A · Temperature, the Curie transition and χ–T behaviour
8 questionsA ferromagnetic material is heated well above a certain characteristic temperature. It then behaves as
Given
A ferromagnet raised above its characteristic (Curie) temperature.
Asked
Its magnetic behaviour at that temperature.
Concept applied
NCERT 5.5.3 states plainly that at high enough temperature a ferromagnet becomes a paramagnet, because the domain structure disintegrates. The individual atomic moments survive — only their cooperation is destroyed.
Formula to use
T > TC : χ ≫ 1 → χ small and positive
Baby steps
- Ferromagnetism depends on domains — groups of about 1011 atoms whose moments are spontaneously aligned.
- Thermal energy competes with the cooperative interaction that holds a domain together.
- Above the Curie point, thermal agitation wins and the domain structure breaks up.
- Each atom still carries its own permanent dipole moment — that has not been removed.
- Randomly oriented permanent atomic moments that partly align in an applied field is precisely the definition of paramagnetism.
- So the material becomes a paramagnet, with χ dropping from ≫ 1 to a small positive value.
Assumption
The material is not chemically altered by the heating; only the magnetic ordering changes.
Shortcut trick
Ferro loses its teamwork, not its members. The atoms keep their moments, so the material lands in the para class, never the dia class and never at χ = 0.
Answer: (B) a paramagnetic material
Statement I: The ferromagnetic property depends on temperature. Statement II: At high temperature the domain structure of a ferromagnetic substance disintegrates. Choose the correct option.
Given
Two statements about the temperature dependence of ferromagnetism.
Asked
Which combination is correct.
Concept applied
Ferromagnetism is the domain phenomenon. So a statement about domains breaking up is not a side-fact but the mechanism behind the temperature dependence.
Formula to use
domain order → ferromagnetism; heat destroys domain order
Baby steps
- Test Statement I: NCERT says explicitly the ferromagnetic property depends on temperature. True.
- Test Statement II: NCERT says the domain structure disintegrates with temperature. True.
- Test the link: does II cause I? Ferromagnetism arises entirely from cooperative domain alignment.
- If the domains break up, the ferromagnetic behaviour must vanish — so the temperature dependence follows directly from the domain picture.
- II is therefore the correct explanation of I.
- Answer: option (A).
Assumption
Standard statement-pair marking scheme with these four options.
Shortcut trick
For statement pairs, do not judge the pair as a unit. Run three separate tests: is I true, is II true, does II cause I. Most lost marks come from merging tests two and three.
Answer: (A) Both statements are true and II explains I
For a paramagnetic sample in a fixed external field, lowering the temperature causes its magnetisation to
Given
Paramagnetic sample, constant applied field B0, temperature lowered.
Asked
Behaviour of the magnetisation M.
Concept applied
In a paramagnet each atom already has a permanent moment; the only obstacle to alignment is random thermal motion. Remove the heat and the field wins more completely, so M rises.
Formula to use
alignment ∝ (field strength) / (thermal agitation)
Baby steps
- Identify the two competing influences: the applied field aligns the moments, thermal motion randomises them.
- The applied field is held fixed in this question.
- Lowering the temperature reduces the randomising influence.
- With less opposition, a greater fraction of the atomic moments point along B0.
- So M increases.
- NCERT adds the endpoint: the magnetisation rises until it reaches the saturation value, at which point every dipole is perfectly aligned and no further increase is possible.
Assumption
The sample stays paramagnetic throughout — no phase change at the temperatures considered.
Shortcut trick
Two levers, same effect: raise B or lower T and magnetisation goes up. Both increase the alignment ratio. The ceiling in either case is saturation.
Answer: (B) increase, approaching a saturation value
Which magnetic behaviour is essentially independent of temperature?
Given
Comparison of the temperature sensitivity of the three classes.
Asked
Which one is essentially temperature independent.
Concept applied
Diamagnetism arises from a field-induced change in electron orbital motion, not from the alignment of pre-existing moments. Since there is no alignment for heat to disrupt, temperature barely matters.
Formula to use
dia: induced effect, no permanent moments to randomise
Baby steps
- Paramagnetism relies on aligning permanent atomic moments against thermal agitation ⇒ strongly temperature dependent.
- Ferromagnetism relies on domain order, which heat destroys at the Curie point ⇒ strongly temperature dependent.
- Diamagnetic atoms have zero resultant moment to begin with.
- The diamagnetic response is induced by the applied field itself, through the speeding up and slowing down of electron orbits.
- There is no ordered arrangement for thermal motion to disturb.
- So diamagnetism is essentially temperature independent.
- NCERT adds that diamagnetism is universal — present in every substance, merely masked when para or ferro effects are also present.
Assumption
Ordinary temperature ranges, excluding the superconducting transition which is a separate phenomenon.
Shortcut trick
Ask what heat can disrupt. Heat disrupts order. Para and ferro depend on order; dia does not. No order, no temperature dependence.
Answer: (C) diamagnetism
For a paramagnetic material, the graph of susceptibility χ against absolute temperature T is best described as
Given
Paramagnetic material; χ plotted against T.
Asked
The shape of the graph.
Concept applied
Susceptibility measures how readily the sample magnetises. As T rises, thermal agitation makes alignment harder, so χ falls. It stays positive throughout — a paramagnet never becomes diamagnetic on heating.
Formula to use
χ > 0 always, and χ decreases as T increases
Baby steps
- Establish the sign: paramagnetic means χ > 0 at every temperature, so the whole curve lies above the T-axis. Eliminate option (D).
- Establish the trend: higher T means more randomisation means weaker response, so χ must fall as T rises. Eliminate options (A) and (B).
- Establish the shape: the fall is not linear — χ drops steeply at low T and flattens at high T, approaching zero without reaching it.
- That is a decreasing curve asymptotic to the T-axis.
- Answer: a curve decreasing steadily towards zero.
- For a ferromagnet the same axes show a very large χ that collapses at the Curie point and then follows a paramagnetic-style curve beyond it.
Assumption
Weak-field (linear) response so that a single χ is meaningful.
Shortcut trick
Graph questions have two independent checks: which side of the axis (fixed by the sign of χ) and which way it slopes (fixed by the physics of alignment). Apply both before looking at curvature — that alone eliminates three of four options.
Answer: (C) a curve decreasing steadily towards zero
On a graph of magnetisation M against magnetising field H, a diamagnetic material is represented by
Given
M plotted against H for a diamagnetic sample.
Asked
The correct representation.
Concept applied
M = χH is a straight line through the origin whose slope is χ. For a diamagnet χ is negative and small, so the line dips gently below the H-axis.
Formula to use
M = χH ⇒ slope of the M–H line = χ
Baby steps
- M = χH is linear with zero intercept, so the graph passes through the origin.
- The slope of that line equals χ.
- For diamagnetic materials χ is negative ⇒ the slope is negative.
- For diamagnetic materials |χ| ≈ 10−5 ⇒ the slope is very small in magnitude.
- So the graph is a straight line through the origin with a small negative slope.
- Compare: paramagnetic gives a small positive slope; ferromagnetic gives a steep non-linear curve that saturates — option (C).
Assumption
Linear response, so a single constant χ describes the material.
Shortcut trick
On an M–H plot the slope is χ. Negative slope = dia, gentle positive = para, steep and saturating = ferro. One glance at the slope classifies the material.
Answer: (A) a straight line through the origin with small negative slope
A paramagnetic sample is said to reach saturation when
Given
A paramagnetic sample in a strong field at low temperature.
Asked
The meaning of saturation.
Concept applied
Magnetisation grows as more atomic moments line up. Once every moment points along the field there is nothing left to align, so M can grow no further — that ceiling is saturation.
Formula to use
Msat = (number of dipoles per unit volume) × (moment of each)
Baby steps
- M is the net moment per unit volume, produced by partial alignment of the atomic dipoles.
- Increasing B or decreasing T aligns a larger fraction of them.
- The fraction cannot exceed one — you cannot align more than all of them.
- At that point M reaches its maximum possible value and stops responding to further increases in field.
- NCERT's wording: the magnetisation increases until it reaches the saturation value, at which point all the dipoles are perfectly aligned with the field.
- Answer: all atomic dipole moments aligned with the field.
Assumption
Sample remains paramagnetic; no structural change at the low temperatures involved.
Shortcut trick
Saturation is a counting limit, not an energy limit. Ask 'is there anything left to align?' When the answer is no, you are at saturation — and this applies to ferromagnets too.
Answer: (B) all its atomic dipole moments are aligned with the field
Which statement is correct about susceptibility and temperature?
Given
Four claims linking χ to temperature across the three classes.
Asked
The correct claim.
Concept applied
Only materials with permanent moments — paramagnets and ferromagnets — are temperature sensitive, and for them cooling helps alignment, so χ rises as T falls.
Formula to use
para and ferro: χ increases as T decreases; dia: χ roughly constant
Baby steps
- Option (A): diamagnetism is an induced effect with no order to disrupt, so χ is essentially temperature independent. False.
- Option (C): raising T increases randomisation, which reduces the response. False.
- Option (D): a ferromagnet loses its ferromagnetism entirely at the Curie point — the strongest temperature dependence of all. False.
- Option (B): cooling a paramagnet reduces thermal randomisation, so a larger fraction aligns and χ rises. True.
- Answer: χ of a paramagnet rises as T falls.
- The same direction of dependence holds for a ferromagnet below its Curie point.
Assumption
Temperatures within the normal range; the paramagnet is not driven to saturation.
Shortcut trick
One sentence covers all four options: cold helps alignment, heat destroys it, and diamagnetism does not care.
Answer: (B) χ of a paramagnet rises as T falls
B · Gauss's law for magnetism and the absence of monopoles
7 questionsThe net magnetic flux through any closed surface is
Given
An arbitrary closed surface in a magnetic field.
Asked
The value of the net magnetic flux through it.
Concept applied
Gauss's law for magnetism. Magnetic field lines form continuous closed loops, so every line that enters a closed surface must also leave it. There are no sources or sinks of B anywhere.
Formula to use
φB = ∑ B·ΔS = 0
Baby steps
- Consider any closed surface S, of any shape, in any position.
- Divide it into small area elements ΔS and compute ΔφB = B·ΔS for each.
- Field lines entering the surface contribute negative flux; lines leaving contribute positive flux.
- Because every magnetic field line is a closed loop, any line that enters must come out again somewhere.
- The negative and positive contributions therefore cancel exactly.
- φB = 0, for every closed surface without exception.
- Reject option (D): the law holds whether or not a magnet is inside — that is precisely what makes it different from the electric case.
Assumption
S is a genuinely closed surface; the result is independent of shape, size and position.
Shortcut trick
The word closed is the trigger. Once you see 'closed surface' plus 'magnetic flux', the answer is zero before you read the rest of the question.
Answer: (A) always zero
A closed surface is drawn so that it encloses only the north pole of a bar magnet. The magnetic flux through this surface is
Given
A closed surface enclosing the north pole of a bar magnet but not the south pole.
Asked
The net magnetic flux through it.
Concept applied
This is the classic trap. Enclosing 'only the N pole' does not enclose an isolated source, because the magnet's field lines pass right through the surface on their way from S to N inside the material. The count still balances.
Formula to use
φB = 0 regardless of which part of the magnet is enclosed
Baby steps
- Sketch a surface cutting through the middle of the magnet, with the N end inside.
- Outside the magnet, lines emerge from the N pole and leave the surface — positive flux.
- But those same lines re-enter the magnet at the S end and travel inside the material from S to N.
- That internal bundle crosses the surface where it slices through the magnet, entering the enclosed region — negative flux.
- The two contributions are equal and opposite, so they cancel.
- φB = zero.
- NCERT makes exactly this point in Example 5.3(e): around both the N-pole and the S-pole, the net flux of the field is zero.
Assumption
The surface is closed and the field lines inside the magnet are included in the count — forgetting them is the source of the wrong answer.
Shortcut trick
Never forget the field lines inside the magnet. Students who imagine only the external lines conclude the flux is positive. The internal S-to-N bundle is what makes the books balance.
Diagram
Answer: (B) zero
Gauss's law for magnetism has zero on its right-hand side because
Given
∑B·ΔS = 0 for magnetism versus ∑E·ΔS = q/ε0 for electrostatics.
Asked
The physical reason for the zero.
Concept applied
The right-hand side of any Gauss law counts the sources enclosed. Electric charge can be isolated, so the electric version has q/ε0. Magnetic poles cannot be isolated, so nothing is ever left over to count.
Formula to use
∑E·ΔS = q/ε0 versus ∑B·ΔS = 0
Baby steps
- In the electric case, the right-hand side is the enclosed charge divided by ε0. A lone positive charge can be trapped inside a surface, giving a non-zero answer.
- For the magnetic case, the analogous source would be an isolated magnetic pole.
- But cutting a magnet always produces two complete magnets — you can never separate an N from its S.
- With no isolated pole available to enclose, the source term is always zero.
- NCERT: there are no sources or sinks of B; the simplest magnetic element is a dipole or a current loop.
- Answer: isolated magnetic poles do not exist.
- Reject option (C): field lines never intersect — that statement is itself false.
Assumption
Monopoles have not been observed. NCERT is careful to say they are 'not known to exist' rather than proven impossible.
Shortcut trick
The zero is a statement about sources, not about field strength. Link it in one sentence: no monopoles ⇒ closed loops ⇒ zero net flux. That chain answers a whole family of questions.
Answer: (B) isolated magnetic poles (monopoles) do not exist
If magnetic monopoles existed, Gauss's law of magnetism would be modified to
Given
Hypothetical existence of magnetic monopoles carrying magnetic charge qm.
Asked
The modified form of the law.
Concept applied
The structure of a Gauss law is fixed: flux equals a constant times the enclosed source. Introduce a magnetic charge and it takes the place of q, with μ0 as the appropriate constant — exactly parallel to the electric version.
Formula to use
∫SB·Δs = μ0qm, with qm the enclosed magnetic charge
Baby steps
- Recall the present law: ∫SB·Δs = 0, the zero arising because no magnetic charge can be enclosed.
- If monopoles existed, a surface could enclose a net magnetic charge qm.
- The right-hand side would then be proportional to qm rather than zero.
- By analogy with the electric law, the constant of proportionality for magnetism is μ0.
- The law becomes ∫SB·Δs = μ0qm.
- This is exactly NCERT's answer to Example 5.4(b).
- Reject option (D): μ0I is Ampere's circuital law, which involves a line integral round a loop, not a surface integral.
Assumption
Magnetic charge would be defined so that μ0 is the natural constant, matching NCERT's stated form.
Shortcut trick
Do not confuse the two integrals. Surface integral of B = Gauss's law for magnetism. Line integral of B = Ampere's law with μ0I. Options (B) and (D) are placed together precisely to test this.
Answer: (B) ∫B·ds = μ0qm
The essential difference between Gauss's law in electrostatics and in magnetism is that
Given
The two Gauss laws side by side.
Asked
The essential difference between them.
Concept applied
Both laws have the same mathematical structure — a closed surface integral of a field. They differ only in what can be enclosed: isolated electric charges exist, isolated magnetic poles do not.
Formula to use
∑E·ΔS = q/ε0 ∑B·ΔS = 0
Baby steps
- Both laws are surface integrals over a closed surface — so option (B) is false.
- It is the magnetic field lines that form closed loops; electric lines begin and end on charges — option (C) has it exactly backwards.
- Magnetic flux is measured in webers, not coulombs — option (D) is false.
- The genuine difference is the source term: an isolated charge can sit inside a surface and produce a net electric flux.
- No isolated magnetic pole can, so the magnetic right-hand side is permanently zero.
- Answer: electric flux can be non-zero because charge can be isolated.
Assumption
Standard SI conventions.
Shortcut trick
When two laws look alike, the difference is almost always in the source term, not the mathematics. Compare right-hand sides first.
Answer: (A) electric flux can be non-zero because charge can be isolated
A closed surface completely encloses a small current-carrying loop. The net magnetic flux through the surface is
Given
A closed surface enclosing a current loop, which is itself a magnetic dipole.
Asked
The net magnetic flux.
Concept applied
A current loop is the simplest magnetic dipole. Gauss's law for magnetism applies to every closed surface and every source, so a dipole inside changes nothing — a dipole is not a net source.
Formula to use
φB = 0 for any closed surface, any enclosed configuration
Baby steps
- The loop produces a dipole field whose lines are closed loops threading through and around the coil.
- Every one of those lines that leaves the closed surface must return through it, since the lines have no ends.
- Therefore the incoming and outgoing flux cancel exactly.
- φB = zero.
- Reject option (A): μ0I is the answer for the line integral of B around a closed path linking the current (Ampere's law), a completely different quantity.
- Reject (C) and (D): the flux does not depend on the strength or size of the enclosed dipole at all.
Assumption
The surface is closed and encloses the whole loop.
Shortcut trick
Two different 'closed' objects, two different laws. Closed surface + flux ⇒ zero. Closed loop + circulation ⇒ μ0I. Identify which one the question names before answering.
Answer: (B) zero
A direct consequence of the non-existence of magnetic monopoles is that
Given
Monopoles do not exist.
Asked
The direct consequence for field lines.
Concept applied
A field line has to start somewhere and end somewhere — unless it has no ends at all. Since there are no poles to start or stop on, magnetic lines must close on themselves.
Formula to use
no sources or sinks of B ⇒ continuous closed loops
Baby steps
- Electric field lines begin on positive charges and end on negative charges, because isolated charges exist to serve as endpoints.
- For magnetism, no isolated pole exists to serve as an endpoint.
- A line with no permitted start and no permitted end must therefore be closed on itself.
- This is exactly what we observe: lines run N to S outside a magnet and S to N inside it, completing the loop.
- Answer: magnetic field lines are continuous and form closed loops.
- This is Point to Ponder 3 in NCERT: the electrostatic lines of force begin on a positive charge and terminate on the negative charge, in contrast.
Assumption
Static magnetic fields produced by magnets and steady currents.
Shortcut trick
Three facts, one chain, learn them as a single unit: no monopoles → closed loops → zero net flux. Any of the three can be asked, and the chain gives the reasoning for all of them.
Answer: (A) magnetic field lines are continuous and form closed loops
C · Validity of magnetic field-line diagrams
6 questionsA diagram shows field lines radiating outward in all directions from a single point. As a magnetic field diagram this is
Given
A diagram of straight lines emanating radially from a point.
Asked
Whether it is a valid magnetic field diagram, and what it really shows.
Concept applied
Radiating lines mean net outward flux through any small surface around the point, which requires a source. Magnetic fields have no sources, so the diagram is impossible for B.
Formula to use
φB = 0 forbids any net outward flux
Baby steps
- Draw a small closed surface around the central point in the diagram.
- Every line crosses it outward and none crosses inward, so the net flux would be positive.
- Gauss's law for magnetism requires zero net flux through every closed surface.
- The diagram therefore cannot represent a magnetic field. Wrong.
- NCERT identifies what it does represent: the electric field of a long positively charged wire.
- The correct magnetic picture for a straight conductor is circles around the wire, from Chapter 4.
Assumption
The diagram is intended as a plane section of the field.
Shortcut trick
The flux test settles every diagram question in one move: can you draw a closed surface with unbalanced lines? If yes, the diagram is not magnetic.
Answer: (B) wrong; it actually shows the electric field of a long charged wire
A diagram shows two sets of field lines crossing each other, with closed loops drawn around empty space. This diagram is
Given
Field lines that intersect, plus closed loops around a region with no current.
Asked
How many errors the diagram contains.
Concept applied
Two independent rules are violated. Intersecting lines make the field direction ambiguous. And a closed loop of static magnetic field line must enclose a region through which current passes — empty space will not do.
Formula to use
unique direction at each point; closed B-loop must enclose current
Baby steps
- Error 1 — the crossing: at an intersection the tangent gives two different directions, so B would not be unique there. Forbidden.
- Error 2 — the loops: NCERT states that magnetostatic field lines can never form closed loops around empty space; a closed loop must enclose a region across which a current is passing.
- So the diagram fails on two counts.
- Answer: option (B).
- Reject (D): electrostatic field lines can never form closed loops at all — neither in empty space nor around charges — so the diagram is not a valid electric picture either.
- Contrast with the toroid, where closed loops are legal because each loop does enclose current-carrying windings.
Assumption
Static fields; the loops are drawn in genuinely current-free space.
Shortcut trick
When an option offers 'wrong on two counts', check for a second violation before rejecting it. NCERT deliberately builds this diagram with a double fault, and the single-fault option is the trap.
Answer: (B) wrong on two counts — the crossing, and closed loops enclosing no current
A diagram shows magnetic field lines as closed loops entirely confined within a toroid. This diagram is
Given
Closed field-line loops drawn inside a toroid.
Asked
Whether the diagram is valid.
Concept applied
Closed loops are not forbidden in themselves — they are forbidden only around empty space. Inside a toroid each loop threads the current-carrying windings, so the requirement is satisfied.
Formula to use
closed B-loop is legal when it encloses a current
Baby steps
- Check the confinement claim: the field of an ideal toroid is entirely inside it, with essentially no external field. Correct.
- Check the closed-loop claim against the rule: a closed magnetostatic loop must enclose a region across which current passes.
- Each circular field line inside the toroid does encircle the current-carrying turns.
- Both conditions are satisfied, so the diagram is right.
- NCERT adds a caution: only a few lines are drawn for clarity, but in reality the entire region enclosed by the windings contains magnetic field.
- Reject option (D): an air-cored toroid behaves the same way; the core changes strength, not the geometry.
Assumption
Ideal closely wound toroid with negligible leakage.
Shortcut trick
The toroid is the standard counter-example to 'closed loops must be wrong'. It also answers Example 5.1(c): a toroid has no N or S pole, because its field is confined and its net magnetic moment is zero.
Answer: (C) right; the loops enclose current-carrying windings
A diagram shows the field of a solenoid as perfectly straight lines that stop abruptly at both ends. This is
Given
Solenoid field lines drawn straight and terminated at the ends.
Asked
Whether the diagram is valid.
Concept applied
Field lines cannot simply stop — they have no ends. They must curve out of one face, travel around outside, and return into the other face. Cutting them off violates Ampere's law.
Formula to use
field lines are closed loops — no line may terminate in space
Baby steps
- Inside a long solenoid the field is genuinely nearly uniform and the lines are nearly straight — that part of the picture is fine.
- The fault is at the ends, where the lines are shown simply stopping.
- A field line that stops would be an endpoint, i.e. a magnetic source, which cannot exist.
- NCERT: field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere's law.
- The lines must curve out at both ends and meet eventually to form closed loops.
- Answer: option (B). Option (C) is a near-miss — a longer solenoid makes the interior more uniform, but the ends must still flare.
Assumption
A finite solenoid; the idealisation of an infinite solenoid has no ends and so no flaring to draw.
Shortcut trick
Any diagram where a magnetic line just ends is wrong, no matter how tidy it looks. Trace each line with your finger — if you cannot get back to where you started, the diagram is faulty.
Answer: (B) wrong — the lines must curve out at the ends and close on themselves
A diagram shows every field line emerging outward from one shaded plate and entering another. This cannot be a magnetic field diagram because
Given
All lines emanating from one plate and terminating on another.
Asked
The reason it fails as a magnetic diagram.
Concept applied
If all lines leave one plate, a surface wrapped around that plate has purely outward flux. That is a net source, which magnetism forbids.
Formula to use
φB through any closed surface must be 0
Baby steps
- Wrap an imaginary closed surface around the upper shaded plate only.
- In the diagram, every line crosses that surface on the way out, and none crosses inward.
- The net flux would therefore be positive, not zero.
- This is impossible for a magnetic field, so the diagram is wrong.
- NCERT identifies the real subject: electrostatic field lines around a positively charged upper plate and a negatively charged lower plate.
- NCERT specifically asks you to grasp the difference between this diagram and the bar-magnet diagram 5.6(e), which looks superficially similar but includes the internal return path.
Assumption
The diagram shows all the lines, not a selected subset.
Shortcut trick
Diagrams (e) and (f) in Figure 5.6 are drawn to look alike on purpose. The distinguishing question is: do the lines close through the object, or do they terminate on it? Closing through ⇒ magnetic. Terminating on ⇒ electric.
Answer: (B) the net flux through a surface around the upper plate would not be zero
A diagram shows perfectly straight, sharply confined field lines between two pole pieces, with no spreading at the edges. This is
Given
Lines between two pole pieces drawn with perfectly sharp edges.
Asked
Whether the diagram is valid.
Concept applied
A perfectly abrupt field boundary would violate Ampere's law. Real fields always bulge outward at the edges — the effect called fringing. Nature does not do sharp corners in field patterns.
Formula to use
abrupt confinement violates Ampere's circuital law
Baby steps
- Take an Amperian loop that lies partly inside the gap and partly outside it.
- With perfectly confined straight lines, the circulation of B around this loop would be non-zero even though the loop encloses no current.
- That contradicts Ampere's law, so the picture cannot be right.
- Physically the lines must bulge outward near the edges — this is fringing, and it is unavoidable.
- Answer: option (B).
- NCERT closes with a useful note: the same is true for electric field lines between charged plates, which is why the parallel-plate capacitor's edge field is a standard correction.
Assumption
Finite pole pieces; only an infinite arrangement could have no edges to fringe at.
Shortcut trick
Three diagram faults recur in this exercise: lines that end, lines that cross, and lines with perfectly sharp edges. Scan any field diagram for these three and you will catch nearly every planted error.
Answer: (B) wrong; some fringing of lines at the edges is inevitable
D · Magnetisation, intensity and susceptibility substitutions
8 questionsA material of susceptibility χ = 0.5 is placed in a magnetising field H = 1600 A m−1. Its magnetisation is
Given
χ = 0.5; H = 1600 A m−1
Asked
The magnetisation M.
Concept applied
Susceptibility is defined as the constant of proportionality between the material's response M and the applied intensity H. Since χ is dimensionless, M automatically comes out in the same units as H.
Formula to use
M = χH
Baby steps
- Write the defining relation: M = χH.
- Substitute the values: M = 0.5 × 1600.
- M = 800 A m−1.
- Check the units: χ is dimensionless, so M inherits the units of H, namely A m−1. ✔
- Check the sign: χ is positive here, so M is parallel to H — the material assists the field.
- Since χ is positive but not enormous, this behaves as a paramagnetic-type response.
Assumption
Linear material, so χ is a constant independent of H.
Shortcut trick
M, H and χ form a three-way set: give any two and the third follows in one line. Both M and H carry units A m−1; χ carries none. If your answer has different units from H, you have used the wrong relation.
Answer: (B) 800 A m−1
Inside a material, H = 1000 A m−1 and M = 3000 A m−1. The magnetic field B is approximately
Given
H = 1000 A m−1; M = 3000 A m−1; μ0 = 4π × 10−7
Asked
The total magnetic field B inside the material.
Concept applied
NCERT partitions the interior field into two contributions: H from external sources (the coil current) and M from the material itself. Add them first, then multiply by μ0 once.
Formula to use
B = μ0(H + M)
Baby steps
- Add the two contributions inside the bracket: H + M = 1000 + 3000 = 4000 A m−1.
- Multiply by μ0: B = 4π × 10−7 × 4000.
- Write 4000 as 4 × 103: B = 4π × 10−7 × 4 × 103.
- Combine: B = 16π × 10−4.
- Evaluate: 16 × 3.1416 = 50.27, so B = 50.27 × 10−4.
- B = 5.03 × 10−3 T.
Assumption
H and M are parallel, so their vector sum is a simple scalar sum — true for para and ferro materials.
Shortcut trick
Add inside the bracket before multiplying. Computing μ0H and μ0M separately doubles the arithmetic and the chances of a slip. Option (A) is μ0H alone; option (B) is μ0M alone — both planted.
Answer: (C) 5.03 × 10−3 T
In a magnetic material B = 1.0 T while H = 2 × 103 A m−1. The magnetisation M is approximately
Given
B = 1.0 T; H = 2 × 103 A m−1; μ0 = 4π × 10−7
Asked
The magnetisation M.
Concept applied
Rearranging B = μ0(H + M) gives M = B/μ0 − H. Numerically the B/μ0 term is enormous compared with H, which tells you at a glance how much the material is contributing.
Formula to use
M = B/μ0 − H
Baby steps
- Compute B/μ0 = 1.0 / (4π × 10−7).
- 4π ≈ 12.57, so the denominator is 1.257 × 10−6.
- B/μ0 ≈ 7.96 × 105 A m−1.
- Subtract H: M = 7.96 × 105 − 0.02 × 105.
- M ≈ 7.94 × 105 ≈ 8 × 105 A m−1.
- Note that H is only about 0.25% of B/μ0 here — the material is doing almost all the work.
Assumption
H and M parallel; this matches Example 5.5(c), where a core of μr = 400 gives M ≈ 8 × 105 A m−1.
Shortcut trick
1/μ0 ≈ 8 × 105 is worth memorising. Then B/μ0 for B = 1 T is instantly 8 × 105, and when H is small the subtraction barely matters.
Answer: (C) 8 × 105 A m−1
A core of relative permeability 400 is placed where H = 2 × 103 A m−1. The magnetic field B inside is
Given
μr = 400; H = 2 × 103 A m−1; μ0 = 4π × 10−7
Asked
The field B inside the core.
Concept applied
Once μr is known, B follows from H in a single multiplication — no need to compute M separately. This is the fastest of the three routes to B.
Formula to use
B = μ0μrH = μH
Baby steps
- Substitute: B = 4π × 10−7 × 400 × 2 × 103.
- Combine the plain numbers: 400 × 2 × 103 = 8 × 105.
- So B = 4π × 10−7 × 8 × 105.
- Combine the exponents: 10−7 × 105 = 10−2.
- B = 4π × 8 × 10−2 = 32π × 10−2 = 100.5 × 10−2.
- B ≈ 1.0 T.
Assumption
The core is linear at this field level — a real ferromagnet would begin to saturate, but NCERT treats μr as constant here.
Shortcut trick
Three equivalent routes to B: μ0(H+M), μ0μrH, or μH. Pick whichever matches the data you are given — converting between them wastes time.
Answer: (C) 1.0 T
For a core of relative permeability 400 in a field H, the magnetisation M equals
Given
μr = 400; magnetising field H.
Asked
M in terms of H.
Concept applied
M = χH, and χ = μr − 1. The subtraction of 1 removes the vacuum's own contribution, leaving only what the material adds.
Formula to use
M = χH = (μr − 1)H
Baby steps
- Start from B = μ0(H + M) and B = μ0μrH.
- Equate: μ0μrH = μ0(H + M).
- Cancel μ0: μrH = H + M.
- Rearrange: M = μrH − H = (μr − 1)H.
- Substitute μr = 400: M = (400 − 1)H = 399 H.
- This is exactly the step NCERT uses in Example 5.5(c) before evaluating M ≈ 8 × 105 A m−1.
Assumption
Linear material with constant μr.
Shortcut trick
Whenever a −1 or +1 appears in this family of formulas, it is the vacuum's contribution. μr = 1 + χ adds it; χ = μr − 1 removes it. Option (A) is what you get by forgetting it.
Answer: (C) 399 H
A substance has μr = 1.00002. Its susceptibility and class are
Given
μr = 1.00002
Asked
χ and the magnetic class.
Concept applied
Subtract 1 to get χ, then read the sign and magnitude against Table 5.2. A μr just above 1 is the paramagnetic signature.
Formula to use
χ = μr − 1
Baby steps
- Compute χ = 1.00002 − 1 = 0.00002.
- Express in scientific notation: χ = 2 × 10−5.
- The sign is positive, so the material reinforces the applied field — diamagnetic is ruled out.
- The magnitude 10−5 is tiny, far from χ ≫ 1 — ferromagnetic is ruled out.
- Small and positive is the definition of paramagnetic.
- Answer: option (A). Note NCERT's Point to Ponder 4: χ = +10−5 versus −10−5 is a miniscule numerical difference that produces radically different behaviour.
Assumption
Room temperature; linear response.
Shortcut trick
Options (A) and (D) share the same χ but different classes; (A) and (B) share the class-sign confusion. Always answer in two moves: compute χ, then classify. Doing both at once is how you land on a mismatched pair.
Diagram
Answer: (A) χ = 2 × 10−5, paramagnetic
Among the three quantities χ, μr and μ, how many are independent?
Given
The three quantities χ, μr and μ describing a magnetic material.
Asked
How many carry independent information.
Concept applied
Two fixed relations connect the three quantities, so specifying any single one determines the other two completely. NCERT states this explicitly at the end of section 5.4.
Formula to use
μr = 1 + χ and μ = μ0μr
Baby steps
- Relation 1 links χ and μr: μr = 1 + χ.
- Relation 2 links μr and μ: μ = μ0μr, with μ0 a universal constant.
- Three quantities minus two independent relations leaves one free parameter.
- So knowing χ gives μr = 1 + χ and then μ = μ0(1 + χ).
- Knowing μ instead gives μr = μ/μ0 and then χ = μr − 1.
- Answer: only one is independent. NCERT: 'Given one, the other two may be easily determined.'
Assumption
Linear isotropic material at a fixed temperature.
Shortcut trick
Treat the three as one quantity wearing three costumes. Convert to whichever form the question's data suits and never carry all three through a calculation.
Answer: (C) only one
The magnetisation of a sample is defined as
Given
A bulk magnetic sample of volume V with net moment mnet.
Asked
The definition of magnetisation M.
Concept applied
Magnetisation is an intensive quantity — it describes how magnetised the material is, independent of how much of it you have. That requires dividing the total moment by the volume.
Formula to use
M = mnet / V
Baby steps
- Each atom in the material carries a magnetic moment.
- In bulk these add vectorially to give a net moment mnet for the whole sample.
- Dividing by the volume V removes the dependence on sample size.
- M = mnet/V — net magnetic moment per unit volume.
- Check the units: A m2 ÷ m3 = A m−1, which matches the units of H as it must. ✔
- Dimensions: [L−1 A].
Assumption
The sample is uniformly magnetised, so a single value of M describes it.
Shortcut trick
Use units as a definition check. If M were moment per unit mass you would get A m2 kg−1, which could never be added to H in B = μ0(H + M). Only 'per unit volume' makes that equation dimensionally legal.
Answer: (B) its net magnetic moment per unit volume
E · The solenoid with a magnetic core
6 questionsAn iron core is inserted into a current-carrying solenoid without changing the current. Inside the solenoid, the magnetic intensity H
Given
Solenoid with fixed current I and fixed turn density n; an iron core is inserted.
Asked
The effect on H.
Concept applied
NCERT's whole reason for defining H is to separate what you supply from what the material adds. H = nI depends only on the coil and the current, so the core cannot change it.
Formula to use
H = nI — depends only on the external current
Baby steps
- Recall the partition: B = μ0(H + M), where H represents external factors and M the material's response.
- For a solenoid, H = nI. Both n (turns per metre) and I (current) are unchanged by inserting a core.
- Therefore H is unchanged.
- What does change is M — the iron becomes strongly magnetised, so M jumps from zero to a large value.
- Consequently B = μ0(H + M) increases dramatically, even though H did not move.
- NCERT's Example 5.5(a) states it directly: 'The field H is independent of the material of the core.'
Assumption
The current is held constant by the supply; the coil geometry is unaltered.
Shortcut trick
H belongs to the coil, M belongs to the material, B is what actually exists. Any question asking what happens 'when a core is inserted' is testing this three-way split, and H is always the unchanged one.
Answer: (C) remains unchanged
A solenoid has 1000 turns per metre and carries 2.0 A. The magnetic intensity H inside is
Given
n = 1000 turns m−1; I = 2.0 A
Asked
The magnetic intensity H.
Concept applied
For a long solenoid the magnetising field is simply the product of turn density and current. No permeability appears — that is exactly the point of the quantity H.
Formula to use
H = nI
Baby steps
- Identify n as turns per metre, not total turns: n = 1000 m−1.
- Substitute: H = 1000 × 2.0.
- H = 2000 A m−1 = 2 × 103 A m−1.
- Check the units: m−1 × A = A m−1. ✔
- No μ0 and no μr enter this step — if either appears in your working, you have computed B or B0 instead.
- This is Example 5.5(a) exactly.
Assumption
Long, closely wound solenoid so the interior field is uniform and end effects are negligible.
Shortcut trick
If a solenoid question quotes turns 'per metre', the answer wants H = nI. If it quotes a total number of turns and a length, divide first. Mixing these up produces answers off by a factor of the length.
Answer: (C) 2000 A m−1
For the same solenoid (n = 1000 m−1, I = 2.0 A) with no core, the magnetic field B0 inside is approximately
Given
n = 1000 m−1; I = 2.0 A; air core; μ0 = 4π × 10−7
Asked
The field B0 with no magnetic core.
Concept applied
With no material present, M = 0, so B reduces to μ0H = μ0nI. This is the baseline against which the core's contribution is measured.
Formula to use
B0 = μ0nI
Baby steps
- With no core, M = 0, so B0 = μ0(H + 0) = μ0H.
- H = nI = 2 × 103 A m−1 from the previous question.
- B0 = 4π × 10−7 × 2 × 103.
- Combine exponents: 10−7 × 103 = 10−4.
- B0 = 8π × 10−4 = 25.1 × 10−4.
- B0 = 2.51 × 10−3 T.
- Compare with the cored value of 1.0 T — the iron multiplies the field by about 400, which is exactly μr.
Assumption
Air behaves as vacuum magnetically, which is standard at this level.
Shortcut trick
B0 and B differ by exactly the factor μr. Once you have one, the other needs a single multiplication or division — never a fresh calculation.
Answer: (A) 2.51 × 10−3 T
A solenoid with n = 1000 m−1 carries 2 A and, with its core, produces B = 1.0 T. The magnetising current IM is about
Given
n = 1000 m−1; I = 2 A; B = 1.0 T; μ0 = 4π × 10−7
Asked
The magnetising current IM.
Concept applied
IM is a bookkeeping device: the extra current you would have to push through a bare coil to reproduce the field that the core is producing for free. It measures the material's contribution in amperes.
Formula to use
B = μ0n(I + IM)
Baby steps
- The core-free coil would need a total current (I + IM) to produce the same B.
- Rearrange: I + IM = B / (μ0n).
- Denominator: μ0n = 4π × 10−7 × 1000 = 1.257 × 10−3.
- I + IM = 1.0 / 1.257 × 10−3 ≈ 795.8 A.
- Subtract the real current: IM = 795.8 − 2 ≈ 794 A.
- Interpretation: you supplied 2 A, and the iron behaved as though someone had quietly added another 794 A.
Assumption
NCERT's printed relation B = μrn(I + IM) contains a typographical slip; the dimensionally correct form is B = μ0n(I + IM), which reproduces the book's own answer of 794 A.
Shortcut trick
IM is almost always huge compared with I, so subtracting I barely changes the answer. Compute B/(μ0n) and you are within a fraction of a percent already — enough to pick the option.
Answer: (B) 794 A
Inserting a core of relative permeability μr into a solenoid, with the current unchanged, multiplies the interior field by a factor of
Given
Same solenoid, same current, core of relative permeability μr inserted.
Asked
The factor by which B increases.
Concept applied
B0 = μ0H without the core and B = μ0μrH with it. Since H is unchanged, the ratio is μr exactly — which is what makes μr a useful number.
Formula to use
B/B0 = μ0μrH / μ0H = μr
Baby steps
- Without the core: B0 = μ0H.
- With the core: B = μ0μrH.
- H is the same in both cases, because the current and coil are unchanged.
- Form the ratio: B/B0 = μr.
- So the field is multiplied by μr.
- Sanity check with Example 5.5: B0 = 2.51 × 10−3 T and B = 1.0 T give a ratio of about 400, which is the quoted μr. ✔
Assumption
The core fills the interior and behaves linearly at this field level.
Shortcut trick
μr literally means 'the factor by which this material multiplies the field'. Option (A), μr − 1, is the factor for the added part (the M contribution) — a genuine quantity, but not what was asked.
Answer: (B) μr
In a cored-solenoid problem, which quantity must be calculated first?
Given
A typical four-part solenoid-with-core problem asking for H, M, B and IM.
Asked
The correct order of attack.
Concept applied
H is the only quantity computable from the given data alone (n and I). Everything else depends on H, so it must come first. Recognising this dependency chain turns a four-part problem into four one-line steps.
Formula to use
H = nI → B = μ0μrH → M = (μr−1)H → IM from B
Baby steps
- H needs only n and I, both given directly. Compute it first.
- B needs H and μr: B = μ0μrH.
- M needs H and μr too: M = (μr − 1)H, or equivalently (B − μ0H)/μ0.
- IM needs B: IM = B/(μ0n) − I.
- So the correct starting point is H.
- Attempting B or M first forces you to work backwards through relations you have not yet evaluated — which is where students stall in this question type.
Assumption
The problem supplies n, I and μr, which is the standard NCERT format.
Shortcut trick
Draw the dependency chain once: H → B → M → IM. Every cored-solenoid problem walks that chain in that order, whatever order the parts are printed in.
Answer: (C) H, since it depends only on the current
F · Cutting, breaking and rejoining magnets
6 questionsA bar magnet of moment m is cut into two equal halves perpendicular to its length. The magnetic moment of each half is
Given
Bar magnet of moment m = qm × 2l, cut transversely into two equal pieces.
Asked
The moment of each piece.
Concept applied
Magnetic moment is pole strength times length. A transverse cut halves the length while leaving the cross-section — and hence the pole strength — unchanged.
Formula to use
m = qm × (2l); transverse cut: qm same, length halved
Baby steps
- Write the original moment: m = qm × 2l.
- A transverse cut splits the magnet across its length, so each piece has length l instead of 2l.
- The cross-sectional area is untouched, so the pole strength qm of each new piece is unchanged.
- New moment: m′ = qm × l.
- Compare with the original: m′ = (qm × 2l)/2 = m/2.
- Each half is a complete magnet with its own N and S pole — no free pole is produced.
Assumption
The cut is clean and the material is uniformly magnetised throughout.
Shortcut trick
Ask which factor in m = qm × length actually changed. Transverse cut ⇒ length halves. Longitudinal cut ⇒ pole strength halves. Either way the moment halves — the answer is the same, only the reason differs.
Answer: (B) m/2
The same magnet is instead cut into two equal halves along its length. The moment of each half is
Given
Bar magnet of moment m cut lengthwise into two equal halves.
Asked
The moment of each half.
Concept applied
Now the length survives but the cross-section halves. Pole strength is proportional to cross-sectional area, so it halves, and the moment halves with it.
Formula to use
qm ∝ area; longitudinal cut halves the area
Baby steps
- Original moment: m = qm × 2l.
- A longitudinal cut leaves the length at 2l for each piece.
- But the cross-sectional area of each piece is half the original.
- Pole strength scales with area, so qm′ = qm/2.
- New moment: m′ = (qm/2) × 2l = m/2.
- NCERT Example 5.1(a): in either case, one gets two magnets, each with a north and a south pole.
Assumption
Uniform magnetisation, so pole strength is proportional to the face area exposed.
Shortcut trick
Both cuts give m/2 — a fact worth memorising, because questions often ask 'which cut gives the larger moment' expecting you to say they are equal. The reasons differ, and that is what a subjective question would probe.
Diagram
Answer: (B) m/2
A bar magnet of moment m is cut transversely into n equal pieces. The moment of each piece and the total number of poles are
Given
Bar magnet of moment m cut transversely into n equal pieces.
Asked
The moment per piece and the total pole count.
Concept applied
Each transverse cut divides the length; n pieces means each has length 2l/n. And every piece, being a complete magnet, contributes exactly two poles.
Formula to use
m′ = qm(2l/n) = m/n; poles = 2 per piece
Baby steps
- Each piece has length 2l/n, with pole strength qm unchanged.
- Moment of each piece: m′ = qm × 2l/n = m/n.
- Every piece is itself a complete bar magnet with one N and one S pole.
- Total poles = 2 × n = 2n.
- Critically, none of these poles is free — each N is permanently paired with an S on the same piece.
- Answer: option (A).
Assumption
Equal transverse cuts; uniform magnetisation.
Shortcut trick
Cutting is a division problem for the moment and a multiplication problem for the poles. Moment ÷ n, poles × 2n. And the free-pole count stays at zero forever — that is the point NCERT is making.
Answer: (A) m/n and 2n poles
A magnet of moment m is cut into two equal halves transversely, and the halves are joined with their axes at right angles (like poles together at the joint). The resultant moment is
Given
Two pieces each of moment m/2, placed with their axes mutually perpendicular.
Asked
The magnitude of the resultant magnetic moment.
Concept applied
Magnetic moment is a vector. Two perpendicular vectors of equal magnitude combine by Pythagoras, not by simple addition.
Formula to use
mR = √(m12 + m22 + 2m1m2cosθ), θ = 90°
Baby steps
- After the transverse cut, each piece has moment m1 = m2 = m/2.
- The angle between the two moment vectors is θ = 90°, so cos θ = 0 and the cross term vanishes.
- mR = √[(m/2)2 + (m/2)2].
- = √[m2/4 + m2/4] = √[m2/2].
- mR = m/√2 ≈ 0.707 m.
- Direction: along the bisector of the two axes, at 45° to each.
Assumption
The two halves are rigidly joined and each retains the moment m/2 it had after the cut.
Shortcut trick
Two-stage problems: cut first, then add as vectors. The common error is using the original m instead of m/2 in the vector sum, which gives m√2 — option (D), planted for exactly that mistake.
Answer: (C) m/√2
A bar magnet is broken into 16 pieces. The number of free (isolated) magnetic poles obtained is
Given
A bar magnet broken into 16 pieces.
Asked
The number of free magnetic poles produced.
Concept applied
Cutting can never isolate a pole. Each fragment, however small, is a complete dipole with a paired N and S. Continue to the atomic scale and each atom is still a tiny current loop — still a dipole.
Formula to use
free poles = 0, always — magnetic monopoles do not exist
Baby steps
- Count the poles present: 16 pieces × 2 poles each = 32 poles in total.
- Now ask how many are free, meaning separated from a partner of opposite type.
- Every N pole created by a break is accompanied by an S pole on the same fragment.
- So the number of isolated poles is zero.
- This is NCERT's point (iv) in the introduction and Point to Ponder 2: slice a magnet in half and you get two smaller magnets.
- Option (B) gives the total pole count — correct arithmetic, wrong question.
Assumption
Ordinary mechanical breaking; no exotic physics.
Shortcut trick
Read the adjective carefully. 'Total poles' ⇒ 2n. 'Free poles' or 'isolated poles' ⇒ always zero. NEET sets both versions with nearly identical wording.
Diagram
Answer: (D) zero
NCERT explains why cutting a magnet never yields a monopole by noting that cutting a bar magnet is like cutting
Given
The bar-magnet-as-solenoid analogy from NCERT 5.2.2.
Asked
What cutting a magnet is analogous to.
Concept applied
If a magnet is really a stack of circulating currents, then cutting it is cutting a coil — and each half of a coil is still a coil, with an anticlockwise face and a clockwise face. Poles come in pairs as a matter of geometry.
Formula to use
bar magnet ≡ solenoid of the same magnetic moment
Baby steps
- NCERT: a bar magnet may be thought of as a large number of circulating currents, in analogy with a solenoid.
- Cut a solenoid transversely and each piece is a shorter solenoid with fewer turns.
- Each shorter solenoid still has one face where the current runs anticlockwise (its N face) and one where it runs clockwise (its S face).
- You cannot obtain a coil with only one face — that is geometrically impossible.
- Hence cutting a magnet cannot yield a single pole. Answer: a solenoid.
- NCERT adds the experimental confirmation: a compass needle deflects identically near a bar magnet and near a current-carrying finite solenoid.
Assumption
Ampere's hypothesis, that all magnetic phenomena can be explained in terms of circulating currents.
Shortcut trick
This analogy is the explanation, not just a coincidence. If a subjective question asks 'why do monopoles not exist in matter', the expected answer is the circulating-current picture, not merely 'because we never find them'.
Diagram
Answer: (A) a solenoid
G · Meissner effect and the origin of diamagnetism
5 questionsThe complete expulsion of magnetic field lines from the interior of a superconductor is called
Given
A superconductor from which magnetic flux is completely expelled.
Asked
The name of the phenomenon.
Concept applied
NCERT names this explicitly: the phenomenon of perfect diamagnetism in superconductors is called the Meissner effect, after its discoverer.
Formula to use
Binside = 0; χ = −1; μr = 0
Baby steps
- A superconductor is a metal cooled to very low temperature, exhibiting both perfect conductivity and perfect diamagnetism.
- Perfect diamagnetism means the field lines are completely expelled, not merely reduced.
- This total expulsion is the Meissner effect.
- Consequences: χ = −1 and μr = 0, the extreme values in Table 5.2.
- The superconductor repels a magnet and, by Newton's third law, is repelled by it — which is what makes magnetic levitation possible.
- Application named by NCERT: magnetically levitated superfast trains.
Assumption
Type-I superconductor below its critical temperature.
Shortcut trick
Attach each name to its phenomenon once: Meissner = flux expulsion in superconductors; Curie point = the temperature where ferro becomes para; hysteresis = magnetisation lagging the field (deleted from your syllabus, but still used as a distractor).
Answer: (B) the Meissner effect
A superconductor is described in NCERT as being simultaneously
Given
The NCERT description of superconductors in section 5.5.1 and Point to Ponder 5.
Asked
The pair of perfect properties.
Concept applied
Two distinct extreme behaviours coexist in the same material: zero electrical resistance and total flux expulsion. NCERT stresses that no classical theory ties the two together.
Formula to use
ρ = 0 (perfect conductivity) and χ = −1 (perfect diamagnetism)
Baby steps
- Property 1 — perfect conductivity: electrical resistance drops to exactly zero below the critical temperature.
- Property 2 — perfect diamagnetism: the magnetic field is completely expelled, χ = −1.
- Both appear together in the same cooled metal.
- Answer: a perfect conductor and a perfect diamagnet.
- NCERT Point to Ponder 5 adds that there exists no classical theory linking these two properties.
- The quantum-mechanical BCS theory of Bardeen, Cooper and Schrieffer explained them — proposed in 1957, recognised with the Nobel Prize in physics in 1972.
Assumption
Below the critical temperature and critical field.
Shortcut trick
NCERT's Points to Ponder are a favourite source of one-line MCQs. The BCS names, the 1957 proposal date and the Nobel recognition are all fair game — and all appear in the printed text.
Answer: (A) a perfect conductor and a perfect diamagnet
Diamagnetism arises in substances whose atoms have
Given
The microscopic origin of diamagnetic behaviour.
Asked
The atomic condition required.
Concept applied
Diamagnetic atoms start with everything cancelled — zero net moment. An applied field then disturbs the electron orbits, speeding some and slowing others, and the leftover moment points against the field.
Formula to use
zero atomic moment + applied field ⇒ induced moment opposing the field
Baby steps
- NCERT: diamagnetic substances are the ones in which the resultant magnetic moment in an atom is zero.
- Apply a field: electrons whose orbital moment lies along the field slow down; those opposite speed up.
- This happens through induced currents, in accordance with Lenz's law (studied in Chapter 6).
- The result is a net moment in the direction opposite to the applied field.
- Opposite moment ⇒ repulsion ⇒ the sample moves from strong field to weak field.
- Answer: zero resultant magnetic moment. Options (A) and (C) describe paramagnets; (D) describes ferromagnets.
Assumption
Non-relativistic classical picture of orbiting electrons, as NCERT presents it.
Shortcut trick
Sort the three classes by their starting atomic state: dia = zero moment to begin with; para = permanent moments, randomly oriented; ferro = permanent moments, already cooperating in domains. Every property of each class follows from its starting state.
Answer: (B) zero resultant magnetic moment
Which statement about diamagnetism is correct?
Given
Claims about how widespread diamagnetism is.
Asked
The correct statement.
Concept applied
The orbital-distortion mechanism operates in every atom that has electrons — which is all of them. It is simply too weak to notice when stronger effects are also present.
Formula to use
|χdia| ≈ 10−5, swamped when para or ferro effects are present
Baby steps
- The induced orbital effect requires only that the atom has orbiting electrons.
- Every substance has orbiting electrons, so every substance is diamagnetic to some degree.
- NCERT: 'Diamagnetism is present in all the substances.'
- But the effect is weak — about one part in 105.
- When the same material also has permanent moments (para) or domains (ferro), those far stronger effects dominate and the diamagnetic contribution is hidden.
- Answer: present in all substances but often masked. Point to Ponder 6 states this directly.
Assumption
Ordinary temperatures, excluding superconducting transitions.
Shortcut trick
Read this as a two-part fact and quote both halves: universal, but weak. Options that say 'only in some materials' fail the first half; options that ignore the masking fail the second.
Answer: (B) It is present in all substances but is often masked
A magnet placed above a cooled superconducting disc floats in mid-air. This happens because the superconductor
Given
A magnet levitating above a superconductor.
Asked
The reason for the levitation.
Concept applied
Perfect diamagnetism means the superconductor develops a magnetisation exactly opposing the applied field. Opposition means repulsion, and a strong enough repulsion balances gravity.
Formula to use
χ = −1 ⇒ M antiparallel to H ⇒ repulsion
Baby steps
- The magnet's field tries to penetrate the superconductor.
- The superconductor responds by expelling it completely — the Meissner effect.
- Expelling the field means developing a magnetisation exactly opposite to the applied field.
- Opposite moments repel, so the superconductor pushes the magnet away.
- NCERT: a superconductor repels a magnet and, by Newton's third law, is repelled by the magnet.
- When this upward repulsion balances the magnet's weight, it floats. Answer: option (B).
- This is the principle NCERT cites for magnetically levitated superfast trains.
Assumption
The superconductor is held below its critical temperature throughout.
Shortcut trick
Repulsion is the diamagnetic signature at every scale: a bismuth rod drifting to the weak-field region, and a magnet floating over a superconductor, are the same physics with different values of χ.
Diagram
Answer: (B) expels the field and therefore repels the magnet
H · Dipole in the field of another dipole
4 questionsNeedle Q lies in the field of needle P. Q is not in equilibrium when its moment is
Given
Needle Q placed in the field BP produced by needle P.
Asked
The orientation in which Q is not in equilibrium.
Concept applied
Equilibrium requires zero torque, and τ = mQBP sin θ. Only θ = 0° and 180° give zero torque; anything in between, and especially 90°, leaves a torque that turns the needle.
Formula to use
τ = mQBP sin θ = 0 requires θ = 0° or 180°
Baby steps
- Parallel (θ = 0°): sin 0° = 0, torque zero ⇒ equilibrium (stable).
- Antiparallel (θ = 180°): sin 180° = 0, torque zero ⇒ equilibrium (unstable, but still equilibrium).
- Perpendicular (θ = 90°): sin 90° = 1, torque is maximum ⇒ definitely not equilibrium.
- So the answer is perpendicular to BP.
- In NCERT's figure these are the configurations PQ1 and PQ2, which the solution lists as not being in equilibrium.
- Option (D) is a distractor about position, not orientation — the two are independent.
Assumption
Q is free to rotate about its centre; P is held fixed.
Shortcut trick
Equilibrium is about the angle between m and B, never about where the needle sits. Two candidates only: 0° and 180°. Anything else has a torque.
Answer: (C) perpendicular to BP
Needle Q is in stable equilibrium in the field of needle P when
Given
Needle Q in the field of needle P.
Asked
The condition for stable equilibrium.
Concept applied
Stability depends on the orientation of mQ relative to the local field BP, not relative to mP itself. Those two differ depending on where Q sits.
Formula to use
U = −mQ·BP, minimum when mQ ∥ BP
Baby steps
- The potential energy of Q is U = −mQ·BP = −mQBP cos θ.
- U is minimum when cos θ = 1, i.e. θ = 0°.
- Minimum energy is the condition for stability.
- So Q is stable when mQ is parallel to BP.
- NCERT states it in exactly these words: equilibrium is stable when mQ is parallel to BP, and unstable when it is antiparallel.
- Option (D) is the trap — parallel to mP is not the same as parallel to BP, as the next question shows.
Assumption
Q is small enough that BP is effectively uniform over its length.
Shortcut trick
Always compare mQ with the local field, never with the distant magnet's moment. This distinction is the entire content of Example 5.2 and it is where most students lose the marks.
Answer: (A) mQ is parallel to BP
Needle Q sits on the perpendicular bisector of needle P. For Q to be in stable equilibrium, mQ must be
Given
Q on the equatorial line of P; stable equilibrium required.
Asked
The orientation of mQ relative to mP.
Concept applied
Two facts must be chained. Stability needs mQ parallel to BP. But on the equatorial line, BP is antiparallel to mP. Combining them flips the answer.
Formula to use
BP = −μ0mP/4πr3 on the bisector
Baby steps
- Step 1 — find the local field direction: on the equatorial line, BP points opposite to mP (the minus sign in Eq. 5.4).
- Step 2 — apply the stability condition: mQ must be parallel to BP.
- Step 3 — chain them: mQ parallel to BP, and BP antiparallel to mP.
- Therefore mQ is antiparallel to mP.
- This corresponds to NCERT's configuration PQ3, which the solution lists as stable.
- The naive answer 'parallel to mP' — option (A) — is what you get by skipping step 1.
Assumption
Short dipoles; Q lies on the true perpendicular bisector.
Shortcut trick
Two-step chain, and the sign flip happens in step 1. On the axis: B is along mP, so stable means parallel to mP. On the equator: B is against mP, so stable means antiparallel to mP. Position determines the answer.
Answer: (B) antiparallel to mP
Needle Q, aligned with the local field, is placed first on the axis of P and then on the equatorial line, at the same distance r. The magnitudes of its potential energy in the two positions are in the ratio
Given
Same needle Q, same distance r, aligned with the local field in both positions.
Asked
|Uaxial| : |Uequatorial|
Concept applied
With Q aligned in both cases, cos θ = 1 throughout, so the energy is proportional to the local field strength alone. The axial field is twice the equatorial field at equal distance.
Formula to use
|U| = mQBP; Baxial = 2Bequatorial
Baby steps
- In both positions Q is aligned with the local field, so |U| = mQBP with cos θ = 1.
- Axial field at distance r: Bax = (μ0/4π)(2mP/r3).
- Equatorial field at the same r: Beq = (μ0/4π)(mP/r3).
- Ratio of fields = 2 : 1.
- Since |U| is proportional to the field, the energy ratio is the same: 2 : 1.
- This is precisely why NCERT identifies PQ6 (on the axis) rather than PQ3 (on the equator) as the configuration of lowest potential energy.
Assumption
Same separation r in both positions; short-dipole approximation; Q aligned with the local field each time.
Shortcut trick
The 2:1 axial-to-equatorial ratio propagates into everything that depends linearly on B — energy, torque, force on a test needle. Establish it once and reuse it rather than recomputing fields.
Answer: (C) 2 : 1
How Tier 2 differs from Tier 1
Tier 1 was mostly plug-and-solve. Tier 2 is mostly decide-then-solve: the marks here are lost in the reasoning step before the arithmetic, not in the arithmetic itself. Three patterns account for nearly all of it.
1. The two-step chain. Q49 is the model case: stability is defined against the local field, and on the equatorial line that field is antiparallel to the source moment. Skip step one and you get a confident wrong answer. The same shape appears in Q35 (H before B before M) and Q39 (cut before adding vectors).
2. The adjective that changes the answer. 'Free poles' versus 'total poles' (Q40). 'Lines concentrated' versus 'lines highly concentrated'. 'Reduced' versus 'completely expelled'. Circle the qualifying word before you choose.
3. The planted intermediate. Options (A) and (B) in Q23 are μ0H and μ0M — both real quantities, neither the answer. Before selecting, ask whether the number you have is the one that was requested.
Where to look first if the score is low. Group C (diagrams) and Group B (Gauss's law) are pure reasoning with no arithmetic to hide behind — errors there are conceptual and will keep costing marks in Ray Optics and Electrostatics too. Group D errors are usually just the missing ±1, which is a five-minute fix.