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A magnet of moment 0.5 J/T is placed at 45° to a field of 0.2 T. Find the torque.
Given: m = 0.5 J/T; B = 0.2 T; θ = 45°
Use: τ = m B sin θ
τ = 0.5 × 0.2 × 0.707 = 0.0707 N m
Answer: 0.071 N m
A magnet of moment 0.4 J/T makes 60° with a field of 0.3 T. Find its potential energy.
Given: m = 0.4 J/T; B = 0.3 T; θ = 60°
Use: U = − m B cos θ
U = − 0.4 × 0.3 × 0.5 = − 0.06 J
Answer: − 0.06 J
A magnet of moment 2 J/T lies along a field of 0.05 T. Find the work done to turn it (a) by 90° and (b) by 180°.
Given: m = 2 J/T; B = 0.05 T
Use: W = mB (cos θ₁ − cos θ₂)
(a) W = mB (1 − 0) = 2 × 0.05 = 0.1 J
(b) W = mB (1 + 1) = 0.2 J
Answer: (a) 0.1 J; (b) 0.2 J
A small magnet of moment 0.6 J/T and moment of inertia 2 × 10⁻⁶ kg m² oscillates in a field of 0.2 T. Find the period.
Given: m = 0.6 J/T; I = 2 × 10⁻⁶ kg m²; B = 0.2 T
Use: T = 2π √(I / mB)
I / mB = 2 × 10⁻⁶ / 0.12 = 1.67 × 10⁻⁵
T = 2π × 4.08 × 10⁻³ = 0.0256 s
Answer: 0.026 s
A solenoid of 100 turns and area 3 × 10⁻⁴ m² carries 2 A. Find its magnetic moment.
Given: N = 100; A = 3 × 10⁻⁴ m²; I = 2 A
Use: m = N I A
m = 100 × 2 × 3 × 10⁻⁴ = 0.06 A m²
Answer: 0.06 A m²
In a hydrogen atom the electron moves at 2.2 × 10⁶ m/s in an orbit of radius 0.53 × 10⁻¹⁰ m. Find the orbital magnetic moment.
Given: v = 2.2 × 10⁶ m/s; r = 0.53 × 10⁻¹⁰ m
Use: m = I A = (e v / 2πr) π r² = e v r / 2
m = 1.6 × 10⁻¹⁹ × 2.2 × 10⁶ × 0.53 × 10⁻¹⁰ / 2 = 9.3 × 10⁻²⁴ A m²
Answer: 9.3 × 10⁻²⁴ A m² (the Bohr magneton)
A short magnet of moment 0.8 J/T. Find its field on the axis at 20 cm from the centre.
Given: m = 0.8 J/T; r = 0.2 m
Use: B = μ₀ 2m / 4π r³
B = 10⁻⁷ × 2 × 0.8 / (0.2)³ = 1.6 × 10⁻⁷ / 8 × 10⁻³ = 2 × 10⁻⁵ T
Answer: 2 × 10⁻⁵ T
For the same magnet find the field at 20 cm on the equatorial line.
Given: m = 0.8 J/T; r = 0.2 m
Use: B = μ₀ m / 4π r³
B = 10⁻⁷ × 0.8 / 8 × 10⁻³ = 10⁻⁵ T
Answer: 1 × 10⁻⁵ T, opposite to m
Find the ratio of the axial to the equatorial field of a short magnet at the same distance.
Given: —
Use: Compare 2m and m
B_axial / B_equatorial = (μ₀ 2m / 4π r³) / (μ₀ m / 4π r³) = 2
Answer: 2
At what distance on its axis is the field of a short magnet (m = 0.5 J/T) equal to 10⁻⁶ T?
Given: m = 0.5 J/T; B = 10⁻⁶ T
Use: r³ = μ₀ 2m / 4π B
r³ = 10⁻⁷ × 2 × 0.5 / 10⁻⁶ = 0.1
r = 0.464 m
Answer: 46 cm
The axial field of a short magnet is 8 × 10⁻⁵ T at 10 cm. Find it at 20 cm.
Given: B₁ = 8 × 10⁻⁵ T at r₁ = 10 cm
Use: B ∝ 1 / r³
B₂ = B₁ (r₁/r₂)³ = 8 × 10⁻⁵ × (1/2)³ = 10⁻⁵ T
Answer: 1 × 10⁻⁵ T
An iron rod of length 10 cm and cross-section 2 × 10⁻⁴ m² has a total magnetic moment of 5 A m². Find the magnetisation.
Given: m = 5 A m²; V = 0.1 × 2 × 10⁻⁴ = 2 × 10⁻⁵ m³
Use: M = m / V
M = 5 / 2 × 10⁻⁵ = 2.5 × 10⁵ A/m
Answer: 2.5 × 10⁵ A/m
A solenoid has 1500 turns per metre and carries 2 A. Find H, and find B when it is filled with a core of relative permeability 500.
Given: n = 1500 m⁻¹; I = 2 A; μr = 500
Use: H = nI; B = μ₀ μr H
H = 1500 × 2 = 3000 A/m
B = 4π × 10⁻⁷ × 500 × 3000 = 1.88 T
Answer: H = 3000 A/m; B = 1.9 T
A paramagnetic material has χ = 0.003. Find the magnetisation and the relative permeability for H = 1000 A/m.
Given: χ = 0.003; H = 1000 A/m
Use: M = χ H; μr = 1 + χ
M = 0.003 × 1000 = 3 A/m
μr = 1.003
Answer: M = 3 A/m; μr = 1.003
In a material B = 0.6 T when H = 200 A/m. Find its relative permeability.
Given: B = 0.6 T; H = 200 A/m
Use: μr = B / (μ₀ H)
μr = 0.6 / (4π × 10⁻⁷ × 200) = 2387
Answer: about 2400
A material has relative permeability 1.0002. What kind of material is it and what is its susceptibility?
Given: μr = 1.0002
Use: χ = μr − 1
χ = 0.0002 = 2 × 10⁻⁴, positive and small → paramagnetic
Answer: 2 × 10⁻⁴; paramagnetic
The susceptibility of a paramagnetic sample is 2.4 × 10⁻⁵ at 300 K. Find it at 200 K.
Given: χ₁ = 2.4 × 10⁻⁵ at 300 K
Use: χ ∝ 1 / T
χ₂ = χ₁ T₁ / T₂ = 2.4 × 10⁻⁵ × 300 / 200 = 3.6 × 10⁻⁵
Answer: 3.6 × 10⁻⁵
A solenoid of 400 turns and length 0.5 m carries 3 A and has an iron core of relative permeability 1000. Find the field in the core.
Given: n = 800 m⁻¹; I = 3 A; μr = 1000
Use: B = μ₀ μr n I
B = 4π × 10⁻⁷ × 1000 × 800 × 3 = 3.02 T
Answer: 3.0 T (an idealised value, since real iron saturates)
A loop of area 5 × 10⁻³ m² is in a field of 0.4 T, with the normal at 60° to the field. Find the flux.
Given: A = 5 × 10⁻³ m²; B = 0.4 T; θ = 60°
Use: Φ = B A cos θ
Φ = 0.4 × 5 × 10⁻³ × 0.5 = 10⁻³ Wb
Answer: 1 × 10⁻³ Wb
A small bar magnet is completely inside a sphere. What is the net flux through the sphere?
Given: —
Use: Gauss's law for magnetism
∮ B · dA = 0 for any closed surface (no monopoles).
Answer: Zero
A bar magnet of moment 0.8 J/T is cut into two equal parts perpendicular to its length. Find the moment of each part.
Given: m = 0.8 J/T
Use: Pole strength is unchanged and the length is halved: m = p × l
m′ = 0.8 / 2 = 0.4 J/T
Answer: 0.4 J/T
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τ = m B sin θ, so m = τ / (B sin θ) = 4.5 × 10⁻² / (0.25 × 0.5) = 0.36 J T⁻¹.
U = − m B cos θ, with mB = 0.32 × 0.15 = 0.048 J.
(a) Stable equilibrium: m parallel to B (θ = 0°), U = − mB = − 0.048 J.
(b) Unstable equilibrium: m antiparallel to B (θ = 180°), U = + mB = + 0.048 J.
A solenoid produces a magnetic field like a bar magnet: its field lines come out of one end (north pole) and enter the other (south pole), and a solenoid in a field experiences a torque m × B like a magnet. Each turn is a current loop of moment IA, so the solenoid is a magnetic dipole whose moment points along its axis.
m = N I A = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.6 J T⁻¹, along the axis.
τ = m B sin θ = 0.6 × 0.25 × sin 30° = 0.075 N m.
mB = 1.5 × 0.22 = 0.33 J. Work done W = mB (cos θ₁ − cos θ₂) with θ₁ = 0.
(a)(i) Normal to the field (θ₂ = 90°): W = 0.33 (1 − 0) = 0.33 J.
(a)(ii) Opposite to the field (θ₂ = 180°): W = 0.33 (1 + 1) = 0.66 J.
(b)(i) τ = mB sin 90° = 0.33 N m. (b)(ii) τ = mB sin 180° = 0. In case (i) the torque tends to restore the magnet to the field direction; in case (ii) it is an unstable equilibrium.
(a) m = N I A = 2000 × 4.0 × 1.6 × 10⁻⁴ = 1.28 J T⁻¹.
(b) The field is uniform, so the net force is zero. Torque τ = m B sin θ = 1.28 × 7.5 × 10⁻² × sin 30° = 0.048 N m, in a direction that tends to align the axis with the field.
(a) On the axis: B = μ₀ 2m / 4π r³ = 10⁻⁷ × 2 × 0.48 / (0.1)³ = 9.6 × 10⁻⁵ T, along the direction of the magnetic moment (from S to N).
(b) On the equatorial line: B = μ₀ m / 4π r³ = 10⁻⁷ × 0.48 / (0.1)³ = 4.8 × 10⁻⁵ T, in the direction opposite to the magnetic moment (from N to S).