NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 5
Chapter Name : Magnetism And Matter

Q5.1 Answer the following questions regarding earth’s magnetism:
(a) A vector needs three quantities for its speci cation. Name the three independent quantities
conventionally used to specify the earth’s magnetic eld.
(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or
smaller dip angle in Britain?
(c) If you made a map of magnetic eld lines at Melbourne in Australia, would the lines seem to go
into the ground or come out of the ground?
(d) 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?
(e) The earth’s eld, it is claimed, roughly approximates the eld due to a dipole of magnetic
moment 8 × 10 located at its centre. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the
earth’s surface oriented in different directions. How is such a thing possible at all?

Answer.(a) The three independent quantities conventionally used for specifying earth’s magnetic
eld are:
(i) Magnetic declination,
(ii) Angle of dip, and
(iii) Horizontal component of earth’s magnetic eld

(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.
Magnetic eld lines emanate from a magnetic north pole and terminate at a magnetic south pole.
Hence, in a map depicting earth’s magnetic eld lines, the eld lines at Melbourne, Australia
would seem to come out of the ground.
(d) If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be
free to move in the horizontal plane while earth's eld is exactly vertical to the magnetic poles. In
such a case, the compass can point in any direction.
(e) Magnetic moment, M = 8 × 10 J T
22 −1

Radius of earth, r = 6.4 × 10 m
6

μo M
Magnetic eld strength, B =
4πr3

where,

Permeability of free space = 4π × 10 −7 −1
μ0 TmA
−7 22
4π×10 ×8×10
∴ B = = 0.3G
3
6
4π×(6.4×10 )

This quantity is of the order of magnitude of the observed eld on earth.
(f) Yes, there are several local poles on earth's surface oriented in different directions. A
magnetised mineral deposit is an example of a local N-S pole.

Page : 200 , Block Name : Exercise

Q5.2 Answer the following questions:
(a) The earth’s magnetic eld varies from point to point in space. Does it also change with time? If
so, on what time scale does it change appreciably?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the
earth’s magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be
responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to
sustain these currents?
(d) The earth may have even reversed the direction of its eld several times during its history of 4
to 5 billion years. How can geologists know about the earth’s eld in such distant past?
(e) The earth’s eld departs from its dipole shape substantially at large distances (greater than
about 30,000 km). What agencies may be responsible for this distortion?
(f) Interstellar space has an extremely weak magnetic eld of the order of 10 T. Can such a
−12

weak eld be of any signi cant consequence? Explain.

Answer. (a) Earth's magnetic eld changes with time. It takes a few hundred years to change by an
appreciable amount. The variation in earth's magnetic eld with the time cannot be neglected.
(b) Earth's core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not
considered as a source of earth's magnetism.
(c) The radioactivity in earth's interior is the source of energy that sustains the currents in the
outer conducting regions of earth's core. These charged currents are considered to
be responsible for earth's magnetism.
(d) Earth reversed the direction of its eld several times during its history of 4 to 5 billion years.
These magnetic elds got weakly recorded in rocks during their solidi cation. One can get clues
about the geomagnetic history from the analysis of this rock magnetism.
(e) Earth's eld departs from its dipole shape substantially at large distances (greater than about
30,000 km) because of the presence of the ionosphere. In this region, earth's eld gets modi ed
because of the eld of single ions. While in motion, these ions produce the magnetic eld
associated with them.
(f) An extremely weak magnetic eld can bend charged particles moving in a circle. This may not
be noticeable for a large radius path. With reference to the gigantic interstellar space, the
de ection can affect the passage of charged particles.

Page : 200 , Block Name : Exercise

Q5.3 A short bar magnet placed with its axis at 30° with a uniform external magnetic eld of 0.25

T experiences a torque of magnitude equal to 4.5 × 10 −2
J . What is the magnitude of magnetic
moment of the magnet?

Answer. Magnetic eld strength, B = 0.25 T
Torque on the bar magnet, T = 4.5 × 10 J −2

Angle between the bar magnet and the external magnetic eld, 0 = 30°
Torque is related to magnetic moment (M) as:
T = MB sin
T
∴ M =
B sin θ
−2
4.5×10 −1
= ∘ = 0.36JT
0.25×sin 30

Hence, the magnetic moment of the magnet is 0.36J T −1

Page : 200 , Block Name : Exercise

Q5.4 A short bar magnet of magnetic moment m = 0.32 J T −1

is placed in a uniform magnetic eld of 0.15 T. If the bar is free to rotate in the plane of the eld,
which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the
potential energy of the magnet in each case?

Answer. Moment of the bar magnet, M = 0.32J T −1

External magnetic eld, B = 0.15 T

(a) The bar magnet is aligned along the magnetic eld. This system is considered as being in stable
equilibrium. Hence, the angle , between the bar magnet and the magnetic eld is 0°.
Potential energy of the system = -MB cos
= -0.32 x 0.15 cos 0°
= −4.8 × 10 J −2

(b) The bar magnet is oriented 180° to the magnetic eld. Hence, it is in unstable equilibrium.
= 180°
Potential energy = - MB cos
= -0.32 x 0.15cos 180°
= 4.8 × 10 J −2

Page : 200 , Block Name : Exercise

Q5.5 A closely wound solenoid of 800 turns and area of cross section 2.5 × 10 m carries a
−4 2

current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its
associated magnetic moment?

Answer. Number of turns in the solenoid, n = 800
Area of cross-section, A = 2.5 x 10 m
−4 2

Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic eld develops along its

axis, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated
as:
M=nIA
= 800 x 3 x 2.5 x 10
−4

= 0.6 J T
−1

Page : 201 , Block Name : Exercise

Q5.6 If the solenoid in above question is free to turn about the vertical direction and a uniform
horizontal magnetic eld of 0.25 T is applied, what is the magnitude of torque on the solenoid
when its axis makes an angle of 30° with the direction of applied eld?

Answer. Magnetic eld strength, B = 0.25 T
Magnetic moment, M = 0.6 T −1

The angle θ, between the axis of the solenoid and the direction of the applied eld is 30⁰.
Therefore, the torque acting on the solenoid is given as:
τ = MB sin θ
= 0.6 x 0.25 sin 30°
= 7.5 x 10⁻² J

Page : 201 , Block Name : Exercise

Q5.7 A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform
magnetic eld of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its
magnetic moment: (i) normal to the eld direction, (ii) opposite to the eld direction?
(b) What is the torque on the magnet in cases (i) and (ii)?

Answer. (a) Magnetic moment, M = 1.5J −1
T

Magnetic eld strength, B = 0.22 T
(i) Initial angle between the axis and the magnetic eld, θ = 0
1
∘

Final angle between the axis and the magnetic eld, θ = 90 .
2
∘

The work required to make the magnetic moment normal to the direction of magnetic eld is
given as:
W = −M B (cos θ2 − cos θ1 )
∘ ∘
= −1.5 × 0.22 (cos 90 − cos 0 )

= −0.33(0 − 1)

= 0.33J

(ii) Initial angle between the axis and the magnetic eld, θ = 0
1
∘

Final angle between the axis and the magnetic eld, θ = 180 .
2
∘

The work required to make the magnetic moment opposite to the direction of the magnetic eld is
given as:

W = −M B (cos θ2 − cos θ1 )
∘
= −1.5 × 0.22 (cos 180 − cos 0 )

= −0.33(−1 − 1)

= 0.66J

(b) For case (i): θ = θ
∘
2 = 90

∴ Torque, τ = M B sin θ
∘
= 1.5 × 0.22 sin 90

= 0.33J

For case (ii): θ = θ 2 = 180
∘

∴ Torque, τ = M B sin θ
∘
= M B sin 180 = 0J

Page : 201 , Block Name : Exercise

Q5.8 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10 m , carrying a
−4 2

current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic eld of
7.5 × 10
−2
T is set up at an angle of 30° with the axis of the solenoid?

Answer. Number of turns in the solenoid, n = 2000
Area of cross-section, A = 1.6 × 10 m
−4 2

Current in the solenoid, I = 4 A

(a) The magnetic moment along the axis of the solenoid is calculated as:
M = nAI
= 2000 × 1.6 × 10 × 4 −4

= 1.28Am
2

(b) Magnetic eld, B = 7.5 × 10 T −2

Angle between the magnetic eld and the axis of solenoid, θ = 30 ∘

Torque, τ = M B sin θ
= 1.28 × 7.5 × 10 sin 30
−2 ∘

= 4.8 × 10 Nm
−2

Since the magnetic eld is uniform, the force on the solenoid is zero. The torque on the solenoid is
−2
4.8 × 10 Nm

Page : 201 , Block Name : Exercise

Q5.9 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 eld of magnitude 5.0 × 10 T. The coil is free to turn about an axis in its
−2

plane perpendicular to the eld direction. When the coil is turned slightly and released, it
oscillates about its stable equilibrium with a frequency of 2.0s . What is the moment of inertia of
−1

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 = Πr = Π × (0.1) m 2 2 2

Current in the coil, I = 0.75 A
Magnetic eld strength, B = 5.0 × 10 T −2

Frequency of oscillation of the coil, v = 2.0s −1

2
∴ Magnetic moment, M = N I A = N I πr
2
= 16 × 0.75 × n × (0.1)
−1
= 0.377J T

Frequency is given by the relation:
1 MB
v = √
2π I

Where,
I = Moment of inertia of the coil
MB
∴ I =
2 2
4π v
−2
0.377×5×10
= 2 2
4π ×(2)

−4 2
= 1.19 × 10 kgm

Hence, the moment of inertia of the coil about its axis of rotation is 1.19 × 10 −4
kgm
2

Page : 201 , Block Name : Exercise

Q5.10 A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its
north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s
magnetic eld at the place is known to be 0.35 G. Determine the magnitude of the earth’s
magnetic eld at the place.

Answer. Horizontal component of earth’s magnetic eld, B = 0.35G H

Angle made by the needle with the horizontal plane = Angle of the dip = δ = 22 ∘

Earth’s magnetic eld strength = B
We can relate B and B as: H

B = B cos θ
H
BH
∴ B =
cos δ
0.35
= ∘
= 0.377G
cos 22

Hence, the strength of earth’s magnetic eld at the given location is 0.377 G.

Page : 201 , Block Name : Exercise

Q5.11 At a certain location in Africa, a compass points 12° west of the geographic north. The north
tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60°

above the horizontal. The horizontal component of the earth’s eld is measured to be 0.16 G.
Specify the direction and magnitude of the earth’s eld at the location.

Answer. Angle of declination, θ = 12 ∘

Angle of dip, δ = 60
∘

Horizontal component of earth’s magnetic eld, B H = 0.16G

Earth’s magnetic eld at the given location = B
We can relate B and B as: H

BH = B cos δ
BH
∴ B =
cos δ
0.16
= ∘
= 0.32G
cos 60

Earth’s magnetic eld lies in the vertical plane, 12 West of the geographic meridian, making an
∘

angle of 60 (upward) with the horizontal direction. Its magnitude is 0.32 G.
∘

Page : 201 , Block Name : Exercise

Q5.12 A short bar magnet has a magnetic moment of 0.48 J T . Give the direction and magnitude
−1

of the magnetic eld produced by the magnet at a distance of 10 cm from the centre of the magnet
on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Answer. Magnetic moment of the bar magnet, M = 0.48J T −1

(a) Distance, d = 10 cm = 0.1 m
The magnetic eld at distance d, from the centre of the magnet on the axis is given by the relation:

μ0 2M
B =
4π d3

Where,
μ = Permeability of free space = 4π × 10
−7 −1
0 TmA
−7
4π×10 ×2×0.48
∴ B =
4π×(0.1)3

= 0.96 × 10 T = 0.96G
−4

The magnetic eld is along the S - N direction.

(b) The magnetic eld at the distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet
is given as:
μ0 ×M
B =
3
4π×d
−7
4π×10 ×0.48
= 3
4π(0.1)

= 0.48 G
The magnetic eld is along the N - S direction.

Page : 201 , Block Name : Exercise

Q5.13 A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic

north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of
the magnet. The earth’s magnetic eld at the place is 0.36 G and the angle of dip is zero. What is
the total magnetic eld on the normal bisector of the
magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null
points, eld due to a magnet is equal and opposite to the horizontal component of earth’s
magnetic eld.)

Answer. Earth’s magnetic eld at the given place, H = 0.36 G
The magnetic eld at the distance d, on the axis of the magnet is given as:
μ0
= H ………….. (i)
2M
B = 1
4π d3

Where,
μ = Permeability of free space
0

M = Magnetic moment
The magnetic eld at the same distance d, on the equatorial line of the magnet is given as:
μ0 M
……….[Using equation (i)]
H
B2 = 3
=
4πd 2

Total magnetic eld, B = B 1 + B2

=H+
H

2

= 0.36 + 0.18 = 0.54 G
Hence, the magnetic eld is 0.54 G in the direction o the earth’s magnetic eld.

Page : 201 , Block Name : Exercise

Q5.14 If the bar magnet in question 5.13 is turned around by 180°, where will the new null points
be located?

Answer. The magnetic eld on the axis of the magnet at the distance d = 14 cm, be written as:
1
μ0 2M
B1 = 3
= H ………..(1)
4π(d1 )

Where,
M = magnetic moment
μ = Permeability of free space
0

H = Horizontal component of the magnetic eld at d 1

If the bar magnet is turned through 180 , then the neutral point lie on the equatorial line.
∘

Hence, the magnetic eld at a distance d , on the equatorial line of the magnet can be written as:
2
μ0 M
B2 =
3
= H …………(2)
4π(d2 )

Equating equations (1) and (2), we get:
2 1
3
= 3
(d1 ) (d2 )

3
d2 1
( ) =
d1 2

1

3
1
∴ d2 = d1 × ( )
2

= 14 x 0.794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.

Page : 201 , Block Name : Exercise

Q5.15 A short bar magnet of magnetic movement 5.25 × 10 3T is placed with its axis −2 −1

perpendicular to the earth’s eld direction. At what distance from the centre of the magnet, the
resultant eld is inclined at 45° with earth’s eld on (a) its normal bisector and (b) its axis.
Magnitude of the earth’s eld at the place is given to be 0.42 G. Ignore the length of the magnet in
comparison to the distances
Involved.

Answer. Magnetic moment of the bar magnet, M = 5.25 × 10 3T −2 −1

Magnitude of earth’s magnetic eld at a place, H = 0.42 G = 0.42 x 10 T
−4

(a) The magnetic eld at a distance R from the centre of the magnet on the normal bisector is
given by the relation:
μ0 M
B = 3
4πR

Where,
μ = Permeability of free space = 4π × 10
−7 −1
0 TmA

When the resultant eld is inclined at 45 with earth’s eld, B = H ∘

μ0 M −4
∴ = H = 0.42 × 10
4πR3
μ0 M
3
R = −4
0.42×10 ×4π
−7 −2
4π×10 ×5.25×10 −5
= = 12.5 × 10
−4
4π×0.42×10

∴ R = 0.05m = 5cm

(b) The magnetic eld at a distance R from the center of the magnet on its axis is given as:
′

μ0 2M
′
B = 3
4πR

The resultant eld is inclined at 45 with the earth’s eld. ∘

′
∴ B = H
μ0 2M
= H
3
4π(R′ )

3 μ0 2M
′
(R ) =
4π×H
−7 −2
4π×10 ×2×5.25×10 −5
= −4
= 25 × 10
4π×0.42×10

′
∴ R = 0.063m = 6.3cm

Page : 202 , Block Name : Exercise

Q5.16 Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising
eld) when cooled?
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the eld in the core be (slightly) greater or (slightly)
less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic eld? If not, is it

more for lower or higher elds?
(e) Magnetic eld lines are always nearly normal to the surface of a ferromagnet at every point.
(This fact is analogous to the static electric eld lines being normal to the surface of a conductor
at every point.) Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of
magnitude as the magnetisation of a ferromagnet?

Answer. (a) Owing to the random thermal motion of molecules, the alignments of the dipoles get
disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic
sample displays greater magnetisation when cooled.
(b) The induced dipole moment in a diamagnetic substance is always opposite to the magnetising
eld. Hence, the internal motion of the atoms ( which is related to the temperature) does not
affect the diamagnetism of the material.
(c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic eld
slightly greater than a toroid whose core is empty.
(d) The permeability of ferromagnetic materials is not independent of the applied magnetic eld.
It is greater for a lower eld and vice versa.
(e) The permeability of the ferromagnetic materials is not less than one. It is always greater than
one. Hence, magnetic eld lines are always nearly normal to the surface of such materials at every
point.
(f) The maximum possible magnetisation of a paramagnetic sample can be of the same order of
the magnitude as the magnetisation of a ferromagnet. This requires high magnetising elds for
saturation.

Page : 202 , Block Name : Additional Exercise

Q5.17 Answer the following questions:
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation
curve of a ferromagnet.
(b) 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?
(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’
Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or
for building ‘memory stores’ in a modern computer?
(e) A certain region of space is to be shielded from magnetic elds. Suggest a method.

Answer. The hysteresis curve (B-H) of a ferromagnetic materials is shown in the following gure.

(a) It can be observed from the given curve that magnetism persists even when the external eld is
removed. This re ects 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 dissipated greater that energy.
(c) The value of magnetism is memory or record of the hysteresis loop cycles of magnetisation.
These bits of information correspond to the cycles of the magnetisation. Hysteresis loops can be
used for storing information.
(d) Ceramic is used for coating magnetic tapes in cassette players and for building memory stores
in modern computers.
(e) A certain region of space can be shielded from magnetic elds if it is surrounded by soft iron
rings. In such arrangements, the magnetic lines are drawn out of the region.

Page : 202 , Block Name : Additional Exercise

Q5.18 A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west
to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic
meridian. The earth’s magnetic eld at the location is 0.33 G, and the angle of dip is zero. Locate
the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic eld due
to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic
eld.)

Answer. Current in the wire, I = 2.5 A
Angle of the dip at the given location on the earth, δ = 0
∘

Earth’s magnetic eld, H = 0.33 G = 0.33 × 10 T −4

The horizontal component of the earth’s magnetic elds is given as:
HH = H cos δ
−4 ∘ −4
= 0.33 × 10 × cos 0 = 0.33 × 10 T

The magnetic eld at the neutral point at a distance R from the cable is given by the relation:
μ0 I
HM =
2πR

Where,
−7 −1
μ0 = Permeability of free space = 4π × 10 TmA
μ0 I
∴ R =
2πHH

−7
4π×10 ×2.5 −3
= −4
= 15.15 × 10 m = 1.51cm
2π×0.33×10

Hence, a set of neutral points parallel to and above the cable are located at a normal distance of
1.5 cm.

Page : 202 , Block Name : Additional Exercise

Q5.19 A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0
A in the same direction east to west. The earth’s magnetic eld at the place is 0.39 G, and the
angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic elds
at points 4.0 cm below the cable?

Answer. Number of horizontal wires in the telephone cable, n = 4
Current in each wire, I = 1.0 A
Earth’s magnetic eld at a location, H = 0.39 G = 0.39 x 10 T −4

Angle of dip at the location, δ = 35 ∘

Angle of declination, θ ∼ 0 ∘

For a point 4 cm below the cable:
Distance, r = 4 cm = 0.4 m
The horizontal component of earth’s magnetic eld can be written as:
Hh = H cos δ − B

Where,
μ0 I
B = Magnetic eld 4 cm due to the current I in the four wires = 4 × 2πr

μ0 = Permeability of free space = 4n × 10 −7
TmA
−1

−7
4π×10 ×1
∴ B = 4 ×
2π×0.04
−4
= 0.2 × 10 T = 0.2G
∘
∴ Hh = 0.39 cos 35 − 0.2

= 0.39 × 0.819 − 0.2 ≈ 0.12G

The vertical component of the earth’s magnetic eld is given as:
Hv = H sin δ
∘
= 0.39 sin 35 = 0.22G

The angle made by the eld with its horizontal component is given as:
Hv
−1
θ = tan
Hh

−1 0.22 ∘
= tan = 61.39
0.12

The resultant eld at the point is given as:
2 2
H1 = √(Hv ) + (Hh )

2 2
= √(0.22) + (0.12) = 0.25G

For a point 4 cm above the cable:
Horizontal component of the earth’s magnetic eld:
Hh = H cos δ + B
∘
= 0.39 cos 35 + 0.2 = 0.52G

Vertical component of the earth’s magnetic eld:
Hv = H sin δ

∘
= 0.39 sin 35 = 0.22G
Hr
−1
=tan
Angle, θ H
h
= tan
−1 0.22

0.52
= 22.9
∘

And the resultant eld:
2 2
H2 = √(Hr ) + (H h)

2 2
= √(0.22) + (0.52) = 0.56T

Page : 202 , Block Name : Additional Exercise

Q5.20 A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of
30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic
meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic eld at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of
90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the
magnetic declination at the places to be zero.

Answer. Number of turns in the circular coil, N = 30
Radius of the circular coil, r = 12 cm = 0.12 m
Current in the coil, I = 0.35 A
Angle of the dip, δ = 45 ∘

(a) The magnetic eld due to the current I, at a distance r, is given as:
μ0 2πN I
B =
4πr

Where,
−7 −1
μ0 = Permeability of free space = 4n × 10 TmA
−7
4π×10 ×2π×30×0.35
∴ B =
4π×0.12
−5
= 5.49 × 10 T

The compass needle points from the west to the east. Hence, the horizontal component of the
earth’s magnetic eld is given as:
BH = B sin δ
−5 ∘ −5
= 5.49 × 10 sin 45 = 3.88 × 10 T = 0.388G

(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an
angle of 90 , the needle will reverse its original direction. In this case, the needle will point from
∘

East to west.

Page : 203 , Block Name : Additional Exercise

Q5.21 A magnetic dipole is under the in uence of two magnetic elds. The
angle between the eld directions is 60°, and one of the elds has a
magnitude of 1.2 × 10 T. If the dipole comes to stable equilibrium at an angle of 15° with this
−2

eld, what is the magnitude of the other eld?

Answer. Magnitude of one of the magnetic elds, B = 1.2 × 10 T
−2
1

Magnitude of the other magnetic eld = B 2

Angle between the two elds, 0 = 60°
At stable equilibrium, the angle between the dipole and eld B , θ = 15 1 1
∘

Angle between the dipole and eld B , θ = θ − θ = 60 − 15 = 45 2 2 1
∘ ∘ ∘

At rotational equilibrium, the torques between both the elds must balance each other.
∴ Torque due to eld B = Torque due to eld B 1 2

M B1 sin θ1 = M B2 sin θ2

Where,
M = Magnetic moment of the dipole
B1 sin θ1
∴ B2 =
sin θ2
−2 ∘
1.2×10 ×sin 15 −3
= ∘
= 4.39 × 10 T
sin 45

Hence, the magnitude of the other magnetic eld is 4.39 × 10 −3
T

Page : 203 , Block Name : Additional Exercise

Q5.22 A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to
a horizontal magnetic eld of 0.04 G normal to the initial direction. Estimate the up or down
de ection of the beam over a distance of 30 cm (m = 9.11 × 10 kg) e
−31

Answer. Energy of an electron beam, E = 18keV = 18 × 10 eV 3

−19
1.6 × 10 C
Charge of an electron, e =
3

E = 18 × 10 × 1.6 × 10 J
3 −19

Magnetic eld, B = 0.04 G
Mass of an electron, m = 9.11 × 10 kg e
−19

Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:
1 2
E = mv
2

2E
v = √
m

3 −19 −15
2×18×10 ×1.6×10 ×10 8
= √ −31
= 0.795 × 10 m/s
9.11×10

The electron beam de ects along a circular path of radius, r.
The force due to the magnetic eld balances the centripetal force of the path.
2
mv
BeV =
r
mv
∴ r =
Be
−31 8
9.11×10 ×0.795×10
= = 11.3m
−4 −19
0.4×10 ×1.6×10

Let the up and down de ection of the electron beam be x = r(1 − cos θ)
Where,
= Angle of declination
d
sin θ =
r

0.3
=
11.3
−1 0.3 ∘
θ = sin = 1.521
11.3
∘
And x = 11.3 (1 − cos 1.521 )

= 0.0039m = 3.9mm

Therefore, the up and down de ection of the beam is 3.9 mm.

Page : 203 , Block Name : Additional Exercise

Q5.23 A sample of paramagnetic salt contains 2.0 × 10 atomic dipoles each of dipole moment
24

1.5 × 10 JT
23
. The sample is placed under a homogeneous magnetic eld of 0.64 T, and cooled
−1

to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the
total dipole moment of the sample for a magnetic eld of 0.98 T and a temperature of 2.8 K?
(Assume Curie’s law)

Answer. Number of atomic dipoles, n = 2.0 × 10 24

Dipole moment of each atomic dipole, M = 1.5 × 10 JT −23 −1

When the magnetic eld, B = 0.64T 1

The sample is cooled to a temperature, T = 4.2 K 1
∘

Total dipole moment of the atomic dipole, M = n × M tot
24 −23
= 2 × 10 × 1.5 × 10
−1
= 30JT

Magnetic saturation is achieved at 15%.
Hence, effective dipole moment,
15 −1
M1 = × 30 = 4.5JT
100

When the magnetic eld, B = 0.98T 2

Temperature, T = 2.8 K 2
∘

Its total dipole moment = M 2

According to Curie's law, we have the ratio of two magnetic dipoles as:
M2 B2 T1
= ×
M1 B1 T2

B2 T1 M1
∴ M2 =
B1 T 2
0.98×4.2×4.5 −1
= = 10.336JT
2.8×0.64

Therefore, 10.336JT is the total dipole moment of the sample for a magnetic eld of 0.98 T and
−1

a temperature of 2.8 K.

Page : 203 , Block Name : Additional Exercise

Q5.24 A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core
of relative permeability 800. What is the magnetic eld B in the core for a magnetising current of
1.2 A?

Answer. Mean radius of a Rowland ring, r = 15 cm = 0.15 m
Number of turns on a ferromagnetic core, N = 3500
Relative permeability of the core material, μ = 800 r

Magnetising current, I = 1.2 A
The magnetic eld is given by the relation:
μ, μ0 I N

B =
2πr

Where,
mu = Permeability of free space = 4π × 10
−7 −1
0 TmA
−7
800×4π×10 ×1.2×3500
B = = 4.48T
2π×0.15

Therefore, the magnetic eld in the core is 4.48 T.

Page : 203 , Block Name : Additional Exercise

Q5.25 The magnetic moment vectors μ and μ associated with the intrinsic spin angular
s 1

momentum S and orbital angular momentum l, respectively, of an electron are predicted by
quantum theory (and veri ed experimentally to a high accuracy) to be given by:
μs = −(e/m)s

μ1 = −[e/2m]1

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-section A, 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 Angular momentum, l
= mvr
Where,
m = Mass of the electron
V = Velocity of the electron
Dividing equation (1) by equation (2), we get:
Therefore, of the two relations, is in accordance with classical physics.

Page : 203 , Block Name : Additional Exercise