NCERT Solutions for Class 12 Physics Chapter 5: Magnetism and Matter
All 7 exercise questions of the latest NCERT textbook (2026–27 reprint) solved like a topper’s answer-sheet — bar magnet as a dipole, torque and potential energy in a uniform field, the solenoid as an equivalent bar magnet, field of a short magnet and magnetic properties of materials — with diagrams linked to the exact NCERT pages, comparison tables, complete revision notes and exam tips.
Chapter at a glance — complete revision notes
5.1 Introduction
- Magnetic phenomena are universal — galaxies, atoms, humans and animals are permeated by magnetic fields of many origins; the Earth’s magnetism predates human evolution. The word magnet comes from Magnesia, a Greek island where magnetic ore deposits were found as early as 600 BC. In Chapter 4 we learned that moving charges/currents produce magnetic fields (Oersted, Ampère, Biot, Savart); here magnetism is studied as a subject in itself.
- Common ideas: (i) the Earth behaves as a magnet with the field pointing approximately from geographic south to north; (ii) a freely suspended bar magnet points north–south — the tip towards geographic north is the north pole, the other the south pole; (iii) like poles repel, unlike poles attract; (iv) north and south poles cannot be isolated — cutting a bar magnet gives two smaller magnets: magnetic monopoles do not exist (unlike isolated electric charges); (v) magnets can be made of iron and its alloys.
5.2 The bar magnet
5.2.1 The magnetic field lines
Iron filings sprinkled over a sheet of glass placed above a short bar magnet show a pattern (Fig. 5.1) revealing the two poles, like the + and − charges of an electric dipole; a similar pattern is seen around a current-carrying solenoid. The properties of magnetic field lines (Fig. 5.2): (i) they form continuous closed loops (unlike electric field lines of a dipole, which begin on + and end on − charges); (ii) the tangent at a point gives the direction of the net field B there; (iii) the larger the number of lines crossing a unit area, the stronger is B (near the poles); (iv) they never intersect. A compass needle placed at various positions maps them. They are not “lines of force” (the force on a moving charge is normal to B).5.2.2 Bar magnet as an equivalent solenoid
The field lines of a bar magnet and of a current-carrying finite solenoid are alike (Fig. 5.2): a bar magnet can be thought of as a large number of circulating (Ampèrian) currents. Cutting a magnet in half is like cutting a solenoid — two weaker solenoids/magnets. The far axial field of a finite solenoid (Fig. 5.3a) equals that of a bar magnet: B = μ₀ 2m/(4π r³), so the magnetic moment m of a bar magnet equals that of the equivalent solenoid (m = NIA).5.2.3 The dipole in a uniform magnetic field
A magnetised needle of moment m in a uniform field B (Fig. 5.3b) feels a torque τ = m × B (magnitude mB sin θ) — a restoring torque that makes it oscillate about B — but no net force. Magnetic potential energy: U_m = −m·B = −mB cos θ (zero at θ = 90°); minimum (−mB) at θ = 0° (stable) and maximum (+mB) at θ = 180° (unstable). Example 5.1 (conceptual): (a) cutting a magnet transversely or lengthwise gives two magnets, each with a north and a south pole; (b) a nail near a magnet feels a force and torque because it acquires an induced moment and the magnet’s field is non-uniform — the force is attractive because the induced pole nearer the magnet is opposite; (c) not every configuration needs poles (a toroid has none: zero net moment); (d) to tell which of two iron bars is magnetised, bring ends together — repulsion shows both are magnets; otherwise lower one end of A on the middle and the end of B — no force at the middle of B means B is the magnet.5.2.4 The electrostatic analogue
Replacing p → m, 1/ε₀ → μ₀, E → B: equatorial field of a short bar magnet at distance r ≫ l: B_E = −μ₀ m/(4π r³) (direction opposite to m); axial field B_A = μ₀ 2m/(4π r³) (along m). Table 5.1 compares electric and magnetic dipoles (energy, torque, fields). Example 5.2: two identical needles — equilibrium is stable when m_Q is parallel to B_P at the position of Q (e.g. configuration Q₃ on the perpendicular bisector with m parallel to B; Q₃ and Q₆ stable, Q₄ and Q₅ unstable) and the lowest potential energy corresponds to the most stable orientation.
5.3 Magnetism and Gauss’s law
- For electric field lines of a dipole (Fig. 5.2c) a closed surface i without net charge has equal lines entering and leaving, while surface ii enclosing a net charge has net outward flux. For the magnetic field lines of Fig. 5.2(a, b), both surfaces have as many lines entering as leaving: net flux zero.
- Magnetic flux through an area element: Δφ_B = B·ΔS (unit weber, Wb = T m²); through a closed surface: φ_B = Σ B·ΔS = 0 — Gauss’s law of magnetism: the net magnetic flux through any closed surface is zero (Fig. 5.5). Compare Gauss’s law for electrostatics, φ_E = q/ε₀. The difference reflects the non-existence of isolated magnetic poles (monopoles): there are no sources or sinks of B; the simplest magnetic element is a dipole or a current loop. (If monopoles existed, the right side would be the enclosed magnetic charge.)
- Example 5.3: in Fig. 5.6 many field-line diagrams are wrong: (a) lines emanating from a point (represent the electric field of a long charged wire; magnetic lines circle the wire); (b) crossing lines/closed loops round empty space are impossible for static fields; (c) lines confined within a toroid — right; (d) straight, completely confined solenoid-end lines violate Ampère’s law (should curve out and meet); (e) bar magnet — right; (f) lines emanating from a plate — the electrostatic field of charged plates; (g) pole pieces with perfectly straight edge lines — some fringing is inevitable. Example 5.4 (conceptual): magnetic field lines are not lines of force on moving charges; Gauss’s law of magnetism would read ∮B·dS = μ₀q_m with monopoles; a bar magnet exerts no torque on itself and an element of wire exerts no force on itself (but on other elements); a system can have magnetic moments although its net charge is zero (paramagnetic atoms).
5.4 Magnetisation and magnetic intensity
- A circulating electron in an atom has a magnetic moment; in a bulk material these add vectorially. Magnetisation M = net magnetic moment per unit volume (vector; unit A m⁻¹; dimensions L⁻¹A).
- For a long solenoid with n turns per unit length carrying I: B₀ = μ₀nI. If the interior is filled with a material of magnetisation M: B = B₀ + B_m = μ₀(H + M), where B_m = μ₀M and magnetic intensity H = B₀/μ₀ = nI (due to external factors such as the current; unit A m⁻¹; independent of the material).
- Magnetic susceptibility χ: M = χ H (dimensionless; small positive for paramagnetic, small negative for diamagnetic materials — M and H opposite in direction for the latter). Hence B = μ₀(1 + χ)H = μ_r μ₀ H = μ H, with relative permeability μ_r = 1 + χ (the analogue of the dielectric constant) and permeability μ = μ₀μ_r; only one of χ, μ_r, μ is independent.
- Example 5.5: solenoid with core μ_r = 400, 1000 turns/m, I = 2 A: H = nI = 2 × 10³ A m⁻¹; M = (μ_r − 1)H = 399 × H ≈ 8 × 10⁵ A m⁻¹; B = μ_r μ₀ H = 400 × 4π × 10⁻⁷ × 2 × 10³ ≈ 1.0 T; magnetising current I_M ≈ 794 A (the extra current needed in the absence of the core to give the same B).
5.5 Magnetic properties of materials
- Classification by χ: diamagnetic (χ negative), paramagnetic (χ small positive), ferromagnetic (χ large positive, μ_r > 1000) (Table 5.2).
5.5.1 Diamagnetism
Diamagnetic substances tend to move from the stronger to the weaker part of an external field — they are repelled by a magnet (Fig. 5.7a: field lines expelled, field inside reduced slightly — about one part in 10⁵). Explanation: atoms have zero net magnetic moment; on applying a field, electrons whose orbital moments are along the field slow down and those opposite speed up (induced currents, Lenz’s law), giving a net moment opposite to the field. Examples: bismuth, copper, lead, silicon, nitrogen (STP), water, sodium chloride. Diamagnetism is present in all substances but usually masked. Superconductors are perfect diamagnets (χ = −1, μ_r = 0): fields are completely expelled — the Meissner effect; used for superconducting magnets (e.g. magnetically levitated trains).5.5.2 Paramagnetism
Weakly magnetised in an external field, tend to move from weak to strong field. Atoms/ions/molecules have permanent dipole moments, randomly oriented by thermal motion (no net M); a strong external field B₀ at low temperature aligns them (Fig. 5.7b: field lines concentrated, enhancement ~ 1 part in 10⁵). Examples: aluminium, sodium, calcium, oxygen (STP), copper chloride. χ and μ_r depend on temperature (Curie’s law); magnetisation increases with field and decreasing T up to saturation.5.5.3 Ferromagnetism
Strongly magnetised in an external field; strong tendency to move from weak to strong field. Atomic dipoles interact (quantum-mechanical exchange effect) so they align spontaneously in domains (~1 mm size, ~10¹¹ atoms); initially random from domain to domain (Fig. 5.8a); on applying B₀ domains orient along B₀ and those along B₀ grow, merging into a single giant domain (Fig. 5.8b). Hard magnetic materials (alnico — an alloy of iron, aluminium, nickel, cobalt and copper; lodestone) retain magnetisation and form permanent magnets (compass needles); soft ferromagnetic materials (soft iron) lose it when the field is removed. Ferromagnetic elements: iron, cobalt, nickel, gadolinium; μ_r > 1000. The property depends on temperature: above the Curie temperature a ferromagnet becomes paramagnetic (domain structure disintegrates).
Summary & “Points to ponder” (NCERT)
- Poles cannot be isolated; U = −m·B; far fields of a bar magnet: axial μ₀2m/4πr³, equatorial μ₀m/4πr³; Gauss’s law for magnetism (net flux zero); H, M, χ, μ_r and μ relations; dia-, para- and ferromagnetism; permanent magnets retain ferromagnetism for long.
- Technological exploitation of magnets predates their scientific explanation by about 2000 years. Magnetic monopoles do not exist (charge is quantised, no analogous unit of magnetic charge) so field lines are continuous closed loops; a minuscule difference in χ (−10⁻⁵ vs +10⁻⁵) separates diamagnetic and paramagnetic behaviour; a superconductor is a perfect diamagnet (BCS theory, 1957, explains it); diamagnetism is universal but weak; other types — ferrimagnetic, anti-ferromagnetic, spin glass — exist.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of magnetic moment of the magnet?
Concept and context. A bar magnet behaves like a magnetic dipole whose strength is measured by its magnetic moment m (a vector pointing from the south pole to the north pole, equal to the moment NIA of the equivalent solenoid; SI unit A m² = J T⁻¹). In a uniform external magnetic field B the net force on the magnet is zero but the two poles (equal and opposite “forces”) form a couple: the torque is τ = m × B, of magnitude τ = m B sin θ, where θ is the angle between the magnetic moment (the axis of the magnet) and B. The torque tends to align m with B and is the restoring torque by which a compass needle oscillates about the field direction (NCERT Fig. 5.3b).
Given: θ = 30° (sin 30° = 0.5); B = 0.25 T; τ = 4.5 × 10⁻² J (N m); m = ?
- Formula: τ = mB sin θ ⇒ m = τ/(B sin θ).
- Substitute: m = (4.5 × 10⁻²)/(0.25 × 0.5) = (4.5 × 10⁻²)/(0.125).
- Calculate: m = 0.36 J T⁻¹.
m = τ / (B sin θ) = 0.36 J T⁻¹ (A m²)
| Quantity | Value |
|---|---|
| Torque τ | 4.5 × 10⁻² N m |
| B sin θ | 0.25 × 0.5 = 0.125 T |
| Magnetic moment m | 0.36 J T⁻¹ = 0.36 A m² |
| Maximum torque (θ = 90°) | mB = 0.09 N m |
- Unit check: N m / T = J T⁻¹ = A m² ✓ (since 1 T = 1 N A⁻¹ m⁻¹).
- Equivalent solenoid: a magnet of moment 0.36 A m² corresponds to, e.g., a solenoid of 1000 turns of area 1.2 × 10⁻⁴ m² carrying 3 A (NIA = 0.36).
Conclusion: the magnetic moment of the bar magnet is 0.36 J T⁻¹ (0.36 A m²).
A short bar magnet of magnetic moment m = 0.32 J T⁻¹ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Concept and context. The potential energy of a magnetic dipole m in a uniform field B is U = −m·B = −mB cos θ (the zero of energy is fixed, by convention, at θ = 90°, when m is perpendicular to B — NCERT Eq. 5.3). The magnet is in equilibrium when the torque τ = mB sin θ vanishes, i.e. at θ = 0° (m parallel to B) or θ = 180° (m antiparallel to B). Equilibrium is stable when the potential energy is a minimum (a small disturbance produces a restoring torque) and unstable when it is a maximum.
Given: m = 0.32 J T⁻¹; B = 0.15 T ⇒ mB = 0.32 × 0.15 = 4.8 × 10⁻² J.
(a) Stable equilibrium
- The magnet is in stable equilibrium when its magnetic moment m is parallel to B (θ = 0°): the north pole points along the field.
- Potential energy: U = −mB cos 0° = −mB = −4.8 × 10⁻² J (the minimum value). Any small rotation increases U and gives a restoring torque tending to bring m back to B.
(b) Unstable equilibrium
- The magnet is in unstable equilibrium when m is antiparallel to B (θ = 180°): the north pole points opposite to the field.
- Potential energy: U = −mB cos 180° = +mB = +4.8 × 10⁻² J (the maximum value). The slightest disturbance produces a torque that turns the magnet away from this position (towards θ = 0°).
| Orientation | θ | Torque mB sin θ | Potential energy −mB cos θ | Nature |
|---|---|---|---|---|
| m ∥ B | 0° | 0 | −4.8 × 10⁻² J | stable (minimum U) |
| m ⟂ B | 90° | 4.8 × 10⁻² N m (max) | 0 (reference) | not in equilibrium |
| m antiparallel B | 180° | 0 | +4.8 × 10⁻² J | unstable (maximum U) |
- Energy difference: the work required to turn the magnet from the stable to the unstable orientation is ΔU = +mB − (−mB) = 2mB = 9.6 × 10⁻² J.
- Analogy: same as the electric dipole in a uniform electric field (U = −p·E): replace p → m.
Conclusion: (a) stable equilibrium: m parallel to B, potential energy −4.8 × 10⁻² J; (b) unstable equilibrium: m antiparallel to B, potential energy +4.8 × 10⁻² J.
A closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Concept and context. A current-carrying solenoid and a bar magnet produce very similar magnetic fields. The magnetic field lines of a finite solenoid (Fig. 5.2b) are closed loops that emerge from one end and enter the other, exactly like those of a bar magnet (Fig. 5.2a). The far axial field of a finite solenoid, B = μ₀·2m/(4πr³), is identical in form to the far axial field of a bar magnet; hence a bar magnet may be regarded as a large number of circulating currents (Ampère’s hypothesis that all magnetic phenomena are due to circulating currents), and its magnetic moment equals that of an equivalent solenoid producing the same field.
(i) Sense in which the solenoid acts like a bar magnet
- Field pattern: the magnetic field lines of the solenoid (inside, along the axis; outside, curving round from one end to the other) have the same shape as those of a bar magnet. One end of the solenoid, where the lines emerge, behaves as the north pole; the other, where the lines enter, as the south pole. (The north end is the one from which, looking at it, the current appears anticlockwise; by the right-hand rule curl the fingers along the current and the thumb points to the north pole.)
- Far field: the axial field at a large distance r on the axis has the same value, B = μ₀ 2m/(4π r³), as that of a bar magnet of the same magnetic moment.
- Behaviour in an external field: a freely suspended solenoid aligns itself with its axis along the north–south direction (like a compass needle), experiences a torque τ = m × B in a uniform field and has a potential energy −m·B.
- Breaking it into two gives two smaller solenoids with weaker fields, just as cutting a magnet gives two smaller magnets: there are no isolated poles.
(ii) Associated magnetic moment — Given: N = 800; A = 2.5 × 10⁻⁴ m²; I = 3.0 A.
- m = NIA = 800 × 3.0 × 2.5 × 10⁻⁴.
- 800 × 3.0 = 2400; 2400 × 2.5 × 10⁻⁴ = 0.60.
- m = 0.60 J T⁻¹ (A m²), directed along the axis of the solenoid, from the south pole end to the north pole end (the direction given by the right-hand thumb rule).
m = NIA = 0.60 J T⁻¹
| Property | Bar magnet | Current-carrying solenoid |
|---|---|---|
| Field lines | closed loops, N → S outside | closed loops, same pattern |
| Poles | N and S at the ends | ends behave as N and S |
| Magnetic moment | m (S → N) | m = NIA along axis |
| Torque in uniform field | m × B | m × B |
| Cutting in half | two smaller magnets | two smaller solenoids |
Conclusion: the solenoid acts like a bar magnet because its field lines and its torque/energy behaviour in an external field are the same — one end is a north pole and the other a south pole; its magnetic moment is NIA = 0.60 J T⁻¹ along its axis.
If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field? (The question in the book refers to “Exercise 5.5” but the solenoid is the one described in Exercise 5.3.)
Concept and context. A solenoid carrying a current has a magnetic moment m = NIA along its axis; like a bar magnet or a current loop, it experiences in a uniform field a torque τ = m × B of magnitude mB sin θ, where θ is the angle between the axis (direction of m) and B, and no net force. When the solenoid is free to rotate about a vertical axis in a horizontal field, this torque tends to rotate it until its axis lies along the field, i.e. along the horizontal field direction (just as a compass needle aligns itself in the north–south direction).
Given: m = 0.60 J T⁻¹ (from Exercise 5.3); B = 0.25 T; θ = 30° (sin 30° = 0.5).
- τ = mB sin θ = 0.60 × 0.25 × 0.5.
- 0.60 × 0.25 = 0.15; 0.15 × 0.5 = 0.075.
- τ = 7.5 × 10⁻² N m (J).
τ = mB sin θ = 7.5 × 10⁻² N m
- Direction: along m × B — vertical (since m and B are both horizontal), in the sense that rotates the axis towards the field direction. The net force on the solenoid is zero in a uniform field.
- Limits: maximum torque mB = 0.15 N m at θ = 90°; zero at θ = 0° (stable) or 180° (unstable).
- Why the vertical axis matters: because the solenoid can turn only about the vertical direction, the horizontal component of the torque is balanced by the suspension; the whole magnetic torque (which is vertical) can rotate the solenoid.
| θ (axis, B) | τ = mB sin θ |
|---|---|
| 0° | 0 |
| 30° (given) | 7.5 × 10⁻² N m |
| 60° | 0.130 N m |
| 90° | 0.15 N m (maximum) |
Conclusion: the torque on the solenoid is 7.5 × 10⁻² N m, tending to turn its axis into alignment with the applied field.
A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?
Concept and context. The work done by an external torque in turning a magnet slowly (without kinetic energy change) from orientation θ₁ to θ₂ equals the change in magnetic potential energy: W = U(θ₂) − U(θ₁), where U(θ) = −mB cos θ (zero at θ = 90°). The torque exerted by the field on the magnet in any orientation is τ = mB sin θ. Initially the magnet is aligned with the field, θ₁ = 0°, in stable equilibrium with U₁ = −mB.
Given: m = 1.5 J T⁻¹; B = 0.22 T ⇒ mB = 1.5 × 0.22 = 0.33 J.
(a) Work required
- (i) m normal to B (θ₂ = 90°): W = U(90°) − U(0°) = 0 − (−mB) = mB = 0.33 J.
- (ii) m opposite to B (θ₂ = 180°): W = U(180°) − U(0°) = (+mB) − (−mB) = 2mB = 0.66 J.
(b) Torque on the magnet
- (i) θ = 90°: τ = mB sin 90° = mB = 0.33 N m (maximum), acting to turn m back towards B.
- (ii) θ = 180°: τ = mB sin 180° = 0; the magnet is in (unstable) equilibrium — though any disturbance would produce a torque.
| Final orientation | Work W = ΔU | Torque mB sin θ |
|---|---|---|
| (i) m ⟂ B (θ = 90°) | 0.33 J | 0.33 N m |
| (ii) m antiparallel to B (θ = 180°) | 0.66 J | 0 |
- Remark: units — 1 J T⁻¹ × 1 T = 1 J; torque is expressed in N m which is dimensionally the same as J but is a different physical quantity (energy vs. torque).
- The work done by the external torque is stored as potential energy of the magnet in the field; if released, the magnet rotates back to the aligned position and converts the stored energy into kinetic energy.
Conclusion: (a) the work required is 0.33 J to turn the magnet normal to the field and 0.66 J to turn it opposite to the field; (b) the torque is 0.33 N m in case (i) and zero in case (ii).
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 30° with the axis of the solenoid?
Concept and context. A current-carrying coil of N turns and area A has a magnetic moment m = NIA along its axis. In a uniform magnetic field the forces on the elements of the coil sum to zero (net force = 0), but they give a torque τ = m × B of magnitude mB sin θ. The solenoid, being free to turn in a horizontal plane about a vertical axis through its centre, will rotate under this torque until its axis lines up with the field, like a compass needle.
Given: N = 2000; A = 1.6 × 10⁻⁴ m²; I = 4.0 A; B = 7.5 × 10⁻² T; θ = 30°.
(a) Magnetic moment
m = NIA = 2000 × 4.0 × 1.6 × 10⁻⁴ = 1.28 J T⁻¹ (≈ 1.3 A m²)
- Direction: along the axis of the solenoid, from its south-pole end to its north-pole end (right-hand thumb rule on the current).
(b) Force and torque
- Force: the field is uniform, so the net force on the solenoid is zero (the forces on the opposite current elements cancel). A net force would appear only in a non-uniform field.
- Torque: τ = mB sin θ = 1.28 × 7.5 × 10⁻² × sin 30° = 1.28 × 7.5 × 10⁻² × 0.5 = 4.8 × 10⁻² N m.
- Direction of the torque: along m × B — vertical, in the sense that rotates the axis of the solenoid towards the field direction (the solenoid turns in the horizontal plane).
| Quantity | Value |
|---|---|
| Magnetic moment m | 1.28 J T⁻¹ |
| Net force in uniform B | 0 |
| Torque τ = mB sin 30° | 4.8 × 10⁻² N m |
| Maximum possible torque (θ = 90°) | mB = 9.6 × 10⁻² N m |
- Equilibrium and oscillation: when released, the solenoid oscillates about the field direction (θ = 0°) with the energy being gradually dissipated by friction and settles with its axis along B — the same principle as a compass needle.
- Comparison with Exercise 5.4: the same relation τ = mB sin θ was used there; here m is larger by a factor of about two (1.28 vs 0.60) because the number of turns is larger.
Conclusion: (a) the magnetic moment is 1.28 J T⁻¹; (b) the net force is zero and the torque is 4.8 × 10⁻² N m, tending to align the solenoid’s axis with the field.
A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Concept and context. The magnetic field of a short bar magnet (or magnetic dipole) at a distance r ≫ l from its centre is obtained from the electric dipole field by the substitutions p → m and 1/ε₀ → μ₀ (NCERT Section 5.2.4). Hence on the axis: B_A = (μ₀/4π)(2m/r³), directed along m (from the south to the north pole); on the equatorial line (normal bisector): B_E = (μ₀/4π)(m/r³), directed opposite to m (parallel to the line from N to S). Thus the axial field is exactly twice the equatorial field at the same distance, and both fall off as 1/r³.
Given: m = 0.48 J T⁻¹; r = 10 cm = 0.10 m ⇒ r³ = 10⁻³ m³; μ₀/4π = 10⁻⁷ T m A⁻¹.
(a) On the axis
- B_A = (μ₀/4π)(2m/r³) = 10⁻⁷ × (2 × 0.48)/10⁻³ = 10⁻⁷ × 960.
- B_A = 9.6 × 10⁻⁵ T (0.96 gauss).
- Direction: along the magnetic moment m, i.e. along the axis from the south pole to the north pole (pointing away from the north-pole end when the point is on the north side).
(b) On the equatorial line (normal bisector)
- B_E = (μ₀/4π)(m/r³) = 10⁻⁷ × 0.48/10⁻³ = 10⁻⁷ × 480.
- B_E = 4.8 × 10⁻⁵ T (0.48 gauss).
- Direction: opposite to the magnetic moment — parallel to the axis, pointing from the north pole towards the south pole direction (i.e. along the direction of N → S).
| Position (r = 10 cm) | Formula | Magnitude | Direction |
|---|---|---|---|
| (a) Axial point | μ₀2m/4πr³ | 9.6 × 10⁻⁵ T | along m (S → N) |
| (b) Equatorial point | μ₀m/4πr³ | 4.8 × 10⁻⁵ T | opposite to m (N → S) |
| Ratio B_A : B_E | — | 2 : 1 | antiparallel to each other |
- Comparison: both are of the same order as the Earth’s magnetic field (≈ 0.3–0.6 gauss) — a compass near such a magnet is strongly affected.
- Analogy with the electric dipole: E_axis = 2p/4πε₀r³ and E_equatorial = −p/4πε₀r³ (Chapter 1) — the same factors of 2 and −1.
- Validity: the formulas assume r ≫ l (length of the magnet); for a “short” magnet and r = 10 cm this holds.
Conclusion: on the axis the field is 9.6 × 10⁻⁵ T directed along m (S → N); on the equatorial line it is 4.8 × 10⁻⁵ T directed opposite to m.