Take a slow breath before you start this one, because Magnetism and Matter has a reputation it does not deserve. Students hear “magnetism” and picture pages of vector derivations. The official CBSE 2026-27 curriculum gives a useful boundary: the bar magnet as an equivalent solenoid, the axial and equatorial field of a dipole, and the torque on a dipole are all marked qualitative treatment only. That means your first job is to understand the physical picture, directions, comparisons and relationships. The worked formula examples on this page are clearly labelled as optional concept checks; they are not presented as proof of what a board paper must assess.
So here is how we will do it. This page teaches the chapter from the physical idea of a magnetic moment. The current NCERT chapter builds its microscopic picture from circulating or orbiting electron motion: those orbital magnetic-moment contributions can cancel or combine to leave a net moment. A current-loop picture is a useful model for several results, but it is not a literal classical description of every electron. Along the way you will get a magnetism and matter formula list class 12, a clear diamagnetic paramagnetic ferromagnetic difference table class 12, an official 2026-27 scope check, and a worksheet of magnetism and matter class 12 practice questions with full worked answers.
Give special attention to the classification of materials and to what temperature does to magnetism. The official scope names magnetisation, para-, dia- and ferromagnetic substances with examples, the effect of temperature, and magnetic field lines. These ideas also connect the chapter into one coherent picture.
What You’ll Learn
Your Game Plan
- Day 1 — get the central idea. Read the first two sections until you can explain magnetic moment and the current-loop analogy in your own words.
- Day 2 — learn the qualitative relationships. Compare the axial and equatorial field directions and magnitudes, then explain how a dipole turns in a uniform field. Treat the numerical examples in these qualitative-only sections as optional concept checks.
- Day 3 — own the materials table. Learn the three L’s mnemonic, then write the dia/para/ferro comparison table from memory on blank paper. Twice.
- Day 4 — temperature. Review how thermal disorder changes paramagnetic alignment and gradually weakens ferromagnetic domain order.
- Day 5 — the worksheet. Attempt all ten questions closed-book, then compare your reasoning with the model answers.
Study Notes
Start Here: Magnetic Moments in Matter
Start with a current loop. A loop of area A carrying current I behaves like a magnetic dipole with magnetic dipole moment m = I A. At the atomic scale, the current chapter uses orbiting-electron motion to introduce orbital magnetic moments. The current-loop picture helps us understand the direction and addition of those moments, although an electron is not literally a tiny classical bead circling on a fixed wire.
Whether an atom has a net magnetic moment depends on how its electron contributions combine. Whether a whole sample behaves as diamagnetic, paramagnetic or ferromagnetic depends on how those microscopic moments respond collectively to an applied field.
There are only three answers, and they are the three material classes you have to learn:
- Orbital magnetic-moment contributions cancel, so an atom has no permanent net magnetic moment in the chapter’s model.
- Atoms do have permanent moments, but without an applied field their orientations are random, so the sample’s average moment is zero.
- Strong interactions can align many neighbouring moments within regions called magnetic domains.
Paramagnetic: atoms have permanent moments that are randomly oriented without a field; an applied field produces partial alignment, so χ is small and positive.
Ferromagnetic: strong interactions produce regions called domains in which many moments align; the response is large and positive. Heating increases disorder, the domain structure gradually disintegrates, and at a high enough temperature the material becomes paramagnetic.
Use this three-part picture as a check. If a question asks why bismuth is repelled, explain its induced moment opposite to the applied field. If it asks why a ferromagnet loses its ferromagnetic order at a high enough temperature, explain that increasing thermal disorder gradually breaks up the ordered domain arrangement until the material behaves as a paramagnet.
Working. For a coil of N turns, m = N I A, where A is the area of one turn.
A = πr² = π × (0.040 m)² = π × 1.6 × 10−3 m² = 5.0265 × 10−3 m²
m = 50 × 0.30 A × 5.0265 × 10−3 m² = 7.54 × 10−2 A·m²
A current loop and a bar magnet can both be described by a magnetic dipole moment. This shared quantity lets us compare their turning response and far-field pattern without claiming that their microscopic structures are identical.
Why it works. A magnetic dipole moment is nothing more than a bookkeeping number that says “here is how strongly this object will try to twist to line up with a magnetic field”. A current loop tries to twist. A bar magnet tries to twist. Nature gives them both the same number and the same unit (A·m²), so we treat them as the same kind of object. If you have already met the electric dipole in Electric Charges and Fields — Class 12 Physics, this will feel familiar: an electric dipole is two opposite charges a small distance apart, and it twists in an electric field in exactly the same way. Magnetism and Matter is, to a large extent, that same chapter wearing a different coat.
The Bar Magnet as a Magnetic Dipole
A bar magnet is a ferromagnetic material with a substantial net alignment of magnetic domains. Their contributions add, so the bar behaves as though it had two poles near its ends: a north-seeking pole N and a south-seeking pole S. The word seeking matters — the north pole of a compass is the end that seeks geographic north.
The two poles are useful labels for reading a field diagram, but the magnet behaves as one dipole. Its magnetic moment m is a vector along the magnet’s axis. The direction is from the south pole towards the north pole inside the magnet. This direction agrees with the internal part of a closed magnetic field line and with the direction in which the north face points.
Reasoning. The magnetic moment points from S to N inside a bar magnet. Here that is from left to right. This is also the direction followed by the internal part of a magnetic field line. Therefore the moment arrow must point right.
Check. Outside the magnet the field emerges from N and returns to S, but the moment direction is read through the magnet, not along the outside return path.
Reasoning. The drawing treats the original poles as separable objects. They are not. Each broken piece becomes a smaller complete magnet with its own north and south ends. Around each piece, field lines still form closed loops. The corrected drawing must therefore show two smaller dipoles, not two isolated poles.
Why it matters. Cutting the magnet illustrates the observed absence of isolated magnetic monopoles. Because isolated magnetic monopoles are not observed, magnetic field lines form closed loops and the net magnetic flux through a closed surface is zero.
What happens when you cut a bar magnet? You do not get a lone north or south pole; you get two smaller complete magnets. The magnetic field lines of each piece remain closed loops. In the classical dipole model used here, an isolated magnetic pole does not appear.
Reasoning. No. An unmagnetised iron bar can be attracted because the nearby magnetic field induces a magnetic response in it. Attraction alone therefore does not establish that both objects are permanent magnets. Repulsion is the decisive observation: if some pair of ends repels, both bars must be magnetised and like poles are facing.
Answer habit. State the observation first, then explain why induction makes attraction ambiguous.
Reasoning. Outside the magnet the line travels from N towards S. It enters the south end, continues through the body of the magnet from S towards N, and returns to its starting region. The answer is therefore a continuous closed loop.
Check. If your line stops at S, starts inside empty space, or crosses another line, the picture cannot represent a magnetic field. Field direction is the tangent at each point, so every point on the loop must have one unambiguous direction.
Magnetic Field Lines and the Rules They Always Obey
Field lines are a drawing trick, not a physical object. We draw them so that the direction of the line at a point is the direction of the magnetic field there, and the crowding of the lines tells you how strong the field is. Where lines bunch up, the field is strong; where they spread out, it is weak. A compass needle placed anywhere on a line will sit along it, pointing the way the arrow points.
Four properties help you read and draw field-line diagrams:
- Magnetic field lines are continuous closed loops. Outside the magnet they run N → S; inside the magnet they continue S → N. There is no start and no finish. Electric field lines, by contrast, can begin or end on isolated electric charges. This different field-line topology reflects the existence of isolated electric charges and the observed absence of isolated magnetic monopoles.
- Two field lines never cross. If they did, the field would have two directions at the crossing point, which is meaningless — a compass cannot point two ways at once.
- The tangent to a field line at any point gives the direction of B there.
- The lines are densest at the poles, which is another way of saying the field is strongest there.
Model answer. The direction of the magnetic field at a point is given by the tangent to the field line at that point. If two field lines crossed, there would be two different tangents and hence two directions for B at the same point. A field has one definite direction at each point, so magnetic field lines cannot intersect.
Bar Magnet as an Equivalent Solenoid (Qualitative Treatment Only)
Stack many current loops side by side along a common axis and you have a solenoid. A bar magnet has a different microscopic structure, but at points outside it the field has the same dipole-like pattern as the field of a current-carrying solenoid: lines emerge from the north face, curve around, enter the south face, and continue through the interior.
For a solenoid of N turns, current I and cross-sectional area A: m = N I A.
One face of the solenoid behaves as a north pole, the other as a south pole. Which is which is settled by the clock rule: viewed head-on, anticlockwise current means you are looking at the N face.
Working. A = πr² = π × (0.015 m)² = 7.0686 × 10−4 m²
m = N I A = 500 × 2.0 A × 7.0686 × 10−4 m² = 0.707 A·m².
Meaning. A bar magnet with the same magnetic moment can produce the same far-away dipole-like external field pattern. The statement compares observable field patterns; it does not claim that the bar and the wire coil have identical microscopic structures.
Model answer. It will turn and settle with its magnetic moment approximately along the local magnetic field, like a suspended bar magnet. A current-carrying solenoid behaves as a magnetic dipole; the field exerts a torque until the dipole reaches its aligned stable orientation.
Magnetic Field of a Bar Magnet: Axial and Equatorial Points (Qualitative Only)
Two special positions around a bar magnet get names because the maths there is clean. Stand on the axis — the straight line through both poles, extended out beyond the ends. That is the axial or end-on position. Now stand on the perpendicular bisector — the line through the centre of the magnet at right angles to its axis. That is the equatorial or broadside-on position.
Equatorial (broadside-on): Beq = (μ0/4π) × m / r³ — direction: antiparallel to m (points opposite to the magnet’s moment).
For CBSE/NCERT textbook calculations, use μ0 ≈ 4π × 10−7 T·m/A, so μ0/4π ≈ 1 × 10−7 T·m/A — a very convenient approximation.
The ratio you must remember: Baxial : Beq = 2 : 1 at the same distance.
Why it works — the intuition, not the proof. On the axis, you are close to one pole and further from the other, and the two contributions add up in the same direction. On the equator you are the same distance from both poles, and their contributions largely cancel — only the sideways components survive, and they point backwards. Adding beats cancelling, so the axial field wins, and it happens to win by exactly a factor of two. Also note the 1/r³: a dipole field dies away much faster than the 1/r² field of a single charge, because the two poles are busy cancelling each other out at long range. This is exactly the same 1/r³ behaviour you met for the electric dipole in Electric Charges and Fields — if you can do that one, you can do this one.
Working. r = 0.20 m, so r³ = 8.0 × 10−3 m³.
B = (μ0/4π) × 2m/r³ = (1 × 10−7) × (2 × 0.40) / (8.0 × 10−3)
= (1 × 10−7) × 100 = 1.0 × 10−5 T
Direction: along m, i.e. pointing away from the N pole along the axis. For scale, that is about a fifth of the Earth’s field — small, but easily enough to swing a compass needle.
Working. B = (μ0/4π) × m/r³ = (1 × 10−7) × 0.40 / (8.0 × 10−3) = (1 × 10−7) × 50 = 5.0 × 10−6 T
Direction: antiparallel to m. And check the ratio: 1.0 × 10−5 ÷ 5.0 × 10−6 = 2. Exactly 2:1, as promised. Use this ratio as a quick error check — if your two answers are not in the ratio 2:1 at the same r, you have slipped somewhere.
Torque on a Magnetic Dipole in a Uniform Magnetic Field (Qualitative Only)
Put a bar magnet into a uniform magnetic field at an angle. The field pulls the north pole one way and pushes the south pole the opposite way with an equal force. Equal and opposite forces on a rigid body, not acting along the same line, is the definition of a couple — so the magnet feels no net force but does feel a twist. It rotates until it lines up with the field, which is precisely what a compass needle is doing all day long.
where θ is the angle between the magnetic moment m and the field B. Unit of τ: N·m.
Three positions worth memorising:
• θ = 0° → τ = 0. Moment along the field. Stable equilibrium.
• θ = 90° → τ = mB, the maximum possible torque.
• θ = 180° → τ = 0 again, but this is unstable — the smallest nudge flips it right round.
Also note: in a uniform field the net force is zero, so the magnet turns on the spot but does not drift. It is only in a non-uniform field that a magnet gets dragged along — which is why a fridge magnet is pulled towards the door, and why paper clips jump to a magnet’s ends and not to its middle.
Working. τ = m B sinθ = 0.072 × 0.25 × sin30° = 0.072 × 0.25 × 0.5
= 9.0 × 10−3 N·m
Sense of rotation: the torque acts to decrease θ, turning the magnet towards alignment with the field.
(a) τmax = mB = 0.072 × 0.50 = 0.036 N·m (this occurs at θ = 90°).
(b) sinθ = τ/τmax = 0.018 / 0.036 = 0.5, so θ = 30°.
Strictly, sinθ = 0.5 also gives θ = 150°, and both are physically possible answers — a good candidate mentions this. If the question says the magnet is “close to alignment”, take 30°.
Magnetisation of Materials: M, H, Susceptibility and Relative Permeability
Everything so far has been about one magnet. Now we ask the question the chapter title promises: what happens when you put matter into a magnetic field? The electron magnetic moments in the material respond, and the sample may develop a net magnetic moment. We describe “how magnetised” the sample is per unit volume, so that differently sized samples of the same uniformly magnetised material can be compared — much as concentration compares amount per unit volume in Solutions — Class 12 Chemistry.
The net magnetic moment developed per unit volume. Unit: A/m (ampere per metre).
2. Magnetising field (magnetic intensity), H
The field you apply, produced by the current in your coil and nothing to do with the sample. Inside a long solenoid, H = nI, where n is turns per metre. Unit: A/m — the same unit as M, which is exactly why comparing them is meaningful.
3. Magnetic susceptibility, χ = M / H
How eagerly the material magnetises when you push it. It is a pure number with no unit, because M and H share a unit. Its sign and size are the whole classification.
4. Relative permeability, μr = 1 + χ
How many times more field you get inside the material than you would in vacuum for the same H. Also dimensionless. The total field inside is B = μ0(H + M) = μ0μrH.
Why it works. H describes the magnetising field and M describes the material’s magnetic response per unit volume. In the simple linear relation used here, χ = M/H. A positive χ means M is along H, a negative χ means M opposes H, and a large magnitude means a strong response. For a linear, isotropic material in SI units, μr = 1 + χ connects that response with relative permeability.
Working. Volume V = length × area = 0.10 m × 5.0 × 10−4 m² = 5.0 × 10−5 m³
M = mnet / V = 4.0 ÷ (5.0 × 10−5) = 8.0 × 104 A/m
Unit check: A·m² ÷ m³ = A/m. Correct.
(a) H = nI = 1000 × 0.50 = 500 A/m
(b) χ = M/H = 0.15 ÷ 500 = 3.0 × 10−4 (no unit)
(c) μr = 1 + χ = 1.0003
(d) χ is positive but tiny, and μr is just barely above 1 — a textbook paramagnetic material. A partly aligned, weakly attracted.
(a) χ = μr − 1 = 4000 − 1 = 3999
(b) M = χH = 3999 × 200 = 8.0 × 105 A/m (7.998 × 105 A/m exactly)
(c) B = μ0μrH = (4π × 10−7) × 4000 × 200 = 1.01 T
Cross-check with the other formula: B = μ0(H + M) = (1.2566 × 10−6) × (200 + 799800) = (1.2566 × 10−6) × 8.0 × 105 = 1.01 T. The two routes agree, as they must.
Notice how strongly the material responds: of the 8.0 × 105 A/m total, the applied H contributes 200 A/m. High permeability is one important reason a suitable ferromagnetic core can greatly increase the magnetic response of a device; practical core choice also depends on losses and the device’s design.
A: negative and tiny → diamagnetic. μr = 1 − 9.5 × 10−6 = 0.9999905, just below 1. It will be weakly repelled.
B: positive and tiny → paramagnetic. μr = 1.000023, just above 1. Weakly attracted.
C: positive and huge → ferromagnetic. μr = 3501. Strongly attracted.
The sign tells you whether the response is opposed to or along the field; the magnitude separates a weak response from a very strong one.
Diamagnetic Paramagnetic Ferromagnetic Difference Table Class 12
This is the highest-yield section in the chapter, so slow down here. The chapter studies three broad classes of magnetic response: diamagnetic, paramagnetic and ferromagnetic. Watching whether a sample is weakly repelled, weakly attracted or strongly attracted by an applied field is a useful conceptual guide to those responses, but it is not a universal one-step laboratory test for every material. The examples come from ordinary chemistry — iron, cobalt and nickel are the classic ferromagnetic three, and if you have already met their reactivity and structure in Metals and Non-metals — Class 10 Science, you will recognise them as old friends.
| Property | Diamagnetic — “Cancelled” | Paramagnetic — “Partly Aligned” | Ferromagnetic — “Domain-Aligned” |
|---|---|---|---|
| Atomic magnetic moment | Zero — electron moments cancel in pairs | Non-zero, but randomly directed | Non-zero and locked parallel to neighbours |
| Behaviour near a magnet | Weakly repelled | Weakly attracted | Strongly attracted |
| Susceptibility χ | Small and negative | Small and positive | Very large and positive (χ >> 1) |
| Relative permeability μr | Slightly less than 1 | Slightly more than 1 | Very much greater than 1 |
| In a non-uniform field | Moves towards the weaker region | Moves towards the stronger region | Moves rapidly towards the stronger region |
| Field lines through a rod | Pushed out — the rod thins them | Slightly drawn in | Strongly crowded into the rod |
| Effect of temperature | No separate quantitative temperature law is needed here | Lower temperature favours stronger field-induced alignment; susceptibility and permeability depend on temperature | Domain order weakens gradually; at high enough temperature the material becomes paramagnetic |
| Keeps its magnetism when the field is removed? | No | No | Hard ferromagnets may retain magnetisation; in soft ferromagnets such as soft iron, magnetisation disappears when the field is removed |
| Examples | Bismuth, copper, lead, silicon, nitrogen at STP, water, sodium chloride | Aluminium, sodium, calcium, oxygen at STP, copper chloride | Iron, cobalt, nickel, gadolinium; alnico is a hard magnetic alloy |
Effect of Temperature on Magnetic Properties
Temperature changes the balance between alignment and random motion. In a paramagnetic substance, the atoms or ions already possess magnetic dipole moments, but constant thermal motion leaves those moments randomly directed when no external field is present. An applied field encourages partial alignment; lowering the temperature reduces the competing thermal disorder, so alignment becomes easier. The current NCERT chapter therefore says that paramagnetic susceptibility and relative permeability depend on temperature and that magnetisation increases when the temperature is lowered, until saturation is reached under suitable conditions.
Paramagnetic: permanent atomic or ionic moments remain mostly random because of thermal motion. Lowering temperature makes field-directed alignment easier, while raising temperature strengthens the randomising influence. Susceptibility and relative permeability therefore depend on sample temperature.
Ferromagnetic: neighbouring moments form ordered domains. As temperature rises, the domain structure disintegrates and magnetisation disappears gradually. At a high enough temperature, the substance behaves as a paramagnet.
Why it works. The field and thermal motion compete. The field favours a common direction; thermal motion constantly changes molecular and atomic orientations. In a paramagnet, cooling helps more moments remain partly aligned with the field. In a ferromagnet, cooperative interactions already organise many moments into domains, so the response is much stronger. Heating progressively disrupts that collective order. Notice the wording: progressively and gradually. A vertical cliff in a memorised graph would misrepresent the current chapter’s explanation.
A useful way to answer a comparison question is to name the magnetic structure before describing the temperature change. For a paramagnet, begin with permanent microscopic moments whose directions are random in the absence of a field. Then add that the applied field produces partial alignment and that lower temperature reduces the competing random motion. For a ferromagnet, begin with domains: neighbouring moments cooperate, and domains can reorient or grow when a field is applied. Then explain that increasing temperature progressively disrupts this ordered structure. Finally state the observable result: the strong ferromagnetic response weakens and, at sufficiently high temperature, the substance behaves as a paramagnet.
Keep that chain separate from diamagnetism. A diamagnetic response is an induced moment opposite to the applied field, not the partial alignment of permanent moments and not the growth of ferromagnetic domains. The three classes can therefore respond differently to the same external field even before temperature is discussed. Temperature does not change the definitions: negative susceptibility still identifies the opposing diamagnetic response, small positive susceptibility identifies a weak paramagnetic response, and a very large positive response with domain behaviour identifies ferromagnetism. This structure-first approach is safer than memorising unrelated slogans because every sentence follows from the material model used in the current chapter.
Reasoning. Both samples contain permanent microscopic moments. The applied field tries to align them; thermal motion disrupts that alignment. Because sample A is cooler, thermal disorder competes less strongly, so more moments can remain partly aligned with the field. Sample A therefore has the larger magnetisation under the stated conditions.
This is a qualitative conclusion from the current source. No undeclared numerical temperature law is needed.
Reasoning. The current NCERT treatment says that the domain structure disintegrates with temperature and that the disappearance of magnetisation is gradual. A better schematic should show ordered domains becoming progressively less effective as thermal disorder grows. At a high enough temperature, the material behaves as a paramagnet; the source does not support presenting this change as a perfectly vertical, all-at-once cliff.
Model answer. In a ferromagnetic substance, neighbouring microscopic moments cooperate and form domains with a common direction. An applied field can orient domains and allow domains already pointing with the field to grow, producing strong magnetisation. Heating increases random thermal motion. As temperature rises, that disorder progressively disturbs the domain arrangement, so the bulk ferromagnetic magnetisation decreases gradually. At a high enough temperature the long-range domain order is lost. The substance then behaves as a paramagnet: microscopic moments are still present, but their directions are largely random unless an external field produces partial alignment.
This answer stays with the supported chain: temperature rises → disorder grows → domain order weakens gradually → paramagnetic behaviour.
Magnetism and Matter Formula List Class 12
This table collects the relationships used on this page. The axial-field, equatorial-field and torque rows support optional mathematical concept checks for topics that the official 2026-27 curriculum labels qualitative treatment only; do not read the table as a prediction of assessment format.
| Quantity | Formula | SI unit | Watch out for |
|---|---|---|---|
| Magnetic moment of a bar magnet | Vector m directed S → N inside the bar | A·m² (or J/T) | Use the dipole direction and closed field-line picture; no pole-strength track is needed |
| Magnetic moment of a coil or solenoid | m = N I A | A·m² | A is the area of one turn |
| Field on the axis (short magnet) | B = (μ0/4π)(2m/r³) | T | r cubed, in metres; B is parallel to m |
| Field on the perpendicular bisector | B = (μ0/4π)(m/r³) | T | Exactly half the axial value; antiparallel to m |
| Torque on a dipole | τ = mB sinθ | N·m | θ is between m and B; max at 90°, zero at 0° |
| Intensity of magnetisation | M = mnet/V | A/m | V is volume, not area or length |
| Magnetising field in a solenoid | H = nI | A/m | n is turns per metre, not total turns |
| Susceptibility and permeability | χ = M/H ; μr = 1 + χ ; B = μ0(H + M) = μ0μrH | both dimensionless; B in T | χ has no unit; its sign classifies the material |
| Temperature trend | Lower T favours paramagnetic alignment; higher T gradually disrupts ferromagnetic domains | Qualitative | Do not invent an undeclared numerical law, constant or instantaneous threshold |
Official 2026-27 Scope Check
Use the current official curriculum, not an old coaching list, to decide the scope. The 2026-27 CBSE Physics curriculum lists bar magnet as an equivalent solenoid, magnetic field lines, the magnetic field due to a dipole along and perpendicular to its axis, torque on a dipole, magnetisation, magnetic properties of materials with examples, and the effect of temperature.
Three listed topics carry the exact note “qualitative treatment only”. Study their physical meaning, directions, comparisons and relationships without turning this note into an unsupported prediction about question type:
- Bar magnet as an equivalent solenoid
- Magnetic field intensity due to a magnetic dipole along its axis and perpendicular to its axis
- Torque on a magnetic dipole in a uniform magnetic field
The practical consequence. Put the three qualitative-only topics into a concept map, then give separate attention to materials, magnetisation and temperature. Before relying on any study page, confirm the current curriculum on the official CBSE Academic website.
How to Write Clear Physics Answers
Good physics writing makes the reasoning easy to follow. Use the following response patterns as study habits, not as a prediction of marks or paper design.
| Question type | Useful response structure | Self-check |
|---|---|---|
| Classify a material from χ or μr | Name the class, then cite sign and magnitude. | Did you use both sign and size? |
| Compare dia / para / ferro | Use a table with the same criterion in each row. | Are examples and temperature effects consistent? |
| Explain a temperature effect | State the change in thermal disorder, then the change in alignment or domain order. | Did you preserve the source’s gradual wording? |
| Show a calculation | Write the relation, substitute SI values, then give unit and direction. | Do units and order of magnitude make sense? |
| Field-line reasoning | Connect the tangent with direction and the curve with a closed loop. | Could two directions exist at one point? |
| Assertion–Reason practice | Test each statement separately, then test the logical link. | Does the reason actually explain the assertion? |
Practice Worksheet
These ten original questions mix calculations with conceptual explanations. Do them on paper with the page scrolled away from the answers. In the three qualitative-only areas, treat calculations as optional mathematical enrichment and prioritise the physical interpretation. Write a full conceptual answer before opening the accordion, then compare your reasoning line by line.
Q1. A rectangular coil of 200 turns and area 1.5 × 10−3 m² carries a current of 0.25 A. Find its magnetic dipole moment.
Show Answer
Direction: perpendicular to the plane of the coil, given by the right-hand rule (curl the fingers along the current; the thumb gives m).
Q2. The coil of Q1 is placed in a uniform magnetic field of 0.40 T. Find the torque on it when its plane is (a) parallel to the field and (b) at 60° to the field.
Show Answer
(a) Plane parallel to B ⇒ m is perpendicular to B, so θ = 90°.
τ = mB sin90° = 0.075 × 0.40 × 1 = 0.030 N·m — this is the maximum torque.
(b) Plane at 60° to B ⇒ θ = 90° − 60° = 30°.
τ = 0.075 × 0.40 × sin30° = 0.075 × 0.40 × 0.5 = 0.015 N·m
Exactly half of the maximum, as it must be since sin30° = ½.
Q3. A short bar magnet produces a certain magnetic field at a point 30 cm away on its axis. At what distance on its perpendicular bisector would the field have the same magnitude?
Show Answer
(μ0/4π)(2m/ra³) = (μ0/4π)(m/re³)
Cancel (μ0/4π) and m: 2/ra³ = 1/re³, so ra³ = 2 re³, giving re = ra / 21/3.
21/3 = 1.2599, so re = 0.30 m ÷ 1.2599 = 0.238 m ≈ 23.8 cm
Sanity check: the equatorial field is weaker at equal distance, so to match the axial field you must move closer. 23.8 cm < 30 cm. Correct.
Q4. A bar of a magnetic material of length 8.0 cm and cross-sectional area 2.5 × 10−4 m² acquires a magnetic moment of 2.4 A·m². Find its intensity of magnetisation.
Show Answer
M = m/V = 2.4 ÷ (2.0 × 10−5) = 1.2 × 105 A/m
Q5. A specimen is placed inside a solenoid wound with 1200 turns per metre carrying 0.40 A, and develops a magnetisation of 0.12 A/m. Find H, χ and μr, and classify the specimen.
Show Answer
χ = M/H = 0.12 ÷ 480 = 2.5 × 10−4 (dimensionless)
μr = 1 + χ = 1.00025
Classification: χ is positive and very small, and μr is just above 1, so the specimen is paramagnetic — a “partly aligned” material, weakly attracted by a magnet.
Q6. Two identical paramagnetic samples are placed in the same applied field. Sample A is cooled while sample B is kept warmer. Which sample should develop the larger field-directed magnetisation, and why?
Show Answer
Q7. A magnetising field of 350 A/m is applied to a core of relative permeability 1200. Find the magnetic field B inside the core.
Show Answer
= (1.2566 × 10−6) × 4.20 × 105 = 0.528 T
With μr = 1200 this is a ferromagnetic core, which is why the field is so much larger than the μ0H = 4.4 × 10−4 T you would get in vacuum.
Q8. Explain, with reference to the atomic picture, why a diamagnetic substance is weakly repelled by a magnet while a paramagnetic substance is weakly attracted.
Show Answer
In a paramagnetic substance each atom already possesses a permanent magnetic moment, but in the absence of a field these moments point in random directions and cancel on average. An applied field exerts a torque on each moment and partially aligns them along the field. The resulting net moment is parallel to the field, so the substance is drawn towards the stronger part of the field — it is weakly attracted, and its susceptibility is small and positive.
Q9. A short bar magnet of magnetic moment 1.2 A·m² is placed on a bench. Find the magnetic field it produces at a point 25 cm from its centre, (a) on its axis and (b) on its perpendicular bisector, giving the direction in each case.
Show Answer
(a) Axial: B = 10−7 × (2 × 1.2)/(1.5625 × 10−2) = 10−7 × 153.6 = 1.536 × 10−5 T, directed parallel to the magnetic moment (i.e. along the axis, away from the N pole).
(b) Equatorial: B = 10−7 × 1.2/(1.5625 × 10−2) = 10−7 × 76.8 = 7.68 × 10−6 T, directed antiparallel to the magnetic moment.
Check the ratio: 1.536 × 10−5 ÷ 7.68 × 10−6 = 2. Good.
Q10. A student breaks a bar magnet into three pieces and draws one piece as north-only, one as south-only and the middle piece with no poles. Correct the drawing in words. Then describe the field-line direction around any one piece.
Show Answer
You have worked through the chapter’s physical picture, comparisons and calculations. Do not try to memorise everything at once. Aim for one more correct explanation than yesterday: one comparison-table row recalled without looking, one direction justified from a field-line diagram, or one optional calculation checked with SI units. Small, verified steps make the chapter manageable.