CLASS 12 PHYSICS CHAPTER 5 – MAGNETISM AND MATTER
NCERT CLASS 12 PHYSICS
CHAPTER 5 – MAGNETISM AND MATTER
PART A – IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
1. What is magnetism?
Magnetism is a physical phenomenon associated with magnetic fields produced by magnets, moving electric charges and electric currents.
A magnet attracts magnetic materials such as iron, nickel and cobalt.
2. What is a bar magnet?
A bar magnet is a rectangular or cylindrical piece of magnetic material that produces a magnetic field around it.
It has two poles:
North pole (N)
South pole (S)
The magnetic strength is generally concentrated near its two ends.
3. What are the magnetic poles of a magnet?
The two regions near the ends of a magnet where its magnetic effect is strongest are called magnetic poles.
They are known as the north pole and the south pole.
4. What happens when two north poles or two south poles are brought near each other?
Like magnetic poles repel each other.
Therefore:
North pole repels north pole.
South pole repels south pole.
5. What happens when a north pole and a south pole are brought near each other?
Unlike magnetic poles attract each other.
Therefore, a north pole attracts a south pole.
6. Can magnetic poles exist independently?
No. Magnetic poles cannot be isolated in ordinary magnets.
If a bar magnet is cut into two pieces, each piece becomes a smaller magnet having its own north and south poles.
This shows that isolated magnetic monopoles have not been observed.
7. What is a magnetic dipole?
A magnetic dipole is a system consisting of two equal and opposite magnetic poles separated by a small distance.
A bar magnet is an example of a magnetic dipole.
8. What is magnetic dipole moment?
Magnetic dipole moment is a quantity that measures the strength and orientation of a magnetic dipole.
For a bar magnet:
m = qₘ × 2l
Where:
qₘ = Pole strength
2l = Magnetic length of the magnet
Its SI unit is A m².
9. What is the direction of magnetic dipole moment of a bar magnet?
The magnetic dipole moment of a bar magnet is directed from its south pole to its north pole along the magnetic axis.
10. What are magnetic field lines?
Magnetic field lines are imaginary curves used to represent the direction and strength of a magnetic field.
The tangent drawn at any point on a magnetic field line gives the direction of the magnetic field at that point.
11. Write the important properties of magnetic field lines.
The important properties are:
Magnetic field lines form continuous closed curves.
Outside a magnet, they travel from the north pole to the south pole.
Inside a magnet, they travel from the south pole to the north pole.
Two magnetic field lines never intersect each other.
The closeness of field lines indicates the strength of the magnetic field.
Magnetic field lines are crowded near the poles, where the field is stronger.
12. Why do two magnetic field lines never intersect each other?
If two magnetic field lines intersected, there would be two different directions of the magnetic field at the same point.
This is impossible because the magnetic field at any point has a unique direction.
Therefore, magnetic field lines never intersect.
13. Why do magnetic field lines form closed curves?
Magnetic field lines form closed curves because isolated magnetic poles do not exist.
Outside a magnet, the field lines travel from north to south, while inside the magnet they travel from south to north.
Thus, they form continuous closed loops.
14. What is the magnetic axis of a bar magnet?
The straight line joining the north and south poles of a bar magnet is called its magnetic axis.
15. What is the magnetic length of a bar magnet?
The distance between the effective north and south poles of a bar magnet is called its magnetic length.
It is slightly smaller than the actual physical length of the magnet.
16. What is a magnetic dipole placed in a uniform magnetic field?
A magnetic dipole placed in a uniform magnetic field experiences a torque that tends to align its magnetic moment with the direction of the magnetic field.
In a uniform magnetic field, the net force on an ideal magnetic dipole is zero.
17. What is the torque acting on a magnetic dipole in a uniform magnetic field?
The torque acting on a magnetic dipole in a uniform magnetic field is:
τ = mB sinθ
Where:
m = Magnetic dipole moment
B = Magnetic field
θ = Angle between magnetic moment and magnetic field
18. When is the torque on a magnetic dipole maximum?
Torque is maximum when the magnetic dipole moment is perpendicular to the magnetic field.
θ = 90°
Therefore,
τmax = mB
19. When is the torque on a magnetic dipole zero?
Torque is zero when the magnetic dipole moment is parallel or antiparallel to the magnetic field.
θ = 0° or 180°
Therefore,
τ = 0
20. What is the potential energy of a magnetic dipole in a uniform magnetic field?
The potential energy of a magnetic dipole in a uniform magnetic field is:
U = −mB cosθ
Where:
m = Magnetic dipole moment
B = Magnetic field
θ = Angle between m and B
21. What is the stable equilibrium position of a magnetic dipole?
A magnetic dipole is in stable equilibrium when its magnetic moment is parallel to the magnetic field.
θ = 0°
At this position, potential energy is minimum.
U = −mB
22. What is the unstable equilibrium position of a magnetic dipole?
A magnetic dipole is in unstable equilibrium when its magnetic moment is antiparallel to the magnetic field.
θ = 180°
At this position, potential energy is maximum.
U = +mB
23. What is the work done in rotating a magnetic dipole in a uniform magnetic field?
The work done by an external agent in slowly rotating a magnetic dipole from angle θ₁ to θ₂ is:
W = mB(cosθ₁ − cosθ₂)
This is equal to the change in magnetic potential energy.
24. What is a magnetic dipole in terms of a current loop?
A current-carrying circular loop behaves like a magnetic dipole.
Its magnetic dipole moment is:
m = IA
For a coil of N turns:
m = NIA
Where:
I = Current
A = Area of the coil
N = Number of turns
25. Why does a current-carrying solenoid behave like a bar magnet?
A current-carrying solenoid produces a magnetic field similar to that of a bar magnet.
One end behaves like a north pole and the other behaves like a south pole.
Its magnetic field lines have a similar pattern to those of a bar magnet.
26. What is the magnetic moment of a current-carrying solenoid?
The magnetic moment of a solenoid is:
m = NIA
Where:
N = Number of turns
I = Current
A = Area of cross-section of the solenoid
Its direction is along the axis of the solenoid, determined by the right-hand rule.
27. What is the magnetic field on the axial line of a short bar magnet?
For a short bar magnet at a distant point on its axial line:
Baxial = (μ₀/4π)(2m/r³)
Where:
m = Magnetic dipole moment
r = Distance from the centre of the magnet
28. What is the magnetic field on the equatorial line of a short bar magnet?
For a short bar magnet at a distant point on its equatorial line:
Bequatorial = (μ₀/4π)(m/r³)
The direction of this field is opposite to the direction of the magnetic dipole moment.
29. What is the relationship between axial and equatorial magnetic fields of a short bar magnet?
At points on the axial and equatorial lines at the same distance from the centre:
Baxial = 2Bequatorial
Thus, the axial magnetic field is twice the equatorial magnetic field in magnitude.
30. What is Gauss's law for magnetism?
Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero.
Mathematically:
∮ B · dA = 0
This means that the total magnetic flux entering a closed surface is equal to the total magnetic flux leaving it.
31. What does Gauss's law for magnetism indicate?
Gauss's law for magnetism indicates that isolated magnetic monopoles do not exist.
Magnetic field lines always form closed loops, so the net magnetic flux through a closed surface is zero.
32. What is magnetic flux?
Magnetic flux is a measure of the total magnetic field passing through a given surface.
For a uniform magnetic field passing through a plane surface:
ΦB = BA cosθ
Where:
B = Magnetic field
A = Area of the surface
θ = Angle between the magnetic field and the normal to the surface
Its SI unit is weber (Wb).
33. What is magnetisation?
Magnetisation is defined as the net magnetic dipole moment per unit volume of a magnetic material.
M = Net magnetic dipole moment/Volume
Its SI unit is A/m.
34. What is magnetic intensity?
Magnetic intensity, also called the magnetic field strength H, describes the magnetic field produced by free currents and is used in the description of magnetic materials.
For a linear magnetic medium:
B = μ₀(H + M)
Where:
B = Magnetic field
H = Magnetic intensity
M = Magnetisation
Its SI unit is A/m.
35. What is magnetic susceptibility?
Magnetic susceptibility is the ratio of magnetisation to magnetic intensity.
χm = M/H
It indicates how easily a material becomes magnetised when placed in an external magnetic field.
It is a dimensionless quantity.
36. What is magnetic permeability?
Magnetic permeability is a measure of the ability of a material to allow magnetic field lines to pass through it.
It is represented by μ.
For a linear, isotropic material:
B = μH
Its SI unit is H/m or T m A⁻¹.
37. What is relative permeability?
Relative permeability is the ratio of the permeability of a material to the permeability of free space.
μr = μ/μ₀
It is a dimensionless quantity.
38. What is the relation between magnetic susceptibility and relative permeability?
For a linear, isotropic material:
μr = 1 + χm
Where:
μr = Relative permeability
χm = Magnetic susceptibility
39. What are diamagnetic materials?
Diamagnetic materials are those materials that develop a weak magnetisation opposite to the direction of an applied magnetic field.
They are weakly repelled by a magnetic field.
Their magnetic susceptibility is small and negative.
Examples:
Bismuth
Copper
Silver
Gold
Water
40. What are paramagnetic materials?
Paramagnetic materials are those materials that develop a weak magnetisation in the direction of an applied magnetic field.
They are weakly attracted by a magnetic field.
Their magnetic susceptibility is small and positive.
Examples:
Aluminium
Platinum
Oxygen
Chromium
41. What are ferromagnetic materials?
Ferromagnetic materials are those materials that are strongly attracted by a magnetic field and can become strongly magnetised.
Their magnetic susceptibility is large and positive.
Examples:
Iron
Cobalt
Nickel
42. What is the difference between diamagnetic, paramagnetic and ferromagnetic materials?
Diamagnetic Materials:
They are weakly repelled by a magnetic field.
Their magnetic susceptibility is small and negative.
Their relative permeability is slightly less than one.
Examples: Bismuth, copper and silver.
Paramagnetic Materials:
They are weakly attracted by a magnetic field.
Their magnetic susceptibility is small and positive.
Their relative permeability is slightly greater than one.
Examples: Aluminium and platinum.
Ferromagnetic Materials:
They are strongly attracted by a magnetic field.
Their magnetic susceptibility is large and positive.
Their relative permeability is much greater than one.
Examples: Iron, cobalt and nickel.
43. What is the effect of temperature on ferromagnetic materials?
When the temperature of a ferromagnetic material increases, its magnetic ordering generally decreases.
At a particular temperature called the Curie temperature, it loses its ferromagnetic property and becomes paramagnetic.
44. What is Curie temperature?
Curie temperature is the temperature above which a ferromagnetic material loses its ferromagnetic behaviour and becomes paramagnetic.
45. What is the difference between a permanent magnet and an electromagnet?
Permanent Magnet:
It retains its magnetism for a long time.
It is generally made from hard magnetic materials.
Its magnetic strength is not easily changed.
Electromagnet:
It produces magnetism when electric current flows through its coil.
Its magnetic strength can be changed by changing the current.
It can be switched on and off.
PART B – NCERT EXERCISE QUESTIONS AND ANSWERS
QUESTION 5.1
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² N m. What is the magnitude of the magnetic moment of the magnet?
Solution:
Given:
Torque, τ = 4.5 × 10⁻² N m
Magnetic field, B = 0.25 T
Angle, θ = 30°
We know:
τ = mB sinθ
Therefore,
m = τ/(B sinθ)
Substituting the values:
m = (4.5 × 10⁻²)/(0.25 × sin30°)
Since:
sin30° = 0.5
m = 0.045/(0.25 × 0.5)
m = 0.045/0.125
m = 0.36 J/T
Answer: The magnetic moment of the magnet is 0.36 J/T or 0.36 A m².
QUESTION 5.2
A short bar magnet of magnetic moment m = 0.32 J T⁻¹ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Solution:
Given:
Magnetic moment, m = 0.32 J/T
Magnetic field, B = 0.15 T
We know:
U = −mB cosθ
(a) Stable equilibrium
A magnetic dipole is in stable equilibrium when its magnetic moment is parallel to the magnetic field.
Therefore:
θ = 0°
U = −mB cos0°
U = −0.32 × 0.15 × 1
U = −0.048 J
(b) Unstable equilibrium
A magnetic dipole is in unstable equilibrium when its magnetic moment is antiparallel to the magnetic field.
Therefore:
θ = 180°
U = −mB cos180°
U = −0.32 × 0.15 × (−1)
U = +0.048 J
Answer:
(a) Stable equilibrium: Magnetic moment parallel to the field; potential energy = −0.048 J.
(b) Unstable equilibrium: Magnetic moment antiparallel to the field; potential energy = +0.048 J.
QUESTION 5.3
A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Solution:
Given:
Number of turns, N = 800
Area, A = 2.5 × 10⁻⁴ m²
Current, I = 3.0 A
A current-carrying solenoid behaves like a bar magnet because:
One end of the solenoid acts as a north pole.
The other end acts as a south pole.
Its magnetic field pattern is similar to that of a bar magnet.
Its magnetic moment is directed along its axis.
The magnetic moment is:
m = NIA
Substituting:
m = 800 × 3.0 × 2.5 × 10⁻⁴
m = 2400 × 2.5 × 10⁻⁴
m = 0.60 A m²
Answer:
The solenoid behaves like a bar magnet.
Its magnetic moment is 0.60 A m² or 0.60 J/T.
QUESTION 5.4
If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field 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 the applied field?
Solution:
Given:
Magnetic moment, m = 0.60 A m²
Magnetic field, B = 0.25 T
Angle, θ = 30°
The torque on a magnetic dipole is:
τ = mB sinθ
Substituting:
τ = 0.60 × 0.25 × sin30°
τ = 0.60 × 0.25 × 0.5
τ = 0.075 N m
Answer: The magnitude of torque is 7.5 × 10⁻² N m.
QUESTION 5.5
A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field 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 field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?
Solution:
Given:
Magnetic moment, m = 1.5 J/T
Magnetic field, B = 0.22 T
Initial angle, θ₁ = 0°
We know:
W = mB(cosθ₁ − cosθ₂)
Also:
τ = mB sinθ
First calculate:
mB = 1.5 × 0.22
mB = 0.33 J
(a) Work done
(i) When the magnetic moment is normal to the field
θ₂ = 90°
W = 0.33(cos0° − cos90°)
W = 0.33(1 − 0)
W = 0.33 J
(ii) When the magnetic moment is opposite to the field
θ₂ = 180°
W = 0.33(cos0° − cos180°)
W = 0.33[1 − (−1)]
W = 0.33 × 2
W = 0.66 J
(b) Torque on the magnet
(i) At 90°
τ = mB sin90°
τ = 0.33 × 1
τ = 0.33 N m
(ii) At 180°
τ = mB sin180°
τ = 0
Answer:
(a) Work done:
(i) 0.33 J
(ii) 0.66 J
(b) Torque:
(i) 0.33 N m
(ii) Zero
QUESTION 5.6
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m², carrying a 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 field of 7.5 × 10⁻² T is set up at an angle of 30° with the axis of the solenoid?
Solution:
Given:
Number of turns, N = 2000
Area, A = 1.6 × 10⁻⁴ m²
Current, I = 4.0 A
Magnetic field, B = 7.5 × 10⁻² T
Angle, θ = 30°
(a) Magnetic moment
We know:
m = NIA
Substituting:
m = 2000 × 4.0 × 1.6 × 10⁻⁴
m = 8000 × 1.6 × 10⁻⁴
m = 1.28 A m²
(b) Force and torque
In a uniform magnetic field, the net force on a magnetic dipole is zero.
Therefore:
F = 0
Torque is:
τ = mB sinθ
Substituting:
τ = 1.28 × 7.5 × 10⁻² × sin30°
τ = 1.28 × 0.075 × 0.5
τ = 0.048 N m
Answer:
(a) Magnetic moment = 1.28 A m².
(b) Net force = 0.
Torque = 4.8 × 10⁻² N m.
QUESTION 5.7
A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial line (normal bisector) of the magnet.
Solution:
Given:
Magnetic moment, m = 0.48 J/T
Distance, r = 10 cm = 0.10 m
Permeability of free space:
μ₀/4π = 10⁻⁷ T m/A
(a) Magnetic field on the axial line
The magnetic field on the axial line is:
Baxial = (μ₀/4π)(2m/r³)
Substituting:
Baxial = 10⁻⁷ × [2 × 0.48/(0.10)³]
Baxial = 10⁻⁷ × [0.96/0.001]
Baxial = 10⁻⁷ × 960
Baxial = 9.6 × 10⁻⁵ T
Direction: Along the magnetic dipole moment, from south to north.
(b) Magnetic field on the equatorial line
The magnetic field on the equatorial line is:
Bequatorial = (μ₀/4π)(m/r³)
Substituting:
Bequatorial = 10⁻⁷ × [0.48/(0.10)³]
Bequatorial = 10⁻⁷ × 480
Bequatorial = 4.8 × 10⁻⁵ T
Direction: Opposite to the magnetic dipole moment, from north to south along the magnet's axis.
Answer:
(a) Axial magnetic field = 9.6 × 10⁻⁵ T, along the magnetic moment.
(b) Equatorial magnetic field = 4.8 × 10⁻⁵ T, opposite to the magnetic moment.
PART C – IMPORTANT DERIVATIONS
1. Derive the expression for the magnetic field on the axial line of a short bar magnet.
Answer:
Consider a short bar magnet having:
Magnetic pole strength = qₘ
Magnetic length = 2l
Magnetic moment = m = 2lqₘ
Let P be a point on the axial line at a distance r from the centre of the magnet.
The distance of P from the north pole is:
r − l
The distance of P from the south pole is:
r + l
Using the magnetic field expression for a pole, the net field along the axial line is:
B = (μ₀/4π)qₘ[1/(r − l)² − 1/(r + l)²]
Taking the LCM:
B = (μ₀/4π)qₘ × [(r + l)² − (r − l)²]/(r² − l²)²
Expanding:
(r + l)² − (r − l)² = 4rl
Therefore,
B = (μ₀/4π)qₘ × 4rl/(r² − l²)²
For a short magnet, r is much greater than l.
Therefore,
(r² − l²)² ≈ r⁴
Thus,
B ≈ (μ₀/4π) × 4qₘrl/r⁴
B ≈ (μ₀/4π) × 4qₘl/r³
But:
m = 2lqₘ
Therefore,
Baxial = (μ₀/4π)(2m/r³)
Hence proved.
2. Derive the expression for the magnetic field on the equatorial line of a short bar magnet.
Answer:
Consider a short bar magnet having:
Magnetic pole strength = qₘ
Magnetic length = 2l
Magnetic moment = m = 2lqₘ
Let P be a point on the equatorial line at a distance r from the centre of the magnet.
The distance of P from each pole is:
d = √(r² + l²)
The magnetic fields due to the two poles have equal magnitudes.
Their components perpendicular to the magnetic axis cancel each other, while the components along the magnetic axis add.
The net magnetic field is:
B = (μ₀/4π) × 2qₘl/(r² + l²)³ᐟ²
For a short magnet, r is much greater than l.
Therefore:
(r² + l²)³ᐟ² ≈ r³
Thus,
B = (μ₀/4π) × 2qₘl/r³
But:
m = 2lqₘ
Therefore,
Bequatorial = (μ₀/4π)(m/r³)
The direction of the field is opposite to the magnetic dipole moment.
Hence proved.
3. Derive the expression for torque acting on a magnetic dipole in a uniform magnetic field.
Answer:
Consider a bar magnet of magnetic dipole moment m placed in a uniform magnetic field B.
Let θ be the angle between the magnetic dipole moment and the magnetic field.
The two poles experience equal and opposite magnetic forces.
These forces form a couple.
The magnitude of each force is:
F = qₘB
If 2l is the magnetic length, the perpendicular distance between the forces is:
d = 2l sinθ
Torque is:
τ = Force × Perpendicular distance
τ = qₘB × 2l sinθ
τ = 2lqₘB sinθ
But:
m = 2lqₘ
Therefore,
τ = mB sinθ
In vector form:
τ = m × B
Hence proved.
4. Derive the expression for potential energy of a magnetic dipole in a uniform magnetic field.
Answer:
Consider a magnetic dipole of magnetic moment m placed in a uniform magnetic field B.
Let θ be the angle between m and B.
The torque acting on the dipole is:
τ = mB sinθ
For a small angular displacement dθ, the work done by an external agent against the magnetic torque is:
dW = mB sinθ dθ
The change in potential energy is equal to the work done by the external agent.
Therefore,
dU = mB sinθ dθ
Integrating:
∫dU = mB∫sinθ dθ
U = −mB cosθ + C
Taking potential energy to be zero at θ = 90°:
C = 0
Therefore,
U = −mB cosθ
In vector form:
U = −m · B
Hence proved.
5. Derive the expression for magnetic moment of a current-carrying solenoid.
Answer:
Consider a solenoid having:
N = Number of turns
I = Current flowing through it
A = Area of cross-section
The magnetic moment of a single current-carrying loop is:
m = IA
For N identical turns, the total magnetic moment is:
m = N × IA
Therefore,
m = NIA
The direction of the magnetic moment is along the axis of the solenoid, determined by the right-hand rule.
Hence proved.
6. State and explain Gauss's law for magnetism.
Answer:
Gauss's law for magnetism states that the net magnetic flux through any closed surface is always zero.
Mathematically:
∮ B · dA = 0
This means:
The total magnetic flux entering a closed surface is equal to the total magnetic flux leaving it.
Magnetic field lines form continuous closed loops.
Isolated magnetic poles have not been observed.
This law is valid for all magnetic fields.
7. Derive the relation between magnetic field, magnetic intensity and magnetisation.
Answer:
For a magnetic material placed in an external magnetic field, the total magnetic field depends on the applied magnetic intensity and the magnetisation produced inside the material.
The relation is:
B = μ₀(H + M)
Where:
B = Magnetic field
H = Magnetic intensity
M = Magnetisation
For a linear magnetic material:
M = χmH
Substituting:
B = μ₀(H + χmH)
Taking H common:
B = μ₀H(1 + χm)
Therefore,
B = μ₀(1 + χm)H
But:
B = μH
Comparing:
μ = μ₀(1 + χm)
Dividing by μ₀:
μr = 1 + χm
Hence proved.
PART D – ADDITIONAL IMPORTANT NUMERICAL QUESTIONS
1. A bar magnet has a magnetic moment of 2 A m² and is placed in a magnetic field of 0.5 T at an angle of 30°. Calculate the torque acting on it.
Solution:
Given:
m = 2 A m²
B = 0.5 T
θ = 30°
Formula:
τ = mB sinθ
Substituting:
τ = 2 × 0.5 × sin30°
τ = 2 × 0.5 × 0.5
τ = 0.5 N m
Answer: The torque is 0.5 N m.
2. A magnetic dipole of moment 0.8 A m² is placed in a magnetic field of 0.4 T. Calculate its potential energy when the angle between the magnetic moment and the field is 60°.
Solution:
Given:
m = 0.8 A m²
B = 0.4 T
θ = 60°
Formula:
U = −mB cosθ
Substituting:
U = −0.8 × 0.4 × cos60°
U = −0.8 × 0.4 × 0.5
U = −0.16 J
Answer: The potential energy is −0.16 J.
3. A solenoid has 500 turns, an area of cross-section 4 × 10⁻⁴ m² and carries a current of 2 A. Calculate its magnetic moment.
Solution:
Given:
N = 500
A = 4 × 10⁻⁴ m²
I = 2 A
Formula:
m = NIA
Substituting:
m = 500 × 2 × 4 × 10⁻⁴
m = 4000 × 10⁻⁴
m = 0.4 A m²
Answer: The magnetic moment is 0.4 A m².
4. A short bar magnet has a magnetic moment of 0.6 A m². Calculate the magnetic field at a point 20 cm away on its axial line.
Solution:
Given:
m = 0.6 A m²
r = 20 cm = 0.20 m
μ₀/4π = 10⁻⁷ T m/A
Formula:
Baxial = (μ₀/4π)(2m/r³)
Substituting:
Baxial = 10⁻⁷ × [2 × 0.6/(0.20)³]
Baxial = 10⁻⁷ × [1.2/0.008]
Baxial = 10⁻⁷ × 150
Baxial = 1.5 × 10⁻⁵ T
Answer: The magnetic field is 1.5 × 10⁻⁵ T.
5. A short bar magnet has a magnetic moment of 0.8 A m². Calculate the magnetic field at a point 10 cm away on its equatorial line.
Solution:
Given:
m = 0.8 A m²
r = 10 cm = 0.10 m
μ₀/4π = 10⁻⁷ T m/A
Formula:
Bequatorial = (μ₀/4π)(m/r³)
Substituting:
Bequatorial = 10⁻⁷ × [0.8/(0.10)³]
Bequatorial = 10⁻⁷ × [0.8/0.001]
Bequatorial = 10⁻⁷ × 800
Bequatorial = 8 × 10⁻⁵ T
Answer: The magnetic field is 8 × 10⁻⁵ T, opposite to the direction of the magnetic moment.
6. A magnetic material has a magnetisation of 400 A/m when placed in a magnetic field of intensity 2000 A/m. Calculate its magnetic susceptibility.
Solution:
Given:
M = 400 A/m
H = 2000 A/m
Formula:
χm = M/H
Substituting:
χm = 400/2000
χm = 0.2
Answer: Magnetic susceptibility = 0.2.
7. A magnetic material has a magnetic susceptibility of 0.5. Calculate its relative permeability.
Solution:
Given:
χm = 0.5
Formula:
μr = 1 + χm
Substituting:
μr = 1 + 0.5
μr = 1.5
Answer: Relative permeability = 1.5.
PART E – IMPORTANT FORMULAS
Magnetic dipole moment of a bar magnet:
m = qₘ × 2l
Magnetic dipole moment of a current loop:
m = IA
Magnetic dipole moment of a coil:
m = NIA
Torque on a magnetic dipole:
τ = mB sinθ
Vector form of torque:
τ = m × B
Potential energy of a magnetic dipole:
U = −mB cosθ
Work done in rotating a magnetic dipole:
W = mB(cosθ₁ − cosθ₂)
Magnetic field on the axial line of a short bar magnet:
Baxial = (μ₀/4π)(2m/r³)
Magnetic field on the equatorial line of a short bar magnet:
Bequatorial = (μ₀/4π)(m/r³)
Relation between axial and equatorial fields:
Baxial = 2Bequatorial
Gauss's law for magnetism:
∮B · dA = 0
Magnetic flux:
ΦB = BA cosθ
Magnetisation:
M = Magnetic dipole moment/Volume
Magnetic susceptibility:
χm = M/H
Magnetic field in a magnetic material:
B = μ₀(H + M)
Relative permeability:
μr = μ/μ₀
Relation between susceptibility and relative permeability:
μr = 1 + χm
Permeability of free space:
μ₀ = 4π × 10⁻⁷ T m A⁻¹
PART F – IMPORTANT SI UNITS
| Physical Quantity | SI Unit |
|---|---|
| Magnetic field | Tesla (T) |
| Magnetic dipole moment | A m² |
| Pole strength | A m |
| Magnetic length | Metre (m) |
| Torque | Newton metre (N m) |
| Potential energy | Joule (J) |
| Magnetic flux | Weber (Wb) |
| Magnetisation | A/m |
| Magnetic intensity | A/m |
| Magnetic susceptibility | No unit |
| Magnetic permeability | H/m |
| Relative permeability | No unit |
| Area | m² |
| Volume | m³ |
PART G – QUICK REVISION POINTS
A bar magnet has two poles: north and south.
Like poles repel and unlike poles attract.
Isolated magnetic monopoles have not been observed.
Magnetic field lines form continuous closed curves.
Outside a magnet, field lines travel from north to south.
Inside a magnet, field lines travel from south to north.
Magnetic field lines never intersect.
Magnetic dipole moment is directed from south to north.
A current-carrying solenoid behaves like a bar magnet.
Magnetic moment of a solenoid is m = NIA.
Torque on a magnetic dipole is τ = mB sinθ.
Potential energy of a magnetic dipole is U = −mB cosθ.
Stable equilibrium occurs when the magnetic moment is parallel to the field.
Unstable equilibrium occurs when the magnetic moment is antiparallel to the field.
The axial magnetic field of a short magnet is twice its equatorial field at the same distance.
Gauss's law for magnetism states that the net magnetic flux through a closed surface is zero.
Magnetisation is magnetic dipole moment per unit volume.
Magnetic susceptibility is the ratio of magnetisation to magnetic intensity.
Diamagnetic materials are weakly repelled by magnetic fields.
Paramagnetic materials are weakly attracted by magnetic fields.
Ferromagnetic materials are strongly attracted by magnetic fields.
Diamagnetic materials have small negative susceptibility.
Paramagnetic materials have small positive susceptibility.
Ferromagnetic materials have large positive susceptibility.
Above the Curie temperature, a ferromagnetic material becomes paramagnetic.
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