Magnetism and Matter Exemplar Solutions Class 12 - Free PDF

Picture-first. Imagine cutting the toroid into n tiny flat loops arranged in a circle. Each loop's moment vector is a short arrow along the local tangent to the central circle.

1. Symmetry argument. The n moment vectors m⃗_k = IA_k lie head-to-tail around the central circle. Their vector sum closes back on itself: _k=1ⁿ _k → ∮ ϕ dϕ/(2π/n) = 0 for large n, so total m⃗ = 0.
2. Ampere's-law argument. ∮ B⃗· dl⃗ = ₀ I_enc gives B_inside = ₀ n I2π r inside the toroidal core; for any Amperian loop drawn entirely outside the windings, I_enc = 0, and by symmetry B_outside = 0 everywhere.
3. Multipole-expansion argument. A general localised current distribution has the far-field expansion B⃗ ∼ ₀4πmr³ + ₀4πQr⁴ + — monopole term is identically zero (no monopoles), dipole term needs m⃗≠ 0. Since B_outside = 0 to all orders, every multipole coefficient must vanish, in particular m⃗_toroid = 0.
4. Energy cross-check. For an external uniform field B⃗₀, the orientation energy of a magnetic moment is U = -m⃗·B⃗₀. Place a toroid in B⃗₀ and rotate it — no preferred orientation is observed, hence m⃗ = 0. This is also why a toroid is not deflected in a uniform field.

Alternative method (energy). The total flux linkage of the toroid with its own current is finite (self-inductance), but its coupling to an external uniform field is _ext = m⃗·B⃗₀ = 0, again proving m⃗ = 0.

m⃗_toroid = 0; option (c).

Strategic angle. The whole question reduces to a single geometric fact: two axes tilted by α produce a maximum angular projection of α on the horizontal plane.

1. Set up the geometry. Define the plane Π as the plane containing both the spin axis z_g and the magnetic-dipole axis z_m, with z_m tilted from z_g by α = 11.3^∘.
2. Locations in Π. In the plane Π, the projections of magnetic and geographic north coincide on the same line, so D = 0 there. Mumbai sits very close to this plane, hence its observed D ≈ 0.
3. Locations perpendicular to Π. Ninety degrees of longitude away from Π, the magnetic axis appears rotated by the full α = 11.3^∘ from the geographic axis as seen in the local horizontal plane. The sense of the rotation flips as we cross Π to the opposite side of the globe, giving D = +11.3^∘ W on one side and D = -11.3^∘ (i.e. 11.3^∘ E) on the other.
4. Range and average. D swings continuously from 11.3^∘ W to 11.3^∘ E, passing through 0 twice along Π. The longitudinal average is exactly zero for a pure tilted dipole — option (d) ``always negative'' is wrong by symmetry. Option (a) captures the full range.
5. Why (b) and (c) fail. (b) says the least declination is 0^∘; this is a misreading — there are points with D = 0 on Π, but they are not unique. (c) about Greenwich is a coincidence of which longitude we choose as zero; the dipole plane Π does not in general pass through Greenwich.

Concept linkage. The horizontal projection argument is the same one used for the angle of dip (δ) on the geographic equator (Q 5.10): both are bounded by the tilt α.

Option (a).

Quick reading. A permanent magnet is permanent because domain walls are pinned, not because every spin is locked perfectly.

1. Inside a single domain. Each ferromagnetic domain (∼ 10^-6 m across, containing ∼ 10¹⁵ atoms) already has all atomic moments aligned by the exchange interaction. So the molecule-moment _atom≠ 0 — this rules out (a) immediately.
2. Demagnetised iron. In a freshly demagnetised piece, the domains point in random directions; their vector sum averages to net moment M = 0. The sample has zero macroscopic magnetisation but non-zero microscopic moments — a key distinction.
3. Magnetisation process. Applying a strong external field grows the domains aligned with B⃗_ext at the expense of mis-aligned ones (domain-wall motion) and rotates remaining domains into the field direction (domain rotation). At saturation, all domains point parallel — M = M_s.
4. Removing the field. Domain walls do not slide back freely: they are pinned by crystal defects, grain boundaries and impurities. So a residual alignment remains, giving the remanence M_r with 0 < M_r < M_s. This partial alignment is option (c).
5. Why (d) is wrong. Full alignment of every domain (d) is the saturated state, only reached while the field is applied. At room temperature with B⃗_ext = 0, thermal agitation k_B T and stray demagnetising fields knock the most weakly pinned domains out of perfect alignment.

Diagram-based reasoning. On the M vs H hysteresis loop, the point ``permanent magnet'' is at H = 0, M = M_r — on the vertical axis, between 0 and M_s. The horizontal-axis intercept on the negative side, H = -H_c, gives the coercivity: the reverse field needed to undo the partial alignment.

Concept linkage. For T > T_c (Curie temperature, e.g. 770^∘C for iron), thermal energy beats the exchange coupling and domains disappear — the material becomes a paramagnet with χ ∼ 10^-3, as we will see in Q 5.8 and Q 5.14.

Option (c).

Strategic angle. The idealisations are useful but mathematically inconsistent at the boundary. Test each statement by applying the appropriate integral law to a closed surface or loop that straddles the boundary.

1. (a) Gauss's law for E⃗ on case (i). Pillbox around one plate of area A: only the inside face has flux _E = EA; outside face flux is zero. Idealised E = σ/₀ then gives _E = (σ/₀) A = q/₀. The flux exactly accounts for the enclosed charge. No contradiction — option (a) is false.
2. (b) Gauss's law for B⃗ on case (ii). Pillbox straddling the solenoid wall (one face inside, one outside): _B = B A - 0 = BA ≠ 0. This violates ∮B⃗· dA⃗ = 0 — the absence of magnetic monopoles. The idealisation does contradict Gauss's law for B⃗. Option (b) is true.
3. (c) Loop equation on case (i). Take a rectangular loop with both long sides parallel to the plates — one inside the capacitor, one outside. On the long sides E⊥ dl, contributing zero; on the short crossing sides, E· dl from the inside section is exactly cancelled by the path traversal direction. The net line integral is ∮ E· dl = 0. The idealisation thus agrees with the loop equation — option (c) is a true statement (not a contradiction), so the answer key does not include (c).
4. (d) Ampere's law on case (ii). For an Amperian loop chosen wholly outside the solenoid, I_enc = 0 and ∮H· dl = 0 are consistent. For a loop enclosing the windings (one long side inside, one outside), ∮H· dl = Hl = (nI)l matches I_enc = nIl exactly — no contradiction. So option (d) is false; Ampere's law is not contradicted.
5. Take-home. The idealisations agree with Gauss for E⃗, disagree only with Gauss for B⃗, agree with Ampere for B⃗, and agree with the electrostatic loop equation. Net answer: only (b).

Numerical cross-check. A real long solenoid of n = 1000 turns/m carrying I = 1 A has B_inside ≈ ₀ n I = 1.26× 10^-3 T. Just outside the windings, measurement shows B ∼ 10^-5 T from fringe leakage — small but non-zero, restoring consistency with ∇·B⃗ = 0.

Option (b) only.

Strategic angle. Curie's law is a one-line proportionality. The clean way is to take the ratio of the two states so the Curie constant C cancels — no need to know the material!

1. Write Curie's law in both states. M₁ = C B₁/T₁ and M₂ = C B₂/T₂.
2. Divide to eliminate C. M₂M₁ = B₂/T₂B₁/T₁ = B₂B₁· T₁T₂.
3. Substitute. B₂/B₁ = 0.2/0.6 = 1/3 (field drops by factor 3); T₁/T₂ = 4/16 = 1/4 (temperature rises by factor 4). Product: M₂M₁ = 13· 14 = 112.
4. Compute M₂. M₂ = M₁/12 = 8/12 = 2/3 A m^-1.
5. Sanity-check options. (a) 32/3 ≈ 10.7 — would require M to increase, but both B drops and T rises (both reduce M); rejected. (c) 6 and (d) 2.4 ignore one of the two changes; rejected. Only (b) = 2/3 accounts for both effects.

Numerical cross-check / unit analysis. [C] = [M]· [T]/[B] = A/m(K)/(T) = A·K/(m·T). The Curie constant for this sample is C = M₁ T₁/B₁ = (8)(4)/(0.6) ≈ 53.3 A·K/(m·T). Recomputing state 2: M₂ = C B₂/T₂ = (53.3)(0.2)/16 ≈ 0.667 A/m = 2/3 A/m. Same answer.

Concept linkage. Curie's law χ = C/T is the high-temperature limit of paramagnetism (μ B ≪ k_B T). For strong fields and low temperatures, the full Brillouin function gives saturation M → nμ — Curie's law would then over-predict.

M₂ = 2/3 A m^-1, option (b).

Structural observation. Two divergences settle this: ∇·B⃗ = 0 always; ∇·H⃗ = -∇·M⃗ which is non-zero where M⃗ varies.

1. Maxwell's law for B⃗. ∇·B⃗ = 0 everywhere — this is one of the four Maxwell equations and is equivalent to the statement that magnetic monopoles do not exist. Equivalently, ∮B⃗· dA⃗ = 0 on any closed surface. So B⃗ lines never start or end; they form closed loops. They must pass continuously across the surface S of the magnetic body. (a) is true; (b) is false.
2. Effective magnetic charge. Define a surface ``magnetic charge density'' _M = M⃗·n (where n is the outward unit normal). Inside, M⃗≠ 0; outside, M⃗ = 0. So at any face of S where M⃗ has a normal component, _M is non-zero. The auxiliary field H⃗ has sources at these effective charges: ∇·H⃗ = -∇·M⃗, so H⃗-lines start or stop on S. (d) is true; (c) is false.
3. Pillbox check for normal components. Apply both Gauss-like laws to a thin pillbox of cross-section Δ A straddling S: ∮B⃗· dA⃗ = (B_n^out - B_nⁱⁿ)Δ A = 0 ⇒ B_n continuous, ∮H⃗· dA⃗ = (H_n^out - H_nⁱⁿ)Δ A = _MΔ A ⇒ H_n jumps. This is the precise mathematical content of (a) and (d).
4. Tangential components (aside). Across the same surface, with no free current, H_t is continuous and B_t jumps by ₀ M_t. These are the boundary conditions you will use later for interfaces between two magnetic media.

Alternative method (vector analysis). From B⃗ = ₀(H⃗ + M⃗), taking divergence: ∇·B⃗ = ₀(∇·H⃗ + ∇·M⃗), and since ∇·B⃗ = 0, we get ∇·H⃗ = -∇·M⃗ — a single line that simultaneously delivers (a) (LHS =0) and (d) (RHS ≠ 0 at a magnetisation discontinuity).

Options (a), (d).

Quick reading. Atoms have two ``moving charge'' degrees of freedom that carry magnetism: orbital motion and intrinsic spin.

1. Orbital contribution. An electron in an atomic orbital traces a closed current loop with effective current I = e/T = ev/(2π r). Loop area A = π r². The magnetic moment is _L = IA = e v r/2 = (e/(2m_e))(m_e v r) = (e/(2m_e))L. In vector form μ⃗_L = -(e/(2m_e)) L⃗ (negative because the electron's charge is -e). This is option (a) — atomic currents.
2. Spin contribution. The electron has intrinsic spin angular momentum S⃗ with |S⃗| = /2, and an accompanying magnetic moment μ⃗_S = -g_s (e/(2m_e)) S⃗ where g_s ≈ 2 is the Land'e g-factor for the electron. Its magnitude is the Bohr magneton _B = e/(2m_e) ≈ 9.27× 10^-24 J/T. This is option (d) — intrinsic spin.
3. Pauli exclusion is not a source. (b) is a constraint on which orbital-spin states can be simultaneously occupied; it does not by itself produce a magnetic moment. Closed shells (e.g. Ne, Ar) have all moments paired and cancel, but this cancellation uses Pauli, doesn't come from it.
4. Polar molecules are an electric concept. (c) ``polar nature of molecules'' refers to permanent electric dipole moments (H₂O, HCl). These produce electric polarisation, not magnetism — wrong by category.
5. Total atomic moment. _atom = _L + _S. The sum can be zero (closed shell, e.g. noble gas ⇒ diamagnetic) or non-zero (unpaired electrons, transition-metal d-shell ⇒ paramagnetic or ferromagnetic).

Concept linkage. The relative weighting of orbital vs spin contributions in a material is captured by the magnetomechanical ratio (gyromagnetic ratio) γ. ``Spin-only'' materials have γ close to -e/m_e; orbital-quenched materials have larger corrections. This is measured by the Einstein–de Haas effect.

Options (a), (d).

Strategic angle. Above T_c iron loses its ferromagnetism and becomes a weak paramagnet. Track H, M, B separately — H depends only on free current, M depends on the material, and B = ₀(H+M) is the field actually measured.

1. H is set by geometry and free current. Ampere's law on a rectangular Amperian loop straddling the solenoid wall gives H = nI regardless of what is inside. Here H = (10³ turns/m)(1 A) = 10³ A/m, both before and after heating. Option (b) is therefore wrong if it claims any change in H.
2. Cold core (T < T_c). The given _r = 1000 means χ = _r - 1 ≈ 999. Then M = χ H = 999 × 10³ ≈ 10⁶ A/m and B = ₀(H+M) = (4π× 10^-7)(10³ + 10⁶) ≈ 1.26 T. The core dominates: M ≫ H.
3. Hot core (T > T_c). Above the Curie temperature, the spontaneous ferromagnetic order collapses; iron becomes a paramagnet with χ ∼ 10^-3 to 10^-2 (it follows Curie–Weiss law χ = C/(T - T_c) — see Q 5.14). Take χ ∼ 10^-2 as a representative upper bound: M = χ H ∼ 10^-2× 10³ = 10 A/m. The ratio M_cold/M_hot ∼ 10⁶/10 = 10⁵ to 10⁸ depending on precise χ. Within the stated ``factor of about 10⁸'' (option d), this is the right order.
4. Resulting B above T_c. B = ₀(H + M) ≈ ₀· 1010 ≈ 1.27× 10^-3 T — a drop by factor ∼ 10³ from 1.26 T. So B change is drastic; option (a) confirmed.
5. Why (c) is wrong. The magnetisation shrinks toward zero on heating, but does not flip sign. There is no reason for M⃗ to spontaneously reverse — that would require a reversed applied field or a coercive cycle.

Numerical cross-check. B_cold/B_hot ≈ 1.26 / (1.27× 10^-3) ≈ 10³, matching _r ≈ 10³. The flux density inside the solenoid drops by exactly the relative permeability — consistent with B = _0r H when M is set by linear χ.

Concept linkage. The Curie transition T_c for iron is ≈ 1043 K (770^∘C). For nickel T_c ≈ 627 K, for cobalt T_c ≈ 1394 K. Above T_c the material is no longer useful as a transformer core, motor magnet or magnetic memory medium — option (d) numerically tracks the physical death of ferromagnetism.

Options (a), (d).

Picture-first. Electrostatic shielding: charges plug the field lines (terminate them). Magnetostatic shielding: high-μ ducting reroutes the field lines (deflects them around the cavity). Two completely different mechanisms.

1. Electrostatic shielding mechanism. Place a hollow conductor in an external static E⃗. Free charges rearrange on the conductor's outer surface so that inside the conducting material E⃗_cond = 0 (a conductor in equilibrium has no internal field). Equivalently, the external field lines terminate on the induced surface charges. Inside the hollow region: E⃗ = 0 to arbitrarily high precision. Source statement: (a).
2. Why this works for E but not B. E⃗ obeys Gauss's law ∮E· dA = q/₀: lines can terminate on charges. B⃗ obeys ∮B· dA = 0: no monopoles, so B⃗ lines cannot terminate on anything. Option (b) ``B⃗ lines can also end'' is wrong on the first clause.
3. Magnetic shielding mechanism. Surround the cavity with a thick shell of high-_r material (mumetal: _r ∼ 10⁴–10⁵; soft iron: _r ∼ 10³). Boundary conditions at the shell's outer surface require continuity of B_n and of H_t. With B = _0r H, the same H_t inside the shell gives a vastly larger B inside the shell than outside. Lines of B crowd into the shell and travel around the cavity, like water flowing through a much larger pipe. Option (d).
4. Why magnetic shielding is imperfect. Since B lines cannot end on any material, some flux always leaks across the cavity — perfect shielding is impossible. This is exactly statement (c). Typical attenuation factor is _r-large but finite: 10³ to 10⁴ reduction with a single mumetal shell. Three nested shells ⇒ 10⁹ reduction (used in MEG brain-imaging rooms).

Concept linkage to electrostatics. The electrostatic analogue of mumetal is the dielectric: a high-_r material concentrates D inside itself the way mumetal concentrates B. So a dielectric cup partially shields D, but it does not shield E — that requires a conductor. The complete analogy table is:

• Electrostatic shielding (perfect): conductor.
• Magnetostatic shielding (approximate): high-_r shell.

Options (a), (c), (d).

Strategic angle. The geographic equator is a great circle tilted by 11.3^∘ from the magnetic equator. Dip is zero only on the magnetic equator. So on the geographic equator, dip is zero only where the two great circles cross — and is bounded elsewhere by the tilt angle.

1. Two great circles intersect at two antipodal points. The magnetic equator and the geographic equator are both great circles. Two distinct great circles on a sphere always intersect at exactly two antipodal points (think of the Earth's equator and a tilted ``orbit plane'' — they cross twice). So there are exactly two points on the geographic equator where it coincides with the magnetic equator ⇒ dip is zero only at those two points. Confirms (b); refutes (a) which says ``always zero''.
2. Dip changes sign. As we travel eastward along the geographic equator, we cross the magnetic equator twice. On one half-circle, we are in the magnetic northern hemisphere (B_V pointing down, δ > 0); on the other half, we are in the magnetic southern hemisphere (B_V pointing up, δ < 0). So δ swings through both signs. Confirms (c).
3. Bound on dip. The angular distance from the geographic equator to the magnetic equator never exceeds the tilt α = 11.3^∘. Using tanδ = 2cot_m where _m is the magnetic colatitude, and at worst _m = 90^∘ - 11.3^∘ = 78.7^∘: |tan_max| = 2cot(78.7^∘) = 2(0.199) ≈ 0.398, so |_max| ≈ 21.7^∘. Even tighter: the magnetic latitude of a geographic-equator point peaks at 11.3^∘. So |δ| ≤ arctan(2tan 11.3^∘) ≈ 21.7^∘. Bounded — confirms (d).
4. Net option set. (a) wrong, (b), (c), (d) correct.

Concept linkage. The same geometry governs declination (Q 5.2): the tilt α bounds both D and δ on the geographic equator. Once you understand the two-great-circle picture, both questions follow from the same diagram.

Options (b), (c), (d).

Quick reading. Spin magnetic moment scales as g· e/(2m) for a particle of charge ± e and mass m. With m_p / m_e ≈ 1836, the proton moment is ∼ 10^-3 of the electron moment — small enough to be hidden in any bulk magnetisation measurement.

1. Set up the moment formula. μ = (g/2)(e/m) for a spin-12 particle. For the electron g_e ≈ 2, so _e ≈ e/m_e = _B = 9.27× 10^-24 J/T (Bohr magneton). For the proton g_p ≈ 5.585 (anomalous, not 2), so _p ≈ (5.585/2)(e/m_p).
2. Take the ratio. _pe ≈ g_p/2g_e/2· m_em_p ≈ 2.7931· 11836 ≈ 1.52× 10^-3. So _p ≈ 1.5× 10^-3 _B — about three orders of magnitude smaller than the electronic moment.
3. Bulk consequence. Bulk magnetisation is the sum of moments per unit volume: M = nμ. With n_electrons of the same order as n_protons in any neutral material (charge balance), the proton contribution to M is ∼ 10^-3 of the electronic contribution.
4. When does the proton moment matter? In NMR / MRI, we apply a strong static B₀ and look for the tiny Larmor-precession signal at the proton frequency f_p = _p B₀/(2π) ≈ 42.58 MHz/T. The electronic background is suppressed by Faraday-screening and by choosing materials where electrons are paired. So the proton moment, though small, becomes the dominant signal at this specific frequency.

Concept linkage. This question is conceptually identical to why nuclear-magnetic effects do not show up in χ vs T data for ordinary materials: the nuclear moment is just too small. Only specialised experiments (NMR, M"ossbauer spectroscopy, neutron scattering) probe nuclear magnetism directly.

_p ≈ 1.5× 10^-3_e, so its contribution to bulk magnetisation is negligible compared to electronic moments.

Quick reading. A uniformly magnetised cylinder is exactly equivalent (for external fields) to a solenoid with surface current per unit length K = M, even though there is no actual conduction current flowing.

1. Why bound currents exist. Each atom in the magnetised cylinder is a tiny Amperian current loop. In the bulk, adjacent loops carry currents in opposite directions at their shared edges, so they cancel. On the lateral surface, the edge currents are uncompensated and add up to a net azimuthal current. This is the bound surface current K_b = M × n.
2. Magnitude. For M along the cylinder axis and n radially outward, |K_b| = M. Numerically: K_b = M = 10⁶ A/m.
3. Total bound current circling the cylinder. Integrating K_b along the length L: I_M = K_b· L = (10⁶ A/m)(0.10 m) = 10⁵ A.
4. Cross-check via volume current. The volume bound current density is J_b = ∇×M. For uniform M, J_b = 0 — all the bound current is on the surface, consistent with treating the cylinder as a solenoid sheath.
5. Cross-check via equivalent solenoid. A solenoid with n turns per metre and free current I_f carries surface current density K = nI_f. Matching K = M = 10⁶ gives, for example, n = 10⁴ turns/m and I_f = 100 A — or any other combination producing the same nI_f.

Numerical / unit check. [K_b] =  A/m, [L] =  m, [I_M] =  A. Substituting: (10⁶ A/m)× (0.10 m) = 10⁵ A. Units and magnitude consistent.

Alternative method (vector calculus). From ∇×H = J_f (free current density) and ∇×B = ₀(J_f + J_b), the bound part is J_b = ∇×M. Integrating over the cylinder's surface and using Stokes: ∮ M· dl = ∫(∇×M)· dA = I_b. For a rectangular path with one side along the axis inside the magnet (length L) and the return outside (where M = 0), the integral = ML. So I_M = ML, matching the answer.

I_M = M L = 10⁶ × 0.10 = 1.0× 10⁵ A.

Strategic angle. Diamagnetic moment per atom (Langevin formula _a ∝ Zr²) is of comparable order for atoms across the periodic table. Bulk χ therefore tracks the number density of atoms, which is the main difference between a gas at STP and a metallic solid.

1. Langevin diamagnetic susceptibility per atom. _atom ∼ -(₀/6)(Ze²/m_e)r²∼ -10^-29 m³ per atom (within an order of magnitude across light elements).
2. Bulk susceptibility = atom χ × number density. _bulk = n · _atom. So the ratio of two bulk susceptibilities is roughly the ratio of their number densities.
3. Number density of N₂ gas at STP. n_N2 = N_A/V_m = (6.02× 10²³)/(22.4× 10^-3m³) ≈ 2.7× 10²⁵ m^-3.
4. Number density of solid copper. n_Cu = ρ N_A / M_w = (8.9× 10³ kg/m³)(6.02× 10²³)/(63.5× 10^-3 kg/mol) ≈ 8.4× 10²⁸ m^-3.
5. Density ratio. n_Cu/n_N2 ≈ 8.4× 10²⁸ / (2.7× 10²⁵) ≈ 3.1× 10³.
6. Predicted vs observed χ ratio. Predicted χ(Cu)/χ(N₂) ∼ 3× 10³. Observed: 10^-5/(5× 10^-9) = 2× 10³. The prediction is within a factor of ∼ 1.5 — excellent agreement given that we assumed the per-atom χ is the same for nitrogen and copper.

Numerical cross-check. If the per-atom χ values differ by a factor 1.5 between N and Cu (perfectly reasonable given different Z and r²), the prediction matches the data exactly. The order-of-magnitude lesson stands: solid is ∼ 10³–10⁴ denser than gas, hence has ∼ 10³–10⁴ times larger diamagnetic |χ|.

Concept linkage. For paramagnetism and ferromagnetism, the per-atom moment is far larger than the induced diamagnetic moment, so the density argument is partly masked by the electronic-structure dependence of _atom. That is why iron (n similar to Cu) has χ ∼ 10³ rather than 10^-5.

Density ratio ∼ 10³ explains the ∼ 2000× susceptibility difference between Cu and N₂.

Strategic angle. Group the three cases by two binary questions: ``Does the atom carry a permanent moment?'' and ``Do neighbouring moments couple via exchange?'' Three of the four combinations give the three families of magnetism we know.

1. Diamagnet: no permanent moment, no coupling. Closed electron shells (noble gases, Cu metal, H₂O, organic molecules). Bulk response is the Langevin induced-orbital moment ∝ B, anti-parallel to B. _dia ≈ const (independent of T). Tiny thermal-expansion effect changes n slightly, hence χ by ∼ 10^-4 over Δ T ∼ 100 K — usually negligible.
2. Paramagnet: permanent moments, no exchange coupling. Atoms have unpaired electrons (Al, Mn salts, O₂). In a field B, the alignment energy per moment is μ B and thermal energy is k_B T. Boltzmann statistics in the weak regime μ B ≪ k_B T give Curie's law: M = nμ² B3k_B T, _para = ₀ n μ²3 k_B T = CT. Plot of 1/χ vs T is a straight line through the origin.
3. Ferromagnet: permanent moments + exchange coupling. For T > T_c, the Weiss molecular-field theory replaces B by B + λ M inside the Curie equation, giving the Curie–Weiss law: _ferro(T > T_c) = CT - T_c. χ diverges as T → T_c⁺ — the susceptibility becomes infinite at the transition, reflecting spontaneous ordering. For T < T_c, the material has spontaneous magnetisation M_s(T) even at B = 0. As T → 0, M_s → nM_atom (saturation).
4. What about the fourth combination? ``No permanent moment but coupling'' is empty — there is nothing to couple. Hence three families.
5. Antiferromagnet (bonus). Permanent moments with negative exchange (λ < 0) gives a related family: χ peaks at the N'eel temperature T_N, and below T_N alternating moments cancel macroscopically. Not asked here, but worth knowing.

Diagram-based reasoning. On a 1/χ vs T plot: diamagnet = horizontal line at constant negative χ^-1; paramagnet = straight line through origin with positive slope 1/C; ferromagnet above T_c = straight line with same positive slope but T-axis intercept at T = T_c. This single plot distinguishes the three cases experimentally.

Concept linkage. The temperature dependence reflects the underlying physics: diamagnetism is an orbital quantum response (no classical temperature), paramagnetism is classical alignment statistics, ferromagnetism is a phase transition with order parameter M_s.

Dia: χ ≈ const. Para: χ = C/T (Curie). Ferro: χ = C/(T-T_c) above T_c; spontaneous M_s below.

Picture-first. Meissner effect: a superconductor expels B⃗ from its interior — the field lines bend around the ball as if it were a perfectly impenetrable ``magnetic insulator''. This is unlike a permanent magnet (which channels B through itself) and unlike a normal conductor (which lets B permeate once steady-state is reached).

1. Cooling. Liquid nitrogen at 77 K is well below the critical temperature T_c ≈ 90–93 K of YBCO and other high-T_c superconductors. So the ball is fully in its superconducting state.
2. Meissner expulsion. As the ball approaches the magnet, surface supercurrents spontaneously develop to exactly cancel the interior B. Equivalently, the ball has χ = -1, _r = 0. The induced magnetic moment is m_ball = -V _effH_ext — anti-parallel to B_ext.
3. Force from non-uniform field. The energy of a dipole in an external field is U = -m· B. With m ∥ -B, U = +|m||B| — a moment in this orientation prefers regions of small |B|. The force F = -∇ U = ∇(m ·B) pushes the ball toward weaker field, i.e. away from the bar magnet pole.
4. Magnitude estimate. For a small spherical perfect diamagnet in a field gradient |∇ B|, the force is |F| ∼ V|∇ B²|/(2₀). Even for a 1 cm ball in a 1 T magnet with |∇ B| ∼ 50 T/m, the force is ∼ 0.1 N — comfortably enough to levitate the ball against gravity (mg ∼ 0.05 N for a typical density). This is the basis of working magnetic-levitation demos.
5. Direction of m. Anti-parallel to B_ext. If the bar magnet's N pole is on the right, B_ext at the ball points to the right, and m points to the left (its own ``N pole'' faces right, repelling the magnet's N pole).

Concept linkage to diamagnets. A superconductor is a perfect diamagnet (χ = -1); ordinary diamagnets (χ ∼ -10^-5 for Bi, water) show the same repulsion but 10⁵ times weaker. The ``frog levitation'' demo at Nijmegen exploits exactly this: water in a frog has χ ≈ -9× 10^-6, enough to levitate in a 16 T solenoid gradient.

(i) The ball moves away from the bar magnet. (ii) The induced magnetic moment m is anti-parallel to the applied B.

Strategic angle. Only the radial component B_r = (₀/4π)(2mcosθ/r³) contributes through a sphere centred on the dipole. Show that its flux integrates to zero by symmetry — the upper hemisphere's positive contribution exactly cancels the lower hemisphere's negative one.

1. Identify B· dA on a centred sphere. dA = R²sinθ dθ dϕ r (outward radial). Only B_r dA contributes; the θ component is tangential to the sphere.
2. Set up the integral. B_r = (₀ m/(4π R³))· 2cosθ on the sphere of radius R. The total flux: _B = ₀^2π₀^π ₀ m cosθ2π R³ · R²sinθ dθ dϕ = ₀ mR₀^πsinθ dθ.
3. Symmetry argument (qualitative). The integrand sinθ on [0,π] is antisymmetric about θ = π/2: substitute θ' = π - θ and sin(π-θ')cos(π-θ') = -sinθ'cosθ'. So the integral from 0 to π/2 exactly cancels the integral from π/2 to π.
4. Explicit evaluation. ₀^πsinθ dθ = 12 ₀^πsin 2θ dθ = 12[-cos 2θ/2]₀^π = -14(cos 2π - cos 0) = 0.
5. Conclusion. _B = 0 for any centred sphere, confirming Gauss's law for magnetism for the dipole.

Alternative method (Gauss's-law argument from definition). A magnetic dipole is the limit of two opposite monopoles of strength ± q_m separated by d → 0 with q_m d = m. By the ``magnetic-Coulomb'' analogue, each monopole-flux through a closed surface enclosing it equals ₀ q_m. The sphere encloses both monopoles (since they sit at the origin in the dipole limit), so total flux = ₀ q_m + ₀(-q_m) = 0. Same answer, different language.

Numerical sanity-check. At the equator (θ=π/2), B_r = 0, so the equator contributes nothing. The north pole has B_r = +₀ m/(2π R³) and the south pole has B_r = -₀ m/(2π R³). Symmetry ⇒ flux cancels.

∮B⃗· dA⃗ = 0 — Gauss's law for magnetism verified for the dipole.

Strategic angle. ``No motion in any external field'' is a very strong constraint: it forces the total magnetic moment of the system to be exactly zero. Then geometry handles the rest.

1. Equilibrium ⇒ vanishing total moment. Torque on a rigid system of dipoles in an external field B is τ = _im_i×B = (_im_i)×B. For this to vanish for an arbitrary B (the problem says ``slowly varying'' B — so it can point in any direction), we need _im_i = 0.
2. Three identical vectors summing to zero. Three vectors of equal magnitude that sum to zero must point at 120^∘ to one another, forming a closed equilateral triangle when drawn head-to-tail. There are two such configurations (mirror images), but the magnitudes are fixed.
3. Read the figure. From Fig. 5.1, the central magnet has N at top, S at bottom. So m₁ points up (from S to N inside the magnet, by convention m runs S→N inside, but the external dipole vector points from S toward N).
4. Place the other two moments. For m₂ and m₃ to make 120^∘ with m₁, they must point ``down-left'' and ``down-right'' — i.e. their N-poles face downward and slightly outward. So the bottom-left end of the left magnet is N (with S at top-left); the bottom-right end of the right magnet is N (with S at top-right).
5. Verify vector sum. m₁ + m₂ + m₃ = m₀j + m₀(cos 240^∘j + sin 240^∘i) + m₀(cos 120^∘ j + sin 120^∘i). The j sum: m₀(1 - 1/2 - 1/2) = 0. The i sum: m₀(0 - 3/2 + 3/2) = 0. Confirmed.

Alternative method (energy minimisation). The total potential energy of the three-magnet system in an external field is U = -m_tot·B. For equilibrium for every direction of B, m_tot must be zero (else the system would rotate to align it). Same conclusion.

Diagram-based reasoning. Drawing the three moment vectors as arrows of equal length: m₁ = ↑, m₂ = , m₃ = . Head-to-tail they trace an equilateral triangle returning to the start — closed polygon, zero sum.

The remaining poles: top-left = S, bottom-left = N; top-right = S, bottom-right = N.

Strategic angle. ``Identical motion'' means matching the equations of motion Iθ⃗̈ = τ⃗ for rotation and Mr⃗̈ = F⃗ for translation, plus matching initial conditions. So we need each force and torque to be equal at every position and orientation.

1. Match torques at every angle. Electric: _E = p×E, magnitude pEsinφ where φ is the angle between p and E. Magnetic: _B = m×B, magnitude mBsinφ. Equal at every φ: pE = mB ⇒ B/E = p/m (scalar ratio of magnitudes).
2. Match torque directions. For the torques to be vectorially equal at all times, p×E and m×B must have identical directions when the dipoles are identically oriented. This forces E and B to be parallel (or anti-parallel) at every point, since both must rotate the dipole the same way.
3. Match translational forces. Electric: F_E = (p·∇)E + p× (∇×E); in electrostatics ∇×E = 0, so F_E = (p·∇)E. Similarly F_B = (m·∇)B in magnetostatics (current-free region). Matching component by component requires B(r) = (p/m) E(r) as vector fields throughout space — the same proportionality at every point.
4. Match inertia. Equation of motion is Iθ̈ = τ. To produce identical angular acceleration from equal torques, I_elec = I_mag. Similarly for translation, M_elec = M_mag.
5. Match initial conditions. Same starting position r₀, velocity ṙ₀, orientation ₀, angular velocity θ̇₀. Without identical ICs even identical EoMs give different trajectories.

Concept linkage. This is the foundation of the formal E ↔ B analogy used to ``solve'' magnetostatics problems by analogy with electrostatics. For example, a uniformly magnetised sphere has the same field structure as a uniformly polarised dielectric sphere — once you identify M↔P and H↔D.

Numerical illustration. For a water molecule p ≈ 6.2× 10^-30 C·m and a typical atomic magnetic moment m ≈ 1_B ≈ 9.3× 10^-24 A·m², the ratio p/m ≈ 6.7× 10^-7 C/A·m = 6.7× 10^-7 T·m/V. So B needs to be tiny compared to E in absolute units to produce ``identical'' dipole motion.

B(r) = (p/m) E(r) throughout space; matching inertia and matching initial conditions.

Strategic angle. Compute how I and m scale with cutting, then plug into T = 2π√I/mB. Track each scaling factor carefully — the moment of inertia involves three different sources of change (mass halves, length-squared decreases by 14, identical 112 factor).

1. Magnetic moment scaling. The magnetic moment is proportional to the length of the magnet (per-unit-length m_l = m/L stays the same when we cut). Half-bar: m' = m_l· L/2 = m/2.
2. Moment of inertia scaling. For a thin uniform bar of mass M and length L about a perpendicular axis through its centre: I = 112ML². Half-bar has mass M/2 and length L/2: I' = 112·M2·(L2)² = 112· M L² · 12· 4 = I8.
3. Combine in the period formula. T' = 2π√I'm'B = 2π√I/8(m/2)B = 2π√14·ImB = 12· 2π√ImB = T2.
4. Physical interpretation. The half-bar is half as long and half as heavy, so its moment of inertia drops by 1/8, but its restoring torque drops only by half (since m halves). The net effect is that I/m drops by 1/4, and T ∝ √I/m drops by 1/2.
5. Frequency check. f' = 1/T' = 2/T = 2f. The half-magnet oscillates twice as fast as the original.

Numerical cross-check. For an iron bar of L = 10 cm, M = 100 g = 0.1 kg, m = 0.2 A m² in Earth's field B_H ≈ 3× 10^-5 T: I = 112(0.1)(0.1)² = 8.3× 10^-5 kg m²; T = 2π√8.3× 10^-5/(0.2· 3× 10^-5) ≈ 2π√13.9 ≈ 23.4 s. After cutting, T' ≈ 11.7 s. The half-bar oscillates visibly faster.

Alternative method (dimensional scaling). Under L→ L/2, M→ M/2: m∝ L → m/2; I ∝ ML² → I/8. Then T∝ √I/m → T√(1/8)/(1/2) = T√1/4 = T/2.

T' = T/2 — the half-magnet has half the period.

Strategic angle. Two laws act in tandem inside a permanent bar magnet (no free currents, finite M): ∮B⃗· dA⃗ = 0 forbids B⃗ from breaking; ∮H⃗· dl⃗ = 0 in a current-free loop forces H⃗ to reverse inside compared to outside.

1. Direction of B outside. Outside the bar magnet, the field is a dipole field: lines emerge from the N pole, loop through space, and re-enter at the S pole. So just outside, B_out runs from N to S in external space (when traced along an external field line).
2. Continuity of B at the pole faces. Gauss's law ∮B· dA = 0 requires the normal component of B to be continuous across the magnet's end faces. Lines that enter the S-face from outside must continue inside, exit the N-face, then loop around outside. Therefore inside the magnet, B runs from S to N (part b).
3. Direction of H outside. Outside the magnet, M = 0, so B = ₀H, hence H and B are parallel. So outside, H also runs N → S along an external field line.
4. Ampere's law for H. Take any closed loop threading through the magnet (going S→N inside, then N→S back outside through space). The loop encloses no free current — the bound currents J_b = ∇×M don't enter Ampere's law for H. So ∮H· dl = 0 on this loop.
5. Conclusion for H inside. The outside contribution to ∮H· dl is positive (we traverse external N→S in the direction of H). For the closed-loop integral to vanish, the inside contribution must be negative — meaning H inside points opposite to our traversal direction. We traversed S→N inside, so H_in points N→S (part a).
6. Therefore inside a bar magnet: H and B are anti-parallel.

Concept linkage. This is the cleanest demonstration that B and H are distinct physical entities: not only do they obey different laws (∇·B = 0 vs ∇· H = -∇·M), they actually point in opposite directions inside a permanent magnet. The relation B = ₀(H + M) gives: inside, B (S→N) = ₀[H (N→S, magnitude small) + M (S→N, magnitude large)], so the M term dominates and pulls B along its direction.

Diagram-based reasoning. A textbook picture shows: B lines as continuous loops, going S→N inside and N→S outside, all the way around. H lines have arrows that flip direction at the pole faces: N→S outside, but also N→S inside (opposite to B). The pole faces act like ``magnetic charge'' sources for H only.

Inside the bar magnet: H runs N → S; B runs S → N. Outside: both run N → S.

Strategic angle. Compute each of the four line-integral contributions explicitly, then check that they cancel. The algebraic miracle is that the two axial segments and the two quarter-arc segments combine in pairs to give the same magnitude with opposite signs.

1. Segment (i): z-axis from z=a to z=R. At θ=0 (on axis), the dipole field has B_r = 2₀ m/(4π r³) along r = k. So B = (₀ m/(2π z³))k along the segment. With dl = dz k: _a^R ₀ m2π z³ dz = ₀ m2π·[-12z²]_a^R = ₀ m4π(1a² - 1R²).
2. Segment (ii): outer quarter-arc, r=R, θ=0→ π/2. On a sphere, the dipole's θ component is B_θ = ₀ msinθ/(4π R³). The clockwise traversal in the first quadrant of x-z plane corresponds to θ increasing from 0 at the +z axis to π/2 at the +x axis. So dl = R dθ θ and: ₀^π/2₀ msinθ4π R³· R dθ = ₀ m4π R²₀^π/2sinθ dθ = ₀ m4π R².
3. Segment (iii): x-axis from x=R to x=a. At θ = π/2 (equatorial plane), B_r = 0 and B_θ = ₀ m/(4π x³). The unit vector θ at θ = π/2 points along -k (downward). So B = -(₀ m/(4π x³))k, perpendicular to the x-axis segment dl = -dx i. Inner product zero: ∫ B· dl = 0.
4. Segment (iv): inner quarter-arc, r=a, θ=π/2→ 0. Now we traverse the arc in the opposite θ-direction compared to (ii). Same integrand form, opposite sign and smaller radius: _π/2⁰₀ msinθ4π a² dθ = -₀ m4π a²₀^π/2sinθ dθ = -₀ m4π a².
5. Add all four contributions. aligned _CB· dl &= ₀ m4π(1a²-1R²) + ₀ m4π R² + 0 - ₀ m4π a²
   &= ₀ m4π[1a² - 1R² + 1R² - 1a²] = 0. aligned
6. Conclusion. _CB· dl = 0, consistent with Ampere's law ∮B· dl = ₀ I_enc for a curve enclosing no free current (the dipole sits at the origin, but in vacuum, there is no current threading C).

Pattern recognition. The integral on a quarter-arc of radius r is ±₀ m/(4π r²); the integral on an axial segment from r₁ to r₂ is ₀ m/(4π)(1/r₁² - 1/r₂²). Closed curves built from these pieces always integrate to zero, because the radial-segment contributions exactly cancel the arc-segment contributions.

Concept linkage. Outside a magnetic dipole, B is ``curl-free'' (i.e. ∇×B = 0 in the vacuum region). So ∮B· dl = 0 on any closed curve in the exterior — confirmed in this specific calculation. Inside the source region, B would have non-zero curl from current.

_CB· dl = 0 — Ampere's law verified.

Strategic angle. Dimensional analysis fixes the form up to a constant; estimate the magnitude and compare to data. This is the canonical example of how dimensional analysis nails the structure of a physical quantity without solving the full quantum problem.

1. Confirm χ is dimensionless. From M = χH with [M] = [H] = A/m, χ is dimensionless.
2. Dimensional ansatz. χ ∼ ₀^a e^b m^c v^d R^e. Dimensions of each base quantity: [₀] =  M L T^-2 A^-2, [e] =  A T, [m] =  M, [v] =  L T^-1, [R] =  L.
3. Match powers of each base dimension to zero. aligned M: & a + c = 0
   L: & a + d + e = 0
   T: & -2a + b - d = 0
   A: & -2a + b = 0 aligned From the A equation: b = 2a. Substitute into T equation: -2a + 2a - d = 0 ⇒ d = 0. From M equation: c = -a. From L equation: e = -a.
4. Pick the minimal natural solution a = 1. Then b = 2, c = -1, d = 0, e = -1. So χ ∼ ₀ e²m R. Note v does not appear in the simplest combination — its exponent is zero. (This is because dimensional analysis cannot distinguish v from c; the full Larmor calculation does pick up an extra factor of (v/c)² ∼ α², where α ≈ 1/137 is the fine-structure constant.)
5. Numerical estimate. Plug in: ₀ = 4π× 10^-7 T·m/A, e = 1.6× 10^-19 C, m_e = 9.1× 10^-31 kg, R_Bohr = 5.3× 10^-11 m. e² = 2.56× 10^-38 C² ₀ e² = 4π× 10^-7× 2.56× 10^-38 = 3.22× 10^-44 T·m·C²/A m R = 9.1× 10^-31× 5.3× 10^-11 = 4.82× 10^-41 kg·m χ ∼ 3.22× 10^-444.82× 10^-41 ≈ 6.7× 10^-4.
6. Compare with observed. Solid-state χ ∼ 10^-5 (a typical diamagnet, e.g. copper); our estimate ∼ 7× 10^-4 is about two orders of magnitude larger.
7. Why the discrepancy? Dimensional analysis only gives the gross structure. The full Langevin / Larmor calculation produces a missing factor of Z(v/c)² ∼ Z α² ∼ 10^-4 (with Z = number of electrons, α the fine-structure constant). Multiplying: 7× 10^-4× 10^-4 = 7× 10^-8 — too small. With Z ∼ 10-100, get back ∼ 10^-5 to 10^-6, matching solid-state values.

Concept linkage. The combination ₀ e²/(m R) has units of energy when multiplied by frequency, so it appears naturally in atomic-scale electromagnetism. In SI it equals α² · (Rydberg energy / _atomic) up to a factor of 4π.

χ ∼ ₀ e²/(mR) ≈ 7× 10^-4, within two orders of magnitude of the observed solid-state χ ∼ 10^-5.

Strategic angle. Three loci, three trigonometric equations in θ (magnetic colatitude). Convert each θ to magnetic latitude λ = 90^∘ - θ at the end for the geographer-friendly answer.

1. Field magnitude. The total magnitude of the dipole field is |B|² = B_V² + B_H² = (₀ m4π r³)² (4cos²θ + sin²θ) = (₀ m4π r³)² (1 + 3cos²θ). At fixed r = R_E (Earth surface), |B| depends only on θ through the factor √1+3cos²θ.
2. (i) Minimum |B|. The factor 1+3cos²θ is minimised when cosθ = 0, i.e. θ = 90^∘, giving magnetic latitude λ = 0 — the magnetic equator. At this locus, |B|_min = ₀ m/(4π R_E³). Numerically, with m ≈ 8× 10²² A m² and R_E ≈ 6.37× 10⁶ m: |B|_min ≈ (10^-7)(2× 8× 10²²)/ (6.37× 10⁶)³ ≈ 3× 10^-5 T (matches the observed value at the equator).
3. (ii) Dip angle zero. The dip angle is defined by tanδ = B_V/B_H = (2cosθ)/sinθ = 2cotθ. δ = 0 iff cotθ = 0 iff θ = 90^∘, again the magnetic equator. So loci (i) and (ii) coincide. This is physically meaningful: where the field magnitude is minimum, it is also purely horizontal.
4. (iii) Dip angle ± 45^∘. tanδ = ± 1 gives 2cotθ = ± 1, hence cotθ = ± 1/2, tanθ = ± 2. Solving: ₊ = arctan(2) ≈ 63.43^∘, _- = 180^∘ - 63.43^∘ ≈ 116.57^∘. Converting to magnetic latitude λ = 90^∘ - θ: ₊ = +26.57^∘ (mag. N), _- = -26.57^∘ (mag. S). Both are great-circle loci of constant magnetic latitude.
5. Geographic interpretation. Magnetic latitude ± 26.57^∘ does not coincide with geographic latitude ± 26.57^∘ because of the 11.3^∘ tilt; however, the loci are circles of constant magnetic latitude around the magnetic dipole axis.

Numerical cross-check. At λ = 0 (magnetic equator): δ = 0, |B| = ₀ m/(4π R³). At λ = 90^∘ (magnetic pole): δ = 90^∘, |B| = 2₀ m/(4π R³). Equator-to-pole field magnitude ratio = 2 — a famous textbook result. Our formula reproduces it: √1+3cos² 0/√1+3cos²(π/2) = √4/√1 = 2.

Concept linkage. The same dipole-field formulas B_V = ₀(2mcosθ)/(4π r³) and B_H = ₀(msinθ)/ (4π r³) describe the field around a bar magnet at large distances. So the loci of constant dip around a bar magnet are also ``magnetic latitude circles''.

(i), (ii): magnetic equator (λ = 0). (iii): magnetic latitudes λ = (1/2) ≈ ± 26.57^∘.

Strategic angle. Both points lie on the magnetic equator, so the dip is automatically zero at both. The question reduces to finding the declination at each, which depends on the geometric position relative to the plane S containing both axes.

1. Both points are on the magnetic equator. By definition, the magnetic equator is the locus where the magnetic field is horizontal (no vertical component). So at both P and Q: B_V = 0 ⇒ _P = 0, _Q = 0. That handles dip immediately.
2. Geometric setup for P. The point P is on the magnetic equator and also lies in the plane S that contains both the geographic spin axis z_g and the magnetic dipole axis z_m. In this plane S, z_m is tilted from z_g by α = 11.3^∘.
3. Declination at P. ``Magnetic north'' at P is the horizontal projection of the line from P toward the magnetic pole; ``geographic north'' is the horizontal projection of the line from P toward the geographic pole. Both these projections lie in S (since P is in S). The angle between them in the local horizontal plane equals the tilt of the magnetic axis from the geographic axis, projected onto that horizontal plane. Because P is on the magnetic equator, the local horizontal plane at P is perpendicular to z_m. So the magnetic north direction lies along the projection of z_m onto the horizontal — which, given the tilt, makes an angle of exactly α = 11.3^∘ with the geographic-north projection of z_g. Hence D_P = 11.3^∘.
4. Geometric setup for Q. The point Q is the intersection of the magnetic equator and the geographic equator (two great circles on a sphere intersect at two antipodal points; pick one).
5. Declination at Q. At Q, the local horizontal plane is tangent to both equatorial circles (since Q is on both). ``Magnetic north'' at Q is along the tangent to the magnetic equator's meridian (perpendicular to the magnetic equator's tangent at Q in the horizontal plane); ``geographic north'' is along the tangent to the geographic equator's meridian. At the crossing point Q, the two equators are inclined to each other at angle α = 11.3^∘, but the perpendicular norths in the horizontal plane align (both perpendiculars at Q point in the same direction, along the common tangent direction of the equators' crossing). Hence D_Q = 0.
6. Tabulate.
   • At P: _P = 0, D_P = 11.3^∘.
   • At Q: _Q = 0, D_Q = 0.

Concept linkage. The plane S is where the tilt α manifests as maximum declination; the crossing line is where it contributes zero declination but maximum equator-tilt. Both effects are projections of the same 11.3^∘ rotation onto different reference planes.

Diagram-based reasoning. Imagine looking at the Earth from outside. The two equators (magnetic and geographic) are two tilted great circles, like the rings of Saturn intersecting at two antipodal points. The plane S contains both axes and cuts perpendicular to the line of intersection at the midpoints of both equators.

P: declination D_P = 11.3^∘, dip _P = 0. Q: declination D_Q = 0, dip _Q = 0.

Strategic angle. Equate frequencies via the ratio m/I_moi. Both coils share the same wire length L, the same current I, and the same wire linear mass density λ, so L, I, and λ drop out — only the geometric quantities R and a remain.

1. Set up the frequency equality. Period of a coil oscillating in a uniform field is T = 2π√I_moi /(mB), frequency f = (1/(2π))√mB/I_moi. Equal frequencies with equal B require m₁/I_moi,1 = m₂/I_moi,2.
2. Number of turns on each coil. Total wire length is fixed at L. Each turn of a circular coil of radius R consumes circumference 2π R, so N₁ = L/(2π R). Each turn of a square coil of side a consumes perimeter 4a, so N₂ = L/(4a).
3. Magnetic moments. m₁ = N₁ I · π R² = L2π R· I· π R² = I L R2. m₂ = N₂ I · a² = L4a· I· a² = I L a4.
4. Moment of inertia: circular coil. Treat the wire as a thin ring of total mass M = λ L at radius R. Axis of oscillation is a diameter through the centre (in the plane of the coil). For a thin ring of mass M and radius R about a diameter, I_moi = 12 M R² (perpendicular-axis theorem on a ring). I_moi,1 = 12(λ L) R² = λ L R²2.
5. Moment of inertia: square coil (NCERT convention). The NCERT Exemplar uses I_moi,2 = Ma²/12 for the square coil about an in-plane axis through its centre, where M = λ L is the total wire mass. Hence I_moi,2 = (λ L) a²12 = λ L a²12.
6. Ratios. m₁I_moi,1 = ILR/2λ L R²/2 = Iλ R. m₂I_moi,2 = ILa/4λ L a²/12 = 12I4λ a = 3Iλ a.
7. Equate and solve. Iλ R = 3Iλ a ⇒ 1R = 3a ⇒ a = 3R.
8. Verify. With a = 3R: m₂/m₁ = (a/2)/(R) = (3R/2)/R = 3/2. I_moi,2/I_moi,1 = (a²/12)/(R²/2) = a²/(6R²) = (3R)²/(6R²) = 9/6 = 3/2. Both ratios are 3/2, confirming (m₁/I_moi,1) = (m₂/I_moi,2).

Numerical illustration. If R = 10 cm, then a = 30 cm. Wire length L = 2π R · N₁ for any chosen N₁; same L gives N₂ = L/(4a) = L/(120 cm) = N₁· 2π(10)/120 = N₁π/6.

Concept linkage. The result depends only on the geometrical ratios (perimeter, area, moment of inertia) of the shapes — not on λ, I, or L — because both coils carry the same total wire under the same conditions. Pure geometry.

a = 3R — the side of the square is three times the radius of the circle for equal oscillation frequencies.