Magnetism and Matter

Magnetic moment is a vector. Two of them add by the parallelogram rule, exactly like two forces. Nothing about magnetism is special here — it is vector addition wearing a magnet costume.

The N-to-S direction of each bar is the direction of its moment. Read that off the figure first, before doing any arithmetic.

1. Reduce the formula: for two equal moments it collapses to 2m cos(θ/2). That is a decreasing function of θ.
2. So the smallest angle wins. No calculation is needed beyond ranking the angles.
3. Rank them: 30° < 60° < 90° < 180°.
4. Check the extremes to be sure: at 180°, 2m cos90° = 0; at 30°, 2m cos15° = 1.93m.

A magnet swinging in the earth's field is a torsional oscillator: T = 2π√(I/MB).

The one thing to notice is that brass is non-magnetic. It contributes nothing to M. So M and B are identical in both runs and cancel in the ratio. Only I changes.

1. Square the period ratio: (10/8)² = 1.5625.
2. That is the factor by which the total inertia grew: I₁ + I' = 1.5625 × 8×10⁻² = 12.5×10⁻² kg m².
3. Subtract the magnet's own share: I' = 12.5×10⁻² − 8×10⁻² = 4.5×10⁻² kg m².
4. The 4 A m² is not used at all — it is there to make you hesitate.

A bar magnet in a uniform field feels equal and opposite forces on its two poles — a couple. Each pole of strength m feels F = mB.

The definition that ties everything together is M = m × 2ℓ. Two equations, two unknowns (m and 2ℓ). Eliminate m.

1. Get pole strength from the force: m = F/B.
2. Get length from the moment: 2ℓ = M/m.
3. Substitute: 2ℓ = M ÷ (F/B) = MB/F.
4. Sanity-check by units: M is A m², B is T = N/(A m), F is N. MB/F = (A m²)(N A⁻¹ m⁻¹)/N = m. ✓

The rule that decides everything: pole strength goes with cross-sectional area; length goes with length.

A cut parallel to the axis slices the bar lengthwise. Length is unchanged; the face area is divided. So m drops, ℓ does not.

A cut perpendicular to the axis chops the bar into shorter bars. Area is unchanged; length is divided. So ℓ drops, m does not.

1. Handle the parallel cuts: area is split five ways, so m' = m/5. Length untouched.
2. Handle the perpendicular cuts: length is split four ways, so ℓ' = ℓ/4. Area untouched.
3. Combine: M' = m'ℓ' = (m/5)(ℓ/4) = mℓ/20 = M/20.
4. Answer: m/5 and M/20.

Two standard results for a short dipole, and the only thing separating them is a factor of 2:

axial — 2M/r³; equatorial — M/r³ (each times μ₀/4π = 10⁻⁷).

"Equatorial" and "broadside-on" are the same thing. So are "axial" and "end-on".

1. Convert first: r = 20 cm = 0.2 m. This is where marks are lost — never leave centimetres in a cube.
2. Cube it: r³ = 0.008 = 8×10⁻³ m³.
3. Substitute: B = 10⁻⁷ × 8 / 8×10⁻³.
4. The 8s cancel: B = 10⁻⁷ × 10³ = 10⁻⁴ Wb m⁻².

The field of a dipole falls as 1/r³. The force between two dipoles is the gradient of an energy, so it picks up one more power: 1/r⁴.

Two arrangements, two constants — end-on carries a 6, broadside-on carries a 3. End-on is the stronger of the two.

1. Pick the end-on constant: 6.
2. Numerator: 6 × 10 × 10 = 600.
3. Denominator: r⁴ = (0.1)⁴ = 10⁻⁴.
4. F = 10⁻⁷ × 600 / 10⁻⁴ = 10⁻⁷ × 6×10⁶ = 0.6 N.

This is Q12's formula used as a proportionality instead of a calculation. Everything except r is unchanged, so:

F ∝ 1/r⁴.

The exponent to carry is 4, not 3. Three is the exponent for the field of one dipole. Four is the exponent for the force between two of them. Both appear in this chapter within a page of each other, which is exactly why the paper tests it.

1. Write the ratio before substituting anything: F₂/F₁ = (r/2r)⁴ = (1/2)⁴.
2. Evaluate: (1/2)⁴ = 1/16.
3. F₂ = 4.8 / 16 = 0.3 N.
4. Guard against the trap: if you had used 1/r³ you would get 4.8/8 = 0.6 N — and 0.6 N is sitting right there as an option, because the setter expected that mistake.

The short-magnet formulas (2M/r³ and M/r³) cannot answer this — length has already been thrown away in deriving them. You must use the exact, finite-length expressions, where ℓ is still visible.

Once you write them down, both statements collapse immediately: the axial denominator shrinks as ℓ grows, and the equatorial denominator grows.

1. Statement A, axial: as ℓ rises, (r² − ℓ²) falls, so the denominator falls and B rises. A claims it decreases — false.
2. Statement B, equatorial: as ℓ rises, (r² + ℓ²) rises, so the denominator rises and B falls. B claims it increases — false.
3. Both false.
4. The tell that you need the exact formulas: the question is about length. If a question varies a quantity, that quantity has to survive into your formula.

The whole question is deciding, for each magnet separately, whether the origin is an axial or an equatorial point. The test: is the origin along the direction of that magnet's moment, or perpendicular to it?

Magnet 1 sits on the x-axis with its moment along x — the origin is straight ahead of it, so axial.

Magnet 2 sits on the y-axis with its moment along x — the origin is sideways from it, so equatorial.

Then the directions: an axial field points along M; an equatorial field points opposite to M.

1. Common factor first: 10⁻⁷ × 9 / (0.03)³ = 10⁻⁷ × 9 / 2.7×10⁻⁵ = 1/30 T.
2. Magnet 1 (axial): B₁ = 2 × 1/30 = 1/15 = 0.0667 T, pointing along +x.
3. Magnet 2 (equatorial): B₂ = 1/30 = 0.0333 T, pointing opposite to its moment. Its moment is along −x, so the field is along +x.
4. Both along +x, so they add: 1/15 + 1/30 = 3/30 = 0.1 T.

Two independent things change, and you must handle them separately.

Magnitude. P was axial; after the turn, P is at 90° to the moment, so P is now equatorial. Equatorial field is half the axial field at the same distance: 5 → 2.5 μT.

Direction. The equatorial field points antiparallel to the moment. Anticlockwise from South is East, so the moment now points East, and the field at P points West.

1. Read the original geometry: axial field is along M, and it is due South, so the moment points South and P lies south of the magnet.
2. Rotate the moment 90° anticlockwise: South → East. (anticlockwise order is E → N → W → S → E.)
3. P has not moved — it is still due south of the magnet, which is now perpendicular to the moment. P is an equatorial point.
4. Magnitude halves to 2.5 μT; direction is antiparallel to the moment (East), so due West.

A rigid body of two dipoles behaves as one dipole whose moment is their vector sum. At equilibrium the resultant lines up with the field — not either individual arm.

So the question reduces to: what angle does the resultant make with the M arm? That is pure trigonometry.

1. Add the two perpendicular moments: magnitude √(M² + 3M²) = 2M.
2. Find where the resultant points relative to the M arm: tanθ = √3M/M = √3, so θ = 60°.
3. At equilibrium, the resultant is along B.
4. Therefore the angle between the M arm and B is that same 60°. (The √3M arm is at 30°, which is the distractor.)

At equilibrium the wire's restoring torque balances the magnetic torque.

The subtlety worth getting right: the wire is twisted by (φ − θ), not by φ. The top turned through φ, but the bottom followed part of the way, through θ. The wire only feels the difference.

Because θ is 30° in both cases, sinθ cancels and the ratio is pure arithmetic.

1. Case 1 net twist: 180 − 30 = 150.
2. Case 2 net twist: 270 − 30 = 240.
3. M₁/M₂ = 150/240.
4. Reduce: divide both by 30 → 5/8. So M₁ : M₂ = 5 : 8.

A neutral point is where the two fields cancel — equal in size, opposite in direction.

Because both poles are like (both north), they push outward against each other, so the cancellation happens between them. If they were unlike, it would lie outside, beyond the weaker one. Decide this first; it fixes the geometry.

The null sits closer to the weaker pole — the weak field needs a short lever to keep up with the strong one.

1. Take the square root of the strength ratio: √(16/4) = √4 = 2.
2. So the distances are in the ratio 2 : 1, with the larger distance next to the stronger pole.
3. Split 30 cm in the ratio 2:1 → 20 cm and 10 cm.
4. The question asks from the weaker pole, which is the shorter leg: 10 cm. (Answering 20 cm would be reading the right split from the wrong end.)

A four-step chain, each step one multiplication:

χ → μ_r → B → Φ.

The only place this goes wrong is the area conversion. 1 cm² = 10⁻⁴ m², not 10⁻² — the square applies to the conversion factor as well.

1. μ_r = 1 + 599 = 600. (Adding the 1 matters far less than remembering to add it at all.)
2. B = 4π×10⁻⁷ × 600 × 1200 = 0.905 T.
3. Convert the area: 0.2 cm² = 0.2×10⁻⁴ = 2×10⁻⁵ m².
4. Φ = 0.905 × 2×10⁻⁵ = 1.81×10⁻⁵ Wb.

Same chain as Q32, stopped two steps earlier. Permeability is an intrinsic property of the material — it does not depend on H at all.

The 1200 A m⁻¹ is deliberately supplied and deliberately unused. Recognising an unused given is a skill in itself.

1. μ_r = 1 + 599 = 600.
2. μ = 4π×10⁻⁷ × 600.
3. 4 × 600 = 2400, so μ = 2400π×10⁻⁷.
4. Normalise the power of ten: 2400×10⁻⁷ = 2.4×10⁻⁴, so μ = 2.4π×10⁻⁴ T m A⁻¹.

Susceptibility compares two per-unit-volume quantities:

χ = I / H, where I is magnetisation = moment per unit volume.

So the moment you are given cannot be used directly — it has to be divided by the volume first. That intermediate step is the whole question.

The volume of a 1 cm cube is 10⁻⁶ m³, because the conversion factor is cubed.

1. Volume: (10⁻² m)³ = 10⁻⁶ m³.
2. Magnetisation: I = 20×10⁻⁶ / 10⁻⁶ = 20 A m⁻¹.
3. Susceptibility: χ = 20 / 60×10³ = 1/3000.
4. 1/3000 = 3.33×10⁻⁴. The 10⁻² options are what you get if you skip the volume division.

Curie's law for a paramagnet: I = C B / T. Magnetisation rises with field and falls with temperature — field aligns the dipoles, heat randomises them.

Since C is a property of the sample and the sample is unchanged, work entirely in ratios and C never appears.

1. Field ratio: 0.02 / 0.06 = 1/3. Field went down, so magnetisation goes down by 3.
2. Temperature ratio: T went from 4 K to 16 K, a factor of 4 up. Since I ∝ 1/T, magnetisation goes down by another factor of 4.
3. Combine: I₂ = 8 × (1/3) × (1/4) = 8/12.
4. = 2/3 A m⁻¹.

The classification is decided entirely by where μ_r sits relative to 1:

• μ_r < 1 (so χ negative) — diamagnetic
• μ_r slightly above 1 — paramagnetic
• μ_r in the hundreds or thousands — ferromagnetic

So compute μ = B/H, compare with μ₀, and read off.

1. μ = 0.0048 / 10000 = 4.8×10⁻⁷ T m A⁻¹.
2. Compare with μ₀ = 4π×10⁻⁷ ≈ 12.57×10⁻⁷.
3. μ_r = 4.8 / 12.57 = 0.38, which is less than 1.
4. χ = 0.38 − 1 = −0.62, negative → diamagnetic.

Claim A. Paramagnetic materials have a small positive χ. Since μ_r = 1 + χ, a positive χ puts μ_r just above 1. Correct. This is not an extra fact to memorise; it follows from the definition.

Claim B. Diamagnetism arises from the field inducing opposing currents in the electron orbits of every atom. It is not a matter of aligning pre-existing dipoles against thermal jostling, so temperature does not enter. Correct. Contrast paramagnetism (χ ∝ 1/T) and ferromagnetism (χ ∝ 1/(T − T_c)), both of which do depend on T.

1. Test A on its own. Paramagnetic → χ > 0 → μ_r = 1 + χ > 1. True.
2. Test B on its own. Diamagnetism is induced, not aligned, so thermal agitation has nothing to disturb. True.
3. Do not let one verdict influence the other — there is no reason a paper cannot make both statements true.
4. Both correct.