26. A circular coil of 20 turns and radius 10 cm carrying current of 0.080 A rests with its plane normal to an external

magnetic field of magnitude 2.0 × 10–2 T. When coil is turned by 30º, find magnitude of field torque on the coil.

Sol. Magnetic dipole moment

M = NiA

= 20 × 0.8 × (0.1)2

= 16 × 10–2 A-m2

= MB sin 30º

= 16 × 10–2 × 2 × 10–2 × 1

2

= 16 × 10–4 N-m

27. A bar magnet of magnetic dipole moment 12 A m2 has magnetic length 48 cm. Calculate the pole strength.

Sol. M = 12 Am2

L = 48 cm = 0.48 m

M = qmL

qm

= M

L=

12

0.48=

1200

48 = 25 Am

28. Whether it is possible that a magnet has

(a) No poles

(b) Two similar poles

(c) Three poles

Sol. (a) Yes, there can be a magnet with no poles. e.g., toroid.

(b) Yes, when two bar magnets are joined as shown.

N S S N

(c) Yes

29. Name parts of earth’s magnetic elements.

Sol. It includes

(a) Angle of declination.

(b) Angle of inclination.

(c) Horizontal component of earth’s magnetic field.



30. Find work done in rotating a magnetic dipole from its stable equilibrium position to unstable equilibrium position.

Sol. Work done in rotating a magnetic dipole in magnetic field.

W = MB (cos i – cos

f)

W = MB (cos 0° – cos 180°)

W = MB (1 – (–1))

W = 2MB

Long Answer Type Questions :

31. What is paramagnetism? How does a paramagnetic material behave in the presence of an external magnetic

field? Explain with the help of appropriate diagram.

Sol. Certain substances when placed in a magnetic field are feebly magnetised in the same direction as the

magnetising field. This is known as paramagnetism.

(a) The relative permeability of paramagnetic substances is slightly greater than one (r > 1). Therefore

external magnetic field prefer to pass through the paramagnetic substances than to air.

Paramagnetic Substance

(b) Paramagnetic substances obey Curie’s law, that their magnetic susceptibility decreases when its absolute

temperature increases according to inverse relation.

1

m

T

On lowering temperature magnetic susceptibility of paramagnetic substances decreases.

(c) Paramagnetic substances corresponds to a polar dielectrics. Both paramagnetic substances and polar

dielectrics have permanent dipole moment.

32. A magnetising field of 1600 A m–1 produces a magnetic flux of 4.8 × 10–5 Wb in an iron bar of cross-sectional

area 0.4 cm2. Calculate its

(a) Magnetic permeability

(b) Magnetic susceptibility

Sol. Magnetic flux density

B = A

=

5

4

4.8 10

0.4 10

= 1.2 Wb/m2

(a) Given, H = 1600 Am–1

= B

H =

1.2

1600 = 7.5 × 10–4 NA–2

(b) = 0 (1 +

m)

0

1m

4

7

7.5 101

4 3.14 10m

596m



33. (a) The magnetic susceptibility of iron is greater than that of aluminium. What do you infer from this?

(b) Explain why a ferromagnetic material is used as a device for storing memory.

Sol. (a) The greater is the value of magnetic susceptibility m

of a substance, greater is the intensity of

magnetisation for given magnetising field. Iron is more easily magnetised than aluminium.

(b) The magnetisation of a ferromagnetic substances are not completely reversible, this can be explain on

the basis of domain theory. When a ferromagnetic substances is kept in a magnetising field, it is

magnetised due to favourable motion of domain boundaries and suitable orientation of domain directions.

When magnetising field is with drawn domains are not completely disorient but retain some memories

of their previous magnetisation. This property of ferromagnetic substances are used to store informations

as in case of computer disc, cassette tapes.

34. Two small magnets are placed horizontally perpendicular to the magnetic meridian. Their north poles are at

30 cm east and 20 cm west from a compass needle. If the compass needle remains undeflected, compare

the magnetic moments of the magnets.

Sol. The compass needle on axis of the magnets of magnetic field at C due to both the magnets are B1 and B

2,

then

Since compass needle remains undeflected therefore

B1 = B

2

0 01 2

3 3

1 2

2 2

4 4

M M

r r

N S S N

N

EW

30 cm 20 cm

2 1

Compass

1 2

3 3(0.2) (0.3)

M M

1

2

8

27

M

M

35. A magnetising field of 1500 A/m produces a magnetic flux density 0.48 Wb/m2 in an iron bar. Calculatemagnetic permeability and susceptibility of iron bar used.

Sol. Given, H = A

1500m

,

B = 0.48 Wb/m2

= 0.48

1500

B

H =

4 T-m3.2 10

A

1r

0

1

4

7

3.2 101

4 10

255 1

254

36. (a) Distinguish between diamagnetic and ferromagnetic material in respect of their (i) Intensity of magnetisation

(ii) Behaviour in non-uniform magnetic field and (iii) susceptibility.

(b) If actual angle of dip at a place is and 1,

2 are apparent angles of dip in two mutually perpendicular

planes, then prove that cot2 = cot2 1 + cot2

2.



Sol. (a)Diamagnetic material Ferromagnetic material

Intensity of magnetization is low, negative. Intensity of magnetization is positive and high.

Behaviour in non uniform magnetic field Diamagnetic material moves from strongerfield to weaker field.

Ferromagnetic material moves from weaker field to stronger magnetic field.

Magnetic susceptibility of material For diamagnetic material it is negative and low.

For ferromagnetic substances it is high and positive.

(b) Angle of dip on meridian one is

1tan

cos

v

H

B

B

1

tantan

cos

1

tancos

tan

...(i)

Angle of dip on the meridian 2 is

90–

Meridian2

Meridian1

Magnetic meridian

BH

BH

sin B

H cos

2tan

sin

v

H

B

B

2

tantan

sin

2

tansin

tan

...(ii)

Squaring and adding equation (i) and (ii)

2 2

2 2

2 2

2 1

tan tansin cos

tan tan

2 2 2

1 2

1 1 1

tan tan tan

2 2 2

1 2cot cot cot Proved.

37. (a) A compass needle having magnetic moment 40 Am2 is pointing towards north at a place where horizontal

component of earth magnetic field is 80 Wb/m2. If torque acting on the needle is 1.6 × 10–3 N-m, then

calculate angle of dip at the place.

(b) What properties should be present in a material which can be used as permanent magnet?

Sol. (a) M = 40 Am2

BH = 80 × 10–6 Wb m–2

= 1.6 × 10–3 N-m

= MB sin

sin = M B

=

3

6

1.6 10

40 80 10

=

3

4

1.6 10

32 10

= 11.6

1032

= 1

2

sin = 30º



(b) (i) Retentivity should be high.

(ii) High coercivity so that magnetism cannot be easily erased.

(iii) High permeability.

(iv) High area under hysteresis loop.

38. A solenoid of 500 turns per meter is carrying current of 3 A. Its core is made of iron, which has a relative permeability

of 5000. Determine the magnitude of intensity of magnetisation I and magnetic field inside the core.

Sol. Magnetic intensity H = ni

= 500 × 3

= 1500 A m–1

r

= 1 + m

m

= r – 1

= 5000 – 1 (r = 5000)

= 4999 5000

r =

0

5000

= 5000 0

I = mH

= 5000 × 1500

6 –17.5 10 A mI

Magnetic field inside core, B = H = 5000 0H = 5000 (4 × 10–7) × 1500 = 3

9.4 TB

39. The magnetic field due to a very small magnet at a point is perpendicular to magnetic moment. Find the angle

made by the line joining centre of magnet to the point and magnetic moment of bar magnet.

Sol. From the diagram,

= 90 –

tan = cot ...(i)

S N M

B

As we know

1tan tan

2

1cot tan

2

cos 1 sin

sin 2 cos

2

tan 2

tan 2 1tan ( 2)



40. What is magnetic shielding? How can a space be shielded from magnetic fields?

Sol. The process of protecting a material from magnetic effect is called magnetic shielding. When we place a soft

iron piece (in the form of a ring) around the material to be protected, the magnetic field lines crowd into the

soft iron ring due to their high , and cannot enter the region within the ring. Thus magnetic field has no effect

on the objects within the ring. For magnetic screening, we can use a soft iron sheet, a hollow soft iron cylinder

or a hollow sphere of iron, apart from soft iron ring.

41. What is pole strength of a bar magnet if its magnetic dipole moment is 4 × 10–2 A-m2 and separation between

two poles in 5 cm?

Sol. Magnetic dipole moment

M = qm

× L

4 × 10–2 = qm

× 5 × 10–2

2

2

4 10

5 10m

q

qm

= 0.8 A m

42. The work done in turning a magnet of magnetic dipole moment M by 90º from its stable equilibrium position

is n times the corresponding work done to turn it through an angle of 60º. Find value of n.

Sol. Work done to turn magnet by 90º is

W1 = MB [1 – cos 90°]

W1 = MB ...(i)

Work done to turn magnet by 60º

W2

= MB (1 – cos 60°) = 1

12

MB

W2

= 2

MB

W1 = nW

2

MB = 2

MBn

2n

43. What do you mean by electromagnet? Compare permanent magnet with electromagnet.

Sol. Electromagnet is a temporary strong magnet and it is obtained by winding of solenoid on a soft iron core which

has high permeability and low retentivity.Soft iron core



Uses of electromagnet :

(i) It is used in electrical devices such as electric bell, electric fan, telegraph etc.

(ii) For lifting and transporting large mass of iron.

(iii) In removing pieces of iron from the wounds.

Comprison between permanent magnet with an electromagnet is :

(i) Permanent magnet cannot be readily demagnetised while electromagnet can be demagnetised by stopping

current through solenoid.

(ii) Strength of magnetic field produced by permanent magnet cannot be charged but that of electromagnet

can be changed by adjusting current through solenoid.

(iii) The polarity of permanent magnet is fixed while that of an electromagnet can be changed by reversing

the direction of current through the solenoid.

44. What do you mean by hysteresis? Explain it. Draw hysteresis loop and show remanence and coercivity in

hysteresis loop.

Sol. The phenomenon of lagging of intensity of magnetisation I or magnetic induction B, behind the magnetic field

intensity H when a magnetic material is subjected to a cycle of magnetisation is called hysteresis.

When an unmagnetised magnetic material is subjected to magnetising field of intensity H�

, material starts

developing magnetic induction inside it. As H�

increases magnetic induction in the material also increases and

finally get saturated. This represent allignment and merger of domains of ferromagnetic material. When H�

is

decreased, B�

also decreases but with lesser rate. When H�

becomes zero, there is some magnetic induction

left in the magnetial ( 0)B �

.

B

OH

e

a

b

c

d

This value of B�

left is called residual magnetism or remanence. Now if H�

is applied in opposite direction

residual magnetism starts reducing till it becomes zero. The value of negative H�

needed to bring residual

magnetism to zero is called coercive force or coercivity.

Ob - represent–remanence,Oc - represent - coercive force.

45. Define terms magnetic permeability and magnetic susceptibility. Derive relation between relative permeability

and susceptibility.

Sol. Magnetic Permeability ()

It is defined as the ratio of the magnitude of total magnetic field (B) inside a material to that of magnitude of

magnetic intensity (H)”

B

H



1. It is a scalar quantity, its unit Wb

A-m and dimensions are [MLT–2A–2]

2. The physical significance of magnetic permeability is that, it measures that extent to which a magnetising

field can penetrate or permeate a given magnetic material.

Magnetic Susceptibility [cm

]

Intensity of magnetisation of a material, I

is dependent on intensity of magnetising field H

as m

I H ,

where cm

is magnetic susceptibility.

1. It is a scalar quantitiy with no dimension, no unit.

2. The physical significance of magnetic susceptibility is that it is the degree of ease with which a magnetic

material can be magnetised. A material with higher value m

can easily be magnetised.

Relation between relative Permeability and magnetic susceptibility

As we known

B = 0H +

0I

B = 0 (H + I)

B = 0 (H +

mH)

B = 0H (1 +

m)

0(1 )

m

B

H = +

= 0 (1 +

m)

0

(1 )m

= +

(1 )r m

= + 0

r

=

SECTION - B

Model Test Paper

Very Short Answer Type Questions : [1 Mark]

1. Name the material which has negative magnetic susceptibility.

Sol. Diamagnetic material.

2. Name the material which is used to form electromagnet.

Sol. Soft iron.

3. What is the value of angle of inclination at south pole?

Sol. = 90º.

4. How does the intensity of magnetisation of a paramagnetic material vary with absolute temperature?

Sol. Intensity of magnetisation I increases with decrease in absolute temperature [T]

1I

T .



5. Give two points to compare the magnetic properties of steel and soft iron.

Sol. (i) Steel has high retentivity, iron has low retentivity.

(ii) Hysteresis loss is more in steel than in iron.

6. A magnetic dipole of dipole moment M is kept in uniform magnetic field aligned to it. If it is rotated by 180º what is

work done needed?

Sol. W = MB [1 – cos ]

W = MB [1 – cos 180°]

W = MB [1 – (–1)]

W = 2 MB

7. What is direction of dipole moment of a magnetic dipole?

Sol. It is directed from south pole to north pole of dipole.

8. Magnetic field lines are endless. What does it mean?

Sol. Magnetic field lines are continuous and form closed loop therefore it do not have ends.

Short Answer Type Questions : [2 Marks]

9 Compare permanent magnet with an electromagnet.

Sol. (i) Permanent magnet cannot be readily demagnetised while electromagnet can be demagnetised by stopping

current through solenoid.

(ii) Strength of magnetic field produced by permanent magnet cannot be changed but that of electromagnet can

be changed by adjusting current through solenoid.

(iii) Two polarity of permanent magnet is fixed while that of an electromagnet can be changed by reversing the

direction of current through the solenoid.

10. What do you mean by electromagnet? Write three uses of an electromagnet.

Sol. Electromagnet is a temporary strong magnet and it is obtained by winding of solenoid on a soft iron core which has

high permeability, low retentivity, and low coercivity.

Soft iron core

Uses of electromagnet :

(i) It is used in electrical devices such as electric bell, electric fan,

telegraph etc.

(ii) For lifting and transporting large mass of iron.

(iii) In removing pieces of iron from the wounds.



11. Define the terms retentivity and coercivity.

Sol. Retentivity : It is the measure of the magnetic field (B) left in a specimen when the magnetising field (H) is

removed.

Coercivity : It is the measure of the magnetising field (H) required to destroy the residual magnetism in the

specimen.

12. What do you mean by soft and hard magnetic material? Give two examples of each.

Sol. On the basis of their retentivity, coercivity, and hysteresis loss ferromagnetic materials are divided into two

categories.

Soft Magnetic Material : They have low coercivity, lower retentivity and small hysteresis loss. They are used in

electromagnets. e.g., soft iron, Mu-metal

Hard Magnetic Material : They have high coercivity, high retentivity and high hysteresis loss. These are used to

make permanent magnet. e.g., steel, alnico.

13. Why diamagnetism is said to be possessed by all magnetic material? Explain.

Sol. Magnetic dipole moment are induced in all materials whenever a magnetic field is applied. Thus, all material have

property of diamagnetism. However, if there is a permanent atomic magnetic moment, (paramagnetism and

diamagnetism) it is much stronger than diamagnetism and then material does not show diamagnetic properties.

14. At certain place the horizontal component of earth’s magnetic field is 3 times its vertical component. Find angle

of inclination at the place.

Sol. Given, 3H v

B B

tan = v

H

B

B

= 3

v

v

B

B

tan = 1

3

= 30º

15. A bar magnet of magnetic dipole moment 5.0 A m2 has poles 20 cm apart. Calculate the pole strength.

Sol. Given, M = 5. 0 A m2

L = 20 cm

= 20 × 10–2 m

= 0.20 m

M = m × L

5 5025 A-m

0.2 2

Mm

L

m = 25 A m.



16. What will be the value of the vertical component and total strength of earth’s magnetic field at a place where dip is

60º? Horizontal component is 0.3 G.

Sol. Given, BH

= 0.3 G

= 0.3 × 10–4 T

= 3 × 10–5 T

= 60º

tanv

H

B

B

5tan60º

3 10

v

B

53 3 10 T

v

B

B = 2 2

v HB B

= 5 2 5 2(3 3 10 ) (3 10 )

= 6 × 10–5 T

17. A circular loop of 300 turns and radius 7 cm carries current of 15 A. Calculate the magnitude of magnetic moment

associated with the loop.

Sol. M = NiA

= 300 × 15 × R2

= 2 222

300 15 7 10 7 107

= 4500 × 22 × 7 × 10–4

= 0.4500 × 154

= 69.30 A-m2

18. Is it possible that a magnet has

(a) No poles

(b) Two similar poles

(c) Three poles

Sol. (a) Yes, there can be a magnet with no pole. e.g., toroid.

(b) Yes, there can be a magnet with two similar poles. It can be obtained by joining two similar poles of two bar

magnet.

(c) Yes, there can be a magnet with three poles.

19. What do you mean by “Earth’s magnetic element”? Name the device which is used to measure angle of dip.

Sol. Magnetic elements of earth at a place are the quantities which describe completely in magnitude as well as in

direction, the magnetic field of earth at that place.



There are following three elements of earth magnetic field

(i) Magnetic declination.

(ii) Magnetic inclination or dip.

(iii) Horizontal component of earth’s magnetic field.

Angle of dip is measured by dip circle.

Long Answer Type Questions : [5 Marks]

20. Three identical specimens of magnetic materials nickel, antimony and aluminium are kept in a non uniform

magnetic field. Draw the modifications in the field lines in each case.

Sol.

Nickel Antimony Aluminium

It is ferromagnetic so field lines

preferred to pass through the

material.

It is diamagnetic so field lines

are repelled away from the

material.

It is paramagnetic so attract

field lines but less than Nickel.

21. Define magnetic dipole moment of a current loop. Write expression for the magnetic moment when an electron

revolves at a speed v around an orbit of radius r in hydrogen atom.

Sol. Magnetic dipole moment of a current loop is equal to the product of current and the area enclosed by the current

loop if there is only one turn.

M = I × A

2

2

evM r

r

2

evrM

( )

2

e mvrM

m

2

hmvr n

4

enhM

m

4

ehM n

m

n = principal quantum number

h = Planck’s constant

m = mass of electron

e = charge of electron

22. Define the terms (a) Intensity of magnetisation (b) Magnetic susceptibility, and magnetic permeability. Derive

relation between relative magnetic permeability and magnetic susceptibility.

Sol. Intensity of Magnetisation (I)



It is equal to the induced dipole moment per unit volume in a magnetic material when it is kept in magnetising field.

Volume

MI

Magnetic Susceptibility [m

]

It is the ratio of magnitude of intensity of magnetisation | |I

to that of magnetic intensity | |H

m

I

H It is unitless, dimensionless constant.

Magnetic Permeability ()

It is the ratio of magnitude of total magnetic flux density in a material to that of magnitude of magnetic intensity

| |H

of magnetising field.

B

H

Relation between relative Permeability and magnetic susceptibility

As we known

B = 0H +

0I

B = 0 (H + I)

B = 0 (H +

mH)

B = 0H (1 +

m)

0(1 )

m

B

H = +

= 0 (1 +

m)

0

(1 )m

= +

(1 )r m

= + 0

r

=

��



Objective Type Questions

(Bar Magnet)

1. Suppose an isolated north pole is kept at the centre of a circular loop carrying a electric current i. The magnetic

field due to the north pole at a point on the periphery of the wire is B. The radius of the loop is a. The force

on the wire is

(1) Nearly 2aiB perpendicular to the plane of the wire

(2) 2aiB in the plane of the wire

(3) aiB along the axis of the wire

(4) Zero

Sol. Answer (1)

dF ldlBi

F

BN

total 2F i Ba

Perpendicular to the plane of the paper.

2. If L be the length of a bar magnet then separation between the two poles is nearly

(1)10

9 L (2)

7

6 L (3)

3

1 L (4)

2

1 L

Sol. Answer (2)

3. Two identical thin bar magnets each of length l and pole strength m are placed at right angle to each other

with north pole of one touching the south pole of the other. Magnetic moment of the system is

(1) 2ml (2) ml (3) 2 ml (4)2

ml

Sol. Answer (3)

S N

S

N

2ml

4. A solenoid of length 10 cm and radius 1 cm contains 200 turns and carries a current of 10 A. The value of

pole strength of each pole is

(1) 2 A m (2) 2 A m (3) 4 A m (4) 10 A m

Sol. Answer (2)

NIA = mL m = NIA/L = 200 (10–2)2 10/0.1 = 2

5. A long magnetic needle of length 2L, magnetic moment M and pole strength m units is broken into two at

the mid point. The magnetic moment and pole strength of each piece will be

(1) M/2, m/2 (2) M, m/2 (3) M/2, m (4) M, m

Sol. Answer (3)

Solutions (Set-2)



M = m(2l)

Pole strength does not change, on breaking the magnet but magnetic moment will become half.

6. A bar magnet of magnetic moment M is cut into two equal parts along its length. The magnetic moment of

either part is

(1) 2M (2) M (3) M/2 (4) Zero

Sol. Answer (3)

M = ml

'2

mM l

M' = M/2

7. A steel wire of length l has a magnetic moment M. It is then bent into a semi-circular arc. The new magnetic

moment is

(1) M (2)2M

(3)

M

(4) 2 M

Sol. Answer (2)

l = r

lr

2r

M = ml

M' = m2r

2'

lM m

2'

MM

8. Points A and B are situated perpendicular to the axis of a small bar magnet at large distances x and 3x from

its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to

(1) 2 : 9 (2) 1 : 9 (3) 27 : 1 (4) 9 : 1

Sol. Answer (3)

0

3

2

4A

MB

x

0

3

2

4 3B

MB

x

1

27

A

B

B

B

9. Figure shows two small identical magnetic dipoles a and b of magnetic moments M each, placed at a

separation 2d, with their axes perpendicular to each other. The magnetic field at the point P mid way between

the dipoles is

aP

b

2d

(1)3

0

4

2

d

M

(2)3

0

4 d

M

(3) Zero (4)3

0

4

5

d

M



Sol. Answer (4)

2d

dd

S

N

N S

B2

0

1 3

2

4

MB

d

0

2 34

MB

d

2 2 0

net 1 2 35

4

MB B B

d

10. Two short magnets of equal dipole moments M are fastened perpendicularly at their centres as shown below

(point P is in the plane of the magnets)

d

P

S

S

N

N

The magnetic field (magnitude) at P, distance d from the centre on the bisector of the right angle is

(1) 3

0

4 d

M

(2)3

0 2

4 d

M

(3) 3

0 2

4 d

M

(4)3

0 22

4 d

M

Sol. Answer (4)

S

S

N

N

M can be broken of both the magnets in terms of their components, one in direction of bisector and other

perpendicular to it

Parallel components will be added whereas perpendicular components will be cancel out

So 0net

3

2 2cos45

4

MB

d

��

0

3

2 2

4

M

d

11. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torqueneeded to maintain the needle in this position will be

(1) W (2) 3W (3)3

2

W(4) 2W

Sol. Answer (2)

– cos0 –i

U MB MB



60°

– cos60f

U MB

–

2

MB

–

f iW U U – 2

2 2

MB MBMB MB W

sin60MB

32 3

2W W

(Bar Magnet in a Magnetic Field, Magnetism and Gauss's Law)

12. Potential energy of a bar magnet of magnetic moment M placed in a magnetic field of induction B such thatit makes an angle with the direction of B is (take = 90° as datum)

(1) –M B sin (2) –M B cos (3) M B (1 – cos ) (4) M B (1 + cos )

Sol. Answer (2)

U = –MB cos

13. A magnetic dipole is placed at right angles to the direction of lines of force of magnetic induction B. If it is rotatedthrough an angle of 180o, then the work done is

(1) 2MB (2) MB (3) –2MB (4) Zero

Sol. Answer (4)

Ui = – MB cos90°

Uf = – MB cos90°

W = 0 – 0 = 0

14. Which of the following statements regarding magnetic lines of force is correct?

(1) Total magnetic flux linked with a closed surface is always zero

(2) They need not be perpendicular to the surface from where they start or where they meet

(3) They may or may not pass through a conductor

(4) All of these

Sol. Answer (4)

15. The time period of a freely suspended magnet is 4 s. If it is broken in length into two equal parts and one

part is suspended in the same way, then its time period will be

(1) 2 s (2) 4 s (3) 0.5 s (4) 0.25 s

Sol. Answer (1)

' 2 s2

TT



16. The magnetic moments of two bar magnets of same size are in the ratio 1 : 2. When they are placed one over

the other with their similar poles together, then their period of oscillation in a magnetic field is 3s. If one of the

magnets is reversed, then the period of oscillation in the same field will be

(1) 3 s (2) 3 3 s (3) 3 s (4) 6 s

Sol. Answer (2)

1T

M

1 2

1 2

–3 3 2 – 1 3 3 s

2 1

M MT

T M M T

17. Magnetostatic screening or shielding can be created by

(1) Super conductor (2) Soft iron ring (3) Both (1) & (2) (4) Neither (1) nor (2)

Sol. Answer (3)

18. Planets producing larger magnetic field have larger

(1) Rotational speed (2) Density (3) Mass (4) Size

Sol. Answer (1)

19. To shield an instrument from external magnetic field, it is placed inside a cabin made of

(1) Wood (2) Plastic

(3) Diamagnetic substances (4) Iron

Sol. Answer (4)

(Earth's Magnetism)

20. A dip needle free to move in a vertical plane perpendicular to the magnetic meridian will remain

(1) Horizontal (2) Vertical

(3) At an angle of 60° to the vertical (4) At an angle of 45° to the horizontal

Sol. Answer (2)

tantan

cos90A

tanA

90A

21. How many neutral points will be obtained when a bar magnet is kept with magnetic moment parallel to earth's

magnetic field?

(1) One (2) Two (3) Four (4) Infinite

Sol. Answer (4)

(Magnetisation and magnetic Intensity, Magnetic properties of Materials)

22. The magnetic induction and magnetising field intensity in a sample of magnetic material are B and H

respectively. The magnetic susceptibility of the material is

(1)H

B

0(2) 1

0

H

B

(3) 1H

B(4) 1

0 B

H



Sol. Answer (2)

0

B I H

0 mB H H

0

–1m

B

H

23. A frog is levitated in a magnetic field produced by current in a vertical solenoid below the frog is

(1) Diamagnetic (2) Paramagnetic (3) Ferromagnetic (4) Ferrimagnetic

Sol. Answer (1)

24. When a ferromagnetic substance is heated to a temperature above its Curie temperature, it

(1) Behaves like a paramagnetic substance (2) Behaves like a diamagnetic substance

(3) Remains ferromagnetic (4) Is permanently magnetised

Sol. Answer (1)

25. Select the incorrect statement

(1) In a diamagnetic substance net magnetic moment of each atom/molecule is zero

(2) In a paramagnetic substance net magnetic moment of each atom/molecule is non-zero

(3) In a ferromagnetic material net magnetic moment of each domain is zero

(4) In a ferromagnetic material net magnetic moment of each domain is non-zero

Sol. Answer (3)

26. Soft iron is used to manufacture electromagnets because their

(1) Magnetic permeability is high and retentivity and coercive force are small

(2) Retentivity is high

(3) Coercive force is high

(4) Area of the hysteresis curve is large

Sol. Answer (1)

27. Relative permeability of superconductors is

(1) 0 (2) 1 (3) –1 (4) 0.5

Sol. Answer (1)

28. The effective radius of a circular coil is R and number of turns is N. The current through it is i ampere. The

work done in rotating the coil by angle of 180° in an external magnetic field B will be (initially plane of coil

is perpendicular to magnetic field)

(1) NiR2B (2) 2NiR2B (3)2

)2(

R

BiN

(4) Zero

Sol. Answer (2)

2 2– – cos180 – – cos0f i

W U U U Ni R Ni R

22 NiR B

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L1_Ch-5_ Sol_ Set-1_ Magnetism and Matter.pmd

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