Chapter 5: Magnetism and Matter
Q1. Define magnetic susceptibility and permeability. How are they related for a paramagnetic material?
Solution
Magnetic susceptibility (χ): Ratio of intensity of magnetization (M) to magnetic field strength (H): χ = M/H
Magnetic permeability (μ): Ratio of B to H: μ = B/H
Relation: μ = μ₀(1 + χ)
For paramagnetic materials, χ is small and positive.
Q2. Classify the following as dia-, para-, or ferromagnetic materials: Aluminium, Sodium, Copper, Iron, Nickel, Cobalt, Water.
Solution
Diamagnetic: Copper, Water
Paramagnetic: Aluminium, Sodium
Ferromagnetic: Iron, Nickel, Cobalt
Q3. What is the earth's magnetic field? What are its components?
Solution
The earth behaves as a magnetic dipole with its south pole near the geographic north. The magnetic field at any point on the earth's surface has three components:
1. Horizontal component (H): Component along the geographic meridian.
2. Declination (θ): Angle between geographic and magnetic meridian.
3. Dip (δ): Angle that the total field makes with the horizontal.
Q4. A bar magnet of magnetic moment 4 A·m² is placed in a uniform magnetic field of 0.2 T at an angle of 30°. Find the torque on it.
Solution
τ = MB sinθ = (4)(0.2) sin 30°
= 0.8 × 0.5 = 0.4 N·m
Q5. Derive the expression for the energy stored in a solenoid.
Solution
Energy = Work done in building up current from 0 to I
U = ½LI²
where L = μ₀n²Al is the self-inductance
n = number of turns per unit length
A = cross-sectional area, l = length
U = ½(μ₀n²Al)I² = ½μ₀n²AlI²
Q6. A bar magnet of magnetic moment 2.0 A·m² and moment of inertia 7.2 × 10−2 kg·m² is freely suspended in a uniform magnetic field of 0.25 T. Find the period of oscillation.
Solution
T = 2π√(I/(MB))
= 2π√(7.2 × 10−2/(2.0 × 0.25))
= 2π√(7.2 × 10−2/0.5)
= 2π√(0.144) = 2π(0.379)
= 2.38 s
Q7. Explain the differences between dia-, para-, and ferromagnetic materials with examples.
Solution
Diamagnetic: Weakly repelled by magnetic field. χ is small and negative. No permanent dipole moments. Examples: Cu, Bi, H₂O
Paramagnetic: Weakly attracted by magnetic field. χ is small and positive. Random dipole moments align with field. Examples: Al, Na, O₂
Ferromagnetic: Strongly attracted by magnetic field. χ is large and positive. Permanent dipoles form domains. Examples: Fe, Ni, Co
Q8. A solenoid has 500 turns per metre and carries a current of 2A. What is the magnetic field inside the solenoid? What is the magnetization if the relative permeability of the core material is 100?
Solution
B = μnI = μ₀μrnI
= (4π × 10−7)(100)(500)(2)
= 4π × 10−7 × 105
= 0.1257 T
H = nI = 500 × 2 = 1000 A/m
M = χH = (μr − 1)H = 99 × 1000
= 9.9 × 104 A/m
Q9. The horizontal component of earth's magnetic field at a place is 0.35 G and the angle of dip is 30°. Find the total magnetic field of earth at that place.
Solution
H = B cosδ
B = H/cosδ = 0.35/cos 30°
= 0.35/0.866 = 0.404 G = 4.04 × 10−5 T
Q10. Compare the properties of soft iron and steel as materials for making permanent magnets.
Solution
Soft Iron:
• High permeability, low retentivity
• Low coercivity (easy to demagnetize)
• Narrow hysteresis loop (low energy loss)
• Used for temporary magnets, electromagnets, transformer cores
Steel:
• High retentivity and high coercivity
• Wide hysteresis loop
• Difficult to magnetize and demagnetize
• Used for permanent magnets
Q11. What is hysteresis? Draw the B-H curve for a ferromagnetic material and explain the terms retentivity and coercivity.
Solution
Hysteresis is the phenomenon where the magnetization (B) of a ferromagnetic material lags behind the applied field (H) during cycles of magnetization.
Retentivity (Br): The value of residual magnetic flux density when H = 0 (after removing the magnetizing field).
Coercivity (Hc): The value of reverse magnetic field needed to reduce B to zero (demagnetize the material).
The area of the B-H loop represents energy dissipated per cycle as heat.
Q12. A short bar magnet of magnetic moment 0.5 A·m² is placed in a uniform magnetic field of 0.25 T. Calculate (i) the maximum torque (ii) the work done to rotate it from unstable to stable equilibrium.
Solution
(i) τmax = MB = 0.5 × 0.25 = 0.125 N·m
(ii) W = −MB(cosθ₂ − cosθ₁)
= −(0.5)(0.25)(cos 180° − cos 0°)
= −(0.125)((−1) − 1)
= 0.125 × 2 = 0.25 J
Q13. At what temperature does a ferromagnetic material become paramagnetic? Define this temperature.
Solution
The temperature above which a ferromagnetic material loses its ferromagnetic properties and becomes paramagnetic is called the Curie temperature (or Curie point).
Examples:
• Iron: Curie temperature = 770°C
• Nickel: Curie temperature = 358°C
• Cobalt: Curie temperature = 1121°C
Above this temperature, the magnetic domains are disrupted by thermal agitation.