Find the best tutors and institutes for Class 12 Tuition
Search in
Answer the following questions: (a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled? (b) Why is diamagnetism, in contrast, almost independent of temperature? (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? (d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields? (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? (f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?
(a) Owing to the random thermal motion of the 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) Each molecule of the diamagnetic material is not a magnetic dipole in itself. Hence, the random thermal motion of the molecules of the diamagnetic material (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 field slightly less 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.
(e) The permeability of a ferromagnetic material is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point. The proof of this fact is based on the boundary conditions of the magnetic fields at the interface of two media.
(f) Yes, the maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.
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 fields. Suggest a method
The hysteresis curve (B-H curve) 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.
8 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 field 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 field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
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 × 10−4 T
The horizontal component of earth’s magnetic field is given as:
HH = H cos
The magnetic field at the neutral point at a distance R from 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.
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 field 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 fields at points 4.0 cm below the cable?
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 × 10−4 T
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:
Hh = Hcosδ − B
Where,
B = Magnetic field at 4 cm due to current I in the four wires
= Permeability of free space = 4π × 10−7 Tm A−1
= 0.2 × 10−4 T = 0.2 G
∴ Hh = 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:
Hv = 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:
Hh = Hcosδ + B
= 0.39 cos 35° + 0.2 = 0.52 G
Vertical component of earth’s magnetic field:
Hv = Hsinδ
= 0.39 sin 35° = 0.22 G
Angle, θ 
= 22.9°
And resultant field:
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 field 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.
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 × 10−7 Tm A−1
= 5.49 × 10−5 T
The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as:
BH = Bsin δ
= 5.49 × 10−5 sin 45° = 3.88 × 10−5 T = 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.
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10–2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field
Magnitude of one of the magnetic fields, B1 = 1.2 × 10−2 T
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
MB1 sinθ1 = MB2 sinθ2
Where,
M = Magnetic moment of the dipole
Hence, the magnitude of the other magnetic field is 4.39 × 10−3 T.
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 × 10 –31 kg). [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.]
Energy of an electron beam, E = 18 keV = 18 × 103 eV
Charge on an electron, e = 1.6 × 10−19 C
E = 18 × 103 × 1.6 × 10−19 J
Magnetic field, B = 0.04 G
Mass of an electron, me = 9.11 × 10−19 kg
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.
A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10–23 J T –1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled 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 field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)
Number of atomic dipoles, n = 2.0 × 1024
Dipole moment of each atomic dipole, M = 1.5 × 10−23 J 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, Mtot = 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.
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 field B in the core for a magnetising current of 1.2 A?
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 × 10−7 Tm A−1
Therefore, the magnetic field in the core is 4.48 T.
The magnetic moment vectors µs and µl associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: µs = –(e/m) S, µl = –(e/2m)l Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
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:
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 12 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 12 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class
