Magnetism and Matter

NEET frequency of this chapter

Priority ranking — 🔥🔥🔥 Magnetic materials (χ, μ, Curie) · 🔥🔥🔥 Bar magnet field (axial + equatorial) · 🔥🔥 Time period of oscillation · 🔥🔥 Cutting a magnet · 🔥 Earth's magnetism (dip, declination).

Aamirah's 5-day sprint

Bar magnet — the fundamentals

• Two poles (N and S) · always occur in pairs (isolated magnetic monopoles have never been observed). Cutting a bar magnet never gives a single pole — each fragment is itself a full dipole.
• Magnetic moment m = pole strength × magnetic length = m_p × 2ℓ. SI unit: A·m². Direction: from S to N inside the magnet.
• Field lines: emerge from N, curve through space, enter S. Inside the magnet they run S → N (closed loops).

Bar magnet as an equivalent solenoid

• A finite solenoid of length 2ℓ, radius a, carrying current I with n turns per unit length has an axial field at large r: B = (μ₀/4π)(2m/r³) where m = n·(2ℓ)·I·A is the equivalent magnetic moment.
• Comparison with bar magnet axial field B = (μ₀/4π)(2m/r³) shows the two are equivalent. This is the theoretical bridge between electromagnetism (moving charges) and static magnetism (magnets).
• Consequence: every bar magnet is essentially a bundle of atomic-scale current loops (electron orbital + spin magnetic moments).

Dipole in a uniform magnetic field

• Torque: τ = m × B · magnitude τ = mB sinθ, where θ is the angle between m and B.
• τ is max when θ = 90° (dipole ⟂ B) · τ = 0 at θ = 0 (stable equilibrium) and θ = 180° (unstable).
• Potential energy: U = −m·B = −mB cosθ. Minimum (−mB) at θ = 0; maximum (+mB) at θ = 180°.
• Work done in rotating from θ₁ to θ₂: W = mB(cosθ₁ − cosθ₂).
• A dipole free to rotate always aligns with B (like a compass needle).

Earth's magnetism (geomagnetism)

• Earth acts approximately like a giant bar magnet tilted ~ 11° from the rotation axis, with its magnetic south near the geographic north (that's why the N pole of a compass points north).
• Three geomagnetic elements at any location:
  • Declination (D): angle between geographic north and magnetic north (horizontal plane).
  • Dip / inclination (δ): angle between the total field and horizontal. At magnetic equator δ = 0°; at magnetic poles δ = 90°.
  • Horizontal component (B_H): horizontal projection of total field B.
• Relations: B_H = B cos δ · B_V = B sin δ · tan δ = B_V/B_H · B = √(B_H² + B_V²).

Magnetic properties of materials (dia / para / ferro)

Curie's law + hysteresis

• Curie's law (paramagnetics only): χ = C/T · where C is the Curie constant. Susceptibility is inversely proportional to absolute temperature.
• Curie temperature (T_C): temperature above which a ferromagnet becomes paramagnetic. Iron: T_C = 1043 K; nickel: 631 K; cobalt: 1394 K; gadolinium: 293 K.
• Hysteresis: for ferromagnets, magnetisation B doesn't simply follow H — it depends on history. The B-H curve traces a loop.
• Retentivity (B_r): residual B when H is reduced to zero. Coercivity (H_c): reverse H needed to bring B to zero.
• Loop area = energy dissipated per cycle per unit volume. Soft ferromagnets (transformer cores) have thin loops (low loss); hard ferromagnets (permanent magnets) have fat loops (high retentivity).

Constants to have on tap

Bar magnet · axial field (at distance r ≫ 2ℓ)

Setup: Bar magnet of length 2ℓ, pole strength m_p. Point P on the axis at distance r from centre. Field at P = vector sum of field from N pole (distance r−ℓ) + field from S pole (distance r+ℓ).

Skeleton:
  B_P = (μ₀/4π) · m_p [1/(r−ℓ)² − 1/(r+ℓ)²]
        = (μ₀/4π) · m_p · [4rℓ/(r²−ℓ²)²]
  For r ≫ ℓ: (r²−ℓ²)² ≈ r⁴ → B_ax = (μ₀/4π)(2m/r³) where m = 2ℓ·m_p.

Bar magnet · equatorial field (at r ≫ 2ℓ)

Setup: Point P on equatorial line (perpendicular bisector) at distance r from centre. Each pole is √(r²+ℓ²) from P; the fields have equal magnitudes but different directions.

Skeleton:
  Only the horizontal components add (vertical ones cancel).
  B_P = 2 · (μ₀/4π) · [m_p/(r²+ℓ²)] · [ℓ/√(r²+ℓ²)]
        = (μ₀/4π) · 2ℓ·m_p/(r²+ℓ²)^(3/2)
  For r ≫ ℓ: (r²+ℓ²)^(3/2) ≈ r³ → B_eq = (μ₀/4π)(m/r³).

Consequence: B_ax : B_eq = 2 : 1 at same r. Direction of B_ax is along m; B_eq is opposite to m.

Torque on a dipole + PE

Setup: Bar magnet in uniform field B, its axis making angle θ with B.

Skeleton:
  Force on N pole = +m_pB (along B). Force on S pole = −m_pB (opposite B). Net force = 0.
  But the two forces form a couple. Perpendicular distance between forces = 2ℓ sinθ.
  Torque τ = (force) × (perp. distance) = m_pB · 2ℓ sinθ = mB sinθ, or τ = m × B.
  PE from τ: U(θ) − U(θ_ref) = ∫_θref^θ τ dθ' = −mB cosθ (choosing U = 0 at θ = π/2). So U = −m·B.

Time period of oscillation

Setup: Bar magnet suspended (or placed on pivot) free to oscillate in a magnetic field B. Small angle displacement.

Skeleton:
  Restoring torque τ = −mB sinθ ≈ −mBθ (small angle).
  Compare with τ = Iα (angular Newton's 2nd): Iα = −mBθ → α = −(mB/I)θ.
  This is SHM with ω² = mB/I. So T = 2π√(I/mB).
  Use: known I → measure T → find B (or vice versa). Classic problem type.

The 5 direction rules you'll need

• Magnetic moment m of a bar magnet: from S pole to N pole (inside the magnet).
• Field along axis (external): same direction as m (points away from N externally).
• Field along equatorial line (external): opposite to m (parallel but reversed).
• Torque τ = m × B: perpendicular to both m and B, tends to rotate m into alignment with B.
• Force on a magnetic dipole in a non-uniform field: F = ∇(m·B). Direction: along ∇B (toward stronger field for m ∥ B).

Bar magnet field at a specific point

Given: m and r (axial or equatorial). Ask: B at point. Method: pick the right formula. Axial: (μ₀/4π)(2m/r³) · Equatorial: (μ₀/4π)(m/r³). If two magnets, add vectorially.

Cutting a bar magnet

Given: bar magnet cut transversely (perpendicular to length) or longitudinally (parallel to length). Method:

• Cut transversely (into 2 shorter magnets): each half has moment m' = m/2 (length halved, pole strength unchanged).
• Cut longitudinally (into 2 thinner magnets): each half has pole strength halved, length unchanged, so m' = m/2 also.
• Time period changes accordingly — since T = 2π√(I/mB), and both I and m change, use the new I and new m.

Time period of oscillation in a field

Given: bar magnet (mass M, length 2ℓ, moment m) oscillating in field B. Ask: T. Method: I = M(2ℓ)²/12 for a thin bar rotating about its centre → T = 2π√(I/mB).

Torque and PE of dipole in field

Given: m, B, angle θ. Ask: τ or U or W (work rotating). Method: τ = mB sinθ · U = −mB cosθ · W = mB(cosθ_i − cosθ_f).

Dip / apparent dip / horizontal component

Given: two of {B, B_H, B_V, δ}. Ask: another. Method: tan δ = B_V/B_H · B_H = B cos δ · B_V = B sin δ. For apparent dip: tan δ' = tan δ / cos α.

Susceptibility identification

Given: sign of χ, temperature dependence, or behaviour in field gradient. Ask: dia/para/ferro. Method:

• χ < 0 · T-independent → diamagnetic
• Small +χ · χ ∝ 1/T → paramagnetic
• Large +χ · drops sharply at some T_C → ferromagnetic

Reading the B-H curve

Given: hysteresis loop. Ask: B_r, H_c, energy loss, soft vs hard classification. Method: retentivity is the y-intercept · coercivity is the x-intercept · loop area is the energy loss per cycle · thin loop = soft, fat loop = hard.

Electric dipole vs Magnetic dipole — perfect parallel

Trick: replace p ↔ m, 1/(4πε₀) ↔ μ₀/(4π), E ↔ B. Every electric-dipole formula converts directly.

Moving Charges & Magnetism (Chapter 4)

• The bar magnet ↔ solenoid equivalence is the theoretical bridge between the two chapters. Solenoid moment m = n(2ℓ)IA has the same axial field formula as a bar magnet.
• The Bohr magneton connects the two: it's the natural unit for atomic magnetic moments arising from electron orbital motion.

Electromagnetic Induction (Chapter 6)

• Flux Φ = B·A links directly to magnetic-field concepts here.
• Hysteresis loss (from this chapter) is an important consideration for transformer-core selection in EMI applications.

Atomic Structure / Quantum

• The Bohr magneton comes from Bohr's model of the H atom: μ_B = e(ℏ)/(2m_e).
• Spin magnetic moment: each unpaired electron contributes ~ 1.73 μ_B via the spin-only formula μ = √(n(n+2)) · used in coordination chemistry (Chemistry Ch 5).