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Magnetism and Matter · NEET 2027 · beyond the rationalised NCERT

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Fifty questions on material that is not in your rationalised NCERT Chapter 5 but still appears in practice papers and older banks: the oscillation period and its cut-magnet variants, the inverse fourth-power force between coaxial magnets, Curie and Curie–Weiss laws, the Bohr magneton, the vibration magnetometer, and the hysteresis terms. Every question is tagged with its syllabus status.

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A · Period of oscillation, T = 2π√(I/mB)

11 questions
Q01Period formulanot in rationalised NCERT

A bar magnet of moment m and moment of inertia I oscillates with small amplitude in a uniform field B. Its time period is

B (horizontal) S N θ T = 2π√(I / mB)
Given
Bar magnet of magnetic moment m, moment of inertia I, in a uniform field B; small angular amplitude.
Asked
The time period of oscillation.
Concept applied
The restoring torque τ = −mB sin θ becomes −(mB)θ for small θ. Comparing with the angular SHM equation τ = −kθ identifies the torsional constant as k = mB, and the period follows from the standard SHM result.
Formula to use
T = 2π√(I / mB)
Baby steps
  1. Restoring torque at angular displacement θ: τ = −mB sin θ.
  2. For small θ, sin θ ≈ θ, so τ ≈ −(mB)θ — torque proportional to displacement and opposite in sign.
  3. This is angular SHM with effective torsional constant k = mB.
  4. Newton's second law for rotation: Iα = −(mB)θ, so α = −(mB/I)θ.
  5. Comparing with α = −ω2θ gives ω2 = mB/I.
  6. T = 2π/ω = 2π√(I/mB).
Assumption
Small angular amplitude so that sin θ ≈ θ; frictionless suspension; uniform field over the magnet.
Shortcut trick
Match it to the pendulum you already know: T = 2π√(length/gravity) has the 'resisting-to-motion' quantity on top and the 'restoring' quantity below. Here I resists rotation and mB restores it — same structure, so I on top, mB below.
Answer: (B) 2π√(I/mB)
Q02Dependence on Bnot in rationalised NCERT

If the magnetic field is made four times stronger, the period of oscillation of the magnet becomes

Given
T = 2π√(I/mB); B → 4B with I and m unchanged.
Asked
The new period.
Concept applied
B sits under a square root in the denominator, so T varies as 1/√B. Multiplying B by a factor divides T by the square root of that factor.
Formula to use
T ∝ 1/√B ⇒ T2/T1 = √(B1/B2)
Baby steps
  1. Write T = 2π√(I/mB), with only B changing.
  2. T2/T1 = √(B1/B2).
  3. Substitute B2 = 4B1: T2/T1 = √(1/4).
  4. √(1/4) = 1/2.
  5. T2 = T1/2 — the period halves.
  6. Physical sense: a stronger field pulls the magnet back harder, so it swings faster and the period drops. ✔
Assumption
Same magnet and same suspension, so I and m are fixed.
Shortcut trick
Under a square root, a factor of 4 becomes a factor of 2. Always halve the exponent: 4× on B gives 2× on 1/T. And check the direction physically — stronger restoring force must mean a shorter period.
Answer: (C) half
Q03Dependence on mnot in rationalised NCERT

Two magnets have the same moment of inertia but magnetic moments in the ratio 4 : 1. In the same field their periods are in the ratio

Given
I1 = I2; m1/m2 = 4; same B.
Asked
T1 : T2
Concept applied
m also sits under the square root in the denominator, so T ∝ 1/√m. A stronger magnet feels a bigger restoring torque and therefore oscillates faster.
Formula to use
T ∝ 1/√m ⇒ T1/T2 = √(m2/m1)
Baby steps
  1. With I and B fixed, T depends only on m.
  2. T1/T2 = √(m2/m1).
  3. Substitute the ratio: √(1/4).
  4. = 1/2.
  5. So T1 : T2 = 1 : 2.
  6. The stronger magnet (magnet 1) has the shorter period, as expected.
Assumption
Identical suspensions and identical field for both magnets.
Shortcut trick
Note the inversion trap: the moments are 4 : 1 but the periods are 1 : 2. Two flips happen — the reciprocal and the square root. Track them one at a time.
Answer: (A) 1 : 2
Q04Period numericnot in rationalised NCERT

A magnet of moment 0.4 A m2 and moment of inertia 1 × 10−4 kg m2 oscillates in a field of 0.25 T. Its period is about

Given
m = 0.4 A m2; I = 1 × 10−4 kg m2; B = 0.25 T
Asked
The time period.
Concept applied
Direct substitution. The safest order is to evaluate the product mB first, then the ratio, then the square root, and only then multiply by 2π.
Formula to use
T = 2π√(I/mB)
Baby steps
  1. Compute mB = 0.4 × 0.25 = 0.1.
  2. Form the ratio: I/mB = 1 × 10−4 / 0.1 = 1 × 10−3.
  3. Take the square root: √(10−3) = 0.0316.
  4. Multiply by 2π: T = 6.283 × 0.0316.
  5. T ≈ 0.20 s.
  6. Option (A) is the answer with the 2π omitted — check that you have applied it.
Assumption
Small amplitude; frictionless suspension; uniform field.
Shortcut trick
Do the square root before multiplying by 2π. Reversing the order forces you to square 2π and drags π2 through the arithmetic for no reason.
Answer: (B) 0.20 s
Q05Moment of inertianot in rationalised NCERT

A uniform bar magnet of mass 60 g and length 10 cm oscillates about a vertical axis through its centre. Its moment of inertia is

Given
Mass Mb = 60 g = 0.06 kg; length L = 10 cm = 0.10 m; axis through centre, perpendicular to length.
Asked
The moment of inertia I.
Concept applied
A bar magnet is treated as a uniform thin rod. About a perpendicular axis through its centre, the standard result is MbL2/12.
Formula to use
I = MbL2/12
Baby steps
  1. Convert to SI: Mb = 0.06 kg, L = 0.10 m.
  2. Square the length: L2 = 0.01 m2 = 1 × 10−2.
  3. Multiply: MbL2 = 0.06 × 0.01 = 6 × 10−4.
  4. Divide by 12: I = 6 × 10−4 / 12.
  5. I = 5 × 10−5 kg m2.
  6. Note the distinction: Mb here is the mass in kg, entirely different from the magnetic moment m in A m2.
Assumption
Uniform thin rod of negligible width; oscillation axis through the centre, perpendicular to the length.
Shortcut trick
The clash of symbols is the real hazard: m is magnetic moment, M is mass. Write them as m and Mb in your working. Mixing them produces answers that look plausible and are badly wrong.
Answer: (A) 5 × 10−5 kg m2
Q06Combined numericnot in rationalised NCERT

The same magnet (mass 60 g, length 10 cm, moment 0.4 A m2) oscillates in a field of 0.25 T. Its period is about

Given
Mb = 0.06 kg; L = 0.10 m; m = 0.4 A m2; B = 0.25 T
Asked
The time period.
Concept applied
A two-stage problem: find I from the geometry, then feed it into the period formula. This is the standard full-length version of the oscillation question.
Formula to use
I = MbL2/12, then T = 2π√(I/mB)
Baby steps
  1. Stage 1 — moment of inertia: I = 0.06 × 0.01/12 = 5 × 10−5 kg m2.
  2. Stage 2 — restoring factor: mB = 0.4 × 0.25 = 0.1.
  3. Ratio: I/mB = 5 × 10−5/0.1 = 5 × 10−4.
  4. Square root: √(5 × 10−4) = 0.02236.
  5. Multiply by 2π: T = 6.283 × 0.02236 ≈ 0.14 s.
  6. Cross-check against Q04: halving I from 10−4 to 5 × 10−5 should divide T by √2, and 0.20/1.414 = 0.14. ✔
Assumption
Uniform thin bar magnet; small amplitude; uniform field.
Shortcut trick
When two questions share data, use the earlier answer as a check rather than recomputing blind. Ratio checks catch arithmetic slips that a fresh calculation would repeat.
Answer: (B) 0.14 s
Q07Frequency formnot in rationalised NCERT

The frequency of oscillation of a bar magnet in a uniform field is

Given
T = 2π√(I/mB).
Asked
The frequency ν.
Concept applied
Frequency is the reciprocal of the period. Inverting a square-root expression flips the fraction inside the root as well as moving the 2π.
Formula to use
ν = 1/T = (1/2π)√(mB/I)
Baby steps
  1. Start from T = 2π√(I/mB).
  2. Take the reciprocal: ν = 1/T = 1/[2π√(I/mB)].
  3. Move the 2π to the front as 1/2π.
  4. Invert the fraction inside the root: 1/√(I/mB) = √(mB/I).
  5. ν = (1/2π)√(mB/I).
  6. Sense check: a stronger field or a stronger magnet increases the frequency, and mB is now in the numerator. ✔
Assumption
Small amplitude SHM.
Shortcut trick
When inverting, two things flip together: the 2π moves from front to denominator, and the fraction inside the root turns upside down. Option (C) flips only one of them — the standard half-done answer.
Answer: (B) (1/2π)√(mB/I)
Q08Two-field comparisonnot in rationalised NCERT

A magnet oscillates with period 2 s in a field B. In a field of 4B, its period becomes

Given
T1 = 2 s in field B; new field 4B; same magnet.
Asked
The new period.
Concept applied
Ratio method — there is no need to know I or m individually, since they cancel when the two cases are divided.
Formula to use
T2 = T1√(B1/B2)
Baby steps
  1. T ∝ 1/√B with I and m fixed.
  2. T2/T1 = √(B1/B2) = √(B/4B) = √(1/4) = 1/2.
  3. T2 = T1 × 1/2.
  4. T2 = 2 × 0.5 = 1.0 s.
  5. Sense check: stronger field ⇒ faster oscillation ⇒ shorter period. 1 s < 2 s. ✔
  6. Options (C) and (D) go the wrong way and can be rejected before any arithmetic.
Assumption
Same magnet, same suspension, small amplitude in both cases.
Shortcut trick
In any two-case problem, form the ratio first and let the unknowns cancel. Then use the physical direction check to eliminate half the options before computing.
Answer: (B) 1.0 s
Q09Small amplitude conditionnot in rationalised NCERT

The formula T = 2π√(I/mB) is valid only when

Given
The derivation of the oscillation period.
Asked
The condition for validity.
Concept applied
The step from τ = −mB sin θ to τ = −(mB)θ requires sin θ ≈ θ, which holds only for small angles. Without it the motion is periodic but not simple harmonic, and the period depends on amplitude.
Formula to use
sin θ ≈ θ requires θ small (in radians)
Baby steps
  1. The exact restoring torque is τ = −mB sin θ, which is not proportional to θ.
  2. SHM requires the restoring torque to be strictly proportional to the displacement.
  3. The approximation sin θ ≈ θ makes it so, but is accurate only for small θ.
  4. At θ = 10° = 0.1745 rad, sin θ = 0.1736 — an error of about 0.5%, acceptable.
  5. At θ = 60° the error exceeds 15% and the formula fails.
  6. So the condition is small angular amplitude.
Assumption
Angles measured in radians; damping negligible.
Shortcut trick
Every SHM period formula in physics carries this same small-amplitude caveat — simple pendulum, torsional pendulum, oscillating magnet. If a question mentions a large amplitude, the standard period formula does not apply.
Answer: (B) the angular amplitude is small
Q10Find m from Tnot in rationalised NCERT

A magnet of moment of inertia 2 × 10−4 kg m2 has period 2 s in a field of 5 × 10−5 T. Its magnetic moment is about

Given
I = 2 × 10−4 kg m2; T = 2 s; B = 5 × 10−5 T
Asked
The magnetic moment m.
Concept applied
Rearranging for m requires squaring the period equation first. This is exactly how a vibration magnetometer is used to measure an unknown magnetic moment.
Formula to use
T = 2π√(I/mB) ⇒ m = 4π2I / (T2B)
Baby steps
  1. Square both sides: T2 = 4π2I/(mB).
  2. Rearrange for m: m = 4π2I/(T2B).
  3. Numerator: 4π2 × 2 × 10−4 = 39.48 × 2 × 10−4 = 7.90 × 10−3.
  4. Denominator: T2B = 4 × 5 × 10−5 = 2 × 10−4.
  5. m = 7.90 × 10−3 / 2 × 10−4 = 39.5.
  6. m ≈ 40 A m2.
Assumption
Small amplitude; the field value quoted is the one acting on the magnet.
Shortcut trick
Remember 4π2 ≈ 39.5, close to 40. In these problems the answer very often lands near a round multiple of 40, which makes it easy to spot the right option once the powers of ten are handled.
Answer: (B) 40 A m2
Q11Effect of demagnetisationnot in rationalised NCERT

A bar magnet is partially demagnetised so that its moment falls to one quarter, with its mass and dimensions unchanged. Its period of oscillation becomes

Given
m → m/4; I unchanged (mass and dimensions unaltered); same field.
Asked
The new period.
Concept applied
Demagnetising changes only the magnetic moment. Since the physical body is untouched, the moment of inertia is exactly the same — a distinction that cutting questions do not share.
Formula to use
T ∝ 1/√m with I fixed
Baby steps
  1. The mass and dimensions are unchanged, so I is unchanged.
  2. Only m changes, falling to m/4.
  3. T2/T1 = √(m1/m2) = √(m / (m/4)) = √4 = 2.
  4. T2 = 2T1 — the period doubles.
  5. Physical sense: a weaker magnet feels a weaker restoring torque and swings more sluggishly. ✔
  6. Contrast with cutting, where both m and I change — the reason cut-magnet answers differ from this one.
Assumption
Demagnetisation is uniform and does not alter the mass distribution.
Shortcut trick
Before applying any ratio, ask which of I and m actually changed. Demagnetising changes m only. Cutting changes both. That single check separates this group from the next.
Answer: (B) twice

B · Cut, stacked and combined magnet variants

8 questions
Q12Transverse cut into twonot in rationalised NCERT

A bar magnet oscillating with period T is cut into two equal halves perpendicular to its length. One half oscillates in the same field with period

Given
Original period T; transverse cut into two equal halves; same field.
Asked
The period of one half.
Concept applied
Both quantities in the formula change. The moment halves, but the moment of inertia falls by a factor of eight, because both mass and length are halved and I depends on ML2.
Formula to use
I = MbL2/12; after transverse cut Mb→Mb/2, L→L/2, m→m/2
Baby steps
  1. New mass: Mb/2. New length: L/2.
  2. New moment of inertia: I′ = (Mb/2)(L/2)2/12 = (MbL2/12) × (1/2)(1/4) = I/8.
  3. New magnetic moment: m′ = m/2 (length halves, pole strength unchanged).
  4. Substitute: T′ = 2π√[(I/8)/((m/2)B)].
  5. Simplify the bracket: (I/8) ÷ (m/2) = (I/8)(2/m) = I/(4m).
  6. T′ = 2π√(I/4mB) = (1/2) × 2π√(I/mB) = T/2.
Assumption
Clean transverse cut; each half remains a uniform bar; same field and suspension.
Shortcut trick
The trap is halving I along with m and concluding T is unchanged. I falls by 8, not 2, because I ∝ ML2 and both factors shrink. Always recompute I from MbL2/12 rather than guessing.
Answer: (B) T/2
Q13n equal partsnot in rationalised NCERT

A magnet of period T is cut transversely into n equal parts. Each part oscillates in the same field with period

Given
Transverse cut into n equal parts; same field.
Asked
The period of each part.
Concept applied
Generalise the previous result. Mass and length each fall by n, so I falls by n3; the moment falls by n. The net effect inside the square root is a factor n2.
Formula to use
I′ = I/n3;  m′ = m/n
Baby steps
  1. New mass Mb/n and new length L/n.
  2. I′ = (Mb/n)(L/n)2/12 = I/n3.
  3. m′ = m/n.
  4. Ratio inside the root: I′/(m′B) = (I/n3) ÷ (mB/n) = I/(n2mB).
  5. T′ = 2π√[I/(n2mB)] = (1/n) × 2π√(I/mB).
  6. T′ = T/n. Check with n = 2: T/2, matching the previous question. ✔
Assumption
Equal transverse cuts; each piece remains a uniform bar of the same cross-section.
Shortcut trick
Remember the two exponents: I ÷ n3 and m ÷ n. Their ratio is n2, and the square root turns that into n. The clean result T′ = T/n is worth memorising directly.
Answer: (A) T/n
Q14Numeric cutnot in rationalised NCERT

A magnet oscillates with period 4 s. It is cut transversely into 4 equal pieces. Each piece oscillates in the same field with period

Given
T = 4 s; cut transversely into n = 4 equal pieces; same field.
Asked
The new period of each piece.
Concept applied
Direct application of T′ = T/n once the general result is established.
Formula to use
T′ = T/n
Baby steps
  1. Identify n = 4 (number of equal transverse pieces).
  2. Apply T′ = T/n.
  3. T′ = 4/4.
  4. T′ = 1.0 s.
  5. Verify from first principles: I′ = I/64 and m′ = m/4, so I′/m′ = I/(16m), and √(1/16) = 1/4. ✔
  6. Option (A) is T/n2 and option (D) is nT — both planted for the wrong exponent.
Assumption
Equal transverse cuts; unchanged field and suspension.
Shortcut trick
Do the first-principles check once for a specific n, then trust T′ = T/n in the exam. Deriving it fresh every time costs 90 seconds you cannot spare.
Answer: (B) 1.0 s
Q15Longitudinal cutnot in rationalised NCERT

A magnet of period T is cut into two equal halves along its length. Each half oscillates in the same field with period

Given
Longitudinal cut into two equal halves; same field.
Asked
The new period.
Concept applied
A longitudinal cut halves the mass but keeps the length, so I halves rather than dropping by eight. The magnetic moment also halves. The two halvings cancel exactly.
Formula to use
I′ = (Mb/2)L2/12 = I/2;  m′ = m/2
Baby steps
  1. New mass: Mb/2. New length: L (unchanged — the cut runs lengthwise).
  2. I′ = (Mb/2)L2/12 = I/2.
  3. New moment: pole strength halves with the cross-section, length unchanged, so m′ = m/2.
  4. Ratio: I′/(m′B) = (I/2) ÷ (mB/2) = I/(mB) — exactly the original ratio.
  5. T′ = 2π√(I/mB) = T, unchanged.
  6. Striking contrast with the transverse cut, which gave T/2.
Assumption
Clean longitudinal cut; uniform magnetisation retained.
Shortcut trick
Two cuts, two different answers: transverse ⇒ T/2; longitudinal ⇒ T unchanged. The question always specifies which. Both cuts halve the magnetic moment, so the difference lies entirely in what happens to I.
Answer: (B) T
Q16Two magnets, like polesnot in rationalised NCERT

Two identical magnets are placed one on top of the other with their like poles together, and the combination oscillates in the same field. The period compared with a single magnet is

Given
Two identical magnets, like poles together (moments parallel); same field.
Asked
The period of the combination relative to one magnet.
Concept applied
Both the moment of inertia and the magnetic moment double, so their ratio — and hence the period — is unchanged.
Formula to use
Itotal = 2I;  mtotal = 2m
Baby steps
  1. Moments of inertia add: Itotal = I + I = 2I.
  2. Magnetic moments are parallel, so they add too: mtotal = m + m = 2m.
  3. Ratio: Itotal/(mtotalB) = 2I/(2mB) = I/(mB).
  4. The 2s cancel exactly.
  5. T′ = 2π√(I/mB) = T, unchanged.
  6. Sensible: the system is twice as hard to turn and twice as strongly restored, so it swings at the same rate.
Assumption
Identical magnets, rigidly bound, oscillating about the same axis.
Shortcut trick
Look for cancellation before computing. Whenever I and m scale by the same factor, T does not move. This applies to the like-poles stack and to the longitudinal cut alike.
Answer: (A) unchanged
Q17Two magnets, unlike polesnot in rationalised NCERT

The same two identical magnets are stacked with their unlike poles together. The combination in a magnetic field will

Given
Two identical magnets stacked with unlike poles together, so their moments are antiparallel.
Asked
The oscillatory behaviour of the combination.
Concept applied
Antiparallel equal moments cancel, giving zero net moment. With no net moment there is no restoring torque, so there is nothing to drive an oscillation.
Formula to use
mtotal = m − m = 0 ⇒ τ = 0 ⇒ T → ∞
Baby steps
  1. Unlike poles together means the two moment vectors point in opposite directions.
  2. Being identical magnets, they cancel exactly: mtotal = 0.
  3. Restoring torque τ = mtotalB sin θ = 0 at every angle.
  4. With no restoring torque there is no SHM — the system stays wherever you put it.
  5. Formally T = 2π√(I/mtotalB) → ∞ as mtotal → 0.
  6. Answer: it does not oscillate at all.
Assumption
The two magnets are exactly identical, so the cancellation is complete. Unequal magnets would leave a residual moment and a finite, long period.
Shortcut trick
An infinite period in a formula is physics, not a maths failure — it means the restoring agency has vanished. Recognise mtotal = 0 as 'no oscillation' rather than trying to compute a number.
Answer: (C) not oscillate at all
Q18Sum and difference methodnot in rationalised NCERT · vibration magnetometer

Two magnets give period 2 s when their like poles are together and 4 s when unlike poles are together. The ratio of their magnetic moments m1/m2 is

Given
Tsum = 2 s (like poles together, moments add); Tdiff = 4 s (unlike poles together, moments subtract).
Asked
The ratio m1/m2.
Concept applied
The classic vibration-magnetometer technique. Both arrangements share the same total moment of inertia, so dividing the two squared periods eliminates I and B and leaves a ratio of moments.
Formula to use
m1/m2 = (Td2 + Ts2) / (Td2 − Ts2)
Baby steps
  1. Like poles together: net moment m1 + m2, so Ts = 2π√[Itot/((m1+m2)B)].
  2. Unlike poles together: net moment m1 − m2, so Td = 2π√[Itot/((m1−m2)B)].
  3. Divide the squares: Td2/Ts2 = (m1+m2)/(m1−m2). Both Itot and B cancel.
  4. Substitute: 16/4 = 4 = (m1+m2)/(m1−m2).
  5. Cross-multiply: 4m1 − 4m2 = m1 + m2, so 3m1 = 5m2.
  6. m1/m2 = 5/3. Equivalently, use the standard formula: (16+4)/(16−4) = 20/12 = 5/3. ✔
Assumption
Both magnets have the same moment of inertia contribution in either arrangement, and m1 > m2 so the difference is positive.
Shortcut trick
The longer period always belongs to the difference arrangement, because the net moment is smaller there. If a question does not label them, the bigger T is Td — and putting them the wrong way round inverts the whole answer.
Answer: (A) 5 : 3
Q19Cut and stacknot in rationalised NCERT

A magnet of period T is cut transversely into two halves, and the halves are placed side by side with like poles together. The period of the combination is

Given
Transverse cut into two halves, then stacked with moments parallel; same field.
Asked
The period of the combination.
Concept applied
Two stages. After cutting, each half has I/8 and m/2. Stacking two of them doubles both, giving I/4 and m — and the ratio determines the period.
Formula to use
each half: I/8, m/2 ⇒ stack: 2(I/8) = I/4, 2(m/2) = m
Baby steps
  1. After the transverse cut, each half has I′ = I/8 and m′ = m/2.
  2. Stacking two halves with like poles together: Itot = 2 × I/8 = I/4.
  3. Moments are parallel and add: mtot = 2 × m/2 = m.
  4. Ratio: Itot/(mtotB) = (I/4)/(mB) = I/(4mB).
  5. T′ = 2π√[I/(4mB)] = (1/2) × 2π√(I/mB) = T/2.
  6. Same as a single half — consistent with the like-poles stacking rule, which never changes the period.
Assumption
The two halves are bound rigidly and oscillate about a common axis; the stacking is side by side so lengths do not add.
Shortcut trick
Chain the two rules rather than starting over: cutting into 2 gives T/2; stacking like poles changes nothing. T → T/2 → T/2. Multi-stage questions are just rules applied in sequence.
Answer: (B) T/2

C · Force between coaxial magnets and field gradients

7 questions
Q20Distance dependenceold bank / JEE

The force between two short bar magnets placed coaxially at a large separation r varies as

S N S N r (centre to centre) attraction F = (μ₀/4π)(6 m₁m₂ / r⁴)
Given
Two short coaxial bar magnets separated by r, with r much larger than either magnet.
Asked
The dependence of the force on r.
Concept applied
The field of a dipole falls as 1/r3. Force on a dipole arises from the field gradient, and differentiating 1/r3 gives 1/r4. One extra power of r comes from the differentiation, not from the field itself.
Formula to use
F = m2(dB1/dr), with B1 ∝ 1/r3
Baby steps
  1. Magnet 1 produces an axial field B1 = (μ0/4π)(2m1/r3).
  2. A dipole in a non-uniform field feels a force F = m2(dB1/dr).
  3. Differentiate: d/dr (r−3) = −3r−4.
  4. So dB1/dr ∝ 1/r4.
  5. Therefore F ∝ 1/r4.
  6. The full result is F = (μ0/4π)(6m1m2/r4) for aligned coaxial dipoles.
Assumption
Both magnets short compared with r; coaxial, aligned arrangement.
Shortcut trick
Count the powers: field 1/r3, force 1/r4. Differentiation always adds one power to the denominator. The same logic gives 1/r4 for two electric dipoles — the analogy carries over intact.
Answer: (C) 1/r4
Q21Doubling the distanceold bank / JEE

Two coaxial short magnets attract with force F. If their separation is doubled, the force becomes

Given
F ∝ 1/r4; r → 2r.
Asked
The new force.
Concept applied
Raise the distance factor to the fourth power and invert. This is the same scaling logic as the field questions, but with exponent 4 rather than 3.
Formula to use
F2/F1 = (r1/r2)4
Baby steps
  1. Write F1 = C/r4.
  2. New separation r2 = 2r.
  3. F2 = C/(2r)4 = C/(16r4).
  4. Compare: F2 = (1/16)F1.
  5. F2 = F/16.
  6. Note how sharply this falls — doubling the gap reduces the force to about 6% of its value, which is why magnets grip strongly only when very close.
Assumption
Both magnets remain short compared with the new separation, and their orientations are unchanged.
Shortcut trick
Options (A) and (B) are the answers for 1/r2 and 1/r3 laws respectively. Confirm the exponent before scaling — force between dipoles is fourth power, not third.
Answer: (C) F/16
Q22Full formulaold bank / JEE

For two short coaxial magnets with their moments aligned, the force between them is

Given
Two short coaxial magnets of moments m1 and m2, separation r, moments aligned.
Asked
The full expression for the force.
Concept applied
Derive rather than memorise: take the axial field of one magnet, differentiate with respect to r, and multiply by the other magnet's moment. The factor 6 emerges from 2 × 3.
Formula to use
F = (μ0/4π)(6m1m2/r4)
Baby steps
  1. Axial field of magnet 1 at distance r: B1 = (μ0/4π)(2m1/r3).
  2. Force on magnet 2 in this non-uniform field: F = m2|dB1/dr|.
  3. Differentiate: dB1/dr = (μ0/4π)(2m1)(−3/r4).
  4. Take the magnitude: |dB1/dr| = (μ0/4π)(6m1/r4).
  5. Multiply by m2: F = (μ0/4π)(6m1m2/r4).
  6. The 6 is simply 2 (from the axial field) × 3 (from differentiating r−3).
Assumption
Short dipoles, r much greater than either magnet's length, moments coaxial and aligned.
Shortcut trick
Never memorise the 6 in isolation — remember it as 2 × 3, where 2 is the axial-field factor and 3 comes from the differentiation. Then you can rebuild the formula even if it slips.
Answer: (B) (μ0/4π)(6m1m2/r4)
Q23Numeric forceold bank / JEE

Two short coaxial magnets, each of moment 1 A m2, are 10 cm apart with moments aligned. The force between them is

Given
m1 = m2 = 1 A m2; r = 0.10 m; μ0/4π = 10−7
Asked
The magnitude of the force.
Concept applied
Direct substitution into the coaxial force formula. The fourth power of the distance is where the arithmetic care is needed.
Formula to use
F = (μ0/4π)(6m1m2/r4)
Baby steps
  1. Fourth power of the distance: r4 = (0.10)4 = 1 × 10−4 m4.
  2. Numerator: 6m1m2 = 6 × 1 × 1 = 6.
  3. Divide: 6 / 10−4 = 6 × 104.
  4. Multiply by μ0/4π: F = 10−7 × 6 × 104.
  5. F = 6 × 10−3 N.
  6. Since the moments are aligned (N of one facing S of the other), the force is attractive.
Assumption
Both magnets short compared with 10 cm; coaxial, aligned arrangement.
Shortcut trick
Raising 0.1 to the fourth power gives 10−4, not 10−3. Counting decimal places under a fourth power is the most common slip here — write it as (10−1)4 and multiply the exponents.
Answer: (A) 6 × 10−3 N
Q24Attraction or repulsionold bank / JEE

Two coaxial magnets are arranged with the north pole of one facing the south pole of the other. The force between them is

Given
Coaxial magnets with N of one facing S of the other — moments pointing the same way along the axis.
Asked
Whether the force is attractive or repulsive.
Concept applied
Unlike poles face each other, so the dominant interaction is attraction. Equivalently, the second dipole sits with its moment aligned to the first's field, which is the minimum-energy configuration — and systems are pulled towards lower energy.
Formula to use
U = −m2·B1; aligned ⇒ U negative and falling with decreasing r
Baby steps
  1. N facing S means the nearest poles are unlike, and unlike poles attract.
  2. Energy view: with the moments aligned, U = −m2B1, which is negative.
  3. B1 grows as r decreases, so U becomes more negative as the magnets approach.
  4. A system moves towards lower potential energy, so they are pulled together.
  5. The force is attractive, at every separation — option (D)'s qualifier is unnecessary.
  6. Reversing one magnet would make the moments antiparallel and the force repulsive.
Assumption
Both magnets free to move along the common axis; short-dipole regime.
Shortcut trick
Two consistent routes, and both are quick: nearest poles (unlike attract) or energy (aligned moments sit lower, so they pull together). Use whichever the question's wording makes easier and they will always agree.
Diagram
S N S N r (centre to centre) attraction F = (μ₀/4π)(6 m₁m₂ / r⁴)
Answer: (A) attractive
Q25Force in a gradientold bank / JEE

A magnetic dipole of moment m placed in a non-uniform field along the x-axis experiences a force of magnitude

Given
Dipole of moment m aligned along a field that varies with position.
Asked
The magnitude of the net force.
Concept applied
In a uniform field the two poles feel equal and opposite forces that cancel. In a non-uniform field the nearer pole sits in a stronger field, so the cancellation is incomplete. What survives is proportional to the rate of change of the field.
Formula to use
F = m (dB/dx)
Baby steps
  1. Model the dipole as poles ±qm separated by 2l along x.
  2. Force on the far pole: −qmB(x); on the near pole: +qmB(x + 2l).
  3. Net force: qm[B(x + 2l) − B(x)].
  4. For small 2l, B(x + 2l) − B(x) ≈ (dB/dx)(2l).
  5. So F = qm(2l)(dB/dx) = m (dB/dx), since m = qm(2l).
  6. If dB/dx = 0 the field is uniform and F = 0 — recovering the familiar in-syllabus result. ✔
Assumption
The dipole is aligned with the field and small enough that the gradient is effectively constant across it.
Shortcut trick
This one formula unifies several in-syllabus facts: why a nail is attracted to a magnet, why a diamagnet drifts to weak field, why a paramagnet drifts to strong field. All are F = m(dB/dx) with different signs of m relative to B.
Answer: (B) m(dB/dx)
Q26Comparison with chargesold bank / JEE

Two point charges interact as 1/r2. Two coaxial short dipoles interact as 1/r4. The extra two powers of r arise because

Given
Comparison of the monopole–monopole and dipole–dipole force laws.
Asked
The origin of the two extra powers of r.
Concept applied
Each dipole is a pair of opposite sources very close together, and their fields largely cancel at a distance. Each such cancellation costs one power of r — one for the source dipole, one for the test dipole.
Formula to use
monopole field 1/r2 → dipole field 1/r3 → dipole–dipole force 1/r4
Baby steps
  1. A single pole would produce a field falling as 1/r2.
  2. A dipole is two opposite poles close together, and their fields nearly cancel at large r, leaving 1/r3 — one power lost to the partial cancellation.
  3. The second dipole responds not to the field itself but to its gradient, since a uniform field gives no net force.
  4. Taking that gradient costs another power, giving 1/r4.
  5. So the two extra powers come from the pairing of opposite poles in each magnet.
  6. Reject option (C): magnetic forces on moving charges do no work, but that is unrelated to this scaling.
Assumption
Short dipoles observed at large separation.
Shortcut trick
Track the exponent through the chain: monopole 2 → dipole field 3 → dipole–dipole force 4. Each step up in structural complexity costs exactly one power of r, and the same pattern holds for electric dipoles.
Answer: (B) each dipole is a pair of opposite poles whose fields partly cancel

D · Curie's law and the Curie–Weiss law

8 questions
Q27Curie's lawnot in rationalised NCERT

Curie's law for a paramagnetic material states that its susceptibility

1/T χ slope = C (Curie constant) χ = C/T plots as a straight line through the origin
Given
A paramagnetic material at absolute temperature T.
Asked
The form of Curie's law.
Concept applied
Thermal agitation opposes alignment, so the response weakens as temperature rises. Curie's law makes that quantitative: χ falls as the reciprocal of the absolute temperature.
Formula to use
χ = C/T, where C is the Curie constant
Baby steps
  1. Paramagnetic atoms have permanent moments that the field tries to align.
  2. Thermal motion randomises them, and the randomising influence grows with T.
  3. The degree of alignment therefore falls as T rises.
  4. Curie's law states this precisely: χ = C/T, i.e. χ is inversely proportional to T.
  5. The rationalised NCERT states the qualitative dependence — magnetisation increases as temperature is lowered — but does not give this equation.
  6. Note T must be the absolute temperature in kelvin, never celsius.
Assumption
Weak field, so the material is far from saturation and the linear relation M = χH holds.
Shortcut trick
Convert to kelvin before using Curie's law, every time. A problem quoting 27°C and 127°C is really 300 K and 400 K, and the ratio 300/400 is nothing like 27/127.
Answer: (B) is inversely proportional to the absolute temperature
Q28Curie's law numericnot in rationalised NCERT

A paramagnetic sample has χ = 6 × 10−3 at 300 K. At 600 K its susceptibility is

Given
χ1 = 6 × 10−3 at T1 = 300 K; T2 = 600 K
Asked
χ2
Concept applied
With χT = C constant, the two states are linked by a simple inverse proportion. No need to compute C explicitly, though it is available if wanted.
Formula to use
χ1T1 = χ2T2 = C
Baby steps
  1. Curie's law gives χT = C, a constant for the material.
  2. So χ1T1 = χ2T2.
  3. χ2 = χ1(T1/T2) = 6 × 10−3 × (300/600).
  4. 300/600 = 0.5.
  5. χ2 = 3 × 10−3.
  6. Optional check: C = χ1T1 = 6 × 10−3 × 300 = 1.8 K, and 1.8/600 = 3 × 10−3. ✔
Assumption
The sample remains paramagnetic and unsaturated over this temperature range.
Shortcut trick
Use the constant-product form χ1T1 = χ2T2 rather than computing C. It is one line, and the direction check is built in — hotter must mean smaller χ.
Answer: (C) 3 × 10−3
Q29Graph of χ vs 1/Tnot in rationalised NCERT

For a paramagnetic material obeying Curie's law, a graph of χ against 1/T is

Given
χ = C/T.
Asked
The shape of the χ versus 1/T graph.
Concept applied
Rewriting χ = C(1/T) makes it a linear relation between χ and the variable 1/T, with slope C and zero intercept. Plotting against the reciprocal is a standard way to linearise an inverse law.
Formula to use
χ = C × (1/T) — compare y = mx
Baby steps
  1. Write Curie's law as χ = C(1/T).
  2. Treat 1/T as the independent variable x and χ as y.
  3. The relation becomes y = Cx, the equation of a straight line through the origin.
  4. So the graph is a straight line through the origin, of slope C.
  5. Against T itself (not 1/T) the graph would instead be a hyperbola-like decreasing curve — option (A) is the answer to that different question.
  6. This linearisation is how the Curie constant is measured experimentally: plot χ against 1/T and take the gradient.
Assumption
Curie's law holds over the range plotted; the sample stays paramagnetic and unsaturated.
Shortcut trick
Read the axis label before choosing the shape. χ vs T ⇒ falling curve. χ vs 1/T ⇒ straight line through the origin. Both appear as options in the same question precisely to test this.
Diagram
1/T χ slope = C (Curie constant) χ = C/T plots as a straight line through the origin
Answer: (B) a straight line through the origin
Q30Curie constant unitsnot in rationalised NCERT

The Curie constant C in the relation χ = C/T has SI unit

Given
χ = C/T, with χ dimensionless and T in kelvin.
Asked
The unit of C.
Concept applied
Rearrange for C and read the units off. Because χ carries no units, C must carry the unit of temperature.
Formula to use
C = χT
Baby steps
  1. Rearrange the law: C = χT.
  2. χ is a dimensionless ratio (M divided by H, both in A m−1).
  3. T is measured in kelvin.
  4. So the unit of C is 1 × K = kelvin.
  5. Check with the earlier numbers: C = 6 × 10−3 × 300 K = 1.8 K. ✔
  6. Dimensionally [C] = [K], with no mass, length or time content.
Assumption
The simple Curie form; the Curie–Weiss version uses the same constant with the same unit.
Shortcut trick
Deriving a constant's unit from its defining equation is always faster and safer than recalling it. Isolate the constant, substitute the known units, and the answer falls out.
Answer: (B) kelvin (K)
Q31Curie–Weiss lawold bank / JEE

Above the Curie temperature TC, the susceptibility of a ferromagnetic material follows

Given
A ferromagnet at temperature T above its Curie temperature TC.
Asked
The applicable susceptibility law.
Concept applied
Above TC the material behaves as a paramagnet, but the residual cooperative interaction between neighbouring moments shifts the reference point from absolute zero to TC.
Formula to use
χ = C/(T − TC) for T > TC
Baby steps
  1. Below TC the material is ferromagnetic and χ is enormous.
  2. Above TC the domains have broken up and it behaves as a paramagnet.
  3. But the exchange interaction between neighbouring atomic moments has not disappeared, and it assists alignment.
  4. The law is modified to χ = C/(T − TC), the Curie–Weiss law.
  5. As T approaches TC from above, the denominator → 0 and χ → ∞, marking the onset of spontaneous magnetisation.
  6. For T ≫ TC the TC becomes negligible and it reduces to ordinary Curie behaviour. ✔
Assumption
T strictly greater than TC; the law fails at and below the transition.
Shortcut trick
The divergence at T = TC is the physical signature to look for. If χ blows up at a non-zero temperature, the law must have T − TC in the denominator — that alone rules out the other three options.
Answer: (B) χ = C/(T − TC)
Q32Curie–Weiss numericold bank / JEE

A ferromagnet with Curie temperature 300 K has χ = 4 × 10−3 at 400 K. At 500 K its susceptibility is

Given
TC = 300 K; χ1 = 4 × 10−3 at T1 = 400 K; T2 = 500 K
Asked
χ2
Concept applied
Apply the Curie–Weiss law twice and divide, so that the constant C cancels. The key is to use (T − TC), not T, in each case.
Formula to use
χ1(T1 − TC) = χ2(T2 − TC) = C
Baby steps
  1. First state: T1 − TC = 400 − 300 = 100 K.
  2. Find the constant: C = χ1 × 100 = 4 × 10−3 × 100 = 0.4 K.
  3. Second state: T2 − TC = 500 − 300 = 200 K.
  4. χ2 = C/(T2 − TC) = 0.4/200.
  5. χ2 = 2.0 × 10−3.
  6. Option (A) is what you get by using plain Curie's law (400/500 × 4 × 10−3 = 3.2 × 10−3) and ignoring TC.
Assumption
Both temperatures lie above TC, so the Curie–Weiss form applies at each.
Shortcut trick
Subtract TC first, then work with those differences as if they were plain temperatures. Here 100 K and 200 K — a factor of 2, so χ halves. Doing the subtraction upfront turns it into an ordinary inverse-proportion problem.
Answer: (B) 2.0 × 10−3
Q33Magnetisation formnot in rationalised NCERT

For a paramagnetic sample obeying Curie's law, the magnetisation depends on the field and temperature as

Given
Paramagnetic sample in field B at temperature T, obeying Curie's law.
Asked
The dependence of M on B and T.
Concept applied
Combine M = χH with χ = C/T. The field promotes alignment while temperature destroys it, so the two appear on opposite sides of the fraction.
Formula to use
M = χH = (C/T)(B/μ0) ⇒ M ∝ B/T
Baby steps
  1. Start from M = χH.
  2. Substitute Curie's law: M = (C/T)H.
  3. For a paramagnet H ≈ B/μ0, so M = CB/(μ0T).
  4. Grouping the constants: M ∝ B/T.
  5. Physical reading: a stronger field aligns more moments (M up), a higher temperature randomises them (M down).
  6. This matches the qualitative statement in NCERT 5.5.2 — as the field increases or the temperature is lowered, the magnetisation increases.
Assumption
Weak field, far from saturation, so the linear relation still holds. At saturation M stops responding to either variable.
Shortcut trick
Check the physics before trusting the algebra: field helps, heat hinders. So B belongs on top and T underneath. Any option with T in the numerator can be rejected on sight.
Answer: (B) M ∝ B/T
Q34Combined changeold bank / JEE

A paramagnetic sample has magnetisation M. The field is doubled and the absolute temperature is halved. The new magnetisation is

Given
M ∝ B/T; B → 2B and T → T/2.
Asked
The new magnetisation.
Concept applied
Two independent proportionalities acting together. Handle each separately and multiply the factors, exactly as in the field-scaling problems.
Formula to use
M′/M = (B′/B) × (T/T′)
Baby steps
  1. Field factor: B′/B = 2, so this doubles M.
  2. Temperature factor: T/T′ = T/(T/2) = 2, so this doubles M again.
  3. Multiply the two factors: 2 × 2 = 4.
  4. M′ = 4M.
  5. Both changes push in the same direction — more field and less heat both improve alignment.
  6. Caveat: this assumes the sample has not reached saturation; once fully aligned, M cannot rise further.
Assumption
Linear regime throughout; the sample remains paramagnetic and unsaturated at the lower temperature.
Shortcut trick
Build a two-column factor table (what helps, what hinders) and multiply down the column. Both changes here help, so the answer must exceed M — which eliminates option (D) instantly.
Answer: (C) 4M

E · The Bohr magneton and quantised atomic moments

7 questions
Q35Bohr magneton formulanot in rationalised NCERT

The Bohr magneton is given by

Given
An electron in the first Bohr orbit, with quantised angular momentum L = h/2π.
Asked
The expression for the Bohr magneton.
Concept applied
Combine the classical gyromagnetic relation m = (e/2me)L with Bohr's quantisation of angular momentum. The smallest allowed L gives the smallest allowed moment.
Formula to use
μB = (e/2me)(h/2π) = eh/4πme
Baby steps
  1. Classical result for an orbiting electron: m = (e/2me)L.
  2. Bohr quantisation: L = nh/2π, with the smallest value at n = 1, so L = h/2π.
  3. Substitute: m = (e/2me)(h/2π).
  4. Multiply the denominators: 2me × 2π = 4πme.
  5. μB = eh/4πme.
  6. Option (D) is the gyromagnetic ratio e/2me alone, which is a ratio, not a moment — its units are C kg−1, not A m2.
Assumption
Bohr's semi-classical model with n = 1; orbital contribution only, ignoring spin.
Shortcut trick
Reconstruct rather than memorise: gyromagnetic ratio × smallest angular momentum. Where the 4π comes from is then obvious — it is the 2 from the ratio times the 2π from quantisation.
Answer: (A) eh/4πme
Q36Bohr magneton valuenot in rationalised NCERT

Substituting e = 1.6 × 10−19 C, h = 6.63 × 10−34 J s and me = 9.1 × 10−31 kg gives μB equal to about

Given
e = 1.6 × 10−19 C; h = 6.63 × 10−34 J s; me = 9.1 × 10−31 kg
Asked
The numerical value of the Bohr magneton.
Concept applied
Straight substitution into μB = eh/4πme. Keeping mantissas and exponents separate is essential with numbers this small.
Formula to use
μB = eh/4πme
Baby steps
  1. Numerator mantissa: 1.6 × 6.63 = 10.61.
  2. Numerator exponent: 10−19 × 10−34 = 10−53. So eh = 10.61 × 10−53 = 1.061 × 10−52.
  3. Denominator: 4π = 12.57, so 4πme = 12.57 × 9.1 × 10−31 = 114.4 × 10−31 = 1.144 × 10−29.
  4. Divide the mantissas: 1.061/1.144 = 0.928.
  5. Divide the exponents: 10−52/10−29 = 10−23.
  6. μB = 0.928 × 10−23 = 9.27 × 10−24 A m2.
Assumption
Standard constants to three significant figures.
Shortcut trick
Split every calculation into a mantissa part and an exponent part, and do them separately. Option (D) is half the correct value — the result of using 2πme instead of 4πme in the denominator.
Answer: (A) 9.27 × 10−24 A m2
Q37nth orbitnot in rationalised NCERT

For an electron in the nth Bohr orbit, the orbital magnetic moment is

Given
Electron in the nth Bohr orbit, L = nh/2π.
Asked
The orbital magnetic moment.
Concept applied
The moment is directly proportional to the angular momentum, and angular momentum is quantised in integer multiples of h/2π. So the moment is quantised in integer multiples of μB.
Formula to use
m = (e/2me)(nh/2π) = nμB
Baby steps
  1. Angular momentum in the nth orbit: L = nh/2π.
  2. Magnetic moment: m = (e/2me)L.
  3. Substitute: m = (e/2me)(nh/2π) = n × eh/4πme.
  4. The bracket is exactly μB.
  5. m = nμB.
  6. So the moment comes in whole-number multiples of the Bohr magneton — which is precisely why μB is called the natural unit of atomic magnetic moment.
Assumption
Bohr model; orbital contribution only, with spin neglected.
Shortcut trick
Option (C), n2μB, is planted because the Bohr radius goes as n2. But the moment follows angular momentum, which goes as n1. Track which quantity you are actually scaling.
Answer: (B) nμB
Q38Orbit comparisonnot in rationalised NCERT

The ratio of the orbital magnetic moments of an electron in the third and first Bohr orbits is

Given
m = nμB; n = 3 and n = 1.
Asked
m3 : m1
Concept applied
With m proportional to n, the ratio of moments is simply the ratio of quantum numbers.
Formula to use
m3/m1 = 3μB/1μB
Baby steps
  1. Third orbit: m3 = 3μB.
  2. First orbit: m1 = 1μB.
  3. Form the ratio: m3/m1 = 3μB/μB.
  4. μB cancels.
  5. Ratio = 3 : 1.
  6. Option (C), 9 : 1, comes from using the n2 radius scaling rather than the n1 moment scaling.
Assumption
Bohr model, orbital moments only.
Shortcut trick
Keep the Bohr scalings straight: radius ∝ n2, angular momentum ∝ n, magnetic moment ∝ n, energy ∝ 1/n2. Most wrong options in this group are built from applying the wrong one.
Answer: (B) 3 : 1
Q39Unit of μBnot in rationalised NCERT

The Bohr magneton has the same unit as

Given
μB = eh/4πme.
Asked
Which quantity shares its unit.
Concept applied
The Bohr magneton is defined as the smallest orbital magnetic moment, so by construction it carries the unit of magnetic moment, A m2.
Formula to use
[μB] = [L2A], unit A m2 = J T−1
Baby steps
  1. μB is defined as the magnetic moment of an electron in the first Bohr orbit.
  2. A magnetic moment has unit A m2, equivalently J T−1.
  3. Verify from the formula: [e][h]/[me] = [AT][ML2T−1]/[M] = [L2A]. ✔
  4. That is the dimension of magnetic moment.
  5. Answer: magnetic moment.
  6. For contrast: magnetic field is in T, flux in Wb, intensity in A m−1 — none of them match.
Assumption
SI units.
Shortcut trick
The name gives it away — a 'magneton' is a unit of magnetic moment. Verifying it dimensionally from eh/me takes ten seconds and confirms it beyond doubt.
Answer: (C) magnetic moment
Q40Directionnot in rationalised NCERT

The orbital magnetic moment of an electron points

+ e− v m (opposite to L) r m = evr/2 = nμB μB = eh/4πme
Given
An electron (negative charge) in a circular orbit.
Asked
The direction of its magnetic moment relative to L.
Concept applied
Conventional current runs opposite to the motion of a negative charge, so the moment defined by that current is antiparallel to the angular momentum defined by the actual motion.
Formula to use
m = −(e/2me)L
Baby steps
  1. The electron physically circulates in one sense around the orbit.
  2. Because its charge is negative, conventional current flows in the opposite sense.
  3. m is set by the conventional current via the right-hand rule.
  4. L = r × p is set by the electron's real motion.
  5. The two are therefore antiparallel.
  6. Both are perpendicular to the orbital plane, so options (C) and (D) are geometrically impossible.
Assumption
e denotes the magnitude of the charge, with the sign carried explicitly.
Shortcut trick
Positive charge ⇒ m parallel to L. Negative charge ⇒ antiparallel. Only the sign of the charge matters — the orbit's size and speed are irrelevant to the direction.
Answer: (B) opposite to its orbital angular momentum
Q41Why atoms are magneticnot in rationalised NCERT

The Bohr magneton is significant because it

Given
μB ≈ 9.27 × 10−24 A m2.
Asked
Its physical significance.
Concept applied
Because orbital moments come in integer multiples of μB, it functions as the natural unit for magnetism at the atomic scale — the yardstick against which atomic moments are quoted.
Formula to use
morbital = nμB, n = 1, 2, 3, …
Baby steps
  1. Orbital moments are quantised as nμB, so μB is the smallest non-zero orbital value — not the largest. Option (A) is wrong.
  2. Atomic moments of real materials are routinely quoted in units of μB, typically a few μB per atom.
  3. That makes it the natural scale for atomic magnetic moments.
  4. A bar magnet's moment is of order 1 A m2 — some 1023 times larger, which is roughly the number of atoms contributing. Option (C) is wrong.
  5. The proton's magnetic moment is about 2000 times smaller, described instead by the nuclear magneton. Option (D) is wrong.
  6. Answer: option (B).
Assumption
Orbital contribution only; electron spin adds a further moment of comparable size.
Shortcut trick
Hold the order-of-magnitude ladder in mind: bar magnet ≈ 1 A m2, atom ≈ 10−23, nucleus ≈ 10−26. That alone answers most 'which is the moment of X' questions by elimination.
Answer: (B) sets the natural scale for atomic magnetic moments

F · The vibration magnetometer

4 questions
Q42Principlenot in rationalised NCERT

A vibration magnetometer works on the principle that a suspended magnet

B (horizontal) S N θ T = 2π√(I / mB)
Given
A magnet suspended horizontally by a torsionless fibre in a magnetic field.
Asked
The operating principle of the instrument.
Concept applied
Displaced from alignment, the magnet feels a restoring torque proportional to the displacement (for small angles), so it oscillates in angular SHM with a period that depends on m and B.
Formula to use
T = 2π√(I/mB)
Baby steps
  1. The magnet rests aligned with the field — its stable equilibrium.
  2. Displace it by a small angle and release: the torque −mB sin θ ≈ −(mB)θ pulls it back.
  3. Restoring torque proportional to displacement means angular SHM.
  4. The period T = 2π√(I/mB) links the measurable T to the quantities m and B.
  5. Measuring T with I known therefore yields m if B is known, or B if m is known.
  6. Reject option (C): no current is induced in the magnet itself; the instrument is purely mechanical.
Assumption
Torsionless suspension, small amplitude, negligible damping.
Shortcut trick
The instrument converts a magnetic unknown into a timing measurement, which is easy to make precisely. That conversion of one measurement into an easier one is the design idea behind most classical instruments.
Answer: (B) executes angular SHM about the field direction
Q43Measuring Bnot in rationalised NCERT

In a vibration magnetometer, the horizontal component of a field is found from the period by

Given
T = 2π√(I/mB), with I and m known.
Asked
The expression for B.
Concept applied
Square both sides to remove the root, then isolate B. This is the standard laboratory route to determining a field from a timing measurement.
Formula to use
B = 4π2I / (mT2)
Baby steps
  1. Start from T = 2π√(I/mB).
  2. Square: T2 = 4π2I/(mB).
  3. Multiply both sides by mB: mBT2 = 4π2I.
  4. Divide by mT2: B = 4π2I/(mT2).
  5. Check the behaviour: a longer period implies a weaker field, and T2 is in the denominator. ✔
  6. Option (B) is the reciprocal — it would wrongly predict a stronger field for a slower oscillation.
Assumption
Small amplitude; I determined from the magnet's mass and dimensions.
Shortcut trick
After any rearrangement, run a one-line physical check on the direction of dependence. Slower swing must mean weaker field. That single check catches inverted answers instantly.
Answer: (A) B = 4π2I/(mT2)
Q44Amplitude requirementnot in rationalised NCERT

In using a vibration magnetometer, the magnet must be given only a small angular displacement because

Given
Standard experimental procedure with a vibration magnetometer.
Asked
The reason for restricting the amplitude.
Concept applied
The period formula assumes sin θ ≈ θ. At large amplitudes that approximation fails, the motion stops being simple harmonic, and the period starts to depend on amplitude — making the measurement meaningless.
Formula to use
τ = −mB sin θ ≈ −(mB)θ only for small θ
Baby steps
  1. The formula T = 2π√(I/mB) is derived using sin θ ≈ θ.
  2. This approximation is good to within about 1% up to roughly 10°–15°.
  3. Beyond that, the restoring torque grows more slowly than the displacement.
  4. The motion is then periodic but not simple harmonic, and the period becomes amplitude dependent.
  5. An amplitude-dependent period would make the measured value of B or m unreliable.
  6. So the amplitude must be kept small, for the SHM condition to hold.
Assumption
Torsionless suspension so the only restoring torque is magnetic.
Shortcut trick
The reason is always mathematical, never mechanical. Options about breaking fibres or demagnetisation sound practical but have nothing to do with why the formula requires small angles.
Answer: (B) the motion is simple harmonic only for small angles
Q45Effect of a shorter magnetnot in rationalised NCERT

In a vibration magnetometer, replacing the magnet with one of half the length and half the mass but the same pole strength changes the period to

Given
New magnet: half the length, half the mass, same pole strength qm; same field.
Asked
The new period.
Concept applied
This is exactly the transverse-cut situation described in different words. Halving both mass and length reduces I by eight; halving the length with qm fixed halves m.
Formula to use
I′ = I/8;  m′ = qm(L/2) = m/2
Baby steps
  1. New moment of inertia: I′ = (Mb/2)(L/2)2/12 = I/8.
  2. New magnetic moment: m′ = qm × (L/2) = m/2, since qm is unchanged.
  3. Ratio: I′/(m′B) = (I/8) ÷ (mB/2) = I/(4mB).
  4. T′ = 2π√[I/(4mB)] = T/2.
  5. T′ = T/2.
  6. Same answer as the transverse cut — the physical situations are identical, only the wording differs.
Assumption
Same field, same suspension, uniform bar in both cases.
Shortcut trick
Questions often disguise a cut as a substitution: 'a magnet of half the length and half the mass' is a transversely cut half. Recognise the disguise and reuse the stored result T′ = T/2.
Answer: (B) half

G · Hysteresis (deleted from both syllabi)

5 questions
Q46Definitiondeleted from NEET & JEE

Hysteresis in a ferromagnetic material refers to

H B retentivity coercivity loop area = energy lost per cycle per unit volume
Given
A ferromagnetic sample taken through a full cycle of magnetising field H.
Asked
The meaning of hysteresis.
Concept applied
Domain walls do not move back reversibly, so the magnetisation depends on the material's history as well as on the present value of H. The B–H curve therefore does not retrace itself, forming a loop.
Formula to use
B is not a single-valued function of H — it depends on history
Baby steps
  1. Increase H from zero: B rises along the initial magnetisation curve to saturation.
  2. Now decrease H back to zero: B does not return along the same path — it stays higher.
  3. At H = 0 a residual field remains, called the retentivity or remanence.
  4. Reversing H eventually drives B to zero, then to saturation the other way, and the cycle closes as a loop.
  5. This lagging of B behind H is hysteresis.
  6. The physical cause is that domain wall motion is not fully reversible.
Assumption
Ferromagnetic material below its Curie temperature.
Shortcut trick
Syllabus note: hysteresis has been removed from both the rationalised NCERT and the JEE Main syllabus. It survives only in older question banks. Know it well enough to recognise the terms, but do not spend serious revision time here.
Answer: (A) the lag of B behind H as H is cycled
Q47Retentivitydeleted from NEET & JEE

The retentivity of a ferromagnetic material is the value of B when

Given
A hysteresis loop for a ferromagnetic sample.
Asked
The definition of retentivity.
Concept applied
Retentivity, also called remanence, is the magnetic flux density that remains once the magnetising field has been switched off. It is the loop's intercept on the B-axis.
Formula to use
retentivity = B at H = 0, after saturation
Baby steps
  1. Drive the sample to saturation with a large H.
  2. Reduce H back to zero.
  3. B does not fall to zero — some domains remain aligned.
  4. The remaining value of B is the retentivity.
  5. Graphically it is where the loop crosses the vertical B-axis.
  6. Retentivity is what makes a permanent magnet possible — hard ferromagnets have high retentivity.
Assumption
The sample was first taken to saturation; retentivity is defined from the saturated state.
Shortcut trick
Locate both quantities on the axes and you will never confuse them: retentivity is the B-axis intercept (H = 0); coercivity is the H-axis intercept (B = 0).
Answer: (B) H is reduced to zero after saturation
Q48Coercivitydeleted from NEET & JEE

Coercivity is the reverse magnetising field required to

Given
A ferromagnetic sample carrying residual magnetism after saturation.
Asked
The definition of coercivity.
Concept applied
Having switched off H, the sample retains magnetism. Coercivity measures how strong a reversed field must be to wipe that residue out completely — a measure of how stubbornly the material holds its magnetisation.
Formula to use
coercivity = |H| at B = 0, on the reverse branch
Baby steps
  1. After saturation and removal of H, the sample retains B equal to its retentivity.
  2. Apply H in the reverse direction.
  3. As reverse H increases, B falls towards zero.
  4. The reverse field at which B first reaches zero is the coercivity.
  5. Graphically it is the loop's intercept on the horizontal H-axis.
  6. High coercivity means a magnet that resists demagnetisation — desirable in permanent magnets, undesirable in transformer cores.
Assumption
The measurement starts from the saturated state.
Shortcut trick
Read the words: retentivity is about what is retained (B when H = 0); coercivity is about how much coercion is needed to erase it (H when B = 0). The names carry the definitions.
Diagram
H B retentivity coercivity loop area = energy lost per cycle per unit volume
Answer: (B) reduce the residual magnetism to zero
Q49Hard vs softdeleted from NEET & JEE

A material suitable for a transformer core should have a hysteresis loop that is

Given
A transformer core magnetised and demagnetised many times per second.
Asked
The desirable loop shape.
Concept applied
The loop area equals the energy lost as heat per cycle per unit volume. A transformer core cycles 50 or 60 times a second, so a large loop would waste a great deal of energy. It also needs to demagnetise easily, requiring low coercivity.
Formula to use
energy lost per cycle per unit volume = area of the B–H loop
Baby steps
  1. A transformer core is taken through a full magnetisation cycle every AC period.
  2. Each cycle dissipates energy equal to the loop area, as heat.
  3. At 50 cycles per second, a broad loop would mean large continuous losses.
  4. So the loop must have small area.
  5. The core must also reverse its magnetisation easily each cycle, which requires low coercivity.
  6. These are the properties of a soft ferromagnet such as soft iron — the material NCERT names for exactly this purpose.
  7. Option (D) is impossible for a ferromagnet: some hysteresis is always present.
Assumption
Standard AC operation; other loss mechanisms such as eddy currents are set aside.
Shortcut trick
Match loop shape to application: broad loop ⇒ hard ⇒ permanent magnet. Narrow loop ⇒ soft ⇒ transformer or electromagnet core. The hard/soft distinction itself is still in the rationalised NCERT, even though the hysteresis loop is not.
Answer: (B) narrow, with low coercivity and small area
Q50Energy lossdeleted from NEET & JEE

The energy dissipated per cycle per unit volume in a ferromagnetic material equals

Given
A ferromagnetic material carried through one complete magnetisation cycle.
Asked
The energy loss per unit volume per cycle.
Concept applied
Work is done on the material as its magnetisation is changed, and because the path is irreversible some of it is not recovered. The unrecovered part is exactly the area enclosed by the loop.
Formula to use
Wcycle = ∮ H dB = area of the hysteresis loop
Baby steps
  1. The work done per unit volume in changing B by dB is H dB.
  2. Over a complete cycle, the total is the closed integral ∮H dB.
  3. Geometrically that closed integral is the area enclosed by the loop.
  4. Because the loop does not retrace itself, this area is non-zero and represents energy converted to heat.
  5. So the material warms up — the origin of hysteresis loss in transformers and motors.
  6. Option (D) is wrong: a soft material has a small loop area, never exactly zero.
Assumption
A complete symmetric cycle; other losses such as eddy currents are excluded.
Shortcut trick
Area on a graph almost always means energy. Under a force–displacement curve it is work; under a P–V curve it is work; enclosed by a B–H loop it is energy lost per cycle. Check the product of the axis units to confirm.
Answer: (B) the area enclosed by the hysteresis loop

Read this before you spend time here

None of the material in this pack is in your rationalised NCERT Chapter 5. It is here because it keeps appearing in practice papers, coaching sheets and older question banks — so meeting it cold in a mock is a wasted mark, and meeting it after this pack is not. But the priority is genuinely lower than Tiers 1 and 2.

Spend your time in this order.

1. Groups A and B (oscillation and cut variants) — worth real effort. These appear most often of anything in this pack, the derivation is short, and the results are clean. Four facts carry the whole group:
  • T = 2π√(I/mB), with I = MbL2/12 for a bar magnet
  • Transverse cut into n parts: I ÷ n³, m ÷ n, so T′ = T/n
  • Longitudinal cut: T unchanged
  • Like poles stacked: T unchanged; unlike poles stacked: no oscillation

2. Group D (Curie's law) — worth a single sitting. χ = C/T, and χ = C/(T − TC) above the Curie point. Always convert to kelvin, and always subtract TC before forming ratios.

3. Groups C, E and F — worth one read each. Learn F ∝ 1/r⁴ and the 6 = 2 × 3 in its formula; learn μB = eh/4πme ≈ 9.27 × 10−24 A m² and that moments go as nμB. That is enough.

4. Group G (hysteresis) — recognition only. Deleted from the rationalised NCERT and from the JEE Main syllabus. Know what retentivity, coercivity and loop area mean so an old paper cannot ambush you. Do not revise it further.

One thing to watch throughout. The symbol clash between magnetic moment m and mass M causes more errors in this pack than any concept does. Write mass as Mb in your working, every time.