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Magnetism and Matter · NEET 2027 · highest-yield types

Tier 1 · 50 Questions

The seven non-negotiable question types from NCERT Class 12 Chapter 5, worth roughly two-thirds of everything asked from this chapter. Every question carries a complete solution: given, asked, concept applied, formula, every baby step, the assumption being made, and the shortcut that saves time in the hall.

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A · Axial and equatorial field of a short bar magnet

10 questions
Q01Axial field · numericNEET + JEE

A short bar magnet of magnetic moment 0.5 J T−1 is placed with its axis along the x-axis. Find the magnitude of the magnetic field at a point 10 cm from its centre on the axis.

S N AXIAL (P) r EQUATORIAL (Q) r m (S to N)
Given
m = 0.5 J T−1; r = 10 cm = 0.10 m; μ0/4π = 10−7 T m A−1
Asked
Magnitude of B at an axial point.
Concept applied
A short bar magnet at large distance behaves as a magnetic dipole. On its axis the field is twice the equatorial field and points along m (from S to N inside the magnet).
Formula to use
Baxial = (μ0/4π) × (2m / r3)
Baby steps
  1. Convert the distance to SI: r = 10 cm = 0.10 m.
  2. Cube it: r3 = (0.10)3 = 1 × 10−3 m3.
  3. Compute the numerator: 2m = 2 × 0.5 = 1.0 J T−1.
  4. Substitute: B = 10−7 × (1.0 / 1 × 10−3).
  5. Divide inside the bracket: 1.0 / 10−3 = 1 × 103.
  6. Multiply: B = 10−7 × 103 = 1.0 × 10−4 T, directed along m.
Assumption
The magnet is 'short', i.e. r ≫ l, so the dipole approximation holds and pole-separation terms are dropped.
Shortcut trick
At r = 10 cm exactly, r3 = 10−3, so Baxial = 10−7 × 2m × 103 = 2m × 10−4. Memorise this: at 10 cm, just double m and slide the decimal four places. Here 2(0.5) = 1 → 1 × 10−4 T.
Answer: (B) 1.0 × 10−4 T
Q02Equatorial field · numericNEET + JEE

For the same magnet (m = 0.5 J T−1), what is the magnitude of the field at a point 10 cm from the centre on the equatorial line?

Given
m = 0.5 J T−1; r = 0.10 m; μ0/4π = 10−7 T m A−1
Asked
Magnitude of B on the normal bisector (equatorial line).
Concept applied
On the equatorial line the dipole field is half the axial value at the same distance, and it points antiparallel to m. The minus sign in NCERT Eq. 5.4 carries that direction, not the magnitude.
Formula to use
Bequatorial = (μ0/4π) × (m / r3)
Baby steps
  1. r3 = (0.10)3 = 1 × 10−3 m3.
  2. Numerator here is just m (no factor of 2): m = 0.5 J T−1.
  3. Substitute: B = 10−7 × (0.5 / 10−3).
  4. 0.5 / 10−3 = 5 × 102.
  5. B = 10−7 × 5 × 102 = 5 × 10−5 = 0.5 × 10−4 T.
  6. Direction: opposite to m, i.e. from N towards S.
Assumption
Short magnet, r ≫ l.
Shortcut trick
Never redo the arithmetic. Once you have Baxial from Q01, just halve it: 1.0 × 10−4 ÷ 2 = 0.5 × 10−4 T.
Answer: (B) 0.5 × 10−4 T
Q03Ratio · conceptualNEET + JEE

At the same distance r from the centre of a short bar magnet, the ratio Baxial : Bequatorial is

Given
Same magnet, same distance r, two different orientations of the field point.
Asked
The ratio of the two field magnitudes.
Concept applied
Both fields fall as 1/r3, so the r-dependence cancels completely. Only the numerical prefactor survives — 2m on the axis versus m on the equator.
Formula to use
Bax/Beq = [(μ0/4π)(2m/r3)] ÷ [(μ0/4π)(m/r3)]
Baby steps
  1. Write both expressions with the same r.
  2. Cancel μ0/4π from numerator and denominator.
  3. Cancel m from numerator and denominator.
  4. Cancel r3 from numerator and denominator.
  5. You are left with 2 / 1 = 2 : 1.
Assumption
Short dipole on both sides; the same magnet is used for both measurements.
Shortcut trick
'Axis is twice.' This single line answers a whole family of questions. Also lock in the direction pair: axial field is along m, equatorial field is against m.
Diagram
S N AXIAL (P) r EQUATORIAL (Q) r m (S to N)
Answer: (C) 2 : 1
Q04Equal-field distanceJEE

The axial field at distance r equals the equatorial field at distance r′ for the same short magnet. Then r′/r is

Given
Baxial(r) = Bequatorial(r′), same magnet of moment m.
Asked
The ratio r′/r.
Concept applied
Set the two dipole expressions equal. The 1/r3 dependence means a factor of 2 in field strength corresponds to a factor of 21/3 in distance — not a factor of 2.
Formula to use
(μ0/4π)(2m/r3) = (μ0/4π)(m/r′3)
Baby steps
  1. Cancel μ0/4π and m from both sides: 2/r3 = 1/r′3.
  2. Cross-multiply: 2 r′3 = r3.
  3. So r′3 = r3/2.
  4. Take the cube root of both sides: r′ = r / 21/3.
  5. Therefore r′/r = 1/21/3 = 2−1/3 ≈ 0.794.
  6. Sanity check: the equatorial field is weaker, so to match the axial field you must move closer. r′ < r ✔
Assumption
Both points are in the far region (r, r′ ≫ l).
Shortcut trick
Whenever a field ratio k is given and B ∝ 1/rn, the distance ratio is k1/n. Here n = 3, k = 2 → 21/3. Direction of the inequality is fixed by asking 'weaker field → must move closer'.
Answer: (B) 2−1/3
Q05Distance scalingNEET + JEE

The axial field of a short bar magnet at a point is B. If the distance of the point from the centre is doubled, the new axial field becomes

Given
B ∝ 1/r3 on the axis; r → 2r.
Asked
The new field in terms of B.
Concept applied
A dipole field falls off as the inverse cube of distance — faster than a point charge's inverse square. Only the r-dependence matters here; m and the constants are unchanged.
Formula to use
B2/B1 = (r1/r2)3
Baby steps
  1. Write B1 = C/r3 where C = (μ0/4π)(2m) is a constant.
  2. New distance r2 = 2r.
  3. B2 = C/(2r)3 = C/(8r3).
  4. Compare: B2 = (1/8) × (C/r3).
  5. So B2 = B/8.
Assumption
The magnet is unchanged and the point stays on the axis; the short-magnet approximation still holds at the larger distance (it holds even better).
Shortcut trick
Cube the distance factor, then invert. Double → 1/8. Triple → 1/27. Halve → 8×. This one line handles every distance-scaling question in this chapter.
Answer: (C) B/8
Q06Find m from BNCERT Ex 5.7

A short bar magnet produces an axial field of 0.96 × 10−4 T at 10 cm from its centre. Its magnetic moment is

Given
Baxial = 0.96 × 10−4 T; r = 0.10 m; μ0/4π = 10−7 T m A−1
Asked
The magnetic moment m.
Concept applied
Same dipole relation, run backwards. Rearranging for m is the single most common 'reverse' variant in NEET.
Formula to use
m = B r3 ÷ [(μ0/4π) × 2]
Baby steps
  1. Start from B = 10−7 × 2m / r3.
  2. Multiply both sides by r3: B r3 = 10−7 × 2m.
  3. Compute B r3 = 0.96 × 10−4 × 10−3 = 0.96 × 10−7.
  4. So 10−7 × 2m = 0.96 × 10−7.
  5. Cancel 10−7 from both sides: 2m = 0.96.
  6. m = 0.48 J T−1.
Assumption
Short magnet; the given point lies exactly on the axis.
Shortcut trick
At r = 10 cm the two powers of ten annihilate each other exactly. The equation collapses to 2m = B × 104 (with B in tesla). Here 0.96 × 10−4 × 104 = 0.96 = 2m.
Answer: (B) 0.48 J T−1
Q07Find r from BNEET

A short bar magnet of moment 0.5 J T−1 gives an axial field of 1 × 10−4 T. The distance of the point from the centre of the magnet is

Given
m = 0.5 J T−1; Baxial = 1 × 10−4 T; μ0/4π = 10−7 T m A−1
Asked
The distance r.
Concept applied
Third rearrangement of the same dipole relation — this time solving for r, which requires a cube root at the end.
Formula to use
r3 = (μ0/4π) × 2m / B
Baby steps
  1. Rearrange B = 10−7 × 2m/r3 to give r3 = 10−7 × 2m / B.
  2. Numerator: 10−7 × 2 × 0.5 = 10−7 × 1 = 1 × 10−7.
  3. Divide by B: r3 = 1 × 10−7 / 1 × 10−4.
  4. r3 = 1 × 10−3 m3.
  5. Cube root: r = (10−3)1/3 = 10−1 m.
  6. r = 0.1 m = 10 cm.
Assumption
Short magnet; answer must satisfy r ≫ l for consistency, which 10 cm does for a typical few-cm magnet.
Shortcut trick
Keep exponents as multiples of 3 so the cube root is painless. 10−3 → 10−1; 8 × 10−6 → 2 × 10−2; 27 × 10−6 → 3 × 10−2.
Answer: (B) 10 cm
Q08Direction · conceptualNEET

The magnetic moment m of a short bar magnet points along +x. At a point on the equatorial line, the magnetic field due to the magnet points along

S N AXIAL (P) r EQUATORIAL (Q) r m (S to N)
Given
m along +x; the field point lies on the perpendicular bisector.
Asked
The direction of B at that equatorial point.
Concept applied
NCERT Eq. 5.4 carries an explicit minus sign: BE = −μ0m/4πr3. The equatorial field is antiparallel to the moment. The physical picture: field lines leave N, sweep around, and are travelling backwards (N → S direction) as they cross the equatorial plane.
Formula to use
BE = − (μ0/4π) m/r3
Baby steps
  1. Identify the direction of m: from S pole to N pole inside the magnet, here +x.
  2. Recall the equatorial expression carries a minus sign relative to m.
  3. Therefore B at the equatorial point is along −m.
  4. Since m is along +x, B is along −x.
  5. Cross-check with the field-line picture: above the magnet the lines curve from the N end back towards the S end, so they are heading in the −x sense there. ✔
Assumption
The point is on the true perpendicular bisector, far from the magnet.
Shortcut trick
Two-word memory: axis-along, equator-against. On the axis B is parallel to m; on the equator it is antiparallel. This is the single highest-yield direction fact in the chapter, and it is exactly where direction-reasoning marks are lost.
Answer: (B) −x
Q09Mixed ratioJEE

For a short bar magnet, the ratio of the axial field at distance r to the equatorial field at distance 2r is

Given
B1 = axial field at r; B2 = equatorial field at 2r; same magnet.
Asked
The ratio B1/B2.
Concept applied
Two independent factors multiply: the 2:1 axial-to-equatorial prefactor, and the 8:1 factor from cubing the distance ratio. Handle them one at a time and multiply.
Formula to use
B1 = k·2m/r3    B2 = k·m/(2r)3
Baby steps
  1. Let k = μ0/4π.
  2. B1 = k × 2m/r3.
  3. B2 = k × m/(2r)3 = k × m/(8r3).
  4. Form the ratio: B1/B2 = (2m/r3) ÷ (m/8r3).
  5. Dividing by a fraction = multiplying by its reciprocal: = (2m/r3) × (8r3/m).
  6. Cancel m and r3: = 2 × 8 = 16.
Assumption
Short magnet at both points.
Shortcut trick
Split-and-multiply: geometry factor (2) × distance factor (23 = 8) = 16. Never try to do both at once — that is where sign and factor slips creep in.
Answer: (D) 16
Q10Combined changeJEE

A short magnet of moment m gives an axial field B at distance r. It is replaced by a magnet of moment 2m and the observation distance is doubled. The new axial field is

Given
m → 2m and r → 2r simultaneously; axial point throughout.
Asked
New field in terms of B.
Concept applied
B ∝ m and B ∝ 1/r3. Apply each proportionality separately, then combine multiplicatively.
Formula to use
B′/B = (m′/m) × (r/r′)3
Baby steps
  1. Moment factor: m′/m = 2m/m = 2. This makes the field 2× bigger.
  2. Distance factor: (r/r′)3 = (r/2r)3 = (1/2)3 = 1/8. This makes the field 8× smaller.
  3. Multiply the two factors: 2 × (1/8) = 1/4.
  4. So B′ = B/4.
  5. Direction is unchanged — still along m on the axis.
Assumption
Both magnets are short; the axial geometry is preserved.
Shortcut trick
Build a two-column factor table (what grows, what shrinks), then multiply the column. Doing it in one line invites the classic error of dividing by 8 and forgetting the 2.
Answer: (B) B/4

B · Torque on a dipole in a uniform field

8 questions
Q11Torque · directNEET + JEE

A magnetic dipole of moment 2 J T−1 is placed at 30° to a uniform magnetic field of 0.5 T. The torque acting on it is

uniform B θ m
Given
m = 2 J T−1; B = 0.5 T; θ = 30°
Asked
Magnitude of the torque τ.
Concept applied
In a uniform field a dipole feels zero net force but a non-zero torque that tries to align m with B. The angle in the formula is always measured between m and B.
Formula to use
τ = m × B  ⇒  τ = mB sin θ
Baby steps
  1. Identify θ as the angle between m and B: θ = 30°.
  2. sin 30° = 0.5.
  3. Substitute: τ = 2 × 0.5 × 0.5.
  4. Multiply the first two: 2 × 0.5 = 1.0.
  5. Multiply by sin θ: τ = 1.0 × 0.5 = 0.50 N m.
Assumption
The field is uniform, so the net force is zero and only the torque acts.
Shortcut trick
Compute mB first (the maximum possible torque), then scale by sin θ. Here mB = 1 N m, so τ is just sin 30° = half of it.
Answer: (B) 0.50 N m
Q12Find m from torqueNCERT Ex 5.1

A short bar magnet placed with its axis at 30° to a uniform field of 0.25 T experiences a torque of 4.5 × 10−2 J. Its magnetic moment is

Given
θ = 30°; B = 0.25 T; τ = 4.5 × 10−2 J (numerically N m)
Asked
The magnetic moment m.
Concept applied
Reverse use of τ = mB sin θ. Note that the torque is quoted in joules here — N m and J are dimensionally identical, and NCERT does write it as J in this exercise.
Formula to use
m = τ / (B sin θ)
Baby steps
  1. sin 30° = 0.5.
  2. Compute the denominator: B sin θ = 0.25 × 0.5 = 0.125.
  3. Substitute: m = 4.5 × 10−2 / 0.125.
  4. Rewrite 0.125 as 1/8, so dividing by it means multiplying by 8.
  5. m = 4.5 × 10−2 × 8 = 36 × 10−2.
  6. m = 0.36 J T−1.
Assumption
The field is uniform over the length of the magnet.
Shortcut trick
Recognise decimal fractions as unit fractions: 0.125 = 1/8, 0.25 = 1/4, 0.2 = 1/5. Dividing becomes multiplying, and the arithmetic goes mental.
Answer: (B) 0.36 J T−1
Q13Maximum torqueNEET

A magnet of moment m in a uniform field B experiences maximum torque when the angle between m and B is

Given
τ = mB sin θ, with m and B fixed.
Asked
The angle for maximum torque and its value.
Concept applied
τ depends on θ only through sin θ, which has its maximum value of 1 at θ = 90°. So the twist is greatest when the magnet sits square across the field.
Formula to use
τmax = mB at θ = 90°
Baby steps
  1. Write τ(θ) = mB sin θ with m, B constant.
  2. The only variable factor is sin θ.
  3. Over 0° ≤ θ ≤ 180°, sin θ is maximum at θ = 90°, where sin 90° = 1.
  4. So τmax = mB × 1 = mB.
  5. Reject the distractors: at θ = 0° and 180°, sin θ = 0, so τ = 0 (both are equilibrium positions).
Assumption
Uniform field, rigid dipole.
Shortcut trick
Torque and energy peak at different angles. Torque is maximum at 90°; potential energy is maximum at 180°. Sketch sin θ and −cos θ side by side once and this never confuses you again.
Answer: (B) 90°, and the torque is mB
Q14Force in uniform fieldNEET + JEE

A bar magnet is placed at an arbitrary angle in a uniform magnetic field. The net force on it is

Given
Bar magnet in a uniform field B, arbitrary orientation.
Asked
The net translational force.
Concept applied
A dipole has two equal and opposite poles. In a uniform field each pole feels an equal-magnitude force in opposite directions, so the forces cancel exactly. Only their separation survives, giving a couple — pure torque, no translation.
Formula to use
Fnet = qmB − qmB = 0  (uniform B)
Baby steps
  1. Model the magnet as poles +qm and −qm separated by 2l.
  2. Force on the north pole: +qmB.
  3. Force on the south pole: −qmB.
  4. Because B is the same at both ends (that is what 'uniform' means), the magnitudes are equal.
  5. Vector sum: qmB + (−qmB) = 0, for every angle θ.
  6. The two forces do not share a line of action, so they form a couple → torque mB sin θ remains.
Assumption
The field is genuinely uniform across the magnet's length. A bar magnet's own field is not uniform, which is why an iron nail near a magnet does feel a net pull.
Shortcut trick
Uniform ⇒ turn only. Non-uniform ⇒ turn and tug. This one line answers Exercise 5.6(b), Example 5.1(b), and every 'force on a magnet' MCQ ever set.
Diagram
θ = 0° · U = −mB · STABLE θ = 180° · U = +mB · UNSTABLE field B points right in both rows
Answer: (C) zero for every angle
Q15Solenoid torqueNCERT Ex 5.6

A closely wound solenoid of 2000 turns and cross-section 1.6 × 10−4 m2 carries 4.0 A. A uniform field of 7.5 × 10−2 T is applied at 30° to its axis. The torque on the solenoid is

S N m = NIA N turns · area A · current I
Given
N = 2000; A = 1.6 × 10−4 m2; I = 4.0 A; B = 7.5 × 10−2 T; θ = 30°
Asked
The torque on the solenoid.
Concept applied
Two-step composite. First convert the coil into an equivalent magnetic dipole using m = NIA, then feed that m into the dipole torque formula. This chaining is the chapter's most common numerical structure.
Formula to use
m = N I A   then   τ = mB sin θ
Baby steps
  1. Step 1 — moment: m = 2000 × 4.0 × 1.6 × 10−4.
  2. 2000 × 4.0 = 8000 = 8 × 103.
  3. m = 8 × 103 × 1.6 × 10−4 = 12.8 × 10−1 = 1.28 J T−1.
  4. Step 2 — torque: τ = 1.28 × 7.5 × 10−2 × sin 30°.
  5. 1.28 × 7.5 × 10−2 = 9.6 × 10−2.
  6. Multiply by sin 30° = 0.5: τ = 4.8 × 10−2 N m.
Assumption
Closely wound solenoid so every turn contributes the same area; field uniform across it; the answer 1.28 J T−1 is option (D) — the classic trap of stopping at step 1.
Shortcut trick
In two-step problems, the intermediate value is almost always planted as a distractor. Before selecting, ask: 'have I finished both steps?' Here 1.28 (the moment) and 9.6 × 10−2 (torque before sin 30°) are both sitting in the options.
Answer: (B) 4.8 × 10−2 N m
Q16Torque ratioJEE

The torques on a magnetic dipole in the same uniform field at 30° and 60° are in the ratio

Given
Same m, same B; θ1 = 30°, θ2 = 60°.
Asked
τ1 : τ2
Concept applied
With m and B fixed, torque is proportional to sin θ alone. The ratio reduces to a ratio of sines.
Formula to use
τ1/τ2 = sin θ1 / sin θ2
Baby steps
  1. τ1 = mB sin 30°, τ2 = mB sin 60°.
  2. Cancel mB: ratio = sin 30° / sin 60°.
  3. sin 30° = 1/2; sin 60° = √3/2.
  4. Ratio = (1/2) ÷ (√3/2) = (1/2) × (2/√3).
  5. The 2s cancel: ratio = 1/√3.
  6. So τ1 : τ2 = 1 : √3 ≈ 1 : 1.73.
Assumption
Identical magnet and identical field for both measurements.
Shortcut trick
Keep sines as halves of 1, √2, √3, 2 (for 30°, 45°, 60°, 90°). The denominators of 2 always cancel in a ratio, so the answer is just the ratio of the numerators: 1 : √3.
Answer: (A) 1 : √3
Q17Torque scalingNEET

A bar magnet experiences a torque of 0.5 N m when placed at 30° to a uniform field. The torque when it is placed at 90° to the same field is

Given
τ30 = 0.5 N m; same magnet, same field; find τ90.
Asked
The torque at 90°, which is also τmax.
Concept applied
You do not need m or B individually. The product mB can be extracted from the first measurement and reused.
Formula to use
mB = τ1/sin θ1   ⇒   τ2 = mB sin θ2
Baby steps
  1. From the first case: 0.5 = mB sin 30° = mB × 0.5.
  2. Solve for the product: mB = 0.5 / 0.5 = 1.0 N m.
  3. At 90°: τ = mB sin 90° = 1.0 × 1.
  4. τ90 = 1.00 N m.
  5. Sense check: 90° is the maximum-torque orientation, so the answer must exceed 0.5 N m. ✔
Assumption
m and B unchanged between the two orientations.
Shortcut trick
Treat mB as a single unknown 'package'. In any two-orientation problem, extract mB from the given case and substitute into the asked case — never try to find m and B separately.
Answer: (D) 1.00 N m
Q18Vector formJEE

A dipole m = 3î A m2 sits in a field B = 2ĵ T. The torque vector on it is

Given
m = 3î; B = 2ĵ
Asked
The torque as a vector, including direction.
Concept applied
Torque is a cross product, so the answer is a vector perpendicular to both m and B. Use the right-hand rule via unit-vector identities rather than trying to visualise.
Formula to use
τ = m × B, with î × ĵ = k̂
Baby steps
  1. Write the cross product: τ = (3î) × (2ĵ).
  2. Pull the scalars out in front: = (3 × 2)(î × ĵ).
  3. = 6 (î × ĵ).
  4. Apply the cyclic identity î × ĵ = k̂.
  5. τ = 6 k̂ N m, i.e. magnitude 6 N m along +z.
  6. Check with the magnitude formula: m and B are perpendicular, so τ = mB sin 90° = 3 × 2 = 6. ✔
Assumption
Uniform field; the dipole is free to rotate about the z-axis.
Shortcut trick
Memorise the cycle î → ĵ → k̂ → î. Going forward round the cycle gives +, going backward gives −. So ĵ × î = −k̂. Order matters: it is m × B, never B × m.
Answer: (A) 6 k̂ N m

C · Work done and potential energy

8 questions
Q19Work 0° to 90°NCERT Ex 5.5

A bar magnet of moment 1.5 J T−1 lies aligned with a uniform field of 0.22 T. The work required to turn it normal to the field direction is

Given
m = 1.5 J T−1; B = 0.22 T; θ1 = 0°, θ2 = 90°
Asked
External work done W.
Concept applied
Work done by an external agent equals the increase in potential energy of the dipole. Since U = −mB cos θ, the work is the difference of two cosine terms.
Formula to use
W = U2 − U1 = −mB(cos θ2 − cos θ1)
Baby steps
  1. Compute mB = 1.5 × 0.22 = 0.33 J.
  2. cos θ1 = cos 0° = 1, so U1 = −0.33 J.
  3. cos θ2 = cos 90° = 0, so U2 = 0 J.
  4. W = U2 − U1 = 0 − (−0.33).
  5. W = 0.33 J.
  6. Positive, as expected — you are pushing the magnet away from its comfortable position.
Assumption
The rotation is quasi-static (no kinetic energy left over at the end) and the field is uniform.
Shortcut trick
From the aligned position, turning to 90° always costs exactly mB, and turning to 180° always costs exactly 2mB. Compute mB once and read both answers off.
Answer: (B) 0.33 J
Q20Work 0° to 180°NCERT Ex 5.5

For the same magnet (m = 1.5 J T−1, B = 0.22 T), the work required to turn it exactly opposite to the field is

Given
m = 1.5 J T−1; B = 0.22 T; θ1 = 0°, θ2 = 180°
Asked
External work done W.
Concept applied
Same energy-difference method. Going from the deepest point of the energy well (−mB) to the very top of the hill (+mB) means climbing a total of 2mB.
Formula to use
W = −mB(cos 180° − cos 0°)
Baby steps
  1. mB = 1.5 × 0.22 = 0.33 J.
  2. cos 180° = −1, so U2 = −mB(−1) = +0.33 J.
  3. cos 0° = +1, so U1 = −0.33 J.
  4. W = U2 − U1 = 0.33 − (−0.33).
  5. W = 0.66 J. Equivalently W = 2mB = 2(0.33).
  6. W = 0.66 J.
Assumption
Quasi-static rotation; uniform field throughout.
Shortcut trick
The 180° flip always costs exactly twice the 90° turn. If a question gives you one, the other needs no calculation at all.
Answer: (B) 0.66 J
Q21PE at stable positionNCERT Ex 5.2

A short bar magnet of moment 0.32 J T−1 is in a uniform field of 0.15 T. Its potential energy in the stable equilibrium orientation is

Given
m = 0.32 J T−1; B = 0.15 T; stable orientation.
Asked
The potential energy U.
Concept applied
Stable equilibrium means m is parallel to B (θ = 0°), where U reaches its minimum value −mB. The negative sign signals a bound, comfortable state.
Formula to use
U = −m·B = −mB cos θ; at θ = 0°, U = −mB
Baby steps
  1. Identify the stable orientation: θ = 0°, m parallel to B.
  2. cos 0° = 1.
  3. U = −mB = −(0.32)(0.15).
  4. 0.32 × 0.15 = 0.048.
  5. U = −0.048 J = −4.8 × 10−2 J.
  6. The sign is essential — dropping it turns a correct answer into option (B).
Assumption
Zero of potential energy is fixed at θ = 90°, which is NCERT's convention (constant of integration set to zero).
Shortcut trick
Stable = most negative. If your stable-equilibrium energy comes out positive, you have used the wrong angle. Minimum energy and maximum stability always travel together.
Diagram
θ = 0° · U = −mB · STABLE θ = 180° · U = +mB · UNSTABLE field B points right in both rows
Answer: (A) −4.8 × 10−2 J
Q22PE at unstable positionNCERT Ex 5.2

For the same magnet (m = 0.32 J T−1, B = 0.15 T), the potential energy in unstable equilibrium is

Given
m = 0.32 J T−1; B = 0.15 T; unstable orientation.
Asked
The potential energy U.
Concept applied
Unstable equilibrium is θ = 180°, m antiparallel to B. Torque is zero there (sin 180° = 0) so it is an equilibrium, but energy is maximum, so any nudge sends it flipping.
Formula to use
U = −mB cos 180° = +mB
Baby steps
  1. Unstable orientation: θ = 180°.
  2. cos 180° = −1.
  3. U = −mB × (−1) = +mB.
  4. mB = 0.32 × 0.15 = 0.048 J.
  5. U = +4.8 × 10−2 J.
  6. Note it is the exact mirror image of the stable value — same magnitude, opposite sign.
Assumption
Same zero-energy convention (θ = 90°) as NCERT.
Shortcut trick
The two equilibrium energies are ±mB. Compute mB once and attach the sign: minus for aligned, plus for anti-aligned. The gap between them is the 2mB you met in Q20.
Answer: (B) +4.8 × 10−2 J
Q23Work 90° to 180°JEE

The work done in rotating a magnetic dipole from the position of zero potential energy to the position of maximum potential energy is

Given
Start at U = 0 (θ = 90°); end at U maximum (θ = 180°).
Asked
Work done W.
Concept applied
Translate the wording into angles first. 'Zero potential energy' is θ = 90° by NCERT's convention; 'maximum potential energy' is θ = 180°. Then it is a simple energy difference.
Formula to use
W = −mB(cos θ2 − cos θ1)
Baby steps
  1. Translate: θ1 = 90° (U = 0), θ2 = 180° (U = +mB).
  2. U1 = −mB cos 90° = 0.
  3. U2 = −mB cos 180° = +mB.
  4. W = U2 − U1 = mB − 0.
  5. W = mB.
  6. Cross-check with the full journey: 0°→90° costs mB, 90°→180° costs mB, total 2mB. ✔
Assumption
NCERT's zero-energy convention at θ = 90°.
Shortcut trick
Learn the three energy landmarks as a ladder: −mB at 0°, 0 at 90°, +mB at 180°. Every work question is just the difference between two rungs, and the rungs are evenly spaced by mB.
Answer: (C) mB
Q24Work between general anglesJEE

A dipole of moment 2 A m2 in a field of 0.4 T is rotated from 60° to 90°. The external work done is

Given
m = 2 A m2; B = 0.4 T; θ1 = 60°, θ2 = 90°
Asked
External work W.
Concept applied
Identical energy-difference method, now with non-special angles. The only new demand is careful handling of cos 60° = 0.5.
Formula to use
W = −mB(cos θ2 − cos θ1)
Baby steps
  1. mB = 2 × 0.4 = 0.8 J.
  2. cos 60° = 0.5 ⇒ U1 = −0.8 × 0.5 = −0.4 J.
  3. cos 90° = 0 ⇒ U2 = 0 J.
  4. W = U2 − U1 = 0 − (−0.4).
  5. W = 0.40 J.
  6. Positive — the dipole is being moved further from alignment, so work must be supplied.
Assumption
Quasi-static rotation; uniform field.
Shortcut trick
W = mB(cos θ1 − cos θ2). Written this way the initial angle comes first and the minus sign disappears — fewer sign slips under time pressure.
Answer: (B) 0.40 J
Q25Work ratioNEET

Starting from the aligned position, the work needed to rotate a dipole to 90° and to 180° are in the ratio

Given
Both rotations begin at θ = 0°; same m and B.
Asked
W90 : W180
Concept applied
Both works are multiples of mB, so mB cancels and only the numerical coefficients remain.
Formula to use
W90 = mB    W180 = 2mB
Baby steps
  1. W90 = −mB(cos 90° − cos 0°) = −mB(0 − 1) = mB.
  2. W180 = −mB(cos 180° − cos 0°) = −mB(−1 − 1) = 2mB.
  3. Form the ratio: mB : 2mB.
  4. Cancel mB from both sides.
  5. Ratio = 1 : 2.
  6. Common wrong answer: 1 : 1, from assuming equal 90° sweeps cost equally. They do not — the second half costs the same as the first only because of where the cosine sits, and here it happens to.
Assumption
Same magnet and field for both rotations; both start from θ = 0°.
Shortcut trick
cos goes 1 → 0 → −1 across 0°, 90°, 180° — equal steps of 1 each. So equal-cost halves, and the totals are in the ratio 1 : 2.
Answer: (B) 1 : 2
Q26Stable to unstableNEET

A magnet of moment 0.5 J T−1 is in a field of 0.2 T. The energy needed to turn it from stable to unstable equilibrium is

Given
m = 0.5 J T−1; B = 0.2 T; stable → unstable.
Asked
The energy (work) required.
Concept applied
Stable is θ = 0° (U = −mB), unstable is θ = 180° (U = +mB). The climb is the full height of the energy landscape, 2mB.
Formula to use
W = Uunstable − Ustable = mB − (−mB) = 2mB
Baby steps
  1. Identify: stable ⇒ θ = 0°; unstable ⇒ θ = 180°.
  2. mB = 0.5 × 0.2 = 0.10 J.
  3. Ustable = −0.10 J.
  4. Uunstable = +0.10 J.
  5. W = 0.10 − (−0.10) = 0.20 J.
  6. Note that 0.10 J (= mB) is planted as option (B) for anyone who forgets to double.
Assumption
Uniform field; quasi-static rotation.
Shortcut trick
'Stable to unstable' is a fixed phrase meaning 2mB, always. Recognise the phrase and skip straight to doubling mB.
Answer: (C) 0.20 J

D · Stable and unstable equilibrium

5 questions
Q27Stable orientationNEET

A magnetic dipole free to rotate in a uniform field settles in stable equilibrium when

Given
Dipole free to rotate in a uniform field.
Asked
The stable orientation.
Concept applied
Equilibrium requires zero torque (θ = 0° or 180°). Stability then requires minimum potential energy, which selects θ = 0°.
Formula to use
τ = mB sin θ = 0 and U = −mB cos θ minimum
Baby steps
  1. Set torque to zero: sin θ = 0 ⇒ θ = 0° or 180°. Both are equilibria.
  2. Evaluate energy at each: U(0°) = −mB; U(180°) = +mB.
  3. The stable one is the minimum-energy state.
  4. −mB < +mB, so θ = 0° wins.
  5. Therefore m is parallel to B.
  6. Physical check: displace it slightly and the restoring torque pushes it back → genuinely stable. ✔
Assumption
No friction, no other torques acting.
Shortcut trick
Two conditions, applied in order: torque zero picks the candidates (0° and 180°); energy minimum picks the winner (0°). Never test only one condition.
Diagram
θ = 0° · U = −mB · STABLE θ = 180° · U = +mB · UNSTABLE field B points right in both rows
Answer: (B) m is parallel to B
Q2890° combinedJEE

At which orientation does a dipole in a uniform field have maximum torque and zero potential energy simultaneously?

Given
τ = mB sin θ; U = −mB cos θ.
Asked
The angle satisfying both conditions at once.
Concept applied
Torque follows sin θ and energy follows −cos θ. These two functions are 90° out of step, so the peak of one coincides with the zero of the other.
Formula to use
τmax at sin θ = 1; U = 0 at cos θ = 0
Baby steps
  1. Maximum torque needs sin θ = 1 ⇒ θ = 90°.
  2. Zero energy needs cos θ = 0 ⇒ θ = 90°.
  3. Both conditions land on the same angle.
  4. Answer: θ = 90°.
  5. Check the others: at 0°, τ = 0 and U = −mB (neither condition met); at 180°, τ = 0 and U = +mB (neither met).
Assumption
NCERT's zero-energy convention at θ = 90° — without it, 'zero potential energy' would be undefined.
Shortcut trick
Sketch sin θ and −cos θ on one axis from 0° to 180° once. Every equilibrium and extremum question in this chapter can then be read straight off the picture instead of recomputed.
Answer: (C) θ = 90°
Q29Assertion–reasonNEET

Assertion: A dipole aligned along the field is in stable equilibrium. Reason: Its potential energy is minimum in that orientation.

Given
Assertion about stability at θ = 0°; reason about minimum energy.
Asked
Which assertion–reason combination is correct.
Concept applied
In mechanics, minimum potential energy is the definition of stable equilibrium. So the reason is not merely a true side-fact; it is precisely why the assertion holds.
Formula to use
U(0°) = −mB = Umin
Baby steps
  1. Test the assertion: at θ = 0°, torque is zero and any small displacement produces a restoring torque. Stable. True.
  2. Test the reason: U(θ) = −mB cos θ is smallest when cos θ is largest, i.e. at θ = 0°, giving U = −mB. True.
  3. Test the link: is minimum energy the cause of stability? Yes — a system displaced from an energy minimum experiences a restoring force or torque by definition.
  4. So both statements are true and the reason explains the assertion.
  5. Answer: option (A).
Assumption
Standard assertion–reason marking scheme with these four choices.
Shortcut trick
For assertion–reason, run three separate tests in order: is A true, is R true, does R cause A. Judging the pair as a whole is how students lose these marks.
Answer: (A) Both true, reason correctly explains assertion
Q30Dipole–dipole configNCERT Ex 5.2

Needle Q is placed in the field of an identical needle P. Among the configurations shown, the one with the lowest potential energy has Q

P Q on equator Q on axis axial field is twice the equatorial field at equal r
Given
Two identical needles; Q sits in the field produced by P.
Asked
The configuration of minimum potential energy.
Concept applied
U = −mQ·BP. To make U as negative as possible you need two things at once: alignment (parallel) and the strongest available field. The axial field is twice the equatorial field at the same distance, so the axis wins.
Formula to use
U = −mQBP cos θ; Baxial = 2Bequatorial
Baby steps
  1. Both parallel options give cos θ = 1, so U = −mQBP in each case — these are the only candidates for a minimum.
  2. The antiparallel option gives U = +mQBP (maximum, not minimum) — reject.
  3. The perpendicular option gives U = 0 and is not even an equilibrium — reject.
  4. Now compare the two parallel cases by field strength: Baxial = 2 × Bequatorial at equal r.
  5. The larger B gives the more negative U.
  6. So the minimum is Q on the axis, parallel to BP — NCERT's configuration PQ6.
Assumption
Both needles are identical and separated by the same distance r in every configuration; short-dipole approximation.
Shortcut trick
Two-filter method: filter 1 is alignment (keeps only the parallel cases), filter 2 is field strength (axis beats equator, 2:1). Applying only filter 1 leaves you stuck between two options, which is exactly the trap.
Answer: (B) on the axis, parallel to BP
Q31Oscillation about equilibriumJEE

A compass needle is displaced slightly from its stable position in a uniform field and released. It will

Given
Small angular displacement from θ = 0°; needle then released.
Asked
The subsequent motion.
Concept applied
The torque τ = −mB sin θ always acts back towards θ = 0°. For small θ, sin θ ≈ θ, so the restoring torque is proportional to the displacement — the signature of simple harmonic motion.
Formula to use
τ = −mB sin θ ≈ −(mB)θ for small θ
Baby steps
  1. At θ = 0° the torque is zero, so it is an equilibrium.
  2. Displace by a small angle θ: torque becomes −mB sin θ, directed back towards θ = 0°.
  3. Restoring torque ⇒ the needle accelerates back through the equilibrium.
  4. It overshoots (it has kinetic energy at θ = 0°), then is pulled back again.
  5. Result: angular oscillation about the field direction, simple harmonic for small amplitudes.
  6. With damping present it eventually settles at θ = 0° — which is what a real compass does.
Assumption
Negligible friction over the timescale considered; small amplitude so sin θ ≈ θ.
Shortcut trick
Restoring torque ⇒ oscillation, always. Note the period formula T = 2π√(I/mB) is not in your rationalised NCERT text — it lives in the Gap Content pack, but the qualitative 'it oscillates' is fully in-syllabus.
Answer: (B) oscillate about the field direction

E · Solenoid and current loop as an equivalent magnet

8 questions
Q32m = NIA · directNCERT Ex 5.3

A closely wound solenoid of 800 turns and cross-section 2.5 × 10−4 m2 carries a current of 3.0 A. Its magnetic moment is

S N m = NIA N turns · area A · current I
Given
N = 800; A = 2.5 × 10−4 m2; I = 3.0 A
Asked
The magnetic moment of the solenoid.
Concept applied
Each turn is a current loop of moment IA. N turns stacked coaxially add their moments arithmetically, giving NIA. The solenoid then behaves exactly like a bar magnet of that moment.
Formula to use
m = N I A
Baby steps
  1. Multiply turns by current: 800 × 3.0 = 2400 = 2.4 × 103.
  2. Multiply by area: m = 2.4 × 103 × 2.5 × 10−4.
  3. Handle the mantissas: 2.4 × 2.5 = 6.0.
  4. Handle the exponents: 103 × 10−4 = 10−1.
  5. m = 6.0 × 10−1 = 0.60 J T−1.
  6. Direction: along the solenoid axis, out of the face where the current appears anticlockwise.
Assumption
Closely wound, so every turn encloses the same area A and the turns are effectively coaxial.
Shortcut trick
Group as (N×I) first, then multiply by A. Keeping powers of ten separate from mantissas prevents the decimal-place errors that make 0.06 or 6.0 look plausible.
Answer: (B) 0.60 J T−1
Q33m = NIA · directNEET

A coil of 500 turns and area 4 × 10−4 m2 carries 2 A. Its magnetic moment is

Given
N = 500; A = 4 × 10−4 m2; I = 2 A
Asked
The magnetic moment.
Concept applied
Same relation. Note that A m2 and J T−1 are the same unit — NEET switches between them freely to test whether you notice.
Formula to use
m = N I A
Baby steps
  1. N × I = 500 × 2 = 1000 = 1 × 103.
  2. m = 1 × 103 × 4 × 10−4.
  3. Mantissas: 1 × 4 = 4.
  4. Exponents: 103 × 10−4 = 10−1.
  5. m = 4 × 10−1 = 0.40 A m2.
  6. Unit check: A × m2 = A m2 ✔ (identical to J T−1).
Assumption
Closely wound coil, all turns of equal area.
Shortcut trick
If the options are all the same digits with shifted decimal points (0.04 / 0.40 / 4.00 / 40.0), the question is testing exponent arithmetic, not physics. Slow down on the powers of ten specifically.
Answer: (B) 0.40 A m2
Q34Circular loopNEET + JEE

A single circular loop of radius 5 cm carries a current of 2 A. Its magnetic moment is approximately

current I, area A, N turns m = NIA right hand: curl fingers along I, thumb gives m
Given
N = 1; r = 5 cm = 0.05 m; I = 2 A
Asked
The magnetic moment.
Concept applied
Same m = NIA, but the area must be computed from the geometry first. For a circle, A = πr2. This extra step is where the marks are lost.
Formula to use
m = N I πr2
Baby steps
  1. Convert the radius: r = 5 cm = 0.05 m = 5 × 10−2 m.
  2. Square it: r2 = 25 × 10−4 = 2.5 × 10−3 m2.
  3. Area: A = π × 2.5 × 10−3 = 7.85 × 10−3 m2.
  4. Moment: m = 1 × 2 × 7.85 × 10−3.
  5. m = 15.7 × 10−3 = 1.57 × 10−2 A m2.
  6. Direction: perpendicular to the loop plane, given by the right-hand rule (curl fingers along I, thumb gives m).
Assumption
Single turn (N = 1) unless stated otherwise; the loop is planar.
Shortcut trick
Square the radius before multiplying by π. Squaring 5 × 10−2 to get 25 × 10−4 keeps everything in clean integers. Option (B) is what you get by using 2πr (circumference) instead of πr2.
Answer: (A) 1.57 × 10−2 A m2
Q35North face identificationNEET

When you look at one end of a current-carrying solenoid and the current appears to flow anticlockwise, that face is

Given
Current appears anticlockwise when viewed from a particular face.
Asked
Which magnetic pole that face represents.
Concept applied
The right-hand rule for a current loop: curl the fingers of the right hand along the current, and the extended thumb points along m, i.e. out of the north face. Anticlockwise as seen by you means m points towards you.
Formula to use
Anticlockwise (as viewed) ⇒ m towards viewer ⇒ NORTH face
Baby steps
  1. Place your right hand so the curled fingers follow the current direction as you see it — anticlockwise.
  2. The thumb then points out of the page, towards you.
  3. m points from the S pole to the N pole inside the magnet, and emerges from the N pole.
  4. Since m points towards you, the face you are looking at is the north pole.
  5. The core material changes the field strength, never which face is north — so option (D) is a distractor.
  6. Mirror rule: clockwise as viewed ⇒ that face is a south pole.
Assumption
Right-handed coordinate convention, as used throughout NCERT.
Shortcut trick
The letters help: draw the current arrows on the letter N going aNticlockwise, and on S going clockwise (S curls like a clock). Crude, but it survives exam pressure.
Answer: (A) a north pole
Q36Cutting a solenoidJEE

A solenoid of N turns carrying current I has magnetic moment m. It is cut into two equal halves along its length axis, and each half carries the same current I. The moment of each half is

Given
Original: N turns, current I, area A, moment m = NIA. Cut into two equal halves, each carrying the same I.
Asked
The magnetic moment of one half.
Concept applied
Cutting a solenoid transversely halves the number of turns. The current and the cross-sectional area are unchanged, so the moment halves. This is the coil version of 'cutting a bar magnet halves its moment'.
Formula to use
m′ = N′ I A with N′ = N/2
Baby steps
  1. Original moment: m = N I A.
  2. After the cut, each half has N′ = N/2 turns.
  3. The current through each half is still I (unchanged, as stated).
  4. The cross-sectional area A is unchanged — a transverse cut does not alter the loop size.
  5. m′ = (N/2) × I × A = (1/2)(NIA).
  6. m′ = m/2.
Assumption
Transverse cut (across the length), current held constant, area unchanged. A cut along the length would halve the area instead, giving the same m/2 by a different route.
Shortcut trick
Ask which of N, I, A actually changed. Only the changed factor scales the moment. Here only N changed, and it halved — so m halves. No formula manipulation needed.
Diagram
S N m, length 2l cut S N S N m/2 each, length l
Answer: (B) m/2
Q37Wire rebentJEE

A wire of fixed length carrying current I is bent first into a single circular turn, then into two turns of a smaller circle. The ratio of the magnetic moments (one turn : two turns) is

Given
Fixed wire length L, same current I, bent into n = 1 then n = 2 turns.
Asked
m1 : m2
Concept applied
Fixing the wire length couples the radius to the number of turns. More turns means a smaller radius, and the area falls as 1/n2 while the turn count only rises as n — so the moment falls as 1/n.
Formula to use
L = n(2πr) ⇒ r = L/2πn ⇒ m = nIπr2 = IL2/4πn
Baby steps
  1. Length constraint: L = n × 2πr, so r = L/(2πn).
  2. Area of one turn: A = πr2 = πL2/(4π2n2) = L2/(4πn2).
  3. Moment: m = nIA = nI × L2/(4πn2) = IL2/(4πn).
  4. So m ∝ 1/n with L and I fixed.
  5. m1/m2 = (1/1) ÷ (1/2) = 2.
  6. Ratio = 2 : 1 — the single large turn has the greater moment.
Assumption
The whole wire is used both times; current is the same in both configurations; turns are coplanar and concentric.
Shortcut trick
Derive once and store the result: m = IL2/4πn. Fewer turns from a fixed wire always means a bigger moment, because area shrinks faster (n2) than turn count grows (n).
Answer: (C) 2 : 1
Q38Bar magnet ≡ solenoidNEET

A bar magnet and a current-carrying solenoid produce identical far axial fields. This shows that

Given
Identical far axial fields for a bar magnet and a solenoid of the same moment.
Asked
The physical conclusion.
Concept applied
This is Ampere's hypothesis, stated in NCERT 5.2.2: all magnetic phenomena can be explained in terms of circulating currents. In a solenoid these are conduction currents in wire; in iron they are the orbital and spin motions of electrons.
Formula to use
Baxial = (μ0/4π)(2m/r3) for both objects
Baby steps
  1. The solenoid's field is unambiguously produced by moving charges — there are no poles inside a copper coil.
  2. The bar magnet produces an indistinguishable field at large distances.
  3. If two objects produce identical fields, no external experiment can tell their sources apart.
  4. The simplest unified explanation is that the magnet's field also arises from circulating currents — at the atomic scale.
  5. Conclusion: magnetism in matter is due to circulating currents.
  6. Reject (A): free poles do not exist. Reject (C): an air-cored solenoid works fine. Reject (D): solenoid field lines are closed loops.
Assumption
Both objects have the same magnetic moment and are observed at r ≫ their size.
Shortcut trick
This idea also explains why cutting never yields a monopole: cut a solenoid and each piece is still a coil with an anticlockwise face and a clockwise face. Geometry, not luck.
Diagram
S N m = NIA N turns · area A · current I
Answer: (B) magnetism in matter can be attributed to circulating currents
Q39Moment of a revolving chargeJEE

A particle of charge q moves in a circle of radius r with speed v. Its magnetic moment is

Given
Charge q in a circular orbit of radius r at speed v.
Asked
The equivalent magnetic moment.
Concept applied
A circulating charge is a current loop. Convert the orbital motion into an equivalent current first (charge passing a point per unit time), then apply m = IA.
Formula to use
I = q/T = qv/2πr   then   m = IA = Iπr2
Baby steps
  1. Time for one revolution: T = 2πr/v.
  2. Equivalent current: I = q/T = qv/(2πr).
  3. Area enclosed: A = πr2.
  4. Moment: m = IA = [qv/(2πr)] × πr2.
  5. Cancel π and one power of r: m = qvr/2.
  6. m = qvr/2, directed perpendicular to the orbital plane.
Assumption
Uniform circular motion at constant speed; single charge; non-relativistic.
Shortcut trick
Two-step chain: orbit → current → moment. The factor of 1/2 always survives because πr2 ÷ 2πr = r/2. Writing m = qvr (option A) is the standard slip — it comes from forgetting the 2π in the period.
Answer: (B) qvr/2

F · Classification from χ, μr and μ

6 questions
Q40Classify from χNEET

A material has magnetic susceptibility χ = −2 × 10−5. The material is

χ = 0 DIA −1≤χ<0 PARA 0<χ≪1 FERRO χ≫1 μr < 1 μr slightly > 1 μr > 1000
Given
χ = −2 × 10−5
Asked
The magnetic classification.
Concept applied
Table 5.2 classification: χ negative and small ⇒ diamagnetic; χ positive and small ⇒ paramagnetic; χ very large and positive ⇒ ferromagnetic. Sign first, magnitude second.
Formula to use
−1 ≤ χ < 0 : dia    0 < χ ≪ 1 : para    χ ≫ 1 : ferro
Baby steps
  1. Read the sign: χ is negative.
  2. A negative χ means M is opposite to H — the material opposes the applied field.
  3. That immediately rules out paramagnetic and ferromagnetic, both of which need χ > 0.
  4. Read the magnitude: 2 × 10−5 is tiny, far from the extreme value χ = −1.
  5. χ = −1 exactly would mean a superconductor (perfect diamagnet); this is not that.
  6. Answer: diamagnetic.
Assumption
Room temperature, linear (weak-field) response so that M = χH applies.
Shortcut trick
One glance at the sign settles dia versus not-dia. Then one glance at the magnitude separates para (≈ 10−5) from ferro (≫ 1). Two glances, no arithmetic.
Answer: (B) diamagnetic
Q41Classify from μrNEET

A substance has relative permeability μr = 0.9999. It is

Given
μr = 0.9999
Asked
The classification.
Concept applied
μr = 1 + χ. Reading off χ from μr converts the question into the standard sign test. μr below 1 means negative χ.
Formula to use
χ = μr − 1
Baby steps
  1. Compute χ: χ = 0.9999 − 1.
  2. χ = −0.0001 = −1 × 10−4.
  3. The sign is negative ⇒ the material opposes the field.
  4. The magnitude is small, so it is ordinary diamagnetism (not the χ = −1 superconducting extreme).
  5. Answer: diamagnetic.
  6. Consistency check: μr < 1 means μ < μ0, so the field inside is reduced — exactly the diamagnetic signature of Fig. 5.7(a). ✔
Assumption
Linear material at room temperature.
Shortcut trick
Use μr = 1 as the dividing line. Below 1 ⇒ diamagnetic. Just above 1 ⇒ paramagnetic. Far above 1 (hundreds or thousands) ⇒ ferromagnetic. You never need to compute χ at all.
Answer: (C) diamagnetic
Q42Classify from χNEET

For a certain material χ = 1200. This material is

Given
χ = 1200
Asked
The classification.
Concept applied
A susceptibility of order 103 means the material's own contribution swamps the applied field by a factor of a thousand. Only cooperative domain alignment can do that.
Formula to use
χ ≫ 1 ⇒ ferromagnetic; μr = 1 + χ ≈ 1201
Baby steps
  1. Sign: positive ⇒ the material reinforces the applied field. Diamagnetic is eliminated.
  2. Magnitude: 1200 is enormous compared with the 10−5 typical of paramagnets.
  3. Paramagnetic is therefore eliminated too.
  4. χ ≫ 1 is the defining ferromagnetic condition in Table 5.2.
  5. Cross-check: μr = 1 + 1200 = 1201, consistent with NCERT's remark that ferromagnetic μr exceeds 1000.
  6. Answer: ferromagnetic.
Assumption
Temperature below the Curie point — above it the material would read as paramagnetic instead.
Shortcut trick
Order of magnitude alone decides: 10−5-ish ⇒ dia or para (sign decides which); 102 to 104 ⇒ ferro. You never need the exact number.
Answer: (C) ferromagnetic
Q43μr from χNEET

A magnetic material has susceptibility χ = 500. Its relative permeability is

Given
χ = 500
Asked
μr
Concept applied
The three quantities χ, μr and μ are interlinked, and only one is independent. The link between the first two is a simple addition of 1 — not a multiplication.
Formula to use
μr = 1 + χ   and   μ = μ0μr
Baby steps
  1. Recall the relation: μr = 1 + χ.
  2. Substitute χ = 500.
  3. μr = 1 + 500.
  4. μr = 501.
  5. Note it is dimensionless, like χ itself.
  6. If μ were also asked: μ = μ0μr = 4π × 10−7 × 501 ≈ 6.3 × 10−4 T m A−1.
Assumption
Linear material, so χ is a constant rather than a function of H.
Shortcut trick
Where the +1 comes from: B = μ0(H + M) = μ0(H + χH) = μ0(1 + χ)H. The 1 is the vacuum's own contribution — it never disappears, however large χ gets. Writing μr = χ is the single most common error in this section.
Answer: (C) 501
Q44SuperconductorNEET + JEE

For a superconductor exhibiting the Meissner effect, the values of χ and μr are

Given
A superconductor showing complete expulsion of magnetic flux.
Asked
Its χ and μr.
Concept applied
Perfect diamagnetism: the induced magnetisation exactly cancels the applied field, so B = 0 inside. That extreme case forces χ to its most negative allowed value, −1.
Formula to use
B = μ0(1 + χ)H = 0 ⇒ χ = −1 ⇒ μr = 0
Baby steps
  1. Meissner effect means the field inside is completely expelled: B = 0.
  2. From B = μ0(1 + χ)H with H ≠ 0, the bracket must vanish.
  3. 1 + χ = 0 ⇒ χ = −1.
  4. Then μr = 1 + χ = 1 − 1 = 0.
  5. And μ = μ0μr = 0 as well.
  6. This is why Table 5.2 sets the diamagnetic lower bound at exactly χ = −1 — nothing can oppose more than total cancellation.
Assumption
Type-I superconductor below its critical temperature and critical field.
Shortcut trick
'Perfect diamagnet' is a fixed phrase meaning the pair (−1, 0). Note that NCERT calls a superconductor both a perfect conductor and a perfect diamagnet, and adds that no classical theory unites the two — BCS theory did, in 1957.
Answer: (B) χ = −1, μr = 0
Q45Permeability comparisonNEET

Which class of material has magnetic permeability less than that of free space?

Given
Comparison of μ with μ0 across the three classes.
Asked
Which class satisfies μ < μ0.
Concept applied
μ = μ0(1 + χ). Whether μ is above or below μ0 is decided entirely by the sign of χ. Only diamagnetic materials have negative χ.
Formula to use
μ = μ0(1 + χ)
Baby steps
  1. Diamagnetic: χ < 0 ⇒ (1 + χ) < 1 ⇒ μ < μ0.
  2. Paramagnetic: χ > 0 but tiny ⇒ (1 + χ) slightly above 1 ⇒ μ > μ0.
  3. Ferromagnetic: χ ≫ 1 ⇒ μ ≫ μ0.
  4. Only the diamagnetic case gives μ below μ0.
  5. Answer: diamagnetic.
  6. Physical picture, Fig. 5.7(a): field lines are pushed out of a diamagnetic bar, so the interior field is weaker than it would have been in vacuum.
Assumption
Linear, isotropic materials at room temperature.
Shortcut trick
Three quantities, one sign. χ < 0 ⇒ μr < 1 ⇒ μ < μ0 ⇒ lines expelled ⇒ repelled by a magnet. Learn the whole chain as one unit rather than five separate facts.
Diagram
DIAMAGNETIC lines pushed out PARAMAGNETIC lines pulled in
Answer: (C) diamagnetic

G · Behaviour of matter in a field and field-line rules

5 questions
Q46Non-uniform fieldNEET

A rod suspended between the poles of a magnet sets itself perpendicular to the field and drifts towards the weaker region. The rod is

STRONG FIELD WEAK FIELD diamagnetic para / ferro
Given
The rod aligns perpendicular to B and moves from the strong-field region to the weak-field region.
Asked
The magnetic classification of the rod.
Concept applied
Diamagnetic materials develop a magnetisation opposite to the applied field, so they are repelled. In a non-uniform field this repulsion pushes them towards the weaker region, and they settle across the field rather than along it.
Formula to use
χ < 0 ⇒ M opposite H ⇒ repelled, strong → weak
Baby steps
  1. Note the direction of drift: strong field → weak field.
  2. Moving away from the strong region means the material is being repelled.
  3. Repulsion requires an induced moment opposite to the applied field, i.e. χ < 0.
  4. Only diamagnetic materials have χ < 0.
  5. The perpendicular alignment is the same effect expressed as a torque — the induced moment opposes B, so the rod is most comfortable lying across it.
  6. Answer: diamagnetic.
Assumption
The field between the pole pieces is non-uniform (it always is near the edges); the rod is free to rotate and translate.
Shortcut trick
Direction of drift is the whole question. Towards weak = diamagnetic. Towards strong = para or ferro. Alignment follows automatically: dia sits across the field, para and ferro sit along it.
Answer: (C) diamagnetic
Q47Field lines in matterNEET

When a bar of a certain material is placed in an external field, the field lines become concentrated inside it and the interior field is enhanced. The material is

DIAMAGNETIC lines pushed out PARAMAGNETIC lines pulled in
Given
Field lines crowd into the bar; the interior field is greater than the applied field.
Asked
Which class or classes this behaviour identifies.
Concept applied
Concentration of lines means Binside > Boutside, which needs μ > μ0 and hence χ > 0. Both paramagnets and ferromagnets qualify; the description does not say by how much, so it cannot single out ferromagnetism.
Formula to use
B = μ0(H + M); M > 0 ⇒ B enhanced
Baby steps
  1. Line crowding ⇒ higher flux density inside ⇒ B enhanced.
  2. Enhancement requires M in the same direction as H, so χ > 0.
  3. Diamagnetic (χ < 0) is eliminated — its lines are expelled.
  4. Superconductor is eliminated — it expels lines completely.
  5. Both paramagnetic (slight enhancement, about 1 part in 105) and ferromagnetic (huge enhancement) fit the description.
  6. Answer: paramagnetic or ferromagnetic.
Assumption
No quantitative figure is given for the enhancement — had the question said 'highly concentrated' or given μr > 1000, ferromagnetic alone would be correct.
Shortcut trick
Watch the adjective. NCERT uses 'slight, about one part in 105' for paramagnets and 'highly concentrated' for ferromagnets. If the question omits the adjective, the answer must include both.
Answer: (B) paramagnetic or ferromagnetic
Q48Strong attractionNEET

A substance is strongly attracted by a magnet and retains its magnetisation after the field is removed. It is

Given
Strong attraction, and magnetisation persists after the external field is switched off.
Asked
The precise classification.
Concept applied
Strong attraction narrows it to ferromagnetic. Within ferromagnets, NCERT distinguishes hard (magnetisation persists → permanent magnets) from soft (magnetisation disappears → electromagnet cores). Retention is the deciding word.
Formula to use
χ ≫ 1, and M ≠ 0 when H = 0 ⇒ hard ferromagnet
Baby steps
  1. Strong attraction ⇒ χ ≫ 1 ⇒ ferromagnetic. Para and dia are eliminated.
  2. Now decide hard versus soft using the retention clue.
  3. Soft ferromagnets (e.g. soft iron) lose magnetisation the moment the field is removed.
  4. Hard ferromagnets (e.g. alnico, lodestone) retain it.
  5. The substance retains magnetisation ⇒ hard ferromagnet.
  6. These are exactly the materials used to make permanent magnets and compass needles.
Assumption
Temperature below the Curie point; otherwise it would behave as a paramagnet.
Shortcut trick
Map the word to the application: retains → hard → permanent magnet; loses → soft → electromagnet or transformer core. The application named in a question is often the fastest route to the classification.
Answer: (C) a hard ferromagnet
Q49Field-line rulesNEET + JEE

Which of the following statements about magnetic field lines is incorrect?

Given
Four candidate statements about magnetic field lines.
Asked
The one that is false.
Concept applied
NCERT 5.2.1 lists exactly four properties. Three of the options restate them correctly; the intersection statement contradicts property (iv) directly.
Formula to use
Property (iv): field lines never intersect — B must be unique at every point
Baby steps
  1. Option (A): closed loops — property (i). True.
  2. Option (B): tangent gives direction of B — property (ii). True.
  3. Option (D): line density measures field strength — property (iii). True.
  4. Option (C): if two lines crossed, the tangent at the crossing point would give two different directions for B.
  5. The magnetic field has one definite value at each point, so that is impossible — and the strength of the field is irrelevant to the argument.
  6. The incorrect statement is (C).
Assumption
Static magnetic fields; the rule applies equally to electric field lines.
Shortcut trick
The reason for the non-intersection rule is always the same one sentence: 'the direction of the field would not be unique.' Learn that sentence — it is the expected justification in every board and NEET answer key.
Answer: (C) Two field lines can intersect where the field is strong
Q50Complete expulsionNEET

Field lines are found to be completely expelled from the interior of a cooled metal sample, so that B = 0 inside. The sample is

Given
Complete expulsion of magnetic flux; B = 0 in the interior; sample is a cooled metal.
Asked
The identification of the sample.
Concept applied
Ordinary diamagnetism reduces the interior field by roughly one part in 105 — a slight effect. Only perfect diamagnetism drives B to exactly zero, and that is the Meissner effect in a superconductor.
Formula to use
B = 0 inside ⇒ χ = −1, μr = 0 (perfect diamagnetism)
Baby steps
  1. Note the word completely — the field is not just reduced, it is zero.
  2. An ordinary diamagnet reduces B by about 1 part in 105, nowhere near zero. Eliminate (A).
  3. Paramagnets and ferromagnets increase the interior field. Eliminate (C) and (D).
  4. Total expulsion requires χ = −1, the extreme end of the diamagnetic range.
  5. The clue 'cooled metal' matches NCERT's description of superconductors as metals cooled to very low temperatures.
  6. Answer: a superconductor showing the Meissner effect.
Assumption
The sample is below its critical temperature and the applied field is below the critical field.
Shortcut trick
Read the adverb. 'Reduced slightly' ⇒ ordinary diamagnet. 'Completely expelled' ⇒ superconductor. One word separates two different answers, and NEET sets both versions.
Answer: (B) a superconductor showing the Meissner effect

How to use this pack

First pass — untimed. Attempt every question, read the full solution even when you were right. The 'Shortcut trick' line is the part worth copying into your formula notebook; the baby steps are there for when a method has slipped.

Second pass — timed. 50 questions in 45 minutes. Anything slower than 55 seconds per question in Tier 1 will cost you in the real paper, because these are the types you must clear fast to buy time elsewhere.

Log your errors by type, not by score. Use the group filter at the top. If more than two errors land in the same group, that group needs a re-read of the concept, not more questions.

The two Tier 1 groups that most often hide silent errors are Group A (direction of the equatorial field) and Group G (which way a sample drifts in a non-uniform field). Both are direction-reasoning, and both are easy to get right for the wrong reason. If you answered those correctly but could not say why in one sentence, treat them as wrong.