Magnetism And Matter

Ans: 1
(a) The three independent quantities conventionally used for specifying earth’s magnetic field are:
(i) Magnetic declination,
(ii)Angle of dip, and
(iii) Horizontal component of earth’s magnetic field
(b)The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.
(c)It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.
Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field 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 field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.
(e)Magnetic moment, M= 8 ×10²² J T⁻¹
Radius of earth, r= 6.4 ×10⁶ m
Magnetic field strength,
Where,
= Permeability of free space =

Thisquantity is of the order of magnitude of the observed field 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.

Ans: 2
(a) Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field 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)Theradioactivity 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 field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.
(e)Earth’s field 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 field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.
(f)An extremely weak magnetic field 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 deflection can affect the passage of charged particles.

Ans: 3
Magnetic field strength, B= 0.25 T
Torque on the bar magnet, T= 4.5 ×10⁻²J
Angle between the bar magnet and the external magnetic field,θ= 30°
Torque is related to magnetic moment (M) as:
T= MB sin θ

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

Ans: 4
Moment of the bar magnet, M= 0.32 J T⁻¹
External magnetic field, B= 0.15 T
(a)The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle θ, between the bar magnet and the magnetic field is 0°.
Potential energy of the system

(b)The bar magnet is oriented 180°to the magnetic field. Hence, it is in unstable equilibrium.
θ= 180°
Potential energy = − MBcos θ


Ans: 5
Number of turns in the solenoid, n = 800
Area of cross-section, A= 2.5 ×10⁻⁴m²
Current in the solenoid, I= 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M= n IA
= 800 ×3 ×2.5 ×10⁻⁴
= 0.6 J T⁻¹

Ans: 6
Magnetic field 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 field is 30°.
Therefore, the torque acting on the solenoid is given as:


Ans: 7
(a)Magnetic moment, M= 1.5 J T⁻¹
Magnetic field strength, B= 0.22 T
(i)Initial angle between the axis and the magnetic field, θ₁= 0°
Final angle between the axis and the magnetic field, θ₂= 90°
The work required to make the magnetic moment normal to the direction of magnetic field is given as:

(ii)Initial angle between the axis and the magnetic field, θ₁= 0°
Final angle between the axis and the magnetic field, θ₂= 180°
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

(b)For case (i):
∴Torque,

Forcase (ii):
∴Torque,


Ans: 8
Number of turns on the solenoid,n = 2000
Area of cross-section of the solenoid, A= 1.6 ×10⁻⁴ m²
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
= 1.28 Am²
(b)Magnetic field, B = 7.5 ×10⁻²T
Angle between the magnetic field and the axis of the solenoid, θ= 30°
Torque,

Sincethe magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is

Ans: 9
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= πr²= π× (0.1)²m²
Current in the coil, I= 0.75 A
Magnetic field strength, B= 5.0 ×10⁻²T
Frequency of oscillations of the coil, v= 2.0 s⁻¹
∴Magnetic moment, M= NIA
= 16 × 0.75 × π× (0.1)²
= 0.377 J T⁻¹

Where,
I= Moment of inertia of the coil

Hence,the moment of inertia of the coil about its axis of rotation is

Ans: 10
Horizontal component of earth’s magnetic field, B_H= 0.35 G
Angle made by the needle with the horizontal plane = Angle of dip =
Earth’smagnetic field strength = B
Wecan relate Band B_Has:

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

Ans: 11
Angle of declination,θ= 12°
Angle of dip,
Horizontal component of earth’s magnetic field, B_H= 0.16 G
Earth’s magnetic field at the given location = B
Wecan relate Band B_Has:

Earth’s magnetic field 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.

Ans: 12
Magnetic moment of the bar magnet, M= 0.48 J T⁻¹
(a) Distance,d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:

Where,
= Permeability of free space =

Themagnetic field is along the S − N direction.
(b) The magnetic field at a distance of 10 cm (i.e., d= 0.1 m) on the equatorial line of the magnet is given as:

Themagnetic field is along the N − S direction.

Ans: 13
Earth’s magnetic field at the given place, H= 0.36 G
The magnetic field at a distance d, on the axis of the magnet is given as:

Where,
= Permeability of free space
M= Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:

Total magnetic field,

Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.

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

Where,
M= Magnetic moment
= Permeability of free space
H= Horizontal component of the magnetic field at d₁
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence,the magnetic field at a distance d₂, on the equatorial line of the magnet can be written as:

Equatingequations (1) and (2), we get:

The new null points will be located 11.1 cm on the normal bisector.

Ans: 15
Magnetic moment of the bar magnet, M= 5.25 ×10⁻²J T⁻¹
Magnitude of earth’s magnetic field at a place, H= 0.42 G = 0.42 ×10⁻⁴T
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:

Where,
= Permeability of free space = 4π×10⁻⁷Tm A⁻¹
When the resultant field is inclined at 45°with earth’s field, B= H

(b) The magnetic field at a distancedfrom the centre of the magnet on its axis is given as:

The resultant field is inclined at 45°with earth’s field.


Ans: 16
(a)Owing to therandom thermal motion of molecules, the alignments of 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 field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.
(c)Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
(d)The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa.
(e)The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field 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 magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.


Ans: 17
The hysteresis curve (B-Hcurve) of a ferromagnetic material is shown in the following figure.

(a) It can be observed from the given curve that magnetisation persists even when the external field is removed. This reflects the irreversibility of a ferromagnet.
(b)The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat energy.
(c)The value of magnetisation is memory or record of hysteresis loop cycles of magnetisation. These bits of information correspond to the cycle of 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 fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

Ans: 18
Current in the wire, I= 2.5 A
Angle of dip at the given location on earth, = 0°
Earth’s magnetic field, H= 0.33 G = 0.33 ×10⁻⁴T
The horizontal component of earth’s magnetic field is given as:
H_H= H cos

The magnetic field at the neutral point at a distance Rfrom the cable is given by the relation:

Where,
= Permeability of free space =

Hence,a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.

Ans: 19
Number of horizontal wires in the telephone cable, n= 4
Current in each wire, I= 1.0 A
Earth’s magnetic field at a location, H= 0.39 G = 0.39 × 10⁻⁴T
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 = 0.04 m
The horizontal component of earth’s magnetic field can be written as:
H_h= Hcosδ− B
Where,
B= Magnetic field at 4 cm due to current Iin the four wires

= Permeability of free space = 4π× 10⁻⁷Tm A⁻¹

= 0.2 × 10⁻⁴T = 0.2 G
∴ H_h= 0.39 cos 35° − 0.2
= 0.39 × 0.819 − 0.2 ≈0.12 G
The vertical component of earth’s magnetic field is given as:
H_v= Hsinδ
=0.39 sin 35° = 0.22 G
The angle made by the field with its horizontal component is given as:

The resultant field at the point is given as:

For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field:
H_h= Hcosδ+ B
= 0.39 cos 35° + 0.2 = 0.52 G
Vertical component of earth’s magnetic field:
H_v= Hsinδ
=0.39 sin 35° = 0.22 G
Angle,θ = 22.9°
And resultant field:


Ans: 20
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 dip, δ= 45°
(a) The magnetic field due to current I, at a distance r, is given as:

Where,
= Permeability of free space = 4π× 10⁻⁷Tm A⁻¹

= 5.49 × 10⁻⁵T
The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as:
B_H= Bsin δ
= 5.49 × 10⁻⁵sin 45° = 3.88 × 10⁻⁵T = 0.388 G
(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 º, the needle will reverse its original direction. In this case, the needle will point from East to West.

Ans: 21
Magnitude of one of the magnetic fields, B₁= 1.2 × 10⁻²T
Magnitude of the other magnetic field = B₂
Angle between the two fields, θ= 60°
At stable equilibrium, the angle between the dipole and field B₁,θ₁= 15°
Angle between the dipole and field B₂,θ₂= θ− θ₁= 60° − 15° = 45°
At rotational equilibrium, the torques between both the fields must balance each other.
∴Torque due to field B₁= Torque due to field B₂
MB₁sinθ₁= MB₂sinθ₂
Where,
M= Magnetic moment of the dipole

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

Ans: 22
Energy of an electron beam, E= 18 keV = 18 × 10³eV
Charge on an electron, e= 1.6 × 10⁻¹⁹C
E= 18 × 10³× 1.6 × 10⁻¹⁹J
Magnetic field, B = 0.04 G
Mass of an electron, m_e= 9.11 × 10⁻¹⁹kg
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:

The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.

Let the up and down deflection of the electron beam be
Where,
θ= Angle of declination

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

Ans: 23
Number of atomic dipoles, n= 2.0 × 10²⁴
Dipole moment of each atomic dipole, M= 1.5 × 10⁻²³J T⁻¹
When the magnetic field, B₁= 0.64 T
The sample is cooled to a temperature, T₁= 4.2°K
Total dipole moment of the atomic dipole, M_tot= n ×M
= 2 × 10²⁴× 1.5 × 10⁻²³
= 30 J T⁻¹
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment,
When the magnetic field, B₂= 0.98 T
Temperature,T₂= 2.8°K
Its total dipole moment = M₂
According to Curie’s law, we have the ratio of two magnetic dipoles as:

Therefore,is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.

Ans: 24
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,
Magnetising current, I= 1.2 A
The magnetic field is given by the relation:
B
Where,
μ₀= Permeability of free space = 4π× 10⁻⁷Tm A⁻¹

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

Ans: 25
The magnetic moment associated with the orbital angular momentum is valid with the classical mechanics.
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:
Magnetic moment
...............................(1)
Where,
e= Charge of the electron
r= Radius of the circular orbit
T= Time taken to complete one rotation around the circular orbit of radius r
Orbital angular momentum, l= mvr
..................................(2)
Where,
m= Mass of the electron
v= Velocity of the electron
r= Radius of the circular orbit
Dividing equation (1) by equation (2), we get: