A · Torque, work and potential energy
13 questionsA short bar magnet of magnetic moment 0.4 J T−1 is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when its potential energy is
Given
m = 0.4 J T−1; B = 0.16 T; stable equilibrium.
Asked
The potential energy.
Concept applied
Stable equilibrium means m parallel to B, i.e. θ = 0°, where U reaches its minimum value −mB.
Formula to use
U = −mB cos θ; at θ = 0°, U = −mB
Baby steps
- Stable equilibrium ⇒ θ = 0° ⇒ cos θ = 1.
- U = −mB.
- Compute mB = 0.4 × 0.16.
- 0.4 × 0.16 = 0.064.
- U = −0.064 J.
- The negative sign is essential — option (B) is the same magnitude with the wrong sign, and is the unstable-equilibrium value.
Assumption
Uniform field; NCERT's zero-energy convention at θ = 90°.
Shortcut trick
Stable is always the most negative energy. Compute mB once, attach a minus for aligned and a plus for anti-aligned. Both values sit in the options every time this is asked.
Answer: (A) −0.064 J
A bar magnet of moment 2 × 104 J T−1 is free to rotate in a horizontal plane where a field of 6 × 10−4 T exists. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from it is
Given
m = 2 × 104 J T−1; B = 6 × 10−4 T; θ1 = 0°, θ2 = 60°
Asked
The external work done.
Concept applied
Work equals the increase in potential energy between the two orientations.
Formula to use
W = mB(cos θ1 − cos θ2)
Baby steps
- Compute mB = 2 × 104 × 6 × 10−4 = 12 J.
- cos θ1 = cos 0° = 1.
- cos θ2 = cos 60° = 0.5.
- W = 12 × (1 − 0.5).
- W = 12 × 0.5 = 6 J.
- Option (C), 12 J, is mB itself — the value before applying the cosine difference.
Assumption
Quasi-static rotation in a uniform field.
Shortcut trick
For a rotation starting from alignment, W = mB(1 − cos θ2). At 60° that is exactly half of mB; at 90° it is mB; at 180° it is 2mB. Three landmarks, no algebra.
Answer: (B) 6 J
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position is
Given
W = √3 J for a rotation from 0° to 60°.
Asked
The torque holding it at 60°.
Concept applied
Extract the product mB from the work equation, then substitute into the torque expression. There is no need to know m or B separately.
Formula to use
W = mB(1 − cos 60°) = mB/2; τ = mB sin 60°
Baby steps
- From the work: √3 = mB(1 − 0.5) = mB/2.
- So mB = 2√3.
- Torque at 60°: τ = mB sin 60° = 2√3 × (√3/2).
- The 2s cancel: τ = √3 × √3.
- τ = 3 J (numerically, in N m).
- General result: τ = √3 W whenever the angle is 60° and the start is aligned.
Assumption
Uniform field; rotation begins from the parallel orientation.
Shortcut trick
This exact question shape recurs constantly — AIEEE 2003 and NEET 2016 Phase 2 both ask it with W left as a symbol. Store the result: τ = √3 W for the 0° → 60° case.
Answer: (B) 3 J
A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position is
Given
W units of work for a 0° → 60° rotation.
Asked
The torque at 60°, in terms of W.
Concept applied
Same structure as the previous question, now purely symbolic. The ratio τ/W depends only on the angle.
Formula to use
W = mB(1 − cos 60°); τ = mB sin 60°
Baby steps
- W = mB(1 − 1/2) = mB/2, so mB = 2W.
- τ = mB sin 60° = 2W × (√3/2).
- The 2s cancel.
- τ = √3 W.
- Check the ratio directly: τ/W = sin 60°/(1 − cos 60°) = 0.866/0.5 = 1.732 = √3. ✔
- This appeared as AIEEE 2003, and again in NEET 2016 Phase 2 with the same numbers.
Assumption
Uniform field; quasi-static rotation from alignment.
Shortcut trick
Note the general ratio τ/W = sin θ / (1 − cos θ). At 60° it is √3; at 90° it is 1; at 120° it is 1/√3. One formula covers every version of this question.
Answer: (A) √3 W
A small bar magnet placed with its axis at 30° to an external field of 0.06 T experiences a torque of 0.018 N m. The minimum work required to rotate it from its stable to its unstable equilibrium position is
Given
θ = 30°; B = 0.06 T; τ = 0.018 N m
Asked
Work from stable to unstable equilibrium.
Concept applied
A two-stage problem. First recover m from the torque data, then apply the stable-to-unstable result W = 2mB.
Formula to use
m = τ/(B sin θ); W = 2mB
Baby steps
- From τ = mB sin θ: 0.018 = m × 0.06 × sin 30° = m × 0.06 × 0.5 = 0.03m.
- So m = 0.018/0.03 = 0.6 A m2.
- Now mB = 0.6 × 0.06 = 0.036 J.
- Stable to unstable is a full 0° → 180° turn, costing W = 2mB.
- W = 2 × 0.036 = 0.072 J = 7.2 × 10−2 J.
- Shortcut: mB = τ/sin θ, so W = 2τ/sin θ = 2(0.018)/0.5 = 0.072 J directly. ✔
Assumption
Uniform field; short magnet.
Shortcut trick
Skip finding m altogether. Since W = 2mB and mB = τ/sin θ, you get W = 2τ/sin θ in one line. JEE Main 2026 (24 Jan) asks the identical question with B = 800 G and τ = 0.016 N m, giving W = 0.064 J.
Answer: (A) 7.2 × 10−2 J
A bar magnet of moment 0.5 A m2 is suspended in a uniform field of 8 × 10−2 T. The work done in rotating it from its most stable to its most unstable position is
Given
m = 0.5 A m2; B = 8 × 10−2 T
Asked
Work from most stable to most unstable position.
Concept applied
Most stable is θ = 0° and most unstable is θ = 180°, so the work is the full 2mB.
Formula to use
W = 2mB
Baby steps
- Identify the endpoints: θ1 = 0° (U = −mB), θ2 = 180° (U = +mB).
- W = U2 − U1 = mB − (−mB) = 2mB.
- Compute mB = 0.5 × 8 × 10−2 = 0.04 J.
- W = 2 × 0.04.
- W = 0.08 J.
- Option (A) is mB — the answer for a 90° turn, planted for anyone who forgets to double.
Assumption
Uniform field; quasi-static rotation.
Shortcut trick
The phrases 'most stable to most unstable', 'parallel to antiparallel' and '0° to 180°' all mean the same thing: 2mB. Recognise the phrase and skip straight to doubling.
Answer: (B) 0.08 J
A bar magnet of moment 5.0 A m2 is placed parallel to a field of 0.4 T. The work required to turn it from the parallel to the antiparallel position is
Given
m = 5.0 A m2; B = 0.4 T; parallel → antiparallel.
Asked
The work done.
Concept applied
Identical to the previous question in different wording — parallel to antiparallel is 0° to 180°.
Formula to use
W = 2mB
Baby steps
- Parallel means θ1 = 0°; antiparallel means θ2 = 180°.
- W = −mB(cos 180° − cos 0°) = −mB(−1 − 1) = 2mB.
- mB = 5.0 × 0.4 = 2 J.
- W = 2 × 2.
- W = 4 J.
- Option (A), 2 J, is mB — again the half-answer.
Assumption
Uniform field throughout the rotation.
Shortcut trick
Three years, three phrasings, one formula. AIPMT 2011, JEE 2020, JEE 2023 and JEE 2024 all reduce to 2mB or a step away from it. This is the single most repeated calculation in the chapter's PYQ history.
Answer: (B) 4 J
A bar magnet of moment 2.0 × 105 J T−1 lies along a uniform field of 14 × 10−5 T. The work done in rotating it slowly through 60° from the field direction is
Given
m = 2.0 × 105 J T−1; B = 14 × 10−5 T; 0° → 60°
Asked
The work done.
Concept applied
Same as AIPMT 2009 with different numbers — a 60° turn from alignment costs exactly half of mB.
Formula to use
W = mB(1 − cos 60°) = mB/2
Baby steps
- Compute mB = 2.0 × 105 × 14 × 10−5.
- Mantissas: 2.0 × 14 = 28. Exponents: 105 × 10−5 = 100 = 1.
- So mB = 28 J.
- W = mB(1 − 0.5) = 28 × 0.5.
- W = 14 J.
- Option (A), 28 J, is mB — the intermediate value.
Assumption
Uniform field; quasi-static rotation from alignment.
Shortcut trick
When the exponents are designed to cancel (105 against 10−5), the examiner wants you to spot it and work with plain numbers. Multiply mantissas, add exponents, and the arithmetic becomes trivial.
Answer: (B) 14 J
A magnetic dipole experiences a torque of 80√3 N m when its moment makes 60° with a uniform field. Its potential energy in that orientation is
Given
τ = 80√3 N m at θ = 60°.
Asked
The potential energy at that orientation.
Concept applied
Extract mB from the torque relation, then substitute into the energy relation at the same angle. Both use the same product mB, so it need never be split.
Formula to use
τ = mB sin θ; U = −mB cos θ
Baby steps
- From the torque: 80√3 = mB sin 60° = mB(√3/2).
- Solve for mB: mB = 80√3 × 2/√3 = 160.
- Now the energy at the same angle: U = −mB cos 60°.
- cos 60° = 0.5.
- U = −160 × 0.5 = −80 J.
- Sign check: at 60° the dipole is still on the aligned side of 90°, so U must be negative. ✔
Assumption
Uniform field; the same orientation for both the torque and the energy.
Shortcut trick
The direct route is U = −τ cot θ, since U/τ = −cos θ/sin θ. Here −80√3 × cot 60° = −80√3/√3 = −80 J in one step.
Answer: (A) −80 J
The magnetic potential energy of a bar magnet of moment m placed perpendicular to a magnetic field B is
Given
m perpendicular to B, i.e. θ = 90°.
Asked
The magnetic potential energy.
Concept applied
NCERT fixes the zero of magnetic potential energy at θ = 90° by taking the constant of integration as zero. So the perpendicular orientation has, by definition, U = 0.
Formula to use
U = −m·B = −mB cos 90° = 0
Baby steps
- U = −m·B = −mB cos θ.
- Perpendicular orientation means θ = 90°.
- cos 90° = 0.
- U = −mB × 0 = zero.
- Note that the torque at this same orientation is maximum, equal to mB.
- NCERT states the convention explicitly: taking the constant of integration to be zero fixes the zero of potential energy at θ = 90°.
Assumption
NCERT's standard zero-energy convention.
Shortcut trick
Zero energy and zero torque are at different angles. U = 0 at 90°; τ = 0 at 0° and 180°. A question naming one is often testing whether you will wrongly supply the other.
Answer: (C) zero
A bar magnet of moment M is placed at right angles to a magnetic induction B. If each pole experiences a force F, the length of the magnet is
Given
Moment M; field B; force F on each pole; magnet perpendicular to B.
Asked
The length of the magnet.
Concept applied
Use the pole model. Pole strength follows from the force on a pole, and length follows from the definition of magnetic moment.
Formula to use
F = qmB and M = qmL
Baby steps
- Force on a pole of strength qm in field B: F = qmB.
- Therefore qm = F/B.
- Magnetic moment is pole strength times length: M = qmL.
- So L = M/qm.
- Substitute qm: L = M/(F/B) = MB/F.
- Dimension check: [L2A][MT−2A−1]/[MLT−2] = [L]. ✔
Assumption
Pole model of the bar magnet with two equal and opposite poles separated by the magnet's length.
Shortcut trick
When a question mentions force on each pole, it is signalling the pole model. Write down F = qmB and M = qmL, and the answer is one substitution away.
Answer: (A) MB/F
A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries 85 μA in a field of 0.85 T. The work done in rotating the coil through 180° against the torque is about
Given
N = 250; length 2.1 cm, width 1.25 cm; I = 85 μA; B = 0.85 T; rotation 180°
Asked
The work done.
Concept applied
Three stages: area from the dimensions, moment from m = NIA, then work from W = 2mB for a 180° turn.
Formula to use
A = lw; m = NIA; W = 2mB
Baby steps
- Area: A = 0.021 × 0.0125 = 2.625 × 10−4 m2.
- Moment: m = 250 × 85 × 10−6 × 2.625 × 10−4.
- 250 × 85 × 10−6 = 2.125 × 10−2.
- m = 2.125 × 10−2 × 2.625 × 10−4 = 5.58 × 10−6 A m2.
- W = 2mB = 2 × 5.58 × 10−6 × 0.85.
- W ≈ 9.5 × 10−6 J.
Assumption
Rotation starts from the aligned position; uniform field; closely wound coil.
Shortcut trick
Convert centimetres to metres before multiplying, not after. Doing it at the end means tracking a factor of 10−4 through three steps, which is where this question is usually lost.
Answer: (A) 9.5 × 10−6 J
A rectangular coil 0.12 m × 0.1 m with 50 turns carries 2 A in a uniform field of 0.2 Wb m−2. The plane of the coil makes 30° with the field direction. The torque required to keep the coil in equilibrium is about
Given
N = 50; A = 0.12 × 0.1 m2; I = 2 A; B = 0.2 T; plane at 30° to B
Asked
The torque required.
Concept applied
The formula τ = NIAB sin θ uses the angle between the coil's normal and B. When the question gives the angle of the plane, you must convert first.
Formula to use
τ = NIAB sin θ, θ measured from the normal
Baby steps
- Plane at 30° to B ⇒ normal at 90° − 30° = 60° to B.
- Area: A = 0.12 × 0.1 = 0.012 m2.
- Moment: m = NIA = 50 × 2 × 0.012 = 1.2 A m2.
- τ = mB sin 60° = 1.2 × 0.2 × 0.866.
- 1.2 × 0.2 = 0.24; × 0.866 = 0.2078.
- τ ≈ 0.20 N m. Option (D), 0.24 N m, is the value before applying sin 60°.
Assumption
Closely wound coil; uniform field; the quoted 30° is the angle of the coil's plane, not its normal.
Shortcut trick
Plane angle and normal angle are complements. If the question says 'plane makes θ with B', use sin(90° − θ) = cos θ. Misreading this is the single commonest error in coil-torque questions.
Answer: (C) 0.20 N m
B · Classification of magnetic materials
10 questionsFour light rods A, B, C, D are suspended by threads. On bringing a magnet near: A is feebly repelled, B is feebly attracted, C is strongly attracted, D is unaffected. Then
Given
A feebly repelled; B feebly attracted; C strongly attracted; D unaffected.
Asked
The classification of each rod.
Concept applied
Two independent clues in each case: the direction of the effect (repel or attract) fixes the sign of χ, and its strength distinguishes paramagnetic from ferromagnetic.
Formula to use
χ < 0 repel · χ small > 0 weak attract · χ ≫ 1 strong attract
Baby steps
- A is repelled ⇒ χ is negative ⇒ diamagnetic. Only diamagnets are repelled.
- B is attracted feebly ⇒ χ small and positive ⇒ paramagnetic.
- C is attracted strongly ⇒ χ very large and positive ⇒ ferromagnetic.
- D is unaffected ⇒ effectively non-magnetic, χ ≈ 0.
- Answer: option (A).
- Note there is only one repelled rod — if a question describes two repelled samples, one of them is a superconductor.
Assumption
Room temperature; the magnet's field is non-uniform, as it always is near a pole.
Shortcut trick
Two-step reading for every such question: direction first (repel = dia), then strength (feeble = para, strong = ferro). AIEEE 2006 asks the same thing with needles N₁, N₂, N₃ and the same logic applies.
Answer: (A) A is diamagnetic, B paramagnetic, C ferromagnetic, D non-magnetic
If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is
Given
A diamagnetic sample near either pole of a bar magnet.
Asked
The nature of the interaction.
Concept applied
The induced moment in a diamagnet always opposes the applied field, whichever way that field points. So the repulsion is independent of which pole is presented.
Formula to use
χ < 0 ⇒ M antiparallel to H, for either polarity
Baby steps
- Bring the north pole close: the field points one way, the induced moment opposes it ⇒ repulsion.
- Bring the south pole close: the field reverses, and the induced moment reverses with it ⇒ still opposing ⇒ still repulsion.
- The sign of χ does not depend on the field direction, so the effect is the same at both poles.
- Answer: repelled by both poles.
- Contrast with para and ferro materials, which are attracted by both poles for the same symmetry reason.
- AIPMT 2003 asks the equivalent question as a drift direction: a diamagnetic material moves from strong to weak field.
Assumption
Ordinary diamagnetic material at room temperature.
Shortcut trick
Induced magnetism is polarity-blind. Whatever a sample does at the north pole, it does at the south pole too. Options that split the behaviour between the two poles are wrong by symmetry alone.
Answer: (B) repelled by both poles
The magnetic susceptibility is negative for
Given
The sign of χ across the three classes.
Asked
Which class has negative χ.
Concept applied
Only diamagnetic materials develop a magnetisation opposing the applied field, which is what a negative χ means.
Formula to use
−1 ≤ χ < 0 (dia) · 0 < χ ≪ 1 (para) · χ ≫ 1 (ferro)
Baby steps
- χ is defined by M = χH, so its sign tells you whether M assists or opposes H.
- Diamagnetic: the induced moment opposes the field ⇒ χ negative.
- Paramagnetic: permanent moments partly align with the field ⇒ χ small and positive.
- Ferromagnetic: domains align strongly with the field ⇒ χ very large and positive.
- So only diamagnetic materials have negative χ.
- The most negative possible value is χ = −1, reached only by a superconductor.
Assumption
Linear response; room temperature.
Shortcut trick
Sign is the fastest classifier in the entire chapter. Negative ⇒ diamagnetic, no exceptions. Magnitude is only needed afterwards, to separate para from ferro.
Answer: (B) diamagnetic materials only
If the magnetic dipole moment of an atom of a diamagnetic, paramagnetic and ferromagnetic material are μd, μp and μf respectively, then
Given
The per-atom magnetic moments of the three classes, in the absence of an applied field.
Asked
The correct comparison.
Concept applied
The classes are distinguished at the atomic level by whether the individual atom carries a permanent moment. Diamagnetic atoms do not; paramagnetic and ferromagnetic atoms do.
Formula to use
μd = 0; μp ≠ 0; μf ≠ 0
Baby steps
- NCERT: diamagnetic substances are the ones in which the resultant magnetic moment in an atom is zero. So μd = 0.
- Paramagnetic atoms possess a permanent magnetic dipole moment of their own, so μp ≠ 0.
- Ferromagnetic atoms also possess a dipole moment, as in a paramagnetic material, so μf ≠ 0.
- The difference between para and ferro is not the per-atom moment but whether neighbouring atoms cooperate to form domains.
- Answer: μd = 0 and μp, μf ≠ 0.
- This is why diamagnetism has to be induced by the applied field, while the other two only need aligning.
Assumption
Moments considered in the absence of an external field.
Shortcut trick
Sort by starting state: dia has nothing to begin with, para has moments in disarray, ferro has moments already organised. Every property of each class follows from its starting state.
Answer: (A) μd = 0 and μp, μf ≠ 0
Which values of relative permittivity εr and relative permeability μr are allowed for a diamagnetic material?
Given
Candidate pairs of εr and μr for a diamagnetic material.
Asked
The physically allowed pair.
Concept applied
Two independent constraints. Relative permittivity of any material is always at least 1. Relative permeability of a diamagnet is always below 1.
Formula to use
εr ≥ 1 always; μr = 1 + χ < 1 for a diamagnet
Baby steps
- Constraint 1: dielectrics reduce the field inside, so εr ≥ 1 for every material. This eliminates options (B) and (D).
- Constraint 2: diamagnetic means χ < 0, so μr = 1 + χ < 1. This eliminates option (A).
- The only pair satisfying both is εr = 1.5 and μr = 0.5.
- Answer: option (C).
- Physically: the material weakens an electric field inside it but also weakens a magnetic field inside it.
- The extreme diamagnetic case is the superconductor, with μr = 0 exactly.
Assumption
Linear, isotropic material.
Shortcut trick
Apply constraints one at a time and cross options off as you go. Two independent constraints on four options usually leaves exactly one survivor — no calculation needed.
Answer: (C) εr = 1.5, μr = 0.5
Match each material with its susceptibility range: (A) Diamagnetic (B) Ferromagnetic (C) Paramagnetic (D) Non-magnetic, against (I) χ = 0 (II) 0 > χ ≥ −1 (III) χ ≫ 1 (IV) 0 < χ < ε
Given
Four material classes and four susceptibility ranges.
Asked
The correct matching.
Concept applied
Straight recall of Table 5.2, with the non-magnetic case added. Work through by sign first, then magnitude.
Formula to use
dia: 0 > χ ≥ −1 · para: 0 < χ < ε · ferro: χ ≫ 1 · non-magnetic: χ = 0
Baby steps
- Diamagnetic is the only class with negative χ, and its floor is −1 ⇒ matches II.
- Ferromagnetic has χ very large and positive ⇒ matches III.
- Paramagnetic has χ small and positive, written as 0 < χ < ε ⇒ matches IV.
- Non-magnetic means no response at all, χ = 0 ⇒ matches I.
- So A-II, B-III, C-IV, D-I — option (A).
- The ε here is NCERT's own notation: a small positive number introduced to quantify paramagnetic materials.
Assumption
NCERT Table 5.2 conventions.
Shortcut trick
Matching questions reward doing the easiest pair first. Diamagnetic is the only negative one, so pin A-II immediately — that alone eliminates three of the four options here.
Answer: (A) A-II, B-III, C-IV, D-I
For paramagnetic substances, which statements are correct? (A) they align along the external field (B) they are attracted strongly (C) susceptibility is a little more than zero (D) they move from strong to weak field
Given
Four candidate statements about paramagnetic behaviour.
Asked
Which are correct.
Concept applied
Paramagnets are attracted, but only weakly, and they drift towards the stronger region of the field. The wrong statements here both belong to other classes.
Formula to use
0 < χ ≪ 1 ⇒ weak attraction, weak → strong drift
Baby steps
- (A) Paramagnetic samples align along the field, since their moments turn to follow it. Correct.
- (B) They are attracted, but weakly — about one part in 105. 'Strongly' describes ferromagnets. Incorrect.
- (C) χ is small and positive, i.e. a little more than zero. Correct.
- (D) They move from weak field towards strong field, not the reverse. Strong-to-weak is diamagnetic. Incorrect.
- So only A and C are correct.
- The two wrong statements are deliberately borrowed — (B) from ferromagnetism and (D) from diamagnetism.
Assumption
Ordinary paramagnetic material at room temperature, far from saturation.
Shortcut trick
In multi-statement questions, check whether each wrong option is describing a different class correctly. Examiners build distractors that way, and spotting the borrowed statement confirms your rejection.
Answer: (A) A and C only
Statement I: For a diamagnetic substance, −1 ≤ χ < 0. Statement II: Diamagnetic substances in an external field tend to move from the stronger to the weaker part of the field. Then
Given
Two statements about diamagnetic materials.
Asked
Their truth values.
Concept applied
Both statements come almost verbatim from NCERT — one from Table 5.2 and one from the opening line of section 5.5.1.
Formula to use
−1 ≤ χ < 0 and drift strong → weak
Baby steps
- Statement I: Table 5.2 gives exactly this range for diamagnetic materials, with −1 as the superconducting extreme. True.
- Statement II: NCERT opens section 5.5.1 by defining diamagnetic substances as those with a tendency to move from stronger to the weaker part of the external field. True.
- The two are also physically linked: negative χ means repulsion, and repulsion means drift away from the strong region.
- Answer: both statements are true.
- Careful: the question here only asks for truth values, not whether II explains I.
- NEET 2018 tested the same physics from an energy angle — a diamagnetic rod pushed up out of the field gains gravitational PE, the work coming from the source maintaining the current in the electromagnet.
Assumption
Non-uniform field, as exists near any real pole piece.
Shortcut trick
Read whether the question asks for truth values only or for explanation as well. The option sets differ, and answering the wrong version is a common way to lose an easy mark.
Answer: (A) Both statements are true
A superconductor is a perfect diamagnet. When it is placed in a magnetic field of magnitude B, the field Bs inside it is such that
Given
A superconductor placed in an external field B.
Asked
The field inside it.
Concept applied
Perfect diamagnetism means χ = −1, so the induced magnetisation cancels the applied field exactly. The interior field is driven to zero — the Meissner effect.
Formula to use
Bs = μ0(1 + χ)H with χ = −1 ⇒ Bs = 0
Baby steps
- Perfect diamagnetism corresponds to the extreme value χ = −1.
- Then μr = 1 + χ = 0.
- Bs = μ0μrH = μ0 × 0 × H.
- Bs = 0: the field is completely expelled from the interior.
- NCERT: here the field lines are completely expelled, χ = −1 and μr = 0.
- This total expulsion is named the Meissner effect and underlies magnetic levitation.
Assumption
Type-I superconductor below its critical temperature and critical field.
Shortcut trick
'Perfect diamagnet' is a fixed phrase carrying three values at once: χ = −1, μr = 0, Binside = 0. Whichever the question asks for, the other two come free.
Answer: (B) Bs = 0
The magnetic moment of an atom of a diamagnetic material, in the absence of an external field, is
Given
A diamagnetic atom with no field applied.
Asked
Its magnetic moment.
Concept applied
Diamagnetism is defined by the complete cancellation of the atom's internal orbital and spin contributions. With nothing left over, the atom has no permanent moment at all.
Formula to use
∑μelectrons = 0 for a diamagnetic atom
Baby steps
- Each electron in an atom contributes an orbital and a spin magnetic moment.
- In a diamagnetic atom these contributions cancel one another completely.
- So the resultant atomic moment is zero when no field is applied.
- Apply a field and a moment is induced, but it is opposite to the field — hence the repulsion.
- NCERT: diamagnetic substances are the ones in which resultant magnetic moment in an atom is zero.
- Option (D), one Bohr magneton, is the smallest non-zero orbital moment — a different quantity entirely.
Assumption
No external field present at the moment of asking.
Shortcut trick
Distinguish permanent from induced. A diamagnetic atom has no permanent moment but does acquire an induced one in a field. Questions specify 'in the absence of a field' precisely to test this.
Answer: (A) zero
C · Temperature, Curie's law and domains
9 questionsAccording to Curie's law, the magnetic susceptibility of a paramagnetic substance at absolute temperature T is proportional to
Given
Paramagnetic substance at absolute temperature T.
Asked
The dependence of χ on T.
Concept applied
Thermal agitation opposes alignment, so raising T weakens the response. Curie's law states this as a simple inverse proportion.
Formula to use
χ = C/T ⇒ χ ∝ 1/T
Baby steps
- Paramagnetic atoms carry permanent moments that the field attempts to align.
- Thermal motion randomises them, and this randomising effect grows with temperature.
- So the fraction aligned — and therefore χ — falls as T rises.
- Curie's law makes it quantitative: χ = C/T, so χ ∝ 1/T.
- T must be the absolute temperature in kelvin.
- Note: this law is not in the rationalised NCERT text, though the qualitative dependence is.
Assumption
Weak field, well below saturation, so the linear relation M = χH holds.
Shortcut trick
Always convert to kelvin first. A question quoting 27°C and 227°C means 300 K and 500 K, and the ratio 300/500 is nothing like 27/227.
Answer: (B) 1/T
Nickel is ferromagnetic at room temperature. If its temperature is raised beyond the Curie temperature, it becomes
Given
Nickel heated above its Curie temperature.
Asked
Its resulting magnetic behaviour.
Concept applied
Heat destroys the cooperative domain structure but not the individual atomic moments. Randomly oriented permanent moments is precisely paramagnetism.
Formula to use
T > TC: domains disintegrate, ferro → para
Baby steps
- Ferromagnetism arises from domains — large groups of atoms whose moments are spontaneously aligned.
- Above the Curie temperature, thermal agitation overcomes the interaction holding a domain together.
- The domain structure disintegrates.
- But each atom keeps its own permanent moment; only the cooperation is lost.
- Randomly oriented permanent moments that partly align in a field is the definition of paramagnetism.
- NCERT states it directly: at high enough temperature, a ferromagnet becomes a paramagnet.
Assumption
No chemical change on heating; only the magnetic ordering is affected.
Shortcut trick
This exact question appears as AIPMT 2006, 2007 and 2008 — three years running, phrased as 'Curie temperature above which...'. The answer is always ferromagnetic becomes paramagnetic, never diamagnetic and never χ = 0.
Answer: (B) paramagnetic
For which type of material does the magnetic susceptibility not depend on temperature?
Given
Comparison of the temperature dependence of χ across the classes.
Asked
Which class is temperature independent.
Concept applied
Temperature disrupts order. Diamagnetism involves no ordered arrangement of permanent moments — the response is induced by the field itself — so heat has nothing to disturb.
Formula to use
χdia ≈ constant; χpara, χferro vary with T
Baby steps
- Paramagnetism depends on aligning permanent moments against thermal agitation ⇒ strongly temperature dependent.
- Ferromagnetism depends on domain order, which collapses at the Curie point ⇒ strongly temperature dependent.
- Diamagnetic atoms have zero resultant moment to begin with.
- Their response is induced by the applied field through changes in electron orbital motion.
- There is no ordered structure for heat to disrupt, so χ is essentially temperature independent.
- Answer: diamagnetic.
Assumption
Ordinary temperature ranges, excluding the superconducting transition.
Shortcut trick
Ask what heat can destroy. Heat destroys order. No order to start with ⇒ no temperature dependence. JEE Main 2023 (12 April) tests the converse and marks 'diamagnetic property depends on temperature' as false.
Answer: (C) diamagnetic
The variation of susceptibility χ with absolute temperature T for a paramagnetic material is best represented by
Given
Paramagnetic material; χ plotted against T.
Asked
The correct graph shape.
Concept applied
Two independent checks settle it: the sign of χ fixes which side of the axis the curve lies on, and Curie's law fixes the direction of the slope.
Formula to use
χ = C/T ⇒ positive, decreasing, asymptotic to zero
Baby steps
- Sign check: paramagnetic means χ > 0 at every temperature, so the curve lies above the T-axis. Eliminate option (D).
- Trend check: χ = C/T falls as T rises. Eliminate options (A) and (C).
- Shape check: the fall is a reciprocal, steep at low T and flattening at high T, approaching but never reaching zero.
- So the graph is a curve above the T-axis falling towards it.
- Plotted against 1/T instead, the same law gives a straight line through the origin of slope C.
- JEE Main 2021 (1 Sept Evening) asks the mirror-image version, matching M–H and χ–T plots to a diamagnetic material.
Assumption
Weak field; the sample remains paramagnetic across the range.
Shortcut trick
Graph questions have two independent filters: which side of the axis (sign), and which way it slopes (physics). Apply both before considering curvature — it usually leaves one option standing.
Diagram
Answer: (B) a curve lying above the T-axis and falling towards it
A paramagnetic material has susceptibility 2.8 × 10−4 at 350 K. Its susceptibility at 300 K is about
Given
χ1 = 2.8 × 10−4 at T1 = 350 K; T2 = 300 K
Asked
χ2
Concept applied
Curie's law gives χT = C, a constant, so the two states are linked by a simple inverse proportion. There is no need to compute C.
Formula to use
χ1T1 = χ2T2
Baby steps
- From χ = C/T, the product χT is the same at both temperatures.
- χ2 = χ1(T1/T2).
- Substitute: χ2 = 2.8 × 10−4 × (350/300).
- 350/300 = 7/6 ≈ 1.1667.
- χ2 = 2.8 × 1.1667 × 10−4.
- χ2 ≈ 3.267 × 10−4.
Assumption
Curie's law holds across this range; the sample stays paramagnetic and unsaturated.
Shortcut trick
Direction check before arithmetic: the sample is being cooled, so alignment improves and χ must rise. That single check eliminates options (A), (C) and (D) instantly.
Answer: (B) 3.267 × 10−4
A paramagnetic sample shows magnetisation 6 A m−1 in a field of 0.4 T at 4 K. In a field of 0.3 T at 24 K, its magnetisation will be
Given
M1 = 6 A m−1 at B1 = 0.4 T, T1 = 4 K; then B2 = 0.3 T, T2 = 24 K
Asked
The new magnetisation M2.
Concept applied
Combining M = χH with χ = C/T gives M ∝ B/T. Apply the two proportionality factors separately and multiply.
Formula to use
M ∝ B/T ⇒ M2 = M1(B2/B1)(T1/T2)
Baby steps
- Field factor: B2/B1 = 0.3/0.4 = 0.75, which reduces M.
- Temperature factor: T1/T2 = 4/24 = 1/6, which also reduces M.
- Multiply the factors: 0.75 × (1/6) = 0.125.
- M2 = 6 × 0.125.
- M2 = 0.75 A m−1.
- Both changes push the same way — weaker field and higher temperature both reduce alignment — so the answer must be well below 6. ✔
Assumption
Linear regime, far from saturation, at both conditions.
Shortcut trick
Build a two-column factor table: what helps, what hinders. Here both changes hinder, so the answer must be much smaller than 6 — which rules out options (C) and (D) before any multiplication.
Answer: (A) 0.75 A m−1
Statement I: Susceptibilities of paramagnetic and ferromagnetic substances increase as temperature decreases. Statement II: Diamagnetism results from orbital motions of electrons developing moments opposite to the applied field. Then
Given
Two statements on the origins and temperature behaviour of magnetism.
Asked
Their truth values.
Concept applied
Statement I is Curie behaviour; Statement II is NCERT's explanation of diamagnetism. Both are correct, though they concern different classes and are not causally linked.
Formula to use
χpara ∝ 1/T; diamagnetism: induced moment opposes B
Baby steps
- Statement I: cooling reduces thermal randomisation, so a larger fraction of moments align and χ rises. True for paramagnets, and for ferromagnets below the Curie point. True.
- Statement II: NCERT explains that when a field is applied, electrons with orbital moment along the field slow down and those opposite speed up, leaving a net moment opposite to the field. True.
- Both statements are therefore correct.
- Answer: both statements are true.
- Note they describe different material classes, so II does not explain I — but this question asks only for truth values.
- Statement II is also the reason diamagnetism is temperature independent: there is no alignment for heat to disturb.
Assumption
Ordinary temperature ranges; ferromagnet below its Curie point.
Shortcut trick
Two true statements about different materials is a common pattern. If the option set includes 'both true but II does not explain I', check the causal link carefully — here it would be the right answer for that version of the question.
Answer: (A) Both statements are true
Assertion: Atoms in a ferromagnetic material possess magnetic dipole moments and interact so as to spontaneously align, forming domains. Reason: At high enough temperature the domain structure disintegrates and magnetisation disappears at the Curie temperature. Then
Given
An assertion about domain formation and a reason about domain destruction by heat.
Asked
The correct combination.
Concept applied
Both statements are true, but they describe opposite processes. The reason explains what destroys domains; it does not explain why they form in the first place.
Formula to use
domains form from exchange interaction; heat destroys them at TC
Baby steps
- Test the assertion: NCERT says ferromagnetic atoms possess dipole moments and interact so as to spontaneously align over a macroscopic volume called a domain. True.
- Test the reason: NCERT says the domain structure disintegrates with temperature, and at high enough temperature a ferromagnet becomes a paramagnet. True.
- Test the link: the assertion is about why domains exist — the interatomic interaction.
- The reason is about what destroys them — thermal energy.
- Destruction does not explain formation, so R does not explain A.
- Answer: option (B).
Assumption
Standard assertion–reason marking scheme.
Shortcut trick
A true reason is not automatically an explanation. Ask the sharp question: does R answer the 'why' of A? Here A asks why domains form and R answers why they break — opposite directions, so the link fails.
Answer: (B) Both true but R does not explain A
In a ferromagnetic material below the Curie temperature, a domain is
Given
A ferromagnetic material below its Curie temperature.
Asked
The definition of a domain.
Concept applied
A domain is a cooperative, macroscopic phenomenon — roughly a millimetre across and containing about 1011 atoms, all of whose moments point the same way without any applied field.
Formula to use
domain ≈ 1 mm across, ≈ 1011 atoms, spontaneously aligned
Baby steps
- In a ferromagnet the atoms interact so as to spontaneously align over a macroscopic volume.
- That volume is called a domain.
- NCERT gives the scale: typical domain size is about 1 mm and it contains roughly 1011 atoms.
- Each domain has its own net magnetisation, even with no applied field.
- Fresh iron appears unmagnetised only because the domains point in random directions and cancel.
- Applying a field does two things at once: the domains rotate towards the field, and those already aligned grow at their neighbours' expense — the point tested by JEE Main 2021 (24 Feb).
Assumption
Temperature below the Curie point, so domains exist at all.
Shortcut trick
Note the scale carefully: a domain is macroscopic, not atomic. Option (A) is the trap for anyone who thinks of magnetism as a per-atom property — which it is for paramagnets, but not for ferromagnets.
Answer: (B) a macroscopic region in which atomic moments are spontaneously aligned
D · Susceptibility, permeability and magnetic intensity
9 questionsAn iron rod of susceptibility 599 is placed in a magnetising field of 1200 A m−1. The permeability of the material is (μ0 = 4π × 10−7 T m A−1)
Given
χ = 599; H = 1200 A m−1; μ0 = 4π × 10−7
Asked
The permeability μ.
Concept applied
Permeability follows from susceptibility in two steps: add 1 to get μr, then multiply by μ0. The value of H is not needed at all — it is a decoy.
Formula to use
μr = 1 + χ; μ = μ0μr
Baby steps
- Compute μr = 1 + 599 = 600.
- μ = μ0μr = 4π × 10−7 × 600.
- 4 × 600 = 2400, so μ = 2400π × 10−7.
- Rewrite: 2400 × 10−7 = 2.4 × 10−4.
- μ = 2.4π × 10−4 T m A−1.
- The given H = 1200 A m−1 plays no part — permeability is a material property, independent of the applied field.
Assumption
Linear material at this field level.
Shortcut trick
Watch for decoy data. NEET routinely supplies a value that is not needed, to see whether you know which quantities the formula actually requires. If a given number never appears in your working, that is often correct rather than a mistake.
Answer: (C) 2.4π × 10−4 T m A−1
The magnetic susceptibility of a rod is 499. With μ0 = 4π × 10−7 H m−1, the absolute permeability of the rod material is
Given
χ = 499; μ0 = 4π × 10−7 H m−1
Asked
The absolute permeability μ.
Concept applied
Identical structure to the previous question. The examiners choose χ values just below a round number so that adding 1 gives a clean multiplier.
Formula to use
μ = μ0(1 + χ)
Baby steps
- μr = 1 + 499 = 500.
- μ = 4π × 10−7 × 500.
- 4 × 500 = 2000, so μ = 2000π × 10−7.
- 2000 × 10−7 = 2 × 10−4.
- μ = 2π × 10−4 H m−1.
- JEE Main 2022 (27 June) asks the same with χ = 99, giving μ = 4π × 10−5 Wb A−1 m−1.
Assumption
Linear material; H m−1 and T m A−1 are the same unit.
Shortcut trick
Values like 99, 499 and 599 are chosen deliberately: +1 turns them into 100, 500 and 600. Seeing a susceptibility ending in 9 is a strong hint that the +1 is the point of the question.
Answer: (A) 2π × 10−4 H m−1
A paramagnetic cube of side 1 cm has magnetic dipole moment 20 × 10−6 J T−1 when a magnetic intensity of 60 × 103 A m−1 is applied. Its susceptibility is
Given
Side = 1 cm = 10−2 m; m = 20 × 10−6 J T−1; H = 60 × 103 A m−1
Asked
The magnetic susceptibility χ.
Concept applied
Three stages. Volume from the side, magnetisation as moment per unit volume, and susceptibility as the ratio M/H.
Formula to use
V = a3; M = m/V; χ = M/H
Baby steps
- Volume: V = (10−2)3 = 10−6 m3.
- Magnetisation: M = m/V = 20 × 10−6 / 10−6 = 20 A m−1.
- Susceptibility: χ = M/H = 20 / (60 × 103).
- 20/60 = 0.333, so χ = 0.333 × 10−3.
- χ = 3.3 × 10−4, dimensionless.
- Sanity check: small and positive, consistent with a paramagnetic material. ✔
Assumption
Uniformly magnetised sample; linear response.
Shortcut trick
Cube the side first. Writing (10−2)3 = 10−6 and not 10−5 is the whole question — option (B) is exactly what an exponent slip produces.
Answer: (A) 3.3 × 10−4
The space inside a current-carrying solenoid is filled with a material of susceptibility 1.2 × 10−5. The fractional increase in the magnetic field inside, relative to air, is
Given
χ = 1.2 × 10−5; solenoid filled with the material, current unchanged.
Asked
The fractional increase in B.
Concept applied
With the current fixed, B increases by the factor μr = 1 + χ. The fractional increase is therefore χ itself — the 1 represents the field that was already there.
Formula to use
ΔB/B0 = (B − B0)/B0 = μr − 1 = χ
Baby steps
- Without the material: B0 = μ0H.
- With the material: B = μ0μrH = μ0(1 + χ)H.
- Increase: ΔB = B − B0 = μ0χH.
- Fractional increase: ΔB/B0 = μ0χH / μ0H = χ.
- So the fractional increase equals 1.2 × 10−5.
- As a percentage it would be 1.2 × 10−3% — which is exactly what JEE Main 2025 (7 April) asks for with magnesium of the same χ.
Assumption
Current held constant so H does not change; linear material.
Shortcut trick
Fractional increase = χ. Percentage increase = 100χ. Option (B) here is the percentage answer. Read which one the question wants — JEE has asked both versions with identical data.
Answer: (A) 1.2 × 10−5
The free space inside a current-carrying toroid is filled with a material of susceptibility 2 × 10−2. The percentage increase in the magnetic field inside the toroid is
Given
χ = 2 × 10−2; toroid filled with the material.
Asked
The percentage increase in B.
Concept applied
Same relation as the previous question, now expressed as a percentage. Multiply the fractional increase by 100.
Formula to use
percentage increase = χ × 100
Baby steps
- Fractional increase in B equals χ.
- χ = 2 × 10−2 = 0.02.
- Percentage increase = 0.02 × 100.
- = 2%.
- Note the geometry is irrelevant — solenoid or toroid, the relation B = μ0(1 + χ)H is the same.
- JEE Main 2021 (25 July) asks the identical question for aluminium with χ = 2.2 × 10−5.
Assumption
Current held constant; linear material; the toroid is uniformly filled.
Shortcut trick
The container does not matter. Solenoid, toroid, or any uniformly filled geometry — if the current is fixed, filling multiplies B by μr. Do not waste time on the toroid geometry.
Answer: (B) 2%
The relationship between magnetic susceptibility χ and magnetic permeability μ is
Given
μ0 the permeability of free space, μr the relative permeability.
Asked
The relation linking μ and χ.
Concept applied
Derive it rather than recall it. Start from B = μ0(H + M), substitute M = χH, and the bracket becomes (1 + χ).
Formula to use
μ = μ0μr = μ0(1 + χ)
Baby steps
- Start from B = μ0(H + M).
- Substitute M = χH: B = μ0(H + χH) = μ0(1 + χ)H.
- But by definition B = μH.
- Comparing the two: μ = μ0(1 + χ).
- Equivalently μr = μ/μ0 = 1 + χ.
- Option (A) drops the 1, which would wrongly give μ = 0 for vacuum where χ = 0.
Assumption
Linear isotropic material.
Shortcut trick
The +1 is the vacuum's own contribution and never disappears, however large χ becomes. Test any candidate relation by setting χ = 0: it must give μ = μ0. Only option (B) survives that test.
Answer: (B) μ = μ0(1 + χ)
A solenoid of 1200 turns is wound uniformly on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at its centre when a current of 2 A flows is
Given
N = 1200 turns; length = 2 m; diameter = 0.2 m; I = 2 A
Asked
The magnetic intensity H at the centre.
Concept applied
H = nI, where n is turns per metre. The diameter is a decoy — H does not depend on the solenoid's cross-section.
Formula to use
H = nI = (N/L)I
Baby steps
- Turns per metre: n = N/L = 1200/2 = 600 m−1.
- H = nI = 600 × 2.
- H = 1200 A m−1.
- The diameter of 0.2 m plays no part — H depends only on turn density and current.
- Note the coincidence that the answer equals the total turn count; that is accidental, not a shortcut.
- Option (D) has units of tesla, so it is B rather than H — a units-level distractor.
Assumption
Long solenoid, closely and uniformly wound, so the interior field is uniform.
Shortcut trick
Check the units in the options. If one option is in tesla and the question asks for magnetic intensity, that option is wrong regardless of its number. H is always in A m−1.
Answer: (B) 1200 A m−1
The magnetic intensity at the centre of a long solenoid is 1.6 × 103 A m−1. If the winding is 8 turns per cm, the current through the solenoid is
Given
H = 1.6 × 103 A m−1; n = 8 turns per cm
Asked
The current I.
Concept applied
Rearrange H = nI. The critical step is converting turns per centimetre into turns per metre before substituting.
Formula to use
I = H/n
Baby steps
- Convert the turn density: 8 turns per cm = 8 × 100 = 800 turns per metre.
- Rearrange: I = H/n.
- I = 1.6 × 103 / 800.
- 1600/800 = 2.
- I = 2 A.
- Skipping the cm → m conversion gives 200 A — option (D), planted for exactly that error.
Assumption
Long solenoid so H = nI applies at the centre.
Shortcut trick
Turns per cm × 100 = turns per metre. Convert before substituting, every time. This single conversion accounts for more lost marks in solenoid questions than the physics does.
Answer: (A) 2 A
The magnetic potential due to a dipole at an axial point 20 cm from its centre is 1.5 × 10−5 T m. Given μ0/4π = 10−7 T m A−1, the magnetic moment is
Given
V = 1.5 × 10−5 T m at r = 0.20 m on the axis; μ0/4π = 10−7
Asked
The magnetic moment m.
Concept applied
The magnetic scalar potential of a dipole on its axis falls as 1/r2, one power slower than the field. Rearranging for m is a single step.
Formula to use
V = (μ0/4π)(m/r2) ⇒ m = V r2 / (μ0/4π)
Baby steps
- Square the distance: r2 = (0.20)2 = 4 × 10−2 m2.
- Numerator: V r2 = 1.5 × 10−5 × 4 × 10−2 = 6 × 10−7.
- Divide by μ0/4π = 10−7.
- m = 6 × 10−7 / 10−7.
- m = 6 A m2.
- Note the potential goes as 1/r2 while the field goes as 1/r3 — using r3 here would give 30 A m2, not among the options.
Assumption
Short dipole; axial point; magnetic scalar potential defined as in the standard treatment.
Shortcut trick
Keep the exponent ladder straight: potential 1/r2, field 1/r3, dipole–dipole force 1/r4. Each differentiation adds one power. Picking the wrong rung is the only real hazard here.
Answer: (A) 6 A m2
E · Bent magnets, combinations and current loops
8 questionsA bar magnet of length l and magnetic dipole moment M is bent into the form of a semicircular arc. Its new magnetic dipole moment is
Given
Magnet of length l and moment M, bent into a semicircle without change of length.
Asked
The new dipole moment.
Concept applied
Magnetic moment is pole strength times the straight-line distance between the poles. Bending preserves the pole strength but shortens that distance from the arc length to the chord.
Formula to use
M = qml; after bending M′ = qm × (chord)
Baby steps
- Original: M = qml, with the poles a straight-line distance l apart.
- After bending, the material length l becomes the arc of a semicircle: l = πR, so R = l/π.
- The poles now sit at opposite ends of a diameter, a straight-line distance 2R apart.
- 2R = 2l/π.
- Pole strength qm is unchanged, so M′ = qm(2l/π) = 2M/π.
- Numerically 2/π ≈ 0.64, so the moment falls to about 64% of its original value.
Assumption
Uniform thin magnet; bending does not disturb the magnetisation.
Shortcut trick
Every bending question is chord ÷ original length. Semicircle gives 2/π; quarter circle gives 2√2/π ≈ 0.90; full circle gives 0. Compute the chord and the rest is a ratio.
Answer: (B) 2M/π
An iron bar of magnetic moment M is bent into an arc subtending 60° at the centre of the circle. Its new magnetic moment is
Given
Magnet of moment M bent into an arc subtending θ = 60° = π/3 radians.
Asked
The new magnetic moment.
Concept applied
Apply the general arc result. The chord of an arc subtending θ is 2R sin(θ/2), and the fixed arc length gives R = L/θ.
Formula to use
M′ = M × 2 sin(θ/2)/θ, θ in radians
Baby steps
- Convert to radians: θ = 60° = π/3.
- Half angle: θ/2 = π/6 = 30°, and sin 30° = 1/2.
- Numerator: 2 sin(θ/2) = 2 × 1/2 = 1.
- Denominator: θ = π/3.
- M′ = M × 1 / (π/3) = 3M/π.
- Check the trend: 3/π ≈ 0.95, larger than the semicircle's 0.64, as a gentler bend should give. ✔
Assumption
θ in radians; uniform arc; pole strength unchanged.
Shortcut trick
Convert degrees to radians before substituting into 2sin(θ/2)/θ. Leaving θ in degrees in the denominator is the standard failure and gives a nonsense answer.
Diagram
Answer: (A) 3M/π
An iron bar of length L has magnetic moment M. It is bent at the middle so that the two arms make 60° with each other. The magnetic moment of the new magnet is
Given
Bar of moment M bent at its midpoint; arms at 60°.
Asked
The resultant moment.
Concept applied
Bending at the midpoint gives two half-magnets of moment M/2 each. Their moment vectors run head to tail along the arms, so the angle between the vectors is 180° minus the angle between the arms.
Formula to use
M′ = 2(M/2)cos[(180°−θ)/2] = M sin(θ/2)
Baby steps
- Each half has moment M/2, since the length halves and the pole strength is unchanged.
- The two moment vectors are separated by 180° − 60° = 120°.
- Combine two equal moments of M/2 at 120°: M′ = 2(M/2)cos 60°.
- cos 60° = 0.5, so M′ = M × 0.5.
- M′ = M/2.
- General result: M′ = M sin(θ/2), with θ the angle between the arms. Check: θ = 180° gives M ✔, θ = 0° gives 0 ✔.
Assumption
Sharp bend at the exact midpoint; each arm retains uniform magnetisation.
Shortcut trick
Note the two different bending questions from the same 2024 paper: an arc of 60° gives 3M/π, but a midpoint bend of 60° gives M/2. Arc and fold are different geometries — read which one is described.
Answer: (B) M/2
Several arrangements of bar magnets are shown, each magnet having dipole moment m. The configuration with the highest net magnetic dipole moment is the one in which the individual moments are
Given
Several configurations of identical bar magnets, each of moment m.
Asked
Which arrangement has the largest net moment.
Concept applied
Magnetic moments add as vectors. The resultant is maximum when all vectors point the same way, since every cross term is then at its most positive.
Formula to use
|∑mi| is maximum when all mi are parallel
Baby steps
- For n identical moments all parallel: the resultant is nm, the largest possible value.
- Alternating directions: adjacent moments cancel in pairs, giving 0 or m depending on whether n is even or odd.
- Closed loop head to tail: the vectors form a closed polygon and sum to zero — the same reason a toroid has no poles.
- Mutually perpendicular: the resultant is m√2 for two, less than 2m.
- So the maximum is the all-parallel arrangement.
- Answer: option (A).
Assumption
Identical magnets, rigidly held, moments acting at a common point for the vector sum.
Shortcut trick
For figure-based configuration questions, redraw the moments as arrows and look for the arrangement where no arrow opposes another. Closed-loop arrangements always give zero and are the most common distractor.
Answer: (A) all parallel and pointing the same way
A conducting wire of length 12a is wound as a coil in the shape of (i) an equilateral triangle of side a and (ii) a square of side a. The magnetic dipole moments in the two cases are respectively
Given
Wire of total length 12a carrying current I, wound as (i) a triangle of side a, (ii) a square of side a.
Asked
The magnetic moments in the two cases.
Concept applied
Fixed wire length means the number of turns is set by the perimeter. Compute turns first, then area, then m = NIA.
Formula to use
N = (total length)/(perimeter); m = NIA
Baby steps
- Triangle: perimeter = 3a, so N = 12a/3a = 4 turns.
- Area of an equilateral triangle of side a: A = (√3/4)a2.
- mtriangle = 4 × I × (√3/4)a2 = √3 I a2.
- Square: perimeter = 4a, so N = 12a/4a = 3 turns.
- Area of a square of side a: A = a2. msquare = 3 × I × a2 = 3 I a2.
- So the pair is √3 I a2 and 3 I a2 — the square wins, since √3 ≈ 1.73 < 3.
Assumption
All the wire is used in each case; turns are coplanar and concentric; same current in both.
Shortcut trick
For a fixed wire length, fewer sides means more turns but the area formula matters more. Work in the fixed order: perimeter → turns → area → moment. Skipping to the area first loses track of the turn count.
Answer: (A) √3 I a2 and 3 I a2
A coil in the shape of an equilateral triangle of side l is suspended between the poles of a magnet with B in the plane of the coil. A current i produces a torque τ. The side l is
Given
Equilateral triangular coil of side l, current i, field B in the plane of the coil, torque τ.
Asked
The side l.
Concept applied
When B lies in the plane of the coil, the normal is perpendicular to B, so sin θ = 1 and the torque is at its maximum value iAB.
Formula to use
τ = iAB with A = (√3/4)l2
Baby steps
- B in the plane of the coil ⇒ angle between the normal and B is 90° ⇒ sin θ = 1.
- So τ = iAB, with A the triangle's area.
- Substitute A = (√3/4)l2: τ = i(√3/4)l2B.
- Rearrange: l2 = 4τ/(√3 iB).
- Take the square root: l = 2(τ/√3 iB)1/2.
- Answer: option (A).
Assumption
Single-turn coil; uniform field; the field lies exactly in the coil's plane.
Shortcut trick
'B in the plane of the coil' means maximum torque, sin θ = 1. The complementary phrase 'B perpendicular to the coil' means zero torque. Translate the geometric phrase into an angle before writing anything else.
Answer: (A) 2(τ/√3 iB)1/2
A charged particle of charge q moves in a circle of radius R with uniform speed v. Its associated magnetic moment is
Given
Charge q in a circular orbit of radius R at speed v.
Asked
The associated magnetic moment.
Concept applied
A circulating charge is a current loop. Convert the orbital motion into an equivalent current, then apply m = IA.
Formula to use
I = q/T = qv/2πR; μ = IA = IπR2
Baby steps
- Period of one revolution: T = 2πR/v.
- Equivalent current: I = q/T = qv/(2πR).
- Area enclosed: A = πR2.
- μ = IA = [qv/(2πR)] × πR2.
- Cancel π and one power of R: μ = qvR/2.
- Direction: perpendicular to the orbital plane, by the right-hand rule applied to the conventional current.
Assumption
Uniform circular motion; single charge; non-relativistic speeds.
Shortcut trick
The chain is always orbit → current → moment, and the 1/2 survives because πR2/(2πR) = R/2. Option (A) is what you get by forgetting the 2π in the period.
Answer: (B) qvR/2
Two identical small bar magnets, each of dipole moment 3√5 J T−1, are placed with their centres 10 cm apart and their axes perpendicular to each other. The magnetic field at the midpoint P between them is closest to
Given
m1 = m2 = 3√5 J T−1; centre-to-centre separation 10 cm, so each magnet is 5 cm from P; axes mutually perpendicular; μ0/4π = 10−7
Asked
The resultant field at the midpoint.
Concept applied
P lies on the axis of one magnet and on the equatorial line of the other, because their axes are perpendicular. Compute the two contributions separately, then add them as perpendicular vectors.
Formula to use
Bax = (μ0/4π)(2m/r3); Beq = (μ0/4π)(m/r3); BR = √(Bax2 + Beq2)
Baby steps
- Distance from each magnet to P: r = 5 cm = 0.05 m, so r3 = 1.25 × 10−4 m3.
- Common factor: (μ0/4π)(m/r3) = 10−7 × 3√5 / 1.25 × 10−4 = 10−7 × 6.708/1.25 × 10−4 ≈ 5.37 × 10−4 T.
- Axial contribution from one magnet: 2 × 5.37 × 10−4 = 1.073 × 10−3 T.
- Equatorial contribution from the other: 5.37 × 10−4 T.
- The two are perpendicular, so BR = √[(1.073×10−3)2 + (5.37×10−4)2] = 5.37×10−4√5.
- √5 ≈ 2.236, giving BR ≈ 1.2 × 10−3 T.
Assumption
Short dipoles compared with 5 cm; P exactly midway; the two contributions are mutually perpendicular.
Shortcut trick
The moment 3√5 is chosen so the √5 from combining a 2:1 pair of perpendicular contributions cancels neatly. An awkward surd in the data usually cancels in the answer — if yours does not, re-check the geometry.
Answer: (A) 1.2 × 10−3 T
F · Oscillation and the vibration magnetometer
7 questionsA thin rectangular magnet suspended freely has a period T. It is broken into two equal halves, each of half the original length, and one piece oscillates in the same field with period T′. The ratio T′/T is
Given
Magnet of period T broken transversely into two equal halves; one half oscillates in the same field.
Asked
T′/T
Concept applied
Both quantities in the period formula change. The moment halves, but the moment of inertia falls by a factor of eight, because both mass and length halve and I ∝ MbL2.
Formula to use
T = 2π√(I/mB), I = MbL2/12
Baby steps
- New mass Mb/2 and new length L/2.
- New moment of inertia: I′ = (Mb/2)(L/2)2/12 = I/8.
- New magnetic moment: m′ = m/2 (length halves, pole strength unchanged).
- Ratio inside the root: I′/(m′B) = (I/8) ÷ (mB/2) = I/(4mB).
- T′ = 2π√[I/(4mB)] = (1/2)T.
- So T′/T = 1/2.
Assumption
Clean transverse break; each half remains a uniform bar; same field and suspension.
Shortcut trick
The trap is halving both I and m and concluding T is unchanged. I falls by 8, not 2, because I depends on MbL2 and both factors shrink. Always recompute I from MbL2/12.
Answer: (C) 1/2
A long thin magnet has a period of 2 s in a vibration magnetometer. It is cut along its length into three equal parts, and the parts are stacked with their like poles together. The period of this combination is
Given
Original period 2 s; cut lengthwise into three equal parts; parts stacked with moments parallel.
Asked
The period of the combination.
Concept applied
A lengthwise cut divides the mass and the pole strength but not the length. Restacking all three parts restores both the total moment and the total moment of inertia, so nothing changes.
Formula to use
lengthwise cut: I → I/3, m → m/3; stacking three restores both
Baby steps
- Each part after the lengthwise cut has mass Mb/3 and the full length L.
- So I′ = (Mb/3)L2/12 = I/3, and m′ = m/3 (pole strength scales with cross-section).
- Stacking three such parts with like poles together: Itotal = 3 × I/3 = I.
- Moments are parallel and add: mtotal = 3 × m/3 = m.
- The ratio I/(mB) is exactly the original, so T is unchanged.
- T = 2 s.
Assumption
Clean lengthwise cuts; the stack oscillates about the same axis; same field.
Shortcut trick
Cut it up and put it back together and you have the original magnet again — the answer should be obvious before any algebra. Look for the physical shortcut first; the formulas are only there to confirm it.
Answer: (C) 2 s
Two bar magnets of the same geometry, with moments M and 2M, are placed with their similar poles on the same side, giving period T1. The polarity of one magnet is then reversed, giving period T2. Then
Given
Moments M and 2M; same geometry so equal moments of inertia; like poles together gives T1, one reversed gives T2.
Asked
The relation between T1 and T2.
Concept applied
The moment of inertia is the same in both arrangements, so only the net magnetic moment changes: the sum in one case and the difference in the other.
Formula to use
T ∝ 1/√(mnet) ⇒ T2/T1 = √(msum/mdiff)
Baby steps
- Similar poles on the same side: moments add, msum = M + 2M = 3M.
- One reversed: moments subtract, mdiff = 2M − M = M.
- The moment of inertia Itotal is unchanged by flipping a magnet over.
- T ∝ 1/√(mnet), so T2/T1 = √(3M/M) = √3.
- T2 = √3 T1.
- Sense check: the reversed arrangement has a smaller net moment and therefore a weaker restoring torque, so it must oscillate more slowly. ✔
Assumption
Same geometry, so both magnets contribute equally to I; small amplitude.
Shortcut trick
The larger period always belongs to the difference arrangement, because the net moment is smaller there. Use that as a direction check before doing any arithmetic.
Answer: (B) T2 = √3 T1
A bar magnet oscillates in the Earth's field with period T. If its mass is quadrupled without changing its dimensions or magnetic moment, its period
Given
Mass × 4; dimensions and magnetic moment unchanged; same field.
Asked
The new period and the nature of the motion.
Concept applied
Moment of inertia is directly proportional to mass at fixed dimensions, and T ∝ √I. Quadrupling the mass therefore doubles the period, and the restoring torque is unaffected so the motion stays simple harmonic.
Formula to use
I ∝ Mb; T ∝ √I ⇒ T ∝ √Mb
Baby steps
- I = MbL2/12, so at fixed L, I is proportional to the mass.
- Quadrupling the mass quadruples I.
- T = 2π√(I/mB), and m and B are unchanged, so T ∝ √I.
- T′/T = √4 = 2, giving T′ = 2T.
- The restoring torque τ = −mB sin θ has not changed, so it is still proportional to θ for small angles.
- The motion therefore remains simple harmonic — just slower.
Assumption
Dimensions and magnetic moment genuinely unchanged, so only I varies.
Shortcut trick
Two-part options need both parts checked. Many students get 2T right and then pick an option that wrongly claims the motion stops being SHM. SHM depends on the restoring torque, not on the inertia.
Answer: (A) becomes 2T and the motion remains simple harmonic
A vibration magnetometer in the magnetic meridian has a small bar magnet with period 2 s in the Earth's horizontal field of 24 μT. A field of 18 μT is then produced opposite to the Earth's field. The new period is
Given
T1 = 2 s in B1 = 24 μT; an opposing field of 18 μT is applied.
Asked
The new period.
Concept applied
The applied field opposes the Earth's field, so the net field is the difference. Then use T ∝ 1/√B.
Formula to use
Bnet = 24 − 18 = 6 μT; T2 = T1√(B1/B2)
Baby steps
- The two fields are antiparallel, so they subtract: B2 = 24 − 18 = 6 μT.
- T ∝ 1/√B, so T2/T1 = √(B1/B2).
- B1/B2 = 24/6 = 4.
- √4 = 2, so T2 = 2 × T1.
- T2 = 2 × 2 = 4 s.
- Sense check: a weaker net field means a weaker restoring torque and therefore a longer period. ✔
Assumption
The applied field is exactly antiparallel to the Earth's horizontal component and uniform over the magnet.
Shortcut trick
The word 'opposite' is the whole question. Subtract to get the net field before touching the period ratio. Adding instead gives 42 μT and a period of about 1.5 s — not among the options, which is itself a check.
Answer: (D) 4 s
Two bar magnets oscillate in the Earth's field with periods 3 s and 4 s. Their moments of inertia are in the ratio 3 : 2. The ratio of their magnetic moments is
Given
T1 = 3 s, T2 = 4 s; I1/I2 = 3/2; same field B.
Asked
m1/m2
Concept applied
Square the period formula and rearrange for m. The field cancels between the two magnets, leaving a relation among I and T only.
Formula to use
m = 4π2I/(T2B) ⇒ m1/m2 = (I1/I2)(T22/T12)
Baby steps
- From T = 2π√(I/mB), squaring gives T2 = 4π2I/(mB), so m = 4π2I/(T2B).
- Form the ratio, with B and 4π2 cancelling: m1/m2 = (I1/I2) × (T22/T12).
- I1/I2 = 3/2.
- T22/T12 = 16/9.
- m1/m2 = (3/2) × (16/9) = 48/18 = 8/3.
- Ratio = 8 : 3.
Assumption
Both magnets in the same field; small amplitude.
Shortcut trick
Note that the period ratio inverts when it moves to the moment: T22 goes with m1. Write the rearranged formula out before substituting rather than juggling the inversion in your head.
Answer: (A) 8 : 3
A magnetic needle of moment 1.0 × 10−2 A m2 and moment of inertia 10−6/π2 kg m2 completes 10 oscillations in 10 s. The magnitude of the magnetic field is
Given
m = 1.0 × 10−2 A m2; I = 10−6/π2 kg m2; 10 oscillations in 10 s
Asked
The magnetic field B.
Concept applied
Get the period from the oscillation count, then rearrange the period formula for B. The π2 in the given moment of inertia is placed there to cancel the 4π2.
Formula to use
T = (total time)/(number of oscillations); B = 4π2I/(T2m)
Baby steps
- Period: T = 10 s / 10 oscillations = 1 s.
- Rearrange T = 2π√(I/mB) for B: B = 4π2I/(T2m).
- Substitute I = 10−6/π2: the numerator becomes 4π2 × 10−6/π2 = 4 × 10−6.
- Denominator: T2m = 1 × 1.0 × 10−2 = 10−2.
- B = 4 × 10−6 / 10−2.
- B = 4 × 10−4 T.
Assumption
Small amplitude; the quoted count is of complete oscillations.
Shortcut trick
When the data contains an odd-looking π2, it is there to cancel the 4π2 in the formula. Spot the cancellation and the arithmetic becomes trivial — NEET 2024 asks the same question with B = 0.049 T and 20 oscillations in 5 s.
Answer: (A) 4 × 10−4 T
G · Field lines, monopoles and forces
6 questionsThe magnetic lines of force inside a bar magnet
Given
The interior of a bar magnet.
Asked
The direction of the field lines inside it.
Concept applied
Magnetic field lines are closed loops. Outside the magnet they run N to S; to close the loop they must run S to N inside the material.
Formula to use
closed loops: N → S outside, S → N inside
Baby steps
- Magnetic field lines form continuous closed loops — NCERT property (i).
- Outside the magnet, the lines emerge from the north pole and enter the south pole.
- To complete the loop, they must travel back from S to N somewhere.
- That return path is through the magnet itself.
- So inside a bar magnet the lines run from south pole to north pole.
- Option (D) is wrong because lines cannot terminate at all — that would require magnetic monopoles.
Assumption
Static bar magnet; lines drawn in the conventional sense.
Shortcut trick
The whole of Gauss's law for magnetism follows from this one picture. If lines close through the magnet, then any closed surface has as many lines entering as leaving — hence zero net flux, whichever part of the magnet is enclosed.
Answer: (B) run from south pole to north pole
Assertion: Magnetic monopoles do not exist. Reason: Magnetic field lines are continuous and form closed loops. The most appropriate answer is
Given
An assertion on monopoles and a reason on closed field lines.
Asked
The correct combination.
Concept applied
The two facts are logically equivalent statements of the same physics. Closed loops mean no line has an endpoint, and a monopole would have to be an endpoint.
Formula to use
no endpoints ⇔ no monopoles ⇔ φB = 0
Baby steps
- Test the assertion: no isolated magnetic pole has ever been observed; cutting a magnet always gives two complete magnets. True.
- Test the reason: NCERT states that magnetic field lines form continuous closed loops. True.
- Test the link: a monopole would be a point from which lines emanate without returning — an endpoint.
- If every line is a closed loop with no endpoints, there is nowhere for a monopole to sit.
- So the closed-loop property does explain the absence of monopoles.
- Answer: option (A).
Assumption
Standard assertion–reason marking scheme.
Shortcut trick
Compare with the domain assertion in Group C, where the reason was true but did not explain the assertion. Run the causal test separately every time — two true statements is not enough to pick option (A).
Answer: (A) Both true and R correctly explains A
Which statements are correct? (A) Electric monopoles do not exist whereas magnetic monopoles exist. (B) Solenoid field lines at the ends cannot be completely straight and confined. (C) Field lines are completely confined within a toroid. (D) Field lines inside a bar magnet are not parallel. (E) χ = −1 is the condition for a perfect diamagnet.
Given
Five statements drawn from NCERT's treatment of field lines and materials.
Asked
Which are correct.
Concept applied
Statement A inverts reality — electric monopoles (charges) do exist while magnetic ones do not. The remaining four are all straight from NCERT.
Formula to use
charges exist; magnetic poles do not; χ = −1 for a perfect diamagnet
Baby steps
- (A) Electric monopoles are simply isolated charges, which certainly exist; magnetic monopoles do not. The statement is exactly backwards. Incorrect.
- (B) Example 5.3(d): solenoid lines at the ends cannot be so completely straight and confined, as that violates Ampere's law. Correct.
- (C) Example 5.3(c): magnetic lines are completely confined within a toroid. Correct.
- (D) Example 5.3(e) notes that not all lines emanate from the north pole; the interior pattern is not simply parallel. Correct.
- (E) Perfect diamagnetism means total flux expulsion, which requires χ = −1. Correct.
- So B, C, D and E are correct — option (B).
Assumption
NCERT's Example 5.3 conventions.
Shortcut trick
In five-statement questions, hunt for the one that is reversed rather than merely wrong. Examiners often build exactly one inverted statement, and finding it usually identifies the answer immediately.
Answer: (B) B, C, D, E only
A current loop placed in a uniform magnetic field experiences
Given
A current loop in a uniform magnetic field.
Asked
The force and torque on it.
Concept applied
A current loop is a magnetic dipole. In a uniform field the forces on opposite sides are equal and opposite and cancel, but they form a couple, so a torque survives.
Formula to use
Fnet = 0; τ = m × B ≠ 0
Baby steps
- The loop behaves as a magnetic dipole of moment m = NIA.
- In a uniform field, forces on opposite sides of the loop are equal in magnitude and opposite in direction.
- Their vector sum is zero, so the net force is zero.
- But the two forces do not act along the same line, so they constitute a couple.
- The resulting torque is τ = mB sin θ, non-zero except at θ = 0° and 180°.
- Answer: a torque but no net force.
Assumption
The field is genuinely uniform across the loop.
Shortcut trick
Uniform ⇒ turn only. Non-uniform ⇒ turn and tug. AIEEE 2005 asks the complementary question — a magnetic needle in a non-uniform field experiences both a force and a torque.
Answer: (B) a torque but no net force
A magnetic needle placed in a non-uniform magnetic field experiences
Given
A magnetic needle in a non-uniform field.
Asked
The force and torque acting on it.
Concept applied
The torque exists in any field; the extra ingredient of non-uniformity is that the two poles now sit in fields of different strength, so the forces no longer cancel.
Formula to use
τ = mB sin θ ≠ 0; F = m(dB/dx) ≠ 0
Baby steps
- Torque: τ = mB sin θ arises in any field, uniform or not, unless the needle is already aligned.
- Force: in a uniform field the equal and opposite pole forces cancel exactly.
- In a non-uniform field, the nearer pole sits where the field is stronger, so its force is larger.
- The cancellation is incomplete and a net force F = m(dB/dx) survives.
- So the needle experiences both a force and a torque.
- This is exactly NCERT's Example 5.1(b): the iron nail near a bar magnet feels both, because the magnet's own field is non-uniform.
Assumption
The needle is not already perfectly aligned, so a torque exists; the field gradient is non-zero.
Shortcut trick
This is the reason magnets stick to things at all. A uniform field could only ever turn an object; the attraction you feel picking up a paperclip comes entirely from the field gradient.
Answer: (C) both a force and a torque
Two identical bar magnets are fixed with their centres a distance d apart. A stationary charge Q is placed at a point P between them. The force on the charge is
Given
A stationary charge Q in the magnetic field produced by two bar magnets.
Asked
The force on the charge.
Concept applied
The magnetic force on a charge is qv × B. With v = 0, the cross product vanishes regardless of how strong the field is.
Formula to use
F = qv × B; v = 0 ⇒ F = 0
Baby steps
- The magnetic force on a charge is F = qv × B.
- The charge is stated to be stationary, so v = 0.
- A cross product with a zero vector is zero.
- Therefore F = 0, however strong the field at P may be.
- The magnitude of B at P is irrelevant — and computing it is exactly the trap the question sets.
- Answer: zero.
Assumption
No electric field present; the charge is genuinely at rest.
Shortcut trick
Scan for the word stationary before doing anything else. A whole geometry of magnets can be set up purely to disguise a one-word answer. Magnetic fields act only on moving charges.
Answer: (C) zero
H · Soft and hard materials, hysteresis
6 questionsElectromagnets are made of soft iron because soft iron has
Given
Soft iron used as the core of an electromagnet.
Asked
The properties that make it suitable.
Concept applied
An electromagnet must switch off on demand. That requires the magnetisation to vanish as soon as the current stops, which means low retentivity, and to be easily reversed, which means low coercivity. High permeability gives the strong field while the current flows.
Formula to use
soft: high μ, low retentivity, low coercivity, narrow loop
Baby steps
- An electromagnet must produce a strong field when energised ⇒ high permeability.
- It must lose that field when de-energised ⇒ low retentivity.
- It must be easy to demagnetise and reverse ⇒ low coercivity.
- Soft iron has exactly this combination, which is why NCERT names it as the classic soft ferromagnet.
- Answer: low retentivity and low coercivity.
- Permanent magnets need the opposite combination — high retentivity and high coercivity — which is what alnico and lodestone provide.
Assumption
Conventional electromagnet operation with a switchable current.
Shortcut trick
Match the property to the job. Electromagnet: must let go ⇒ soft. Permanent magnet: must hold on ⇒ hard. JEE Main 2023 (8 April) asks the same as an assertion–reason, where the answer is that R correctly explains A.
Answer: (B) low retentivity and low coercivity
Two hysteresis loops are given: material A has a narrow loop and material B a broad one. For a transformer core and for a permanent magnet, the proper choices are
Given
Material A with a narrow hysteresis loop; material B with a broad loop.
Asked
Which material suits a transformer core and which a permanent magnet.
Concept applied
Loop area is the energy lost per cycle. A transformer cycles fifty times a second, so it needs the smallest possible loop. A permanent magnet is never cycled and needs a large retentivity and coercivity, which means a broad loop.
Formula to use
loop area = energy lost per cycle per unit volume
Baby steps
- A transformer core is magnetised and demagnetised every AC cycle.
- Each cycle dissipates energy equal to the loop area, so a broad loop would waste energy continuously as heat.
- The core must also demagnetise easily, needing low coercivity.
- Narrow loop ⇒ material A for the transformer core.
- A permanent magnet must retain its magnetisation and resist demagnetisation, needing high retentivity and high coercivity.
- Broad loop ⇒ material B for the permanent magnet. Answer: option (C).
Assumption
Standard AC operation; other loss mechanisms such as eddy currents set aside.
Shortcut trick
Narrow loop = soft = switches, wastes little. Broad loop = hard = holds, stores a lot. Note that hysteresis has since been deleted from both the NEET and JEE Main syllabi, so this is recognition material only.
Answer: (C) A for the transformer core, B for the permanent magnet
The coercivity of a small magnet is 3 × 103 A m−1. The current that must be passed through a solenoid of length 10 cm with 100 turns, so that the magnet inside is demagnetised, is
Given
Coercivity Hc = 3 × 103 A m−1; solenoid length 0.10 m; N = 100 turns.
Asked
The current required for demagnetisation.
Concept applied
Coercivity is a magnetic intensity, so it equates directly to H = nI. Demagnetising means supplying exactly that H.
Formula to use
Hc = nI ⇒ I = Hc/n, with n = N/L
Baby steps
- Turns per metre: n = N/L = 100/0.10 = 1000 m−1.
- Set the solenoid's intensity equal to the coercivity: nI = Hc.
- I = Hc/n = 3 × 103 / 1000.
- I = 3 A.
- JEE Main 2024 (8 April) asks the same with Hc = 5 × 103 A m−1, L = 30 cm and N = 150, giving n = 500 and I = 10 A.
- JEE Main 2019 (9 January) runs it backwards: L = 0.2 m, N = 100, I = 5.2 A gives Hc = 500 × 5.2 = 2600 A m−1.
Assumption
Long solenoid so H = nI; the magnet sits inside where the field is uniform.
Shortcut trick
Coercivity is quoted in A m−1, the units of H — that is your signal to use H = nI and nothing else. No μ0 ever enters these problems, and if it appears in your working you have taken a wrong turn.
Answer: (A) 3 A
For permanent magnets (P) and magnets used in a transformer (T), the best matching of properties is
Given
Two applications: a permanent magnet and a transformer core.
Asked
The matching of magnetic properties.
Concept applied
The two applications have opposite requirements. One must hold its magnetisation indefinitely; the other must reverse it fifty times a second with minimal loss.
Formula to use
hard: high retentivity + high coercivity; soft: low both
Baby steps
- A permanent magnet must keep its magnetisation with no field applied ⇒ high retentivity.
- It must resist being demagnetised by stray fields ⇒ high coercivity.
- A transformer core must lose its magnetisation each half-cycle ⇒ low retentivity.
- It must reverse easily with minimal energy loss ⇒ low coercivity and a narrow loop.
- So P needs high-high and T needs low-low — option (A).
- In NCERT's language: permanent magnets are hard ferromagnets such as alnico; transformer cores are soft ferromagnets such as soft iron.
Assumption
Conventional AC transformer operation.
Shortcut trick
One question, two applications, opposite answers. Ask 'must it hold on or let go?' Hold on ⇒ hard ⇒ high-high. Let go ⇒ soft ⇒ low-low. That framing prevents mixing the pairs up.
Answer: (A) P: high retentivity, high coercivity; T: low retentivity, low coercivity
On an experimentally measured B–H curve for a ferromagnet, the retentivity, coercivity and saturation are read respectively as
Given
A measured B–H hysteresis loop.
Asked
Where retentivity, coercivity and saturation are read from the graph.
Concept applied
Each quantity is an intercept or an extreme value on the loop. Locating them by axis makes them impossible to confuse.
Formula to use
retentivity = B-intercept; coercivity = H-intercept; saturation = maximum B
Baby steps
- Retentivity: the flux density remaining after the magnetising field is removed ⇒ the value of B when H = 0, the vertical intercept.
- Coercivity: the reverse field needed to drive B to zero ⇒ the value of H when B = 0, the horizontal intercept.
- Saturation: the plateau the curve reaches at large H ⇒ the maximum value of B.
- So the reading is: B at H = 0, H at B = 0, maximum B — option (A).
- The enclosed area is a fourth quantity: the energy lost per cycle per unit volume.
- JEE Main 2018 (15 April) tests the same curve by asking for the demagnetising current from the coercivity.
Assumption
A complete symmetric loop measured from the saturated state.
Shortcut trick
Anchor each term to an axis: retentivity on the B-axis, coercivity on the H-axis. The names help too — retentivity is what is retained, coercivity is the coercion needed to erase it.
Diagram
Answer: (A) B at H = 0, H at B = 0, the maximum B
Soft iron is suitable for making an electromagnet because it has
Given
Soft iron as an electromagnet core.
Asked
The property combination that makes it suitable.
Concept applied
Two requirements pulling in different directions: the core must strongly amplify the field while the current flows (high permeability), yet retain nothing once it stops (low retentivity).
Formula to use
B = μ0μrH with μr ≫ 1, and M → 0 when H → 0
Baby steps
- While energised, the core must multiply the field by a large factor ⇒ high permeability, μr > 1000 for soft iron.
- When de-energised, the magnetism must disappear so the electromagnet can be switched off ⇒ low retentivity.
- Soft iron satisfies both, which is why NCERT names it as the archetypal soft ferromagnetic material.
- Answer: high permeability and low retentivity.
- Option (C) describes a permanent magnet: strong and persistent, which is the wrong behaviour for a switchable device.
- This same pairing appears as an assertion–reason in JEE Main 2023 (8 April), where the reason correctly explains the assertion.
Assumption
Conventional switchable electromagnet.
Shortcut trick
High permeability and low retentivity are not contradictory — one describes the response while a field is applied, the other what remains after it is removed. Students reject option (A) thinking it is inconsistent; it is not.
Answer: (A) high permeability and low retentivity
What the examiners' record actually shows
These 68 questions are drawn from the full NEET/AIPMT record 2001–2024 and the JEE Main/AIEEE record 2003–2026. Reading them in sequence, four patterns stand out.
1. One calculation dominates everything. Torque and work on a dipole account for roughly a fifth of all questions ever set from this chapter, and most reduce to τ = mB sin θ or W = 2mB. AIPMT 2011, AIPMT 2012, AIEEE 2003, NEET 2016 Ph-2, JEE 2020, JEE 2022, JEE 2023, JEE 2024, JEE 2025 and JEE 2026 are all the same two formulas in different clothes. If Aamirah can do these in under forty seconds each, a large fraction of the chapter's marks are secured.
2. The classification questions never change. Dia/para/ferro identification has been asked in nearly identical form since AIPMT 2001, and NEET 2024's match-the-list is the same content in a newer format. The sign of χ settles almost every one of them.
3. NEET and JEE have diverged since 2024. NEET 2024 asked two bending-geometry questions in one paper — an arc of 60° and a midpoint fold of 60° — and a numerical oscillation question. None of these three is derivable from the rationalised NCERT text alone; all three are in your Tier 3 and Gap Content packs. JEE Main has moved towards assertion–reason and multi-statement formats, which reward precise recall of NCERT's own sentences rather than calculation.
4. Deleted topics still appear in old papers. Earth's magnetism (dip, declination, tangent galvanometer) fills many pre-2020 questions and is now removed from both syllabi — NEET 2019, NEET 2017, AIPMT 2012 and several JEE 2022 questions on dip are outside your scope and are deliberately excluded here. Hysteresis is likewise deleted, but included in Group H for recognition only, since it appears throughout older banks.
On the wording. Every question here has been rewritten in plain language while preserving the physics, the numbers and the answer. Year and session tags were verified against published past-paper indexes rather than reconstructed from memory. Where a question recurs across years, the repeat sessions are named in the solution so you can see the pattern.
Suggested use. Attempt Group A cold and time it — that group alone predicts a large share of the chapter's marks. Then use the type filter to work through whichever group your mock analysis flags. Do not attempt all sixty-eight in one sitting; the value here is in the year-to-year patterns, not in the volume.