Exercises
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 J. What is the magnitude of magnetic moment of the magnet?
Answer
τ = mB sin θ
⇒ m = τ/(B sin θ)
⇒ m = (4.5 × 10⁻²)/(0.25 × 0.5)
⇒ m = 0.36 J T⁻¹
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?
Answer
Potential energy U = –mB cos θ; mB = 0.32 × 0.15 = 0.048 J.
(a) Stable equilibrium: magnetic moment parallel to the field (θ = 0°). U = –mB
⇒ U = –4.8 × 10⁻² J (minimum energy).
(b) Unstable equilibrium: magnetic moment antiparallel to the field (θ = 180°). U = +mB
⇒ U = +4.8 × 10⁻² J (maximum energy).
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?
Answer
Each turn of a current-carrying solenoid is a small current loop, i.e., a magnetic dipole. Their moments add along the axis, so the solenoid as a whole has a north pole at one end and a south pole at the other. Its external field lines match those of a bar magnet, and when freely suspended it aligns along the north–south direction like a compass.
m = NIA
⇒ m = 800 × 3.0 × 2.5 × 10⁻⁴
⇒ m = 0.60 J T⁻¹, directed along the axis (sense given by the right-hand rule).
4. If the solenoid in Exercise 5.5 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 applied field?
Answer
The solenoid meant here is the one in Exercise 5.3 (the reference to 5.5 is a misprint), with m = 0.60 J T⁻¹.
τ = mB sin θ
⇒ τ = 0.60 × 0.25 × sin 30°
⇒ τ = 0.15 × 0.5
⇒ τ = 7.5 × 10⁻² N m
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)?
Answer
mB = 1.5 × 0.22 = 0.33 J. Work W = U_final – U_initial = mB(cos θ₁ – cos θ₂), with θ₁ = 0°.
(a)(i) θ₂ = 90°: W = 0.33 × (1 – 0)
⇒ W = 0.33 J
(a)(ii) θ₂ = 180°: W = 0.33 × (1 – (–1))
⇒ W = 0.66 J
(b) τ = mB sin θ.
(i) θ = 90°: τ = 0.33 × 1
⇒ τ = 0.33 N m (tending to bring the moment back along the field)
(ii) θ = 180°: τ = 0.33 × sin 180°
⇒ τ = 0 (unstable equilibrium)
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?
Answer
(a) m = NIA
⇒ m = 2000 × 4.0 × 1.6 × 10⁻⁴
⇒ m = 1.28 J T⁻¹, along the axis of the solenoid.
(b) Force: in a uniform field the forces on the two "poles" are equal and opposite, so the net force is zero.
Torque: τ = mB sin θ
⇒ τ = 1.28 × 7.5 × 10⁻² × sin 30°
⇒ τ = 0.096 × 0.5
⇒ τ = 4.8 × 10⁻² N m, tending to align the axis with the field.
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 lines (normal bisector) of the magnet.
Answer
r = 0.10 m, μ₀/4π = 10⁻⁷ T m A⁻¹.
(a) Axis: B = (μ₀/4π)(2m/r³)
⇒ B = 10⁻⁷ × (2 × 0.48)/(0.10)³
⇒ B = 10⁻⁷ × 960
⇒ B = 9.6 × 10⁻⁵ T, directed along the magnetic moment, i.e., from the S-pole to the N-pole direction.
(b) Equatorial line: B = (μ₀/4π)(m/r³)
⇒ B = 10⁻⁷ × 0.48/(0.10)³
⇒ B = 4.8 × 10⁻⁵ T, directed opposite to the magnetic moment, i.e., from the N-pole to the S-pole direction.
Important Extra Questions
State Gauss's law for magnetism. What does it imply?
Answer
The net magnetic flux through any closed surface is zero: ∮B·dS = 0. It implies that isolated magnetic poles (monopoles) do not exist; magnetic field lines form continuous closed loops.
Define magnetisation and magnetic susceptibility.
Answer
Magnetisation M is the net magnetic moment per unit volume of a material. Susceptibility χ = M/H, the ratio of magnetisation to magnetic intensity; it is dimensionless and μᵣ = 1 + χ.
How do diamagnetic, paramagnetic and ferromagnetic materials differ in susceptibility?
Answer
Diamagnetic: χ small and negative (–1 ≤ χ < 0), weakly repelled. Paramagnetic: χ small and positive, weakly attracted. Ferromagnetic: χ large and positive (χ ≫ 1), strongly attracted.
Write the relation between B, H and M inside a material.
Answer
B = μ₀(H + M). For a linear material M = χH, so B = μ₀(1 + χ)H = μ₀μᵣH = μH, where μ = μ₀μᵣ is the permeability.
Why does a superconductor show perfect diamagnetism?
Answer
Below its critical temperature, a superconductor expels magnetic field lines completely from its interior (Meissner effect), so B = 0 inside, χ = –1 and μᵣ = 0, which is perfect diamagnetism.