Chapter 5 – Magnetism And Matter Questions and Answers: NCERT Solutions for Class 12 Physics

1. Answer the following questions carefully about magnetism on Earth:
a) What are the three independent quantities which are conventionally used to describe the magnetic field on the Earth’s surface specifically?

Ans: The three independent quantities conventionally utilized for specifying Earth’s magnetic field are:
I. Magnetic declination
II. The horizontal component of Earth’s magnetic field and
III. Magnetic inclination or angle of dip.

b) At one specific point in a village of south India, the angle of dip is nearly 18∘18∘. Compare this dip angle with the one you expect somewhere in Britain?

Ans: The angle that magnetic field lines would make with the surface of the Earth is called the dip angle. Thus, it depends on how far the point is located with respect to the North Pole or the South Pole.
The dip angle is maximum at the poles and minimum at the equator. The angle of dip would be greater at the location in Britain as it would be closer to the North Pole (and for a fact, it is about 70∘70∘) as compared to that at a location in southern India because this location is closer to the equator.

c) Let us make a map of the magnetic field lines at Melbourne in Australia which is somewhat close to the South Pole. Which way would the lines seem to go? Would they go into the ground, or would they come out of the ground (ground means the surface of the Earth)?

Ans: To understand the Earth’s magnetism in a simpler way, the Earth is considered to have a giant bar magnet inside itself. This magnet is aligned such that its north pole points to a location close to the geographic South Pole and its South Pole points to the geographic North Pole.
Hence, the magnetic field lines emanate from close to the south pole of the Earth and terminate at the North Pole. Now, in a map depicting Earth’s magnetic field lines, the field lines at Melbourne, Australia, which is close to the Earth’s South Pole, would come out of the surface of the Earth.

d) A free-to-move compass is kept such that its needle can point in the vertical plane. Which direction will it point to, if the compass is located right at the geomagnetic north pole or at the geomagnetic south pole?

Ans: The magnetic field lines originate and terminate at the geomagnetic south and north poles, respectively.
Hence, a compass on the geomagnetic North Pole or the South Pole that too aligned in the vertical plane would point upwards and downwards at the geomagnetic south and North Pole, respectively.
If it were kept horizontally, the needle would be perpendicular to the field lines of the Earth, and hence would be free to move in every direction.

e) The Earth’s field is claimed to roughly approximate the field due to a dipole of magnetic moment 8×1022J/T8×1022J/T that is located at the Earth’s centre. How can we check the order of magnitude of this number?

Ans: The magnetic strength is given by
B=μ0M4πr3B=μ0M4πr3
Here,
MM
is the magnetic moment, and is given as
MM
= 8×1022J/T8×1022J/T,
rr
is the radius of Earth,
r=6.4×106mr=6.4×106m
μ0μ0
is the permeability of the free space =
4π×10−7Tm/A4π×10−7Tm/A
Hence,
B=4π×10−7×8×10224π(6.4×106)3B=4π×10−7×8×10224π(6.4×106)3
⇒B=0.3G=3G×(10−1)⇒B=0.3G=3G×(10−1)
The order of magnitude of the magnetic field is −1−1.

f) Scientists claim the existence of several local poles on the Earth’s surface oriented in different directions, besides the main geomagnetic N-S poles. Justify the possibility of this.

Ans: Every small magnet has its own N-S poles. Similarly, a lot of magnetised minerals have their N-S poles. The field strength of these poles is very small at large distances, hence they don’t have a global effect, but could affect experiments in the lab-scale lengths.

2. Answer the following questions carefully about magnetism on Earth:
a) Does the Earth’s magnetic field vary in time as it varies from point to point in space. If it does, then on what time scale does it change significantly?

Ans: Earth’s magnetic field dynamically changes with time. To observe any appreciable change, it takes (in seconds) of the order of nearly 9 or above, i.e., a few hundred years.
The change in the Earth’s magnetic field with time should be neglected if we consider multiple decades’ time scales.

b) Why do Geologists not regard the iron in the Earth’s core as a source of the Earth’s magnetism?

Ans: Earth’s core contains molten iron. This form of iron is not magnetized. Hence, this iron can’t be considered a source of Earth’s magnetism.

c) It is considered that the charged currents in the outer regions of the Earth’s core are responsible for Earth’s magnetism. What could be the ‘battery’ (or the source of energy) to sustain these currents?

Ans: The radioactivity in Earth’s interior emits a lot of energy. Moreover, this energy sustains the charged currents in the conducting regions outside the Earth’s core which are considered to cause the Earth’s magnetism. However, this is not guaranteed as this is a topic in active research.

d) During the past 4 to 5 billion years. The Earth may have reversed the direction of its magnetic field several times. How do geologists know about something from such a distant past?

Ans: Direction of the Earth’s magnetic field has flipped several times during the last 4 to 5 billion years. During this time, the Earth has also cooled down, solidifying a lot of the molten metals into solid rocks.
Some of these rocks have had ferrous metals, and the magnetic fields got weakly recorded in them. Hence the geologists get clues about the geomagnetic history from the analysis of such rocks.

e) The Earth’s field departs from its dipole shape significantly at large distances (nearly >30,000 km). Which agencies could be responsible for this distortion?

Ans: Earth’s field usually departs from its dipole shape quite significantly at large distances because of the presence of the moving charges in the ionosphere. While in motion, these charged particles or ions generate the magnetic field.

f) Can the weak magnetic field of interstellar space have any significant consequence? Explain. The field in the interstellar space is of the order of
10−1210−12
T.
Note: The above exercise is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown and could be a great field of research. For details, one should consult a good text on geomagnetism.

Ans: We know that an extremely weak magnetic field can bend the trajectory (or the path) of moving charged particles. If the field is weak, then there will be a minor change in the direction of charged particles.
A constant field would make a moving charge go in circles. Since the interstellar field is very weak, the deviation in the path of the charged particle may not be noticeable.
For instance, if a charged particle is moving with a constant speed and is affected by the interstellar magnetic field perpendicular to its path, then it will experience a force that will make it go in a circle of a radius so large it can barely be noticed in small length scales.
However, with reference to the gigantic interstellar space, the deflection can affect the passage of charged particles since the distance scales under consideration are huge.

3. A short bar magnet placed with its axis at
30∘30∘
with a uniform external magnetic field of
0.25T0.25T
experiences a torque of magnitude equal to
4.5×10−2J4.5×10−2J
. What is the magnitude of magnetic moment of the magnet?

Ans: Provided in the question,
Magnetic field strength
B=0.25TB=0.25T
Torque on the bar magnet,
T=4.5×10−2JT=4.5×10−2J
Angle between the given bar magnet and the external magnetic field,
θ=30∘θ=30∘
Torque is related to magnetic moment (M) as:
T=MBsin(θ)T=MBsin⁡(θ)
⇒M=4.5×10−20.25×sin30∘=0.36J/T⇒M=4.5×10−20.25×sin⁡30∘=0.36J/T
Clearly, the magnetic moment of the magnet is
0.36J/T0.36J/T

4. A short bar magnet of magnetic moment
M=0.32J/TM=0.32J/T
is placed in a uniform magnetic field of
0.15T0.15T
. If the bar is free to rotate in the plane of the field, which orientation and would correspond to its
a) Stable?

Ans: It is provided that moment of the bar magnet,
M=0.32J/TM=0.32J/T
External magnetic field,
B=0.15TB=0.15T
It is considered as being in stable equilibrium, when the bar magnet is aligned along the magnetic field. Therefore, the angle
θθ
, between the bar magnet and the magnetic field is
0∘0∘
.
Potential energy of the system
=−MBcos(θ)=−MBcos⁡(θ)
⇒−MBcos(θ)=−0.32×0.15×cos(0)=−4.8×10−2J⇒−MBcos⁡(θ)=−0.32×0.15×cos⁡(0)=−4.8×10−2J
Hence the potential energy is
=−4.8×10−2J=−4.8×10−2J

b) Unstable equilibrium? What is the potential energy of the magnet in each case?

Ans: It is provided that moment of the bar magnet,
M=0.32J/TM=0.32J/T
External magnetic field,
B=0.15TB=0.15T
When the bar magnet is aligned opposite to the magnetic field, it is considered as being in unstable equilibrium,
θ=180∘θ=180∘
Potential energy of the system is hence
=−MBcos(θ)=−MBcos⁡(θ)
⇒−MBcos(θ)=−0.32×0.15×cos(180∘)=4.8×10−2J⇒−MBcos⁡(θ)=−0.32×0.15×cos⁡(180∘)=4.8×10−2J
Hence the potential energy is
=4.8×10−2J=4.8×10−2J

5. A closely wound solenoid of
800800
turns and area of cross section
2.5×10−4m22.5×10−4m2
carries a current of
3.0A3.0A
. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Ans: It is provided that number of turns in the solenoid,
n=800n=800
Area of cross-section,
A=2.5×10−4m2A=2.5×10−4m2
Current in the solenoid,
I=3.0AI=3.0A
A current-carrying solenoid is analogous to a bar magnet because a magnetic field develops along its axis, i.e., along its length joining the north and south poles.
The magnetic moment due to the given current-carrying solenoid is calculated as:
M=nIA=800×3×2.5×10−4=0.6J/TM=nIA=800×3×2.5×10−4=0.6J/T
Thus, the associated magnetic moment
=0.6J/T=0.6J/T

6. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of
0.25T0.25T
is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of
30∘30∘
with the direction of applied field?

Ans: Given is the magnetic field strength,
B=0.25TB=0.25T
Magnetic moment,
M=0.6/TM=0.6/T
The angle,
θθ
between the axis of the turns of the solenoid and the direction of the external applied field is
30∘30∘
Hence, the torque acting on the solenoid is given as:
τ=MBsin(θ)τ=MBsin⁡(θ)
⇒τ=0.6×0.25sin(30∘)⇒τ=0.6×0.25sin⁡(30∘)
⇒τ=7.5×10−2J⇒τ=7.5×10−2J
Hence the magnitude of torque is
=7.5×10−2J=7.5×10−2J

7. A bar magnet of magnetic moment
1.5J/T1.5J/T
lies aligned with the direction of a uniform magnetic field of
0.22T0.22T
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 field direction, (ii) opposite to the field direction?

Ans: Provided that,
Magnetic moment,
M=1.5J/TM=1.5J/T
Magnetic field strength,
B=0.22TB=0.22T

(i) Initial angle between the magnetic field and the axis is,
θ1=0∘θ1=0∘
Final angle between the magnetic field and the axis is,
θ2=90∘θ2=90∘
The work that would be required to make the magnetic moment perpendicular to the direction of magnetic field would be:
W=−MB(cosθ2−cosθ1)W=−MB(cos⁡θ2−cos⁡θ1)
⇒W=−1.5×0.22(cos90∘−cos0∘)⇒W=−1.5×0.22(cos⁡90∘−cos⁡0∘)
⇒W=−0.33(0−1)⇒W=−0.33(0−1)
⇒W=0.33J⇒W=0.33J

(ii) Initial angle between the magnetic field and the axis,
θ1=0∘θ1=0∘
Final angle between the magnetic field and the axis,
θ2=180∘θ2=180∘
The work that would be required to make the magnetic moment opposite (180 degrees) to the direction of magnetic field is given as:
W=−MB(cosθ2−cosθ1)W=−MB(cos⁡θ2−cos⁡θ1)
⇒W=−1.5×0.22(cos180∘−cos0∘)⇒W=−1.5×0.22(cos⁡180∘−cos⁡0∘)
⇒W=−0.33(−1−1)⇒W=−0.33(−1−1)
⇒W=0.66J⇒W=0.66J

b) What is the torque on the magnet in cases (i) and (ii)?

Ans: For the first (i) case,
θ=θ1=90∘θ=θ1=90∘
Hence the Torque,
τ⃗ =M⃗ ×B⃗ τ→=M→×B→
And its magnitude is:
τ=MBsin(θ)τ=MBsin⁡(θ)
⇒τ=1.5×0.22sin(90∘)⇒τ=1.5×0.22sin⁡(90∘)
⇒τ=0.33Nm⇒τ=0.33Nm
Hence the torque involved is
=0.33Nm=0.33Nm
For the second-(ii) case:
θ=θ1=180∘θ=θ1=180∘
And its magnitude of the torque is:
τ=MBsin(θ)τ=MBsin⁡(θ)
⇒τ=1.5×0.22sin(180∘)⇒τ=1.5×0.22sin⁡(180∘)
⇒τ=0Nm⇒τ=0Nm
Hence the torque is zero.

8. A closely wound solenoid of
20002000
turns and area of cross-section
1.6×10−4m21.6×10−4m2
, carrying a current of
4.0A4.0A
, is suspended through its center allowing it to turn in a horizontal plane.
a) What is the magnetic moment associated with the solenoid?

Ans: Given is the number of turns on the solenoid,
n=2000n=2000
Area of cross-section of the solenoid,
A=1.6×10−4m2A=1.6×10−4m2
Current in the solenoid,
I=4AI=4A
The magnetic moment inside the solenoid at the axis is calculated as:
M=nAI=2000×1.6×10−4×4=1.28Am2M=nAI=2000×1.6×10−4×4=1.28Am2

b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of
7.5×10−2T7.5×10−2T
is set up at an angle of
30∘30∘
with the axis of the solenoid?

Ans: Provided that,
Magnetic field,
B=7.5×10−2TB=7.5×10−2T
Angle between the axis and the magnetic field of the solenoid,
θ=30∘θ=30∘
Torque,
τ=MBsin(θ)τ=MBsin⁡(θ)
⇒τ=1.28×7.5×10−2sin(30∘)⇒τ=1.28×7.5×10−2sin⁡(30∘)
⇒τ=4.8×10−2Nm⇒τ=4.8×10−2Nm
Given the magnetic field is uniform, and the force on the solenoid is zero. The torque on the solenoid is
4.8×10−2Nm4.8×10−2Nm

9. A circular coil of
1616
turns and radius
10cm10cm
carrying a current of
0.750.75
A rests with its plane normal to an external field of magnitude
5.0×10−2T5.0×10−2T
. 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/s2.0/s
. What is the moment of inertia of the coil about its axis of rotation?

Ans: It is provided that,
The number of turns in the given circular coil solenoid,
N=16N=16
Radius of the coil,
r=10cm=0.1mr=10cm=0.1m
Cross-section of the coil,
A=πr2=π×(0.1)2m2A=πr2=π×(0.1)2m2
Current in the coil,
I=0.75AI=0.75A
Magnetic field strength,
B=5.0×10−2TB=5.0×10−2T
Frequency of oscillations of the coil,
υ=2.0/sυ=2.0/s
Therefore, magnetic moment,
M=NAI=NIπr2M=NAI=NIπr2
⇒M=16×7.5×π×0.12⇒M=16×7.5×π×0.12
⇒M=0.3777J/T⇒M=0.3777J/T
Frequency is given by the relation:
ν=12πMBI−−−−√ν=12πMBI
where,
I=I=
Moment of inertia of the coil
⇒I=MB4π2ν2⇒I=MB4π2ν2
⇒I=0.377×5×10−24π2×22⇒I=0.377×5×10−24π2×22
⇒I=1.19×10−4kgm2⇒I=1.19×10−4kgm2
Clearly, the moment of inertia of the coil about its axis of rotation
1.19×10−4kgm21.19×10−4kgm2

10. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at
22∘22∘
with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be
0.35G0.35G
. Determine the magnitude of the earth’s magnetic field at the place.

Ans: It is provided that,
The horizontal component of earth’s magnetic field,
BH=0.35GBH=0.35G
The angle made by the needle with the horizontal plane (angle of dip)
=δ=22∘=δ=22∘
Earth’s magnetic field strength is
BB
We can relate
BB
and
BHBH
as:
BH=BcosδBH=Bcos⁡δ
⇒B=BHcosδ⇒B=BHcos⁡δ
⇒B=0.35cos22∘=0.377G⇒B=0.35cos⁡22∘=0.377G
Clearly, the strength of earth’s magnetic field at the given location is
0.377G0.377G

11. At a certain location in Africa, a compass points
12∘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∘60∘
above the horizontal. The horizontal component of the earth’s field is measured to
0.16G0.16G
. Specify the direction and magnitude of the earth’s field at the location.

Ans: Provided that,
Angle of declination,
θ=12∘θ=12∘
Angle of dip,
δ=60∘δ=60∘
Horizontal component of earth’s magnetic field,
BH=0.16GBH=0.16G
The magnetic field of the Earth at the given location
=B=B
We can relate
BB
and
BHBH
as:
B=BHcosδB=BHcos⁡δ
⇒B=0.16cos60∘=0.32G⇒B=0.16cos⁡60∘=0.32G
Magnetic field of Earth lies in the vertical plane,
θ=12∘θ=12∘
west of the geographic meridian, making an angle of
δ=60∘δ=60∘
(upward) with the horizontal direction. Its magnitude is
0.32G0.32G

12. A short bar magnet has a magnetic moment of
0.48J/T0.48J/T
. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of
10cm10cm
from the center of the magnet on
a) the axis,

Ans: Provided that the magnetic moment of the given bar magnet,
MM
is
0.48J/T0.48J/T
Given distance,
d=10cm=0.1md=10cm=0.1m
The magnetic field at d-distance, from the centre of the magnet on the axis is given by the relation:
B=μ04π2Md3B=μ04π2Md3
here,
μ0=μ0=
Permeability of free space
=4π×10−7Tm/A=4π×10−7Tm/A
Substituting these values,
BB
becomes as follows:
⇒B=4π×10−74π2×0.480.13⇒B=4π×10−74π2×0.480.13
⇒B=0.96×10−4T=0.96G⇒B=0.96×10−4T=0.96G
The magnetic field is
0.96G0.96G
along the South-North direction.

b) the equatorial lines (normal bisector) of the magnet.

Ans: The magnetic field at a point which is
d=10cm=0.1md=10cm=0.1m
away on the equatorial of the magnet is given as:
B=μ04πMd3B=μ04πMd3
⇒B=4π×10−74π0.480.13⇒B=4π×10−74π0.480.13
⇒B=0.48×10−4T=0.48G⇒B=0.48×10−4T=0.48G
The magnetic field is
0.48G0.48G
along the North-South direction.

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
14cm14cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36G0.36G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e.,14cm14cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field).

Ans: Provided that,
The magnetic field of Earth at the given place,
H=0.36GH=0.36G
The magnetic field at a
dd
-distance, on the axis of the magnet is given as:
B1=μ04π2Md3=HB1=μ04π2Md3=H
Here,
μ0=μ0=
Permeability of free space
=4π×10−7Tm/A=4π×10−7Tm/A
M=M=
The magnetic moment
The magnetic field at the same distance
dd
, on the equatorial line of the magnet is given as:
B2=μ04πMd3B2=μ04πMd3
⇒B2=H/2⇒B2=H/2
(comparing with
B1B1
)
Therefore, the total magnetic field,
B=B1+B2B=B1+B2
⇒B=H+H/2⇒B=H+H/2
⇒B=0.36+0.18=0.54⇒B=0.36+0.18=0.54
Clearly, the magnetic field is 0.54 G along the direction of earth’s magnetic field.

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

Ans: According to what is given, the magnetic field on the axis of the magnet at a distance
d1=14cmd1=14cm
, can be written as:
B1=μ04π2Md13=HB1=μ04π2Md13=H
here,
MM
is the magnetic moment
μ0μ0
is the permeability of free space
HH
is the horizontal component of the given magnetic field at d1d1
When the bar magnet is turned through
180∘180∘
, then the neutral point will lie on the equatorial line.
Also, the magnetic field at a distance
d2d2
on the equatorial line of the magnet can be written as:
B2=μ04πMd23=HB2=μ04πMd23=H
Equating
B1B1
and
B2B2
we get:
2d13=1d232d13=1d23
⇒d2=d1(12)1/3⇒d2=d1(12)1/3
⇒14×0.794=11.1cm⇒14×0.794=11.1cm
Thus, the new null point will be located
11.1cm11.1cm
on the normal bisector.

15. A short bar magnet of magnetic moment 5.25×10−2J/T5.25×10−2J/T is placed with its axis perpendicular to the earth’s field direction. Magnitude of the earth’s field at the place is given to be 0.42G0.42G
. Ignore the length of the magnet in comparison to the distances involved. At what distance from the center of the magnet, the resultant field is inclined at 45∘45∘ with earth’s field on
a) its normal bisector and

Ans: Provided that,
The magnetic moment of the bar magnet,
M=5.25×10−2J/TM=5.25×10−2J/T
The magnitude of the Earth’s magnetic field at a place,
G=0.42G=0.42×10−4TG=0.42G=0.42×10−4T
The magnetic field at the distance of R from the centre of the magnet on the normal bisector is given by the relation:
B=μ04πMR3B=μ04πMR3
Here,
MM
is the above-mentioned magnetic moment
μ0μ0
is the permeability of free space
When the resultant field is inclined at
45∘45∘
with earth’s field,
B=HB=H
μ04πMR3=H=0.42×10−4μ04πMR3=H=0.42×10−4
⇒R3=4π×10−74π5.25×10−20.42×10−4=12.5×10−5⇒R3=4π×10−74π5.25×10−20.42×10−4=12.5×10−5
⇒R=5×10−2m=5cm⇒R=5×10−2m=5cm
Clearly, at a distance of 5cm from the center of the magnet, the resultant field is inclined at
45∘45∘
with earth’s field on its normal bisector.

b) its axis

Ans: Provided that,
The magnetic moment of the bar magnet,
M=5.25×10−2J/TM=5.25×10−2J/T
The magnitude of the Earth’s magnetic field at a place,
G=0.42G=0.42×10−4TG=0.42G=0.42×10−4T
The given magnetic field at
RR
distanced from the center of the magnet on a point on its axis is given as:
B′=μ04π2MR3B′=μ04π2MR3
The resultant field is inclined at
45∘45∘
with earth’s field
B′=HB′=H
⇒μ04π2M(R′)3=H⇒μ04π2M(R′)3=H
⇒(R′)3=4π×10−74π2×5.25×10−20.42×10−4=2.5×10−4⇒(R′)3=4π×10−74π2×5.25×10−20.42×10−4=2.5×10−4
⇒R=6.3×10−2m=6.3cm⇒R=6.3×10−2m=6.3cm
Clearly, at a distance of 6.3cm from the center of the magnet, the resultant field is inclined at
45∘45∘
with earth’s field on its axis.

16. Answer the following questions:
a) Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?

Ans: The thermal motion of molecules is random, and the randomness increases with increasing temperature. Considering this fact, the alignments of dipoles get disrupted at high temperatures.
On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetization when its temperature is lowered i.e., it is cooled.

b) Why is diamagnetism, in contrast, almost independent of temperature?

Ans: In presence of a magnetizing field, the induced dipole moment in a diamagnetic substance is always opposite to the magnetizing field.
Hence, the change in temperature that leads to a change in the internal motion of the atoms does not affect the diamagnetism of a material.

c) 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?

Ans: It is known that Bismuth is a diamagnetic substance. This means, the magnetic field due to the toroid will be the magnetizing field for the bismuth core which will be opposite to the induced magnetic field of Bismuth.
Hence the total field generated by the toroid will be slightly less than the empty-core-toroid.

d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

Ans: The permeability of ferromagnetic materials is dependent of the applied magnetic field. As observed from hysteresis curve, it is greater for a lower field and vice versa.

e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point). Why?

Ans: The permeability of ferromagnetic material is greater than 1 not less than 1. Therefore, magnetic field lines are always nearly normal to the surface of such materials at every point.

f) Would the maximum possible magnetization of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?

Ans: The maximum possible magnetization of a paramagnetic sample can be of the same order of magnitude as the magnetization of a ferromagnet. This requires high magnetizing fields for saturation.

17. Answer the following questions:
a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetization curve of a ferromagnet.

Ans: The B-H curve i.e., the Hysteresis curve of a ferromagnetic material is as shown in the figure below:

It can be seen from the above-given curve that magnetization-B persists even when the external field-H is removed. This shows the irreversibility of a ferromagnet, i.e., the magnetization will not drop by reducing the magnetization field just the same way it was increased by increasing the magnetization field.

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 magnetization, which piece will dissipate greater heat energy?

Ans: The dissipated heat energy is in proportion to the area inside the hysteresis loop. For a carbon steel piece, the hysteresis curve area is large. Thus, it dissipates 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.

Ans: The information of magnetization corresponds to the cycle of magnetization. Also, it can be seen that Hysteresis loops can be used for storing such information.
The value of magnetization is memory or record of hysteresis loop cycles of magnetization.

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?

Ans: Ceramic.
Ceramic is usually used for coating magnetic tapes in memory storage devices like cassette players and also for building memory stores in today’s computers.

e) A certain region of space is to be shielded from magnetic fields. Suggest a method.

Ans: A region of space can be shielded from magnetic fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

18. A long straight horizontal cable carries a current of
2.5A2.5A
in the direction
10∘10∘
south of west to
10∘10∘
north of east. The magnetic meridian of the place happens to be
10∘10∘
west of the geographic meridian. The earth’s magnetic field at the location is
0.33G0.33G
, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Ans: Provided that,
Current in the wire,
I=2.5AI=2.5A
The angle of dip at the location,
δ=0∘δ=0∘
The Earth’s magnetic field,
H=0.33G=0.33×10−4TH=0.33G=0.33×10−4T
The horizontal component of earth’s magnetic field is given as:
HH=Hcosδ=0.33×10−4×cos0∘=0.33×10−4THH=Hcos⁡δ=0.33×10−4×cos⁡0∘=0.33×10−4T
The magnetic field at the neutral point at a distance R from the cable is given by the relation:
HH=μ02πIRHH=μ02πIR
here,
μ0μ0
= Permeability of free space
=4π×10−7Tm/A=4π×10−7Tm/A
⇒R=μ02πIHH⇒R=μ02πIHH
⇒R=4π×10−72π2.50.33×10−4=15.15×10−3m=1.51cm⇒R=4π×10−72π2.50.33×10−4=15.15×10−3m=1.51cm
Clearly, a set of neutral points lie on a straight line parallel to the cable at a perpendicular distance of
1.51cm1.51cm
above the plane of the paper.

19. A telephone cable at a place has four long straight horizontal wires carrying a current of
1.0A1.0A
in the same direction east to west. The earth’s magnetic field at the place is
0.39G0.39G
, and the angle of dip is
35∘35∘
. The magnetic declination is nearly zero. What are the resultant magnetic fields at points
4.0cm4.0cm
below and above the cable?

Ans: It is provided that,
The number of horizontal wires in the telephone cable,
n=4n=4
Current in each wire,
I=1.0AI=1.0A
Earth’s magnetic field at any location,
H=0.39G=0.39×10−4TH=0.39G=0.39×10−4T
The angle of dip at the location,
δ=35∘δ=35∘
and the angle of declination,
θ∼0∘θ∼0∘
For a point that is 4cm below the cable:
Distance,
r=4cm=0.04mr=4cm=0.04m
The horizontal component (parallel to Earth’s Surface) of Earth’s magnetic field is:
HH=H.cosδ−BHH=H.cos⁡δ−B
Here,
BB
is the magnetic field at 4 cm due to current I in the four wires and
B=4μ02πIrB=4μ02πIr
Here,
μ0μ0
is the permeability of free space
=4π×10−7Tm/A=4π×10−7Tm/A
⇒B=44π×10−72π10.01⇒B=44π×10−72π10.01
⇒B=0.2×10−4T=0.2G⇒B=0.2×10−4T=0.2G
∴HH=0.39×cos35∘−0.2≈0.12G∴HH=0.39×cos⁡35∘−0.2≈0.12G
The vertical component (perpendicular to Earth’s surface) of earth’s magnetic field is given as:
Hv=H.sinδHv=H.sin⁡δ
⇒Hv=0.39×sin35∘=0.22G⇒Hv=0.39×sin⁡35∘=0.22G
The angle between the field with its horizontal component is given as:
θ=tan−1HvHHθ=tan−1HvHH
⇒θ=tan−10.220.12=61.39⇒θ=tan−10.220.12=61.39
The resultant field at the point is obtained as:
H1=HH2+Hv2−−−−−−−−−√H1=HH2+Hv2
⇒H1=0.222+0.122−−−−−−−−−−−√=0.25G⇒H1=0.222+0.122=0.25G
For a point that is 4 cm above the cable,
Horizontal component of earth’s magnetic field:
HH=H.cosδ−BHH=H.cos⁡δ−B
⇒HH=0.39cos35∘+0.2=0.52⇒HH=0.39cos⁡35∘+0.2=0.52
Vertical component of earth’s magnetic field:
Hv=H.sinδHv=H.sin⁡δ
⇒Hv=0.39sin35∘=0.22⇒Hv=0.39sin⁡35∘=0.22
The angle
θ=tan−1HvHHθ=tan−1HvHH
⇒θ=tan−10.220.52=22.9∘⇒θ=tan−10.220.52=22.9∘
And the resultant field is:
H1=HH2+Hv2−−−−−−−−−√H1=HH2+Hv2
⇒H1=0.222+0.522−−−−−−−−−−−√=0.56G⇒H1=0.222+0.522=0.56G
Clearly, the resultant magnetic field below the cable is
0.25G0.25G
and above the cable is
0.56G0.56G

20. A compass needle free to turn in a horizontal plane is placed at the center of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of
45∘45∘ with the magnetic meridian. When the current in the coil is 0.35A0.35A, the needle points west to east.
a) Determine the horizontal component of the earth’s magnetic field at the location.

Ans: Provided that,
The number of turns in the given circular coil,
N=30N=30
The radius of the given circular coil,
r=12cm=0.12mr=12cm=0.12m
Current in the coil,
I=0.35AI=0.35A
Angle of dip,
δ=45∘δ=45∘
The magnetic field due to the current I , at a distance r , is given as:
B=4μ02πIrB=4μ02πIr
here,
μ0μ0
= Permeability of free space
=4π×10−7Tm/A=4π×10−7Tm/A
⇒B=4π×10−74π2π×30×0.350.12⇒B=4π×10−74π2π×30×0.350.12
⇒B=54.9×10−4T=0.549G⇒B=54.9×10−4T=0.549G
The compass needle points West to East. Hence, the horizontal component of earth’s magnetic field is given as:
BH=B.sinδBH=B.sin⁡δ
⇒BH=0.549sin45∘=0.388G⇒BH=0.549sin⁡45∘=0.388G

b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90∘90∘ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Ans: If the direction of the current flowing in the coil is reversed and if the coil is also rotated about its vertical axis by an angle of 90∘90∘, the needle will rearrange and reverse its original direction. In the given case, the needle would point from East to West.

21. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60∘60∘, and one of the fields has a magnitude of1.2×10−2T1.2×10−2T. If the dipole comes to stable equilibrium at an angle of 15∘15∘ with this field, what is the magnitude of the other field?

Ans: Provided that,
Magnitude of one of the magnetic fields,
B1=1.2×10−2TB1=1.2×10−2T
Magnitude of the other magnetic field is
B2B2
Angle between the above-mentioned two fields,
θ=60∘θ=60∘
At the state of stable equilibrium, the angle between the dipole and field
B1B1
is
θ1=15∘θ1=15∘
Angle between the dipole and field
B2B2
is
θ2=θ−θ1=60∘−15∘=45∘θ2=θ−θ1=60∘−15∘=45∘
At a rotational equilibrium, the torques experienced by the dipole, due to both the fields must balance each other.
Therefore, torque due to field
B1=B1=
Torque due to field
B2B2
MB1sinθ1=MB2sinθ2MB1sin⁡θ1=MB2sin⁡θ2
Where,
M=M=
Magnetic moment of the dipole
⇒B2=B1sinθ1sinθ2⇒B2=B1sin⁡θ1sin⁡θ2
⇒B2=4.39×10−3T⇒B2=4.39×10−3T
Clearly, the magnitude of the other magnetic field is
4.39×10−3T4.39×10−3T

22. A monoenergetic ( 18keV18keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04G0.04G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30cm(me=9.11×10−31kg)30cm(me=9.11×10−31kg)
. Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.

Ans: Provided that,
Energy of an electron beam,
E=18keV=18×103eVE=18keV=18×103eV
Charge on an electron,
e=1.6×10−19Ce=1.6×10−19C
∴E=18×103×1.6×10−19V=2.88×10−15V∴E=18×103×1.6×10−19V=2.88×10−15V
The magnetic field,
B=0.04GB=0.04G
The mass of an electron,
me=9.11×10−31kgme=9.11×10−31kg
Distance till where the electron beam travels,
d=30cm=0.3md=30cm=0.3m
We can write the kinetic energy carried by the electron beam as:
E=12mv2E=12mv2
⇒v=2Em−−−√⇒v=2Em
⇒v=2×2.88×10−159.11×10−31−−−−−−−−−−−−−−√=0.795×108m/s⇒v=2×2.88×10−159.11×10−31=0.795×108m/s
The electron beam deflects and gets in a circular path of radius,rr
The force experienced due to the magnetic field balances the centripetal force of the path.
BeV=mv2rBeV=mv2r
⇒r=mvBe⇒r=mvBe
⇒r=9.11×10−31×0.795×1080.4×10−4×1.6×10−19=11.3m

⇒r=9.11×10−31×0.795×1080.4×10−4×1.6×10−19=11.3m
Let the up-down deflection of the beam be
x=r(1−cosθ)x=r(1−cos⁡θ)
Where,θθ
is the angle of declination given by,
θ=sin−1(d/r)=1.521∘θ=sin−1(d/r)=1.521∘
And
x=11.3(1−cos1.521∘)=0.0039mx=11.3(1−cos⁡1.521∘)=0.0039m
Clearly, the up and down deflection of the bean
=3.9mm=3.9mm

23. A sample of paramagnetic salt contains 2.0×10242.0×1024 atomic dipoles each of dipole moment 1.5×10−23J/T1.5×10−23J/T. The sample is placed under a homogeneous magnetic field of 0.64T0.64T , and cooled to a temperature of 4.2K4.2K The degree of magnetic saturation achieved is equal to 1515. What is the total dipole moment of the sample for a magnetic field of  0.98T0.98T and a temperature of 2.8K2.8K ? (Assume Curie’s law)

Ans: Provided that,
The number of atomic dipoles,
n=2.0×1024n=2.0×1024
Dipole moment for each atomic dipole,
M=1.5×10−23J/TM=1.5×10−23J/T
The given magnetic field,
B1=0.64TB1=0.64T
The sample is then cooled to a temperature,
T1=4.2KT1=4.2K
Total dipole moment of the atomic dipole,
Mtot=n×M=2×1024×1.5×10−23=30J/TMtot=n×M=2×1024×1.5×10−23=30J/T
Magnetic saturation is achieved at
1515
Hence, effective dipole moment,
M1=1510030=4.5J/TM1=1510030=4.5J/T
Now when the magnetic field is
B2=0.98TB2=0.98T
Temperature,
T2=2.8KT2=2.8K
Its total dipole moment
=M2=M2
According to Curie’s law, the ratio of the two magnetic dipoles at different temperatures is:
M2M1=B2B1T1T2M2M1=B2B1T1T2
⇒M2=B2T1M1B1T2⇒M2=B2T1M1B1T2
⇒M2=10.336J/T⇒M2=10.336J/T
Clearly, it can be seen that,
10.336J/T10.336J/T
is the total dipole moment of the sample for a magnetic field of
0.98T0.98T
when its temperature is
2.8K2.8K

24. A Rowland ring of mean radius 15cm350015cm3500 turns of wire wound on a ferromagnetic core of relative permeability 800800. What is the magnetic field B⃗ B→
in the core for a magnetizing current of 1.2A1.2A?

Ans: Provided that the mean radius of a Rowland ring is,
r=15cm=0.15mr=15cm=0.15m
And the number of turns on a ferromagnetic core,
N=3500N=3500
Relative permeability of the core material,
μr=800μr=800
Given magnetizing current is,
I=1.2AI=1.2A
The magnetic field generated by this current is given by the relation:
B=μrμ02πIrNB=μrμ02πIrN
where,
μ0μ0
is the permeability of free space
=4π×10−7Tm/A=4π×10−7Tm/A
⇒B=800×4π×10−72π1.20.153500=4.48T⇒B=800×4π×10−72π1.20.153500=4.48T
Clearly, the magnetic field in the core is obtained as
4.48T4.48T

25. The magnetic moment vectors μsμs and μlμl associated with the intrinsic spin angular momentum S⃗ S→ and orbital angular momentum l⃗ l→, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by μs=−(e/m)Sμs=−(e/m)S μl=−(e/2m)lμl=−(e/2m)l Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Ans: According to the definition of magnetic moment- μlμl and orbital angular momentum- ll Magnetic moment associated with the motion of the electron is: μl=iA= (e/T).πr2μl=iA=−(e/T).πr2 And the corresponding angular momentum is: l=mvr=m(2πr/T)rl=mvr=m(2πr/T)r Where rr is the radius of the orbit, which the mass of an electron is mm and its charge (−e)(−e) completes in time TT Dividingμlbylμlbyl , one would get: μll=−eT.πr2×Tm2πr2=−e2mμll=−eT.πr2×Tm2πr2=−e2m ⇒μ⃗ l=−(e2m)l⃗ ⇒μ→l=−(e2m)l→ Evidently, it can be seen that μ⃗ lμ→l and l⃗ l→will be antiparallel (both being normal to the plane of the orbit). In contrast, μss=(em)μss=(em) and it is derived on the basis of quantum mechanics and is verified experimentally.