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Chapter test · Physics · Class 12 · Ch 4 + Ch 5

Moving Charges and Magnetism + Magnetism and Matter — 80-Question Test

Aamirah Fathima · NEET 2027 · every question lost on this paper, rebuilt in full · ‡ marks gap content (cyclotron, earth's magnetism, hysteresis)

203marks / 320
55correct
17attempted, wrong
8left blank
117marks lost (incl. −17)
Target was 220+, danger line 190: this lands between the two, 17 marks short. Section G (galvanometer, ammeter, voltmeter) was 7 / 7 — the shunt-versus-series drill has worked. The losses are concentrated in three places: directions, formula factors, and assertion–reason.

Weak areas from this paper

Assertion–reason has not transferred to Physics recurring

Q15, Q23, Q65, Q74 — four A–R items attempted, three wrong, in both directions: a false assertion accepted (Q15), a true one rejected (Q23, Q74), and a false reason accepted (Q65). Her Biology A–R is now 8/8 plus 10/10 on the drill, so this is a habit that is not being carried across subjects.

Fix to drill. Write the same three lines in the Physics margin: A: T/F — reason, R: T/F — reason, then 'A because R' aloud. In Physics, test R with one formula line (B_H = B_E cos I, μ_r = 1 + χ, no E field inside a dee).

Direction questions answered without the triad recurring

Q2, Q3, Q35, Q36 — 20 marks. Q2's answer (West) is not even one of the two possible directions, so no cross product was worked out; Q3 lost the sign of the electron; Q35 used the wire rule on charged beams; Q36 put the null point outside the wires.

Fix to drill. At the top of every Physics paper write x̂ → ŷ → ẑ and, beside it, null point → nearer the smaller current. Apply the sign of q only at the last step.

Factors inside formulas: fractions, powers, r vs d recurring

Q7, Q17, Q20, Q30, Q40, Q57 — 30 marks. Each answer was a clean multiple of the right one: sin for cos, 60/180 for 60/360, power 1 for 3/2, 1/r inside the conductor, diameter for radius, and a period ratio left unsquared.

Fix to drill. One value per line with its unit before substituting: θ = …, fraction = …/360, r = … m, (bracket)^{3/2} = …, then compute. No two-step arithmetic in the head.

Magnetic materials and hysteresis: eight blanks, four of them here attempt rate

Q73, Q77, Q78, Q80 (plus Q64, Q60, Q14, Q24) — 32 marks left untouched. Q64, Q73, Q77, Q78 and Q80 are one-line recall that needs no calculation at all.

Fix to drill. A single card: χ negative → diamagnetic → repelled → strong to weak · χ small positive → paramagnetic · χ large → ferromagnetic, becomes paramagnetic above T_c · permanent magnet = high retentivity + high coercivity · electromagnet core = high permeability + low retentivity.

The twenty-five at a glance

QStatusMarkedKeyTopicError family
Q1wrongBDLorentzAngle between v and B misread
Q2wrongADLorentzCross-product sign
Q3wrongDBLorentzSign of the charge applied too early
Q7wrongBDHelical motioncos vs sin (pitch vs radius)
Q8wrongBACircular motionWhich quantity is held constant
Q14blank—BCyclotronBlank — cyclotron energy
Q15wrongADCyclotronAssertion–Reason: where the energy is gained
Q17wrongCABiot–Savart arcArc fraction measured against the wrong circle
Q20wrongCABiot–Savart loopExponent 3/2 dropped
Q23wrongDABiot–Savart loopAssertion–Reason link denied
Q24blank—CBiot–Savart compositeBlank — finite sides need the sin terms
Q30wrongCAAmpèreInside or outside decided too late
Q31wrongACAmpèreSolid graph given for a hollow conductor
Q35wrongCBParallel wiresWires vs beams
Q36wrongADParallel wiresNull point placed on the wrong side
Q40wrongCAMagnetic momentRadius, diameter and the area formula
Q57wrongBCDipole oscillationPeriod ratio not squared
Q60blank—ADipole fieldBlank — length hidden inside the formula
Q64blank—DEarth's fieldBlank — declination vs dip
Q65wrongACEarth's fieldAssertion–Reason: components swapped
Q73blank—DMaterialsBlank — which way a diamagnet moves
Q74wrongDBMaterialsAssertion–Reason: two true statements judged as one
Q77blank—BHysteresisBlank — permanent magnet material
Q78blank—DHysteresisBlank — electromagnet core
Q80blank—BMaterialsBlank — above the Curie temperature

Question by question

Lorentz Force and Motion in a Magnetic Field

Q1Q2Q3Q7Q8
Q1Attempted · wrong Angle between v and B misread
An electron moves with speed 2 × 10⁶ m/s perpendicular to a uniform magnetic field of 0.5 T. The magnitude of the magnetic force on it is
  1. 3.2 × 10⁻¹³ N
  2. zero her answer
  3. 8 × 10⁻¹⁴ N
  4. 1.6 × 10⁻¹³ N ✓ key
Given
  • q = 1.6 × 10⁻¹⁹ C
  • v = 2 × 10⁶ m/s
  • B = 0.5 T
  • θ = 90°
Concept
The magnetic force is F = qvB sin θ. Here the electron enters perpendicular to B, so θ = 90° and sin θ = 1 — the force is maximum, not zero. Zero force happens only when the velocity is parallel (or antiparallel) to B, where sin θ = 0.
Formula
F = qvB sin θ
Baby steps
  1. sin 90° = 1
  2. F = 1.6 × 10⁻¹⁹ × 2 × 10⁶ × 0.5
  3. F = 1.6 × 10⁻¹³ N
Answer
(D) 1.6 × 10⁻¹³ N
Why not the others
A · rule out3.2 × 10⁻¹³ N doubles the charge.
B · rule outZero needs v ∥ B; here v ⊥ B.
C · rule out8 × 10⁻¹⁴ N halves the product.
D · keepqvB = 1.6 × 10⁻¹⁹ × 2 × 10⁶ × 0.5.
Shortcut
Write θ on the diagram before touching the formula: ⊥ → sin = 1 (maximum), ∥ → sin = 0 (zero).
Trap (as built)
Zero is the answer only when v ∥ B. Doubling or halving comes from mixing up magnitude of charge.
from: ILTS Q131/Q133 (angle between v and B)
Where it went wrong
Marked (B): zero.
'Zero' is the answer to the other standard question — the one where the particle moves ALONG the field (as in Q10 of this paper, which she got right). The word 'perpendicularly' in the stem was passed over.
Q2Attempted · wrong Cross-product sign
A proton moves towards East in a uniform magnetic field directed vertically upward. The magnetic force on it is directed towards (Take x̂ = East, ŷ = North, ẑ = vertically up.)
  1. West her answer
  2. vertically downward
  3. North
  4. South ✓ key
Given
  • v along +x̂ (East)
  • B along +ẑ (up)
  • q positive
Concept
Set the axes as the paper does: x̂ = East, ŷ = North, ẑ = up. The cycle x̂ → ŷ → ẑ → x̂ gives a plus sign; going backwards gives a minus. Here v ∝ x̂ and B ∝ ẑ, so v × B ∝ x̂ × ẑ, which runs backwards in the cycle and equals −ŷ. The charge is positive, so the force is along −ŷ = South.
Diagram
x̂(East)ŷ(North)ẑ(up)forward = +x̂ × ŷ = +ẑ ŷ × ẑ = +x̂ẑ × x̂ = +ŷbackwards → put a minus signProton East, B up (Q2)v × B ∝ x̂ × ẑ = −ŷ → SOUTHElectron +x, B +y (Q3)x̂ × ŷ = +ẑ, then q = −e flips it→ force along −zWrite the cycle first, read the cross product, apply the sign of q LAST
Animated · the still frame carries the full message.
Formula
F = q (v × B)
Baby steps
  1. v × B ∝ x̂ × ẑ
  2. x̂ × ẑ runs backwards round the cycle x̂ → ŷ → ẑ, so it is −ŷ
  3. −ŷ = South; q is positive, so F points South
Answer
(D) South
Why not the others
A · rule outWest would be −x̂; the cross product has no x̂ component.
B · rule out'Downward' would need a force along −ẑ.
C · rule outNorth is +ŷ — the sign-flipped answer.
D · keepx̂ × ẑ = −ŷ, and q is positive.
Shortcut
Draw the three-letter cycle in the margin of every direction question. Forward = +, backwards = −.
Trap (as built)
North is the sign-flipped answer: x̂ × ẑ = −ŷ, not +ŷ. Write the triad in the margin.
from: ILTS Q95, Q105 (cross-product sign)
Where it went wrong
Marked (A): West.
West is neither of the two possible answers (±ŷ) — the cross product was not worked out at all, only guessed. Two minutes of writing x̂ ŷ ẑ at the top of the page would have protected Q2, Q3, Q35 and Q36 together (20 marks).
Q3Attempted · wrong Sign of the charge applied too early
An electron moves along +x in a uniform magnetic field along +y. The magnetic force on the electron is along
  1. +y
  2. −z ✓ key
  3. −x
  4. +z her answer
Given
  • v along +x̂
  • B along +ŷ
  • q = −e
Concept
x̂ × ŷ = +ẑ. That is the direction for a positive charge. An electron carries −e, so the force is reversed: along −z.
Diagram
x̂(East)ŷ(North)ẑ(up)forward = +x̂ × ŷ = +ẑ ŷ × ẑ = +x̂ẑ × x̂ = +ŷbackwards → put a minus signProton East, B up (Q2)v × B ∝ x̂ × ẑ = −ŷ → SOUTHElectron +x, B +y (Q3)x̂ × ŷ = +ẑ, then q = −e flips it→ force along −zWrite the cycle first, read the cross product, apply the sign of q LAST
Animated · the still frame carries the full message.
Formula
F = q (v × B)
Baby steps
  1. x̂ × ŷ = +ẑ
  2. q = −e reverses it: F = −e(v × B) points along −ẑ
  3. Force along −z
Answer
(B) −z
Why not the others
A · rule out+y is neither v nor F here.
B · keepx̂ × ŷ = +ẑ, reversed by q = −e.
C · rule out−x is not perpendicular to both v and B in the right sense.
D · rule out+z is the answer for a proton.
Shortcut
Do it in two steps: (1) work out v × B for a positive charge, (2) flip it if the particle is an electron. Never mix the two steps.
Trap (as built)
+z is the force on a proton. The negative charge flips the direction — check the sign of q last.
from: ILTS Q105, Q109 (charge sign)
Where it went wrong
Marked (D): +z.
+z is the proton answer. The cross product was right and the sign of the charge was dropped.
Q7Attempted · wrong cos vs sin (pitch vs radius)
A proton enters a uniform magnetic field of 0.5 T with speed 6 × 10⁵ m/s at an angle of 30° to the field. The pitch of its helical path is close to (mp = 1.67 × 10⁻²⁷ kg)
  1. 0.63 cm
  2. 3.9 cm her answer
  3. 13.6 cm
  4. 6.8 cm ✓ key
Given
  • v = 6 × 10⁵ m/s at 30° to B
  • B = 0.5 T
  • m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C
Concept
Split v into a part along B and a part across it. The part along B (v cos θ) is untouched by the magnetic force and carries the particle forward — it fixes the pitch. The part across B (v sin θ) makes the circle and fixes the radius. Pitch = v cos θ × T with T = 2πm/qB.
Diagram
Bv30°v cos30° → PITCH (along B)v sin30° → circle (radius)pitch = v∥ × T, T = 2πm/qBThe component ALONG B carries the particle forward — that is the pitch
Animated · the still frame carries the full message.
Formula
T = 2πm/qB ; pitch = (v cos θ) × T
Baby steps
  1. T = 2π × 1.67 × 10⁻²⁷ / (1.6 × 10⁻¹⁹ × 0.5) = 1.31 × 10⁻⁷ s
  2. v∥ = v cos 30° = 5.2 × 10⁵ m/s
  3. pitch = 5.2 × 10⁵ × 1.31 × 10⁻⁷ = 6.82 × 10⁻² m ≈ 6.8 cm
Answer
(D) ≈ 6.8 cm
Why not the others
A · rule out0.63 cm is the radius r = mv sin θ/qB.
B · rule out3.9 cm uses v sin 30° — that component makes the circle.
C · rule out13.6 cm is twice the pitch.
D · keepv cos 30° × T = 6.8 cm.
Shortcut
Label the two components on the sketch: along B → pitch (cos), across B → radius (sin).
Trap (as built)
3.9 cm uses sin 30° — that component makes the circle, not the forward drift. 0.63 cm is the radius.
from: ILTS Q119 (cos vs sin)
Where it went wrong
Marked (B): 3.9 cm.
sin 30° was used for the pitch. Both components appear in the same problem, so the label on the sketch is what keeps them apart.
Q8Attempted · wrong Which quantity is held constant
A proton and an α-particle enter a uniform magnetic field perpendicularly with the same kinetic energy. The ratio of the radii of their paths rp : rα is
  1. 1 : 1 ✓ key
  2. 2 : 1 her answer
  3. 1 : √2
  4. 1 : 2
Given
  • mα = 4mp
  • qα = 2qp
  • same K
Concept
r = mv/qB. With equal kinetic energy, v is not the same for the two particles: v = √(2K/m), so r = √(2mK)/qB, i.e. r ∝ √m / q. For the α-particle, m is 4× and q is 2×, so r_p : r_α = (√1/1) : (√4/2) = 1 : 1.
Formula
r = mv/qB = √(2mK)/qB
Baby steps
  1. r ∝ √m / q at fixed K and B
  2. rp/rα = √(m/4m) × (2q/q)
  3. = (1/2) × 2 = 1
Answer
(A) 1 : 1
Why not the others
A · keep√m/q is the same for both at equal K.
B · rule out2 : 1 is the equal-speed answer inverted.
C · rule out1 : √2 uses √m only.
D · rule out1 : 2 uses r ∝ m/q, valid only at equal speeds.
Shortcut
Read the fixed quantity first: same speed → r ∝ m/q; same energy → r ∝ √m/q; same momentum → r ∝ 1/q.
Trap (as built)
1 : 2 uses r ∝ m/q, which is true only for equal speeds, not equal energies.
from: Two-quantity ratio (ILTS Q127 pattern)
Where it went wrong
Marked (B): 2 : 1.
The equal-speed rule was applied to an equal-energy question. Both rules are needed in this chapter, so write the three-line table above in the margin.

Cyclotron ‡

Q14Q15
Q14Left blank Blank — cyclotron energy
A cyclotron with dees of radius 0.5 m and magnetic field 1.0 T accelerates protons. The maximum kinetic energy of the protons is about (mp = 1.67 × 10⁻²⁷ kg)
  1. 6 MeV
  2. 12 MeV ✓ key
  3. 48 MeV
  4. 24 MeV
Given
  • R = 0.5 m
  • B = 1.0 T
  • m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C
Concept
At the rim of the dees, v = qBR/m, so K = ½mv² = q²B²R²/2m. Everything in the formula is given; only the unit conversion (J → MeV, divide by 1.6 × 10⁻¹³) remains.
Diagram
D₁D₂gaposcillatoracross gapB ⊥ plane
Animated · the still frame carries the full message.
Formula
vmax = qBR/m ; K = q²B²R²/2m
Baby steps
  1. q²B²R² = (1.6 × 10⁻¹⁹)² × 1 × 0.25 = 6.4 × 10⁻³⁹
  2. K = 6.4 × 10⁻³⁹ / (3.34 × 10⁻²⁷) = 1.92 × 10⁻¹² J
  3. In MeV: 1.92 × 10⁻¹² / 1.6 × 10⁻¹³ ≈ 12 MeV
Answer
(B) ≈ 12 MeV
Why not the others
A · rule out6 MeV halves the result.
B · keepq²B²R²/2m = 1.92 × 10⁻¹² J ≈ 12 MeV.
C · rule out48 MeV uses the diameter as R.
D · rule out24 MeV drops the ½ in ½mv².
Shortcut
Memorise the one line K = q²B²R²/2m; it is the only cyclotron numerical NEET asks.
Trap (as built)
24 MeV drops the ½ in K = ½mv²; 48 MeV uses the diameter (1 m) as R.
from: Factor-of-2 trap
Q15Attempted · wrong Assertion–Reason: where the energy is gained
Assertion (A): A charged particle in a cyclotron gains kinetic energy while it moves inside the dees.
Reason (R): There is no electric field inside the dees; the particle is accelerated only while crossing the gap.D₁D₂gaposcillatoracross gapB ⊥ plane
  1. Both A and R are true and R is the correct explanation of A her answer
  2. Both A and R are true but R is NOT the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true ✓ key
Given
  • Oscillator voltage is applied across the gap
  • Dees are metal boxes
Concept
Inside a dee there is no electric field (the metal shields it), so only the magnetic force acts there — and the magnetic force does no work. The particle merely turns inside the dees. All the energy is gained while crossing the gap, where the oscillator's electric field acts. So A is false and R is true.
Diagram
D₁D₂gaposcillatoracross gapB ⊥ plane
Animated · the still frame carries the full message.
Formula
Energy gain = qV per gap crossing
Baby steps
  1. Inside a dee the field is shielded, so only the magnetic force acts
  2. The magnetic force does no work → no energy gained inside
  3. A is false; R is true
Answer
(D) A is false but R is true
Why not the others
A · rule outWould require energy gain inside the dees.
B · rule outA is false, so 'both true' fails.
C · rule outR is true — the dees are field-free.
D · keepA false, R true.
Shortcut
In every cyclotron question ask: turning (dees, magnetic) or speeding up (gap, electric)?
Trap (as built)
The particle only turns inside the dees; the gap is where energy is added.
from: AR verdict flips (MM B1 ILTS-09 Q15, Q44)
Where it went wrong
Marked (A): Both A and R are true and R is the correct explanation of A.
Option A accepted both statements. R contradicts A directly — if there is no electric field inside the dees, the particle cannot gain energy there. Reading R as a check on A settles this pair in seconds.

Biot–Savart Law

Q17Q20Q23Q24
Q17Attempted · wrong Arc fraction measured against the wrong circle
A wire carrying 4 A is bent into a circular arc of radius 2 cm that subtends 60° at the centre. The magnetic field at the centre is
  1. 2.09 × 10⁻⁵ T ✓ key
  2. 1.26 × 10⁻⁴ T
  3. 4.19 × 10⁻⁵ T her answer
  4. 1.05 × 10⁻⁵ T
Given
  • I = 4 A
  • R = 0.02 m
  • θ = 60° = π/3 rad
Concept
An arc contributes the same fraction of the full-loop field as it does of the full circle: B = (θ/360°) × μ₀I/2R. For 60°, that fraction is 60/360 = 1/6.
Diagram
60°arc of radius Rfraction = 60/360 = 1/6 of the FULL circleB = (1/6) × μ₀I/2R = μ₀Iθ/4πR60/180 = 1/3 compares it with a SEMICIRCLE ✗A fraction of an arc is always measured against 360°
Animated · the still frame carries the full message.
Formula
B = (θ/2π) × μ₀I/2R = μ₀Iθ/4πR
Baby steps
  1. Fraction of full circle = 60/360 = 1/6
  2. Full loop: μ₀I/2R = 1.26 × 10⁻⁴ T
  3. B = (1/6) × that = 2.09 × 10⁻⁵ T
Answer
(A) 2.09 × 10⁻⁵ T
Why not the others
A · keep(1/6) × μ₀I/2R = 2.09 × 10⁻⁵ T.
B · rule out1.26 × 10⁻⁴ T is the full loop.
C · rule out4.19 × 10⁻⁵ T uses 60/180 — a semicircle.
D · rule out1.05 × 10⁻⁵ T uses 1/12.
Shortcut
Write the fraction first, as a fraction of 360, and only then multiply the full-loop field.
Trap (as built)
4.19 × 10⁻⁵ T measures the arc against a semicircle (60/180) instead of the full circle (60/360).
from: ILTS Q92 (marked exactly double)
Where it went wrong
Marked (C): 4.19 × 10⁻⁵ T.
Exactly double the correct value — the same 60/180 slip as ILTS Q92. The semicircle is a tempting reference because the half-circle formula (μ₀I/4R) is memorised, but the reference for an arc is always the full circle.
Q20Attempted · wrong Exponent 3/2 dropped
For a circular current loop of radius R, the ratio of the magnetic field at an axial point at distance √3 R from the centre to the field at the centre is
  1. 1 : 8 ✓ key
  2. 1 : 2
  3. 1 : 4 her answer
  4. 1 : 16
Given
  • x = √3 R on the axis
Concept
On the axis of a loop, B = μ₀IR²/2(R² + x²)^{3/2}. With x = √3 R, R² + x² = 4R², and (4R²)^{3/2} = 8R³, so the field is one-eighth of the centre value μ₀I/2R.
Formula
B(x) = μ₀IR² / 2(R² + x²)3/2
Baby steps
  1. B(0) = μ₀I/2R
  2. R² + x² = 4R², (4R²)3/2 = 8R³
  3. B(x)/B(0) = R³/8R³ = 1/8
Answer
(A) 1 : 8
Why not the others
A · keepR³/8R³ = 1 : 8.
B · rule out1 : 2 would need (R²+x²)^{1/2}.
C · rule out1 : 4 uses power 1 instead of 3/2.
D · rule out1 : 16 squares the 4 without the 3/2.
Shortcut
Do the bracket first (4R²), then apply the power 3/2 = 'square root, then cube': √4 = 2, 2³ = 8.
Trap (as built)
1 : 4 uses power 1 instead of 3/2.
from: Exponent slip
Where it went wrong
Marked (C): 1 : 4.
1 : 4 treats the denominator as (R² + x²) — the 3/2 power was lost. Writing (4R²)^{3/2} = 8R³ as its own line prevents it.
Q23Attempted · wrong Assertion–Reason link denied
Assertion (A): The magnetic field at the centre of a circular current loop is perpendicular to the plane of the loop.
Reason (R): For every current element of the loop, dl × r̂ at the centre points along the axis of the loop.
  1. Both A and R are true and R is the correct explanation of A ✓ key
  2. Both A and R are true but R is NOT the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true her answer
Given
  • Circular loop, field point at the centre
Concept
At the centre of a loop, every current element dl is tangential and its r̂ points inward along the radius; dl × r̂ is therefore perpendicular to the plane of the loop — the same direction for every element, so all the contributions add along the axis. That is exactly why the field at the centre is perpendicular to the plane. Both statements are true and R is the reason.
Formula
dB = (μ₀/4π) I dl × r̂ / r²
Baby steps
  1. dl is tangential and r̂ is radial, both in the plane
  2. Their cross product is perpendicular to the plane, the same way for every element
  3. So all dB add along the axis → A follows from R
Answer
(A) Both true, R explains A
Why not the others
A · keepR is the mechanism that produces A.
B · rule outDenies a real link.
C · rule outR is a correct statement of the Biot–Savart geometry.
D · rule outA is true — the field at the centre is axial.
Shortcut
Say 'A because R' aloud. If R describes the mechanism, choose option A.
Trap (as built)
When R is the mechanism that produces A, the link is real — she has denied such links before.
from: AR link denied (HHD 60Q Q53 pattern; MCM Q120)
Where it went wrong
Marked (D): A is false but R is true.
Marked 'A false, R true' — so the assertion itself was rejected, although the axial field at the centre of a loop is standard. The same over-correction appeared in Q65 and Q74 on this paper: the three-step defence is being used in Biology but not in Physics.
Q24Left blank Blank — finite sides need the sin terms
A square loop of side 10 cm carries a current of 5 A. The magnetic field at its centre isOside a, current I
  1. 1.41 × 10⁻⁵ T
  2. 8 × 10⁻⁵ T
  3. 5.66 × 10⁻⁵ T ✓ key
  4. 2.83 × 10⁻⁵ T
Given
  • a = 0.1 m, I = 5 A
  • distance of centre from each side d = a/2
  • each side subtends 45° on either side
Concept
Each side of the square is a finite wire: B = (μ₀I/4πd)(sin θ₁ + sin θ₂) with d = a/2 and both angles 45°. One side gives (μ₀I/4πd)·√2; four sides give B = 2√2 μ₀I/πa.
Diagram
Oside a, current I
Animated · the still frame carries the full message.
Formula
One side: B = (μ₀I/4πd)(sin 45° + sin 45°)
Baby steps
  1. d = 0.05 m; sin 45° + sin 45° = √2
  2. One side: 10⁻⁷ × 5/0.05 × √2 = 1.41 × 10⁻⁵ T
  3. Four sides, same direction: B = 4 × that = 5.66 × 10⁻⁵ T (= 2√2 μ₀I/πa)
Answer
(C) 5.66 × 10⁻⁵ T
Why not the others
A · rule out1.41 × 10⁻⁵ T is one side only.
B · rule out8 × 10⁻⁵ T treats each side as an infinite wire.
C · keep4 × (10⁻⁷ × 5/0.05) × √2 = 5.66 × 10⁻⁵ T.
D · rule out2.83 × 10⁻⁵ T is half the total.
Shortcut
Infinite wire → μ₀I/2πd. Finite wire → keep (sin θ₁ + sin θ₂). For a square, that bracket is always √2 and the answer is 2√2 μ₀I/πa.
Trap (as built)
8 × 10⁻⁵ T treats each side as an infinite wire (μ₀I/2πd). A side is finite: keep the sin terms.
from: ILTS Q94 (infinite-wire formula)

Ampère's Law

Q30Q31
Q30Attempted · wrong Inside or outside decided too late
A long solid wire of radius a carries a current uniformly distributed over its cross-section. The ratio of the magnetic field at distance a/2 from the axis to that at 3a is
  1. 3 : 2 ✓ key
  2. 1 : 6
  3. 6 : 1 her answer
  4. 2 : 3
Given
  • r₁ = a/2 (inside)
  • r₂ = 3a (outside)
Concept
For a solid wire, B rises linearly inside (B ∝ r) up to the surface value B_s at r = a, then falls as 1/r outside. At r = a/2 the field is B_s/2; at r = 3a it is B_s/3. The ratio is 1/2 : 1/3 = 3 : 2.
Diagram
rBSOLID wire (Q30)RB ∝ rB ∝ 1/rrBHOLLOW cylinder (Q31)RB = 0Read 'solid' or 'hollow' first: a hollow cylinder encloses no current inside
Animated · the still frame carries the full message.
Formula
Inside: B ∝ r ; outside: B ∝ 1/r ; peak Bs at r = a
Baby steps
  1. B₁ = (1/2)Bs
  2. B₂ = (1/3)Bs
  3. B₁ : B₂ = 1/2 : 1/3 = 3 : 2
Answer
(A) 3 : 2
Why not the others
A · keep(1/2)B_s : (1/3)B_s = 3 : 2.
B · rule out1 : 6 inverts the pair.
C · rule out6 : 1 applies the 1/r law to the inside point.
D · rule out2 : 3 is the reciprocal.
Shortcut
Mark each radius as IN or OUT before computing: in → multiply by r/a, out → multiply by a/r.
Trap (as built)
6 : 1 treats the inside point with the outside 1/r law. Decide inside or outside first.
from: ILTS Q101 (blank)
Where it went wrong
Marked (C): 6 : 1.
6 : 1 comes from using B ∝ 1/r for the point at a/2, which lies inside the conductor where only the enclosed current counts.
Q31Attempted · wrong Solid graph given for a hollow conductor
A long, thin hollow metal cylinder of radius R carries a steady current along its length. The magnetic field B varies with distance r from the axis as
  1. rises linearly from zero to r = R, then falls as 1/r her answer
  2. constant and non-zero for r < R, then falls as 1/r
  3. zero for r < R, then jumps at r = R and falls as 1/r ✓ key
  4. zero everywhere, inside and outside
Given
  • Hollow: all current flows on the surface
Concept
Ampère's law uses the current enclosed. A hollow cylinder carries all its current on the surface, so a loop drawn inside encloses nothing and B = 0 there. At r = R the whole current is suddenly enclosed, so B jumps to μ₀I/2πR and then falls as 1/r.
Diagram
rBSOLID wire (Q30)RB ∝ rB ∝ 1/rrBHOLLOW cylinder (Q31)RB = 0Read 'solid' or 'hollow' first: a hollow cylinder encloses no current inside
Animated · the still frame carries the full message.
Formula
∮B · dl = μ₀ Ienclosed
Baby steps
  1. Inside: an Amperian loop encloses no current → B = 0
  2. At r = R the whole current is enclosed
  3. Outside: B = μ₀I/2πr
Answer
(C) Zero inside, jump at R, then 1/r
Why not the others
A · rule outThe linear rise is the SOLID-wire graph.
B · rule outA constant non-zero field inside would need current spread through the interior.
C · keepZero inside, jump at R, 1/r outside.
D · rule outOutside, the full current is enclosed, so B cannot be zero.
Shortcut
Circle the word 'solid' or 'hollow' in the stem, then ask what the Amperian loop encloses.
Trap (as built)
The linear rise is the SOLID-conductor graph. Read 'solid' or 'hollow' first.
from: ILTS Q112 + Q128 (same graph for both)
Where it went wrong
Marked (A): rises linearly from zero to r = R, then falls as 1/r.
Both this and ILTS Q112/Q128 were answered with the solid-conductor graph. Keep the two sketches side by side (below) until the difference is automatic.

Parallel Currents

Q35Q36
Q35Attempted · wrong Wires vs beams
Two parallel beams of electrons move in the same direction. They
  1. attract or repel depending on their speed only
  2. repel each other ✓ key
  3. attract each other her answer
  4. neither attract nor repel
Given
  • Both beams are negatively charged
  • Both move the same way
Concept
Two parallel wires attract when their currents are parallel, because a wire is electrically neutral and only the magnetic force acts. Two parallel beams of electrons also carry net charge, and the electric repulsion between like charges is larger than the magnetic attraction by a factor of about c²/v². So the beams repel.
Diagram
−−−−v−−−−velectric repulsion (strong)magnetic attraction (weak)F(magnetic) / F(electric) ≈ v²/c² → the beams REPELWires are neutral, so only the magnetic force acts; beams are charged, so the electric force wins
Animated · the still frame carries the full message.
Formula
Fmagnetic / Felectric ≈ v²/c²
Baby steps
  1. Electric force: like charges → repulsion (strong)
  2. Magnetic force: parallel currents → attraction (weak)
  3. Electric force wins → repel
Answer
(B) They repel
Why not the others
A · rule outSpeed changes the size of the magnetic force, not the outcome.
B · keepElectric repulsion dominates → repel.
C · rule outAttraction is the answer for neutral wires.
D · rule outBoth forces exist, so 'neither' is impossible.
Shortcut
Ask first: is the object neutral? Wire → yes → magnetic only. Beam → no → electric force wins.
Trap (as built)
Parallel current-carrying WIRES attract because a wire is neutral. Beams are charged, so the electric force dominates.
from: ILTS Q113 (blank)
Where it went wrong
Marked (C): attract each other.
'Attract' is the wire rule copied onto a beam. This question was also left blank on ILTS Q113, so it is a gap rather than a slip.
Q36Attempted · wrong Null point placed on the wrong side
Two long parallel wires are placed at the 0 cm and 12 cm marks of a scale, carrying currents I and 2I in the same direction. The magnetic field is zero at the
  1. −12 cm mark her answer
  2. 6 cm mark
  3. 8 cm mark
  4. 4 cm mark ✓ key
Given
  • Wire 1 (I) at 0 cm
  • Wire 2 (2I) at 12 cm
  • same direction
Concept
Between two wires carrying current in the same direction, their fields oppose, so the null point lies between them. Setting μ₀I/2πx = μ₀(2I)/2π(12 − x) gives 12 − x = 2x, x = 4 cm — measured from the weaker current.
Diagram
I0 cm2I12 cmnull point4 cmI/x = 2I/(12 − x) → 12 − x = 2x → x = 4 cmthe null point always lies NEARER the smaller current
Animated · the still frame carries the full message.
Formula
μ₀I/2πx = μ₀(2I)/2π(12 − x)
Baby steps
  1. Same-direction currents → null point lies between the wires
  2. 12 − x = 2x → x = 4 cm
  3. Null point at the 4 cm mark (nearer the weaker current)
Answer
(D) 4 cm mark
Why not the others
A · rule out−12 cm is outside the pair; for same-direction currents no null exists there.
B · rule out6 cm is the midpoint, valid only for equal currents.
C · rule out8 cm is 4 cm from the stronger wire.
D · keepx = 4 cm from the wire carrying I.
Shortcut
Remember one line: the null point is always nearer the smaller current. That alone eliminates three options.
Trap (as built)
8 cm puts the null nearer the stronger wire. The null point always sits nearer the smaller current.
from: ILTS Q93 (blank)
Where it went wrong
Marked (A): −12 cm mark.
−12 cm places the null outside the wires, which only happens with opposite currents. Deciding 'between or outside' from the direction of the currents is the first step here.

Magnetic Moment and Oscillation

Q40Q57
Q40Attempted · wrong Radius, diameter and the area formula
A circular coil of 100 turns and radius 5 cm carries 0.5 A. Its magnetic dipole moment is
  1. 0.393 A m² ✓ key
  2. 0.125 A m²
  3. 1.57 A m² her answer
  4. 15.7 A m²
Given
  • N = 100
  • r = 0.05 m
  • I = 0.5 A
Concept
The magnetic moment of a coil is m = NIA, and for a circular coil A = πr². With N = 100, I = 0.5 A and r = 0.05 m: A = 7.85 × 10⁻³ m², so m = 0.393 A m².
Diagram
Bnormal30°coil (edge-on)
Animated · the still frame carries the full message.
Formula
m = NIA = NIπr²
Baby steps
  1. A = π(0.05)² = 7.85 × 10⁻³ m²
  2. m = 100 × 0.5 × 7.85 × 10⁻³ = 0.393 A m²
Answer
(A) 0.393 A m²
Why not the others
A · keepNIπr² = 100 × 0.5 × 7.85 × 10⁻³.
B · rule out0.125 A m² drops π.
C · rule out1.57 A m² uses the diameter (0.1 m) as r.
D · rule out15.7 A m² uses the circumference 2πr as the area.
Shortcut
Write r = … m on its own line before squaring. Area needs r², not d² and not 2πr.
Trap (as built)
1.57 A m² uses the diameter as r; 15.7 A m² uses the circumference 2πr as the area; 0.125 A m² drops π.
from: π and r/d slips
Where it went wrong
Marked (C): 1.57 A m².
1.57 A m² is exactly 4× the answer — the radius was doubled before squaring, i.e. the diameter was used. This is the same family as the radius/diameter slip in ILTS Q116.
Q57Attempted · wrong Period ratio not squared
A magnet of moment of inertia 9 × 10⁻² kg m² oscillates in a uniform field with period 6 s. A non-magnetic brass bar placed on it makes the period 10 s. The moment of inertia of the brass bar is
  1. 0.06 kg m²
  2. 0.25 kg m² her answer
  3. 0.16 kg m² ✓ key
  4. 0.0324 kg m²
Given
  • I₁ = 9 × 10⁻² kg m²
  • T₁ = 6 s, T₂ = 10 s
  • brass adds inertia only
Concept
For a magnet oscillating in a uniform field, T = 2π√(I/mB), so T ∝ √I and I ∝ T². Adding a non-magnetic brass bar changes only the moment of inertia: (I₁ + I′)/I₁ = (T₂/T₁)² = (10/6)² = 2.78, so I₁ + I′ = 0.25 kg m² and I′ = 0.25 − 0.09 = 0.16 kg m².
Formula
T = 2π√(I/mB) → T ∝ √I
Baby steps
  1. (T₂/T₁)² = 100/36 = 2.778
  2. I₁ + I′ = 2.778 × 0.09 = 0.25 kg m²
  3. I′ = 0.25 − 0.09 = 0.16 kg m²
Answer
(C) 0.16 kg m²
Why not the others
A · rule out0.06 kg m² forgets to square the ratio.
B · rule out0.25 kg m² is the TOTAL moment of inertia, not the bar's.
C · keep0.25 − 0.09 = 0.16 kg m².
D · rule out0.0324 kg m² squares in the wrong direction.
Shortcut
Two steps, both written down: (1) square the period ratio, (2) subtract the magnet's own I.
Trap (as built)
Forgetting to square the period ratio gives 0.06 kg m²; not subtracting I₁ gives 0.25 kg m².
from: MM B1 ILTS-09 Q3 (blank)
Where it went wrong
Marked (B): 0.25 kg m².
0.25 kg m² is the total — the last subtraction was skipped. The question asks for the brass bar alone.

Bar Magnet and Earth's Magnetism ‡

Q60Q64Q65
Q60Left blank Blank — length hidden inside the formula
The magnetic moment of a bar magnet is kept fixed while its length is increased. At a fixed distance from its centre, the field on the axial line and the field on the equatorial line, respectively,
  1. increases and decreases ✓ key
  2. both remain unchanged
  3. both decrease
  4. decreases and increases
Given
  • m fixed, ℓ increases, r fixed
Concept
For a magnet of finite length, B_axial = (μ₀/4π)·2mr/(r² − ℓ²)² and B_equatorial = (μ₀/4π)·m/(r² + ℓ²)^{3/2}. Keep m and r fixed and increase ℓ: the axial denominator (r² − ℓ²) gets smaller, so the axial field rises; the equatorial denominator (r² + ℓ²) gets larger, so that field falls.
Diagram
AXIAL pointB = (μ₀/4π) · 2mr /(r² − ℓ²)²ℓ ↑ → (r² − ℓ²) ↓ → B ↑EQUATORIAL pointB = (μ₀/4π) · m /(r² + ℓ²)^{3/2}ℓ ↑ → (r² + ℓ²) ↑ → B ↓With m and r fixed, the sign inside the bracket decides which field rises
Animated · the still frame carries the full message.
Formula
Bax = (μ₀/4π) 2mr/(r² − ℓ²)² ; Beq = (μ₀/4π) m/(r² + ℓ²)3/2
Baby steps
  1. Axial: (r² − ℓ²) shrinks → B increases
  2. Equatorial: (r² + ℓ²) grows → B decreases
Answer
(A) Axial increases, equatorial decreases
Why not the others
A · keepAxial ↑ (minus sign), equatorial ↓ (plus sign).
B · rule outThey cannot both stay the same — ℓ appears in both.
C · rule outOnly the equatorial field decreases.
D · rule outThis reverses the two.
Shortcut
The short-magnet formulas hide ℓ. When the question varies the LENGTH, write the full expressions and look only at the sign inside the bracket.
Trap (as built)
The short-magnet formulas hide ℓ — use the full expressions when length is the variable.
from: MM B1 ILTS-09 Q15 (marked 'A true')
Q64Left blank Blank — declination vs dip
The angle of declination at a place is the angle between
  1. the magnetic axis and the rotation axis of the earth
  2. the horizontal component and the vertical component
  3. the earth's total field and the horizontal
  4. the geographic meridian and the magnetic meridian ✓ key
Given
  • Definitions of the magnetic elements
Concept
Declination D is measured in the horizontal plane: the angle between the geographic meridian (true north) and the magnetic meridian (the compass direction). Dip I is measured in the vertical plane: the angle the total field makes with the horizontal.
Diagram
geographic northmagnetic northdeclination DDECLINATIONangle between the twomeridians (horizontal plane)DIP (inclination)angle of the total fieldwith the horizontalDeclination is measured in the horizontal plane; dip in the vertical plane
Animated · the still frame carries the full message.
Formula
Declination D: in the horizontal plane ; dip I: in the vertical plane
Baby steps
  1. Declination compares true north with magnetic north
  2. Dip is the angle of BE below the horizontal
Answer
(D) Geographic vs magnetic meridian
Why not the others
A · rule outThat is roughly the tilt of the earth's magnet, not declination.
B · rule outThe two components define the dip through tan I = B_V/B_H.
C · rule outThis is the definition of dip.
D · keepGeographic meridian vs magnetic meridian.
Shortcut
Two words: declination = horizontal (left–right), dip = vertical (up–down).
Trap (as built)
The angle between the total field and the horizontal is dip, not declination.
from: Definition swap
Q65Attempted · wrong Assertion–Reason: components swapped
Assertion (A): At a magnetic pole of the earth, a freely suspended horizontal compass needle can stay in any direction.
Reason (R): At the magnetic poles, the vertical component of the earth's field is zero.horizontalBHBEBVI (dip)
  1. Both A and R are true and R is the correct explanation of A her answer
  2. Both A and R are true but R is NOT the correct explanation of A
  3. A is true but R is false ✓ key
  4. A is false but R is true
Given
  • At a magnetic pole, dip I = 90°
Concept
At a magnetic pole the field points straight down, so the dip is 90° and the horizontal component B_H = B_E cos 90° = 0 — that is why a horizontal compass needle has no directing torque and can rest in any direction. R claims the vertical component is zero, which is the situation at the magnetic equator. So A is true and R is false.
Diagram
horizontalBHBEBVI (dip)
Animated · the still frame carries the full message.
Formula
BH = BE cos I ; BV = BE sin I
Baby steps
  1. I = 90° → BH = 0, so a horizontal needle has no directing torque → A true
  2. BV = BE (maximum), not zero → R false
Answer
(C) A is true but R is false
Why not the others
A · rule outR is false, so this cannot be right.
B · rule outR is false.
C · keepA true, R false.
D · rule outA is true — the needle is free to rest anywhere.
Shortcut
Write both components before judging R: at the poles B_H = 0, B_V = B_E; at the equator B_V = 0, B_H = B_E.
Trap (as built)
R swaps the components: at the poles the HORIZONTAL component vanishes.
from: Component swap; AR verdicts
Where it went wrong
Marked (A): Both A and R are true and R is the correct explanation of A.
Option A accepted R because it 'sounded like' the pole condition. One line of B_H = B_E cos I with I = 90° settles it. Together with Q15, Q23 and Q74, this is three-quarters of the assertion–reason block lost.

Magnetic Materials and Hysteresis ‡

Q73Q74Q77Q78Q80
Q73Left blank Blank — which way a diamagnet moves
A small diamagnetic rod is placed in a non-uniform magnetic field. It tends to
  1. stay at rest wherever it is placed
  2. move from the weaker to the stronger part of the field
  3. align itself along the field and move to the strongest region
  4. move from the stronger to the weaker part of the field ✓ key
Given
  • Diamagnetic: induced moment opposite to B
Concept
A diamagnetic substance is magnetised opposite to the applied field (χ is small and negative), so it is repelled and drifts towards the weaker part of a non-uniform field. Paramagnetic and ferromagnetic substances are magnetised along the field and move towards the stronger part.
Diagram
0−1DIAMAGNETIC−1 ≤ χ < 0 · μr < 1 · repelledPARAMAGNETICsmall positive χ · μr > 1FERROMAGNETICχ ≫ 1 · μr ≫ 1moves strong → weak fieldmove weak → strong fieldabove the Curie temperature a ferromagnet becomes PARAMAGNETIC (domains break up)χ sign decides everything: negative = repelled, positive = attracted
Animated · the still frame carries the full message.
Formula
Diamagnets are repelled by fields
Baby steps
  1. Opposite magnetisation → repulsion
  2. They move towards weaker field regions
Answer
(D) Stronger → weaker
Why not the others
A · rule outA non-uniform field always exerts a net force.
B · rule outThat is paramagnetic behaviour.
C · rule outA diamagnet aligns across the field, not along it.
D · keepRepelled → strong to weak.
Shortcut
One sentence: negative χ → repelled → strong to weak.
Trap (as built)
Moving to the stronger region is paramagnetic/ferromagnetic behaviour.
from: Class swap
Q74Attempted · wrong Assertion–Reason: two true statements judged as one
Assertion (A): The relative permeability of a paramagnetic material is slightly greater than one.
Reason (R): Diamagnetic behaviour of a material is independent of temperature.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is NOT the correct explanation of A ✓ key
  3. A is true but R is false
  4. A is false but R is true her answer
Given
  • Two independent statements
Concept
A: a paramagnetic material has a small positive susceptibility, so μ_r = 1 + χ is slightly greater than 1 — true. R: diamagnetism does not depend on temperature — also true, and it is an NCERT line. But R is about a different class of material, so it cannot be the explanation of A. The answer is 'both true, R not the explanation'.
Diagram
0−1DIAMAGNETIC−1 ≤ χ < 0 · μr < 1 · repelledPARAMAGNETICsmall positive χ · μr > 1FERROMAGNETICχ ≫ 1 · μr ≫ 1moves strong → weak fieldmove weak → strong fieldabove the Curie temperature a ferromagnet becomes PARAMAGNETIC (domains break up)χ sign decides everything: negative = repelled, positive = attracted
Animated · the still frame carries the full message.
Formula
μr = 1 + χ
Baby steps
  1. A: χ small positive → μr > 1 → true
  2. R: diamagnetism is temperature-independent → true
  3. R is about a different class, so it cannot explain A
Answer
(B) Both true, R not the explanation
Why not the others
A · rule outR says nothing about paramagnets.
B · keepBoth true, no causal link.
C · rule outR is a true NCERT statement.
D · rule outA is true — μ_r for a paramagnet is just above 1.
Shortcut
Judge A alone, then R alone. Only if both are true do you ask whether R explains A.
Trap (as built)
Judge A and R separately first; she marked A false on this pair before.
from: MM B1 ILTS-09 Q44 (marked A false)
Where it went wrong
Marked (D): A is false but R is true.
'A false, R true' rejects a standard fact. This exact pair was marked the same way on the MM ILTS-09 paper (Q44), so it is a repeat, not a one-off.
Q77Left blank Blank — permanent magnet material
Materials suitable for making permanent magnets should have
  1. high permeability and low retentivity
  2. high retentivity and high coercivity ✓ key
  3. low retentivity and low coercivity
  4. high retentivity and low coercivity
Given
  • A permanent magnet must keep its magnetisation and resist demagnetisation
Concept
A permanent magnet must (i) stay strongly magnetised when the field is removed — high retentivity, and (ii) resist being demagnetised by stray fields — high coercivity. Steel and alnico qualify; soft iron does not.
Diagram
HBPQSRR: region enclosed by the loop
Animated · the still frame carries the full message.
Formula
Retentivity = B left at H = 0 ; coercivity = reverse H needed to make B = 0
Baby steps
  1. High retentivity keeps it strongly magnetised
  2. High coercivity stops stray fields from demagnetising it
  3. e.g. steel, alnico
Answer
(B) High retentivity, high coercivity
Why not the others
A · rule outHigh permeability with low retentivity describes an electromagnet core.
B · keepHigh retentivity + high coercivity.
C · rule outLow on both counts is a poor magnet of any kind.
D · rule outLow coercivity means it demagnetises easily.
Shortcut
Permanent = keeps it and defends it: retentivity high, coercivity high.
Trap (as built)
Low coercivity + high permeability describes an electromagnet core (soft iron).
from: Permanent vs electromagnet swap
Q78Left blank Blank — electromagnet core
The core of an electromagnet is made of soft iron because soft iron has
  1. negative susceptibility
  2. high coercivity
  3. low permeability and high retentivity
  4. high permeability and low retentivity ✓ key
Given
  • The core must magnetise strongly and lose it when the current stops
Concept
An electromagnet must magnetise strongly while the current flows (high permeability) and lose that magnetisation the moment the current stops (low retentivity, so a narrow hysteresis loop and little energy wasted per cycle). Soft iron has exactly these properties.
Diagram
HBPQSRR: region enclosed by the loop
Animated · the still frame carries the full message.
Formula
Electromagnet: high μ, low retentivity, low hysteresis loss
Baby steps
  1. High permeability → strong field for a given current
  2. Low retentivity → switches off cleanly
Answer
(D) High permeability, low retentivity
Why not the others
A · rule outNegative susceptibility is diamagnetic — it would be repelled.
B · rule outHigh coercivity is a permanent-magnet property.
C · rule outThat is the opposite of what a core needs.
D · keepHigh permeability, low retentivity.
Shortcut
Permanent magnet keeps its magnetism; electromagnet must let it go. The words retentivity and coercivity go with 'permanent'.
Trap (as built)
High retentivity is what a PERMANENT magnet needs.
from: Permanent vs electromagnet swap
Q80Left blank Blank — above the Curie temperature
When a ferromagnetic material is heated above its Curie temperature, it becomes
  1. diamagnetic
  2. paramagnetic ✓ key
  3. more strongly ferromagnetic
  4. a superconductor
Given
  • Curie temperature Tc
Concept
Thermal agitation breaks up the ordered domains of a ferromagnet. Above the Curie temperature T_c the domains no longer hold together and the material behaves as a paramagnet, with χ falling as C/(T − T_c).
Diagram
0−1DIAMAGNETIC−1 ≤ χ < 0 · μr < 1 · repelledPARAMAGNETICsmall positive χ · μr > 1FERROMAGNETICχ ≫ 1 · μr ≫ 1moves strong → weak fieldmove weak → strong fieldabove the Curie temperature a ferromagnet becomes PARAMAGNETIC (domains break up)χ sign decides everything: negative = repelled, positive = attracted
Animated · the still frame carries the full message.
Formula
Above Tc: χ ∝ 1/(T − Tc)
Baby steps
  1. Thermal agitation destroys domain alignment
  2. The material behaves as a paramagnet
Answer
(B) Paramagnetic
Why not the others
A · rule outDiamagnetism would need the magnetisation to oppose the field.
B · keepFerromagnet → paramagnet above T_c.
C · rule outHeating weakens, never strengthens, ferromagnetism.
D · rule outSuperconductivity appears at very LOW temperatures.
Shortcut
Heat destroys order: ferro → para at T_c. The atoms still have moments, so it never becomes diamagnetic.
Trap (as built)
It does not become diamagnetic — its atoms still carry permanent moments.
from: Class swap

What to carry into the next paper

HabitWould have saved
Three-line A–R defence in Physics too; test R against one formula.Q15, Q23, Q65, Q74 — 20 marks
Triad x̂ → ŷ → ẑ at the top of the page; sign of q applied last; null point nearer the smaller current.Q2, Q3, Q36 — 15 marks
One value per line with units: θ, fraction of 360, r in metres, bracket to the power 3/2, ratio squared.Q7, Q17, Q20, Q30, Q40, Q57 — 30 marks
Ask what the Amperian loop encloses; circle 'solid' or 'hollow'. Ask whether the object is neutral (wire) or charged (beam).Q31, Q35 — 10 marks
One recall card for magnetic materials and hysteresis; attempt every one-line question.Q64, Q73, Q77, Q78, Q80 — 20 marks