Class 12 Physics | Unit III — Magnetic Effects
Chapter 5: Magnetism and Matter
Bar Magnet • Earth's Magnetism • Magnetisation • Dia/Para/Ferromagnetism • Hysteresis
1. Bar Magnet as a Magnetic Dipole
Magnetic Dipole Moment (M): M = m×2l where m = pole strength (A·m) and 2l =
magnetic length (distance between poles). Direction: S to N pole (inside magnet). Unit: A·m².
Axial field (end-on position): Baxial = μ0/(4π) × 2Mr
/ (r²−l²)² ≈ μ02M / (4πr³) for r >> l
Equatorial field (broadside-on): Beq = μ0M / (4π(r²+l²)3/2) ≈ μ0M / (4πr³) for r >> l
Note: Baxial = 2Beq at same distance (exact analogy with electric dipole).
Equatorial field (broadside-on): Beq = μ0M / (4π(r²+l²)3/2) ≈ μ0M / (4πr³) for r >> l
Note: Baxial = 2Beq at same distance (exact analogy with electric dipole).
- Axial field direction: along M (from S to N, extended).
- Equatorial field direction: opposite to M.
- Magnetic length ≈ 0.84 × geometric length for a bar magnet.
1.1 Torque and PE on Dipole in Uniform B
Torque: τ = MB sinθ (τ = M × B, vector)
PE: U = −MB cosθ (U = −M·B)
Stable equilibrium: θ = 0° (M along B). Unstable: θ = 180° (M against B).
PE: U = −MB cosθ (U = −M·B)
Stable equilibrium: θ = 0° (M along B). Unstable: θ = 180° (M against B).
1.2 Time Period of Oscillation in Magnetic Field
T = 2π√(I/MB)
I = moment of inertia of magnet, M = magnetic dipole moment, B = magnetic field
I = moment of inertia of magnet, M = magnetic dipole moment, B = magnetic field
⚠️ NEET: Baxial = μ02M/4πr³;
Beq = μ0M/4πr³. Baxial = 2Beq. τ =
MBsinθ. T = 2π√(I/MB). Analogy: magnetic dipole ↔ electric dipole (M↔p;
B↔E; μ0/4π ↔ 1/4πε0).
2. Earth's Magnetism
The Earth behaves like a magnetic dipole. The geographic North Pole of Earth is actually the
magnetic South Pole (that is why the N-pole of a compass needle is attracted toward it).
Earth's geographic and magnetic axes are tilted by about 11.3°.
2.1 Key Terms
| Term | Symbol | Definition | Unit |
|---|---|---|---|
| Magnetic Declination | α | Angle between geographic north and magnetic north (compass) at a place | Degrees |
| Magnetic Dip (Inclination) | δ | Angle between Earth's total magnetic field (BH) and horizontal plane | Degrees |
| Horizontal Component | BH | Component of Earth's field in horizontal plane | Tesla (T) |
| Vertical Component | BV | Component of Earth's field in vertical plane | Tesla (T) |
| Total intensity | Be | Be = √(BH²+BV²) | Tesla (T) |
BH = Be cosδ
BV = Be sinδ
tanδ = BV/BH
BV = Be sinδ
tanδ = BV/BH
- At magnetic equator: δ = 0°, BV = 0, BH = Be (maximum).
- At magnetic poles: δ = 90°, BH = 0, BV = Be (maximum).
- Average BH in India ≈ 4×10−5 T. Earth's Be ≈ 0.3−0.6 × 10−4 T.
⚠️ NEET: tanδ = BV/BH. BH =
Becosδ. Dip δ=0 at equator; δ=90° at poles. Geographic North → Magnetic
South. Compass needle: N-pole points geographic north (= magnetic south of Earth).
3. Magnetisation and Magnetic Intensity
Magnetisation M⃗ = mnet/V (net magnetic dipole moment per unit volume).
Unit: A/m.
Magnetic intensity H = B/μ0 − M or B = μ0(H + M)
Magnetic susceptibility: χm = M/H (dimensionless)
Relative permeability: μr = 1 + χm
B = μ0μrH = μ H where μ = μ0μr = absolute permeability
Magnetic intensity H = B/μ0 − M or B = μ0(H + M)
Magnetic susceptibility: χm = M/H (dimensionless)
Relative permeability: μr = 1 + χm
B = μ0μrH = μ H where μ = μ0μr = absolute permeability
4. Classification of Magnetic Materials
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| Cause | Orbital motion (all paired e−) | Unpaired electrons, weak alignment | Domain formation (strong alignment) |
| Susceptibility χm | Small, negative (−1≤χm<0) | Small, positive (0<χm<1) | Very large, positive (χm>>1) |
| Relative permeability μr | Slightly < 1 | Slightly > 1 | Very large (>>1) |
| In external field | Repelled; field lines avoid material | Weakly attracted; field lines enter | Strongly attracted; field lines concentrated |
| Temp dependence | Largely independent | χm ∝ 1/T (Curie's law) | Ferromagnetic above Tc → becomes paramagnetic |
| Examples | Bi, Cu, Pb, H2O, NaCl | Al, Pt, Mn, O2, Na, CuCl2 | Fe, Ni, Co (and their alloys) |
| Retentivity | None | None | High (permanent magnet) |
4.1 Curie's Law and Curie Temperature
Curie's Law (paramagnetic): χm = C/T where C = Curie constant.
Curie temperature Tc: Above Tc, ferromagnetic → paramagnetic. Tc(Iron) = 1043 K; Tc(Nickel) = 631 K; Tc(Cobalt) = 1388 K.
Curie temperature Tc: Above Tc, ferromagnetic → paramagnetic. Tc(Iron) = 1043 K; Tc(Nickel) = 631 K; Tc(Cobalt) = 1388 K.
5. Permanent Magnets and Electromagnets
5.1 Hysteresis Loop (B-H curve for ferromagnets)
- Retentivity/Remanence: Value of B when H = 0 (after removing field). High retentivity → strong permanent magnet.
- Coercivity: Reverse H needed to reduce B to zero. High coercivity → harder to demagnetise.
- Hysteresis loss: Energy dissipated per cycle = area enclosed by B-H loop (as heat). Larger loop = more loss.
| Application | Required Properties | Suitable Material |
|---|---|---|
| Permanent magnets | High retentivity + High coercivity | Steel, Alnico (Al-Ni-Co), Cobalt |
| Electromagnets | High permeability + Low coercivity + Low retentivity | Soft iron (largest B for given H, easily demagnetised) |
| Transformer cores | Low hysteresis loss + High permeability | Soft iron, silicon steel |
| Recording tapes/hard disk | High coercivity + Moderate retentivity | Iron oxide, CrO2 coating |
⚠️ NEET: Diamagnetic: χm negative. Paramagnetic: Curie law
χm=C/T. Ferromagnetic: domains, very large χm, becomes paramagnetic above
Tc. Permanent magnets: high coercivity + high retentivity. Electromagnets: low coercivity (soft
iron). Hysteresis loss = area of B-H loop.
🎓 NEET Previous Year Questions
Q1. [NEET 2022] The angle of dip at a place where BH =
0.4×10−4 T and BV = 0.4×10−4 T is:
Answer tanδ = BV/BH =
0.4/0.4 = 1. δ = tan−1(1) = 45°.
Q2. [NEET 2021] Which of the following is diamagnetic?
Answer Bismuth (Bi). Diamagnetic: all
e− paired (Bi, Cu, NaCl, H2O). Paramagnetic: Al, Mn, O2. Ferromagnetic: Fe,
Ni, Co.
Q3. [NEET 2020] The magnetic susceptibility of a paramagnetic material at
−73°C is 0.0060. Find susceptibility at 27°C.
Answer Curie's law: χmT =
constant. χ1T1 = χ2T2. T1=200 K,
T2=300 K. χ2 = 0.006×200/300 = 0.004.
Q4. [NEET 2019] For permanent magnets, the desired properties are:
Answer High retentivity and high
coercivity. High retentivity → doesn't lose magnetism easily; High coercivity
→ hard to demagnetise by external fields. Steel/Alnico are typical permanent magnet materials.
Q5. [NEET 2018] The axial field of a bar magnet of moment M at distance r is equal to
the equatorial field at what distance?
Answer Baxial(r) =
2μ0M/4πr³. Beq(r') = μ0M/4πr'³.
Set equal: 2/r³ = 1/r'³ ⇒ r'³ = r³/2 ⇒ r' =
r/21/3 = r/(³√2).
💡 Rapid Revision
- Baxial = μ02M/4πr³ | Beq = μ0M/4πr³ | Baxial=2Beq
- τ=MBsinθ | U=−MBcosθ | T=2π√(I/MB)
- tanδ=BV/BH | BH=Becosδ | dip 0° at equator; 90° at poles
- Diamagnetic: χm<0 | Paramagnetic: χm=C/T>0 | Ferromagnetic: χm>>1
- Permanent magnet: high coercivity + high retentivity (steel)
- Electromagnet: low coercivity + high μr (soft iron)
CLASS 12 PHYSICS | NCERT SOLUTIONS
Chapter 5 — Magnetism and Matter
20+ NCERT Exercise & Exemplar Questions — Step-by-Step Solutions
Key formulas: Bax=μ02M/4πr³ |
Beq=μ0M/4πr³ | tanδ=BV/BH |
T=2π√(I/MB) | τ=MBsinθ | χm=C/T (Curie law)
📝 NCERT Exercise Questions
3 MarksQ1 (Ex 5.1). A short bar magnet of moment 0.32
J/T is placed in uniform external B of 0.15 T. Find torques for angles 30°, 60°, 90°.
✓ Solution
τ = MB sinθ; M = 0.32 J/T, B = 0.15 T
θ=30°: τ = 0.32×0.15×sin30° = 0.048×0.5 = 0.024 N·m
θ=60°: τ = 0.048×sin60° = 0.048×0.866 = 0.0416 N·m
θ=90°: τ = 0.048×1 = 0.048 N·m (maximum)
τ = MB sinθ; M = 0.32 J/T, B = 0.15 T
θ=30°: τ = 0.32×0.15×sin30° = 0.048×0.5 = 0.024 N·m
θ=60°: τ = 0.048×sin60° = 0.048×0.866 = 0.0416 N·m
θ=90°: τ = 0.048×1 = 0.048 N·m (maximum)
3 MarksQ2 (Ex 5.2). A bar magnet of moment 1.5 J/T
has field on axial line at 20 cm = 1.28×10−4 T. Verify axial formula.
✓ Solution
Bax = μ0/(4π) × 2M/r³ = 10−7 × 2×1.5/(0.20)³
Bax = μ0/(4π) × 2M/r³ = 10−7 × 2×1.5/(0.20)³
= 10−7 × 3/(8×10−3) =
3×10−7/(8×10−3) = 3.75×10−5 ×
…
More precisely: 10−7×2×1.5/(0.008) = 3×10−7/0.008 =
3.75×10−5 T ≈ given value. (Exact: r=0.20m=20cm,
(0.2)³=0.008 m³. B = 2×10−7×1.5/0.008 =
3.75×10−5 T.)3 MarksQ3 (Ex 5.3). A short bar magnet has equatorial
field of 1.2×10−4 T at 10 cm. Find M and axial field at same distance.
✓ Solution
Beq = μ0M/(4πr³) ⇒ M = Beq×r³/(10−7)
Beq = μ0M/(4πr³) ⇒ M = Beq×r³/(10−7)
M = 1.2×10−4×(0.10)³/10−7 =
1.2×10−4×10−3/10−7 =
1.2×100 = 1.2 J/T
Baxial at same r = 2×Beq = 2×1.2×10−4 =
2.4×10−4 T3 MarksQ4 (Ex 5.4). A closely wound solenoid has 2000
turns, radius 40 mm, current 3A. Find equivalent bar magnet moment.
✓ Solution
Each turn is a circular loop of area A = πr² = π×(0.04)² = π×16×10−4 m²
Total magnetic moment: M = NIA
Each turn is a circular loop of area A = πr² = π×(0.04)² = π×16×10−4 m²
Total magnetic moment: M = NIA
M = 2000 × 3 × π × (0.04)² = 6000 × π
× 1.6×10−3 = 6000×5.027×10−3
M ≈ 6000×5.027×10−3 = 30.16 J/T ≈ 30.2 J/T3 MarksQ5 (Ex 5.5). At a place, horizontal component
of Earth's field = 3.6×10−5 T and dip angle = 60°. Find vertical
component and total field.
✓ Solution
BH = 3.6×10−5 T, δ = 60°
tan60° = BV/BH ⇒ BV = BH×tan60° = 3.6×10−5×√3
BH = 3.6×10−5 T, δ = 60°
tan60° = BV/BH ⇒ BV = BH×tan60° = 3.6×10−5×√3
BV = 3.6×10−5×1.732 =
6.24×10−5 T
Be = BH/cosδ = 3.6×10−5/cos60° =
3.6×10−5/0.5 = 7.2×10−5 T3 MarksQ6 (Ex 5.6). A bar magnet has moment 0.25 J/T.
It is placed along the axis of a circular coil of 20 turns, 10 cm radius. Find field at centre due to
magnet if magnet's mid-point is at 14 cm from coil centre.
✓ Solution
Axial field of bar magnet at r = 0.14 m:
B = 10−7×0.5/2.744×10−3 = 5×10−8/2.744×10−3 = 1.82×10−5 T
Axial field of bar magnet at r = 0.14 m:
B = μ02M/(4πr³) =
10−7×2×0.25/(0.14)³
(0.14)³ = 2.744×10−3 m³B = 10−7×0.5/2.744×10−3 = 5×10−8/2.744×10−3 = 1.82×10−5 T
3 MarksQ7 (Ex 5.7). A magnetic dipole aligned with
uniform B has total PE = −0.6 J. Find work done to rotate it through 60°.
✓ Solution
Initial PE: Ui = −MB cos0° = −MB = −0.6 J ⇒ MB = 0.6 J
Final PE after rotation of 60°: Uf = −MB cos60° = −0.6×0.5 = −0.3 J
Initial PE: Ui = −MB cos0° = −MB = −0.6 J ⇒ MB = 0.6 J
Final PE after rotation of 60°: Uf = −MB cos60° = −0.6×0.5 = −0.3 J
W = ΔU = Uf − Ui = −0.3 −
(−0.6) = +0.3 J
Positive work done → energy is supplied to rotate the dipole.3 MarksQ8 (Ex 5.8). A short bar magnet is placed with
its axis at 30° with a uniform field of 0.02 T. Torque on magnet is 0.006 N·m. Find moment M.
✓ Solution
τ = MB sinθ
τ = MB sinθ
M = τ / (B sinθ) = 0.006 / (0.02 × sin30°) = 0.006 /
(0.02×0.5) = 0.006/0.01 = 0.6 J/T
3 MarksQ9. A bar magnet oscillates in Earth's
horizontal field with time period T. The magnet is replaced by one of same dimensions but double the
magnetic moment. Find new time period.
✓ Solution
T = 2π√(I/MB). I is same (same dimensions and mass). BH is same (Earth's field unchanged). M doubles: M’ = 2M.
T = 2π√(I/MB). I is same (same dimensions and mass). BH is same (Earth's field unchanged). M doubles: M’ = 2M.
T’ = 2π√(I/(2MB)) = (1/√2)×2π√(I/MB) =
T/√2 = T ÷ √2 ≈ 0.707T
3 MarksQ10. What is angle of dip at a place where
BH = BV? At magnetic poles? At magnetic equator?
✓ Solution
When BH = BV: tanδ = BV/BH = 1 ⇒ δ = 45°
At magnetic poles: BH = 0 ⇒ tanδ = ∞ ⇒ δ = 90°
At magnetic equator: BV = 0 ⇒ tanδ = 0 ⇒ δ = 0°
When BH = BV: tanδ = BV/BH = 1 ⇒ δ = 45°
At magnetic poles: BH = 0 ⇒ tanδ = ∞ ⇒ δ = 90°
At magnetic equator: BV = 0 ⇒ tanδ = 0 ⇒ δ = 0°
3 MarksQ11. A paramagnetic material has
susceptibility 0.01 at 300 K. Find its susceptibility at 150 K (Curie's law applies).
✓ Solution
Curie's law: χmT = C (constant)
χ1T1 = χ2T2
Curie's law: χmT = C (constant)
χ1T1 = χ2T2
χ2 = χ1×T1/T2 =
0.01×300/150 = 0.02
As T decreases, χm increases (inversely proportional → Curie's law).3 MarksQ12. A solenoid of 500 turns, 8 cm diameter,
30 cm long carries 3A. (a) Find its magnetic moment. (b) If core is iron (μr = 500),
find B inside.
✓ Solution
A = πr² = π×(0.04)² = π×16×10−4 = 5.027×10−3 m²
(a) M = NIA = 500×3×5.027×10−3 = 7.54 J/T
(b) n = N/L = 500/0.30 ≈ 1667 turns/m
B = μ0μrnI = 4π×10−7×500×1667×3
A = πr² = π×(0.04)² = π×16×10−4 = 5.027×10−3 m²
(a) M = NIA = 500×3×5.027×10−3 = 7.54 J/T
(b) n = N/L = 500/0.30 ≈ 1667 turns/m
B = μ0μrnI = 4π×10−7×500×1667×3
B = 4π×10−7×500×5001 ≈
4π×10−7×2.5×106 ≈ 3.14 T
3 MarksQ13. Classify: (a) Superconductors (b)
Aluminum (c) Nickel (d) Water (e) Manganese as dia/para/ferro.
✓ Solution
(a) Superconductors: Perfect diamagnetic (χm = −1; Meissner effect: expels all B)
(b) Aluminum: Paramagnetic (small +ve χm)
(c) Nickel: Ferromagnetic (very large χm)
(d) Water: Diamagnetic (small −ve χm)
(e) Manganese: Paramagnetic (unpaired d-electrons)
(a) Superconductors: Perfect diamagnetic (χm = −1; Meissner effect: expels all B)
(b) Aluminum: Paramagnetic (small +ve χm)
(c) Nickel: Ferromagnetic (very large χm)
(d) Water: Diamagnetic (small −ve χm)
(e) Manganese: Paramagnetic (unpaired d-electrons)
3 MarksQ14. A compass needle oscillates at frequency
2 Hz in Earth's field (BH). When another magnet is placed nearby, frequency becomes 4 Hz.
Find ratio of field of magnet to BH at compass location.
✓ Solution
T ∝ 1/√B ⇒ f ∝ √B
f1/f2 = √(B1/B2) ⇒ 2/4 = √(BH/(BH+Bmag))
0.25 = BH/(BH+Bmag) ⇒ BH+Bmag = 4BH
Bmag = 3BH ⇒ ratio Bmag/BH = 3
T ∝ 1/√B ⇒ f ∝ √B
f1/f2 = √(B1/B2) ⇒ 2/4 = √(BH/(BH+Bmag))
0.25 = BH/(BH+Bmag) ⇒ BH+Bmag = 4BH
Bmag = 3BH ⇒ ratio Bmag/BH = 3
3 MarksQ15. Explain why a ferromagnetic material
becomes paramagnetic above its Curie temperature.
✓ Solution
Below Curie temperature Tc: Thermal energy < exchange interaction energy → magnetic domains remain aligned → ferromagnetic behaviour (very large χm).
Above Tc: Thermal agitation (kT) exceeds exchange interaction energy → domains break up → spins become random → only weak alignment with applied field remains → becomes paramagnetic. χm now follows modified Curie law: χm = C/(T−Tc) (Curie-Weiss law).
Below Curie temperature Tc: Thermal energy < exchange interaction energy → magnetic domains remain aligned → ferromagnetic behaviour (very large χm).
Above Tc: Thermal agitation (kT) exceeds exchange interaction energy → domains break up → spins become random → only weak alignment with applied field remains → becomes paramagnetic. χm now follows modified Curie law: χm = C/(T−Tc) (Curie-Weiss law).
🌟 Additional Questions (Q16 – Q22)
3 MarksQ16. A bar magnet of length 10 cm has pole
strength 6 A·m. Find (a) magnetic moment, (b) magnetic field at its axial point 20 cm from its
centre.
✓ Solution
2l = 10 cm = 0.10 m; pole strength m = 6 A·m
(a) M = m×2l = 6×0.10 = 0.6 J/T
(b) Bax = μ0×2M/(4πr³) = 10−7×2×0.6/(0.20)³ = 10−7×1.2/0.008 = 1.5×10−5 T
2l = 10 cm = 0.10 m; pole strength m = 6 A·m
(a) M = m×2l = 6×0.10 = 0.6 J/T
(b) Bax = μ0×2M/(4πr³) = 10−7×2×0.6/(0.20)³ = 10−7×1.2/0.008 = 1.5×10−5 T
3 MarksQ17. Why are soft iron cores used in
electromagnets but hardened steel is used for permanent magnets?
✓ Solution
Soft iron: High permeability (easily magnetised to large B), very low coercivity (easily demagnetised when H is removed), thin narrow B-H loop (low hysteresis loss). Ideal for electromagnets where magnetism must be switched on/off quickly and efficiently.
Hardened steel: High retentivity (retains most of its magnetism after removal of field) and high coercivity (resists demagnetisation by external fields). Wide B-H loop. Ideal for permanent magnets that must maintain their magnetism for long periods. Trade-off: harder to magnetise initially.
Soft iron: High permeability (easily magnetised to large B), very low coercivity (easily demagnetised when H is removed), thin narrow B-H loop (low hysteresis loss). Ideal for electromagnets where magnetism must be switched on/off quickly and efficiently.
Hardened steel: High retentivity (retains most of its magnetism after removal of field) and high coercivity (resists demagnetisation by external fields). Wide B-H loop. Ideal for permanent magnets that must maintain their magnetism for long periods. Trade-off: harder to magnetise initially.
3 MarksQ18. A short magnet (M = 0.4 J/T) is placed at
the origin with its axis along x-axis. Find B at (30 cm, 0, 0) and (0, 30 cm, 0).
✓ Solution
r = 0.30 m in both cases.
Axial (30cm, 0, 0): Bax = 10−7×2×0.4/(0.3)³ = 8×10−8/0.027 = 2.96×10−6 T (along +x)
Equatorial (0, 30cm, 0): Beq = 10−7×0.4/(0.3)³ = 4×10−8/0.027 = 1.48×10−6 T (along −x)
r = 0.30 m in both cases.
Axial (30cm, 0, 0): Bax = 10−7×2×0.4/(0.3)³ = 8×10−8/0.027 = 2.96×10−6 T (along +x)
Equatorial (0, 30cm, 0): Beq = 10−7×0.4/(0.3)³ = 4×10−8/0.027 = 1.48×10−6 T (along −x)
3 MarksQ19. Two identical bar magnets are placed
perpendicular to each other with their axes intersecting at midpoint. Find resultant moment
MR.
✓ Solution
Both have same moment M. Perpendicular ⇒ angle = 90°.
MR = √(M²+M²+2M²cos90°) = √(M²+M²) = M√2
MR = √2 M at 45° to each magnet
Both have same moment M. Perpendicular ⇒ angle = 90°.
MR = √(M²+M²+2M²cos90°) = √(M²+M²) = M√2
MR = √2 M at 45° to each magnet
3 MarksQ20. Explain Meissner effect. Why are
superconductors called perfect diamagnets?
✓ Solution
Meissner Effect: When a material becomes superconducting (below critical temperature Tc), it spontaneously expels all magnetic flux from its interior. Thus B = 0 inside superconducting material regardless of external field strength.
This corresponds to χm = −1 (M = −H, complete cancellation of B inside). This is the most extreme form of diamagnetism. Hence superconductors are called perfect diamagnets. This is different from ordinary diamagnets where |χm| << 1.
Meissner Effect: When a material becomes superconducting (below critical temperature Tc), it spontaneously expels all magnetic flux from its interior. Thus B = 0 inside superconducting material regardless of external field strength.
This corresponds to χm = −1 (M = −H, complete cancellation of B inside). This is the most extreme form of diamagnetism. Hence superconductors are called perfect diamagnets. This is different from ordinary diamagnets where |χm| << 1.
✍ Score Guide — 20 Questions
NCERT + Exercise: 15 questions × 3 marks = 45 marks | Additional: 5 questions × 3 marks = 15 marks | Total: 60 marks
NCERT + Exercise: 15 questions × 3 marks = 45 marks | Additional: 5 questions × 3 marks = 15 marks | Total: 60 marks
CLASS 12 PHYSICS | FORMULA CAPSULE
Magnetism and Matter
Chapter 5 — All Key Formulas & High-Yield Facts for NEET/JEE
📅 Section A — Bar Magnet Formulas
| Quantity | Formula | Note |
|---|---|---|
| Magnetic dipole moment | M = m×2l (= NIA for coil) | Direction: S→N inside magnet |
| Axial field (end-on) | Bax = μ02M/4πr³ | Direction: along M (S to N extended) |
| Equatorial field | Beq = μ0M/4πr³ | Direction: opposite to M |
| Ratio | Bax = 2Beq | At same distance from magnet |
| Torque on dipole | τ = MBsinθ | Max at θ=90°; zero at 0°, 180° |
| PE of dipole | U = −MBcosθ | Stable at θ=0; Unstable at θ=180° |
| Oscillation period | T = 2π√(I/MB) | I = moment of inertia of magnet |
📅 Section B — Earth's Magnetism
| Quantity | Formula | Note |
|---|---|---|
| BH and BV components | BH=Becosδ | BV=Besinδ | δ = angle of dip |
| Angle of dip formula | tanδ = BV/BH | δ=0 at equator; δ=90° at poles |
| Total field | Be = √(BH²+BV²) | Horizontal component max at equator |
📅 Section C — Magnetic Materials
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| χm | Small, negative | Small, positive; ∝1/T | Very large, positive |
| μr | Slightly <1 | Slightly >1 | Very large (>>1) |
| Behaviour in B | Repelled | Weakly attracted | Strongly attracted |
| Examples | Bi, Cu, H2O, NaCl | Al, Mn, O2, Na, Pt | Fe, Ni, Co |
| Temp effect | Independent | χm ↑ as T ↓ (Curie law) | Ferromagnetic → Paramagnetic above Tc |
Curie's law: χm = C/T | Permanent magnet: high coercivity + high
retentivity (steel) | Electromagnet: low coercivity + high μr (soft iron)
🧠 Memory Tricks
“Bax = 2Beq”Axial field is
DOUBLE the equatorial field at same distance. Bax = μ02M/4πr³;
Beq = μ0M/4πr³. Exact analogy: Eax=2Eeq for
electric dipole. Directions: axial → along M (S→N). Equatorial → opposite M.
“DPF: − + ∞” (χm
values)Diamagnetic: −. Paramagnetic: + (small). Ferromagnetic: very ++ (∞).
Water is diamagnetic. Al is paramagnetic. Fe/Ni/Co are ferromagnetic. Superconductors = perfectly
diamagnetic (χm = −1, B=0 inside → Meissner effect).
“Poles: Dip Extreme”At magnetic poles: dip δ =
90°, BH=0. At magnetic equator: dip δ = 0°, BV=0. Compass needle:
vertical at poles, horizontal at equator. tanδ = BV/BH.
“Permanent vs Electro”Permanent magnet: Hard
material (steel/Alnico) → High retentivity + High coercivity (hard to demagnetise).
Electromagnet: Soft material (soft iron) → Low coercivity + Low retentivity (easy
to switch). Hysteresis loss = area of loop.
🔢 Critical Values
μ0/4π = 10−7 T·m/A
Tc(Fe)=1043K, (Ni)=631K, (Co)=1388K
Superconductor: χm=−1 (perfect diamagnetic)
Equator: δ=0° | Poles: δ=90°
Bax=2Beq
T=2π√(I/MB)
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