Formula
Field of magnetic dipole
Axial: B = (μ₀/4π)(2m/r³). Equatorial: B = -(μ₀/4π)(m/r³). Falls off as 1/r³ (analogous to electric dipole).
-- NCERT Class 12 Physics, Ch. 5, p. 141Magnetic Field of Dipole
8 MCQs1 revision card9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismPYQ coverage: NEET 2023, 2024Official key: NTA-verifiedLast updated: 25 Sep 2026
Magnetic Field of Dipole, explained for NEET
The factor-of-two swap is what costs marks here. At the same distance from the same bar magnet, the axial field is exactly twice the equatorial field — and a question that says "on the perpendicular bisector" while you reach for the 2m expression takes four marks off you in under a minute. The distractor is always sitting in the options.
NCERT Class 12 Physics, Chapter 5, page 139 gives both far-field results for a dipole of moment m at distance r:
- Axial (on the magnet's own axis, produced end-on): B = (μ₀/4π)(2m/r³)
- Equatorial (on the perpendicular bisector, broadside-on): B = (μ₀/4π)(m/r³)
Two things are worth fixing in memory. First, the numerator: axial carries the 2, equatorial does not. Second, the direction: the axial field points parallel to m, while the equatorial field points antiparallel to m. A question can test the sign without asking for a number at all.
Both expressions are far-field approximations — they hold when r is much larger than the magnet's own length (2l). If a problem gives you a magnet of length comparable to r, these results do not apply, and NEET stems signal this with phrases like "short bar magnet" or "a point far from the magnet."
The r⁻³ dependence is the other high-value feature. Halving the distance multiplies the field by 8, not by 2 or 4. Ratio questions exploit this constantly, and they combine it with the axial/equatorial factor: an axial point at distance r and an equatorial point at the same r give a ratio of exactly 2, but an axial point at r against an equatorial point at 2r gives 2 × 8 = 16.
Watch-out: "equatorial" refers to the magnet's equator — the plane through its centre perpendicular to its axis. It has nothing to do with the Earth's magnetic equator, which is a separate idea in the same chapter.
Can you answer these Magnetic Field of Dipole MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
For a short bar magnet of magnetic moment m, the magnetic field at an axial point a distance r from its centre is given by
Show answer and why every option is right or wrong
Answer: C. C is correct. The axial (end-on) field of a short dipole carries the factor 2 in the numerator and falls as the cube of the distance, as stated in NCERT Class 12 Physics, Chapter 5, page 139.
Why A is wrong: A is wrong because it is the equatorial expression — the factor 2 is missing, which is exactly the axial/equatorial swap this topic's PYQ pattern exploits.
Why B is wrong: B is wrong because it combines the equatorial numerator with an inverse-square distance law; dipole fields fall as r⁻³, not r⁻².
Why D is wrong: D is wrong because the axial numerator 2m is correct but the distance dependence is not — an r⁻² law belongs to a monopole-like source, which does not exist magnetically.
MCQ 2Easy RecallPractice
The magnetic field at a point on the equatorial line of a short bar magnet is directed
Show answer and why every option is right or wrong
Answer: B. B is correct. The equatorial (broadside-on) field of a dipole opposes the moment vector, whereas the axial field is along it — the direction contrast set out with the two field expressions in NCERT Class 12 Physics, Chapter 5, page 139.
Why A is wrong: A is wrong because parallel-to-m describes the axial field, not the equatorial one; this is the direction half of the axial/equatorial confusion.
Why C is wrong: C is wrong because the equatorial field is antiparallel to m, not perpendicular to it — a perpendicular field would arise nowhere on the bisector for a dipole.
Why D is wrong: D is wrong because a radially outward field would require a magnetic monopole; the dipole field has a definite direction set by m, not by the radial direction.
MCQ 3Easy RecallPractice
The axial and equatorial expressions for the field of a bar magnet are valid under which condition?
Show answer and why every option is right or wrong
Answer: A. A is correct. Both results are far-field approximations, valid when r is large compared with the magnet's own length — the condition NCERT Class 12 Physics, Chapter 5, page 139 attaches to the short-dipole formulas.
Why B is wrong: B is wrong because the approximation runs the other way: close to the magnet, the two poles cannot be treated as a point dipole and the r⁻³ result fails.
Why C is wrong: C is wrong because r comparable to the magnet length is precisely the regime where the far-field expansion is not justified.
Why D is wrong: D is wrong because the dipole result is not exact at all distances; the exact axial field of a finite magnet of half-length l involves (r² − l²) in the denominator and reduces to the r⁻³ form only for r ≫ l.
MCQ 4Direct ApplicationPractice
A short bar magnet of magnetic moment 0.80 A·m² is placed in vacuum. The magnetic field at a point on its axis, 2.0 × 10⁻¹ m from its centre, is (take μ₀/4π = 1.0 × 10⁻⁷ T·m/A)
Show answer and why every option is right or wrong
Answer: B. B is correct. B_ax = (μ₀/4π)(2m/r³) = 1.0 × 10⁻⁷ × (2 × 0.80)/(8.0 × 10⁻³) = 2.0 × 10⁻⁵ T, using the axial expression from NCERT Class 12 Physics, Chapter 5, page 139.
Why A is wrong: A is wrong because it drops the factor 2 from the axial numerator — this is the equatorial value at the same distance, the standard distractor for this pattern.
Why C is wrong: C is wrong because it applies the factor 2 twice, or equivalently uses 4m in the numerator; only one factor of 2 distinguishes axial from equatorial.
Why D is wrong: D is wrong because it squares 2.0 × 10⁻¹ m (4.0 × 10⁻²) instead of cubing it — an r⁻² substitution in an r⁻³ formula.
MCQ 5Direct ApplicationPractice
For a short bar magnet, the magnitude of the field at an axial point a distance r from the centre, divided by the magnitude of the field at an equatorial point at the same distance r, equals
Show answer and why every option is right or wrong
Answer: C. C is correct. At equal r the two expressions differ only by the factor 2 in the axial numerator, so B_ax/B_eq = 2 — read directly off the pair of formulas in NCERT Class 12 Physics, Chapter 5, page 139.
Why A is wrong: A is wrong because it inverts the ratio, placing the factor 2 with the equatorial field instead of the axial field.
Why B is wrong: B is wrong because the two fields are not equal at the same distance; the axial field is stronger by exactly the factor 2.
Why D is wrong: D is wrong because 4 would require the numerators to differ by 4m versus m; the dipole result gives 2m versus m.
MCQ 6Direct ApplicationPractice
The magnetic field at an equatorial point of a short bar magnet is B₀ at a distance r. If the distance is halved, the field at the new equatorial point becomes
Show answer and why every option is right or wrong
Answer: C. C is correct. The dipole field varies as r⁻³, so halving r multiplies the field by 2³ = 8, following from the equatorial expression in NCERT Class 12 Physics, Chapter 5, page 139.
Why A is wrong: A is wrong because it treats the field as varying as r⁻¹, which would be the long-straight-wire behaviour, not a dipole's.
Why B is wrong: B is wrong because it applies an inverse-square law; the dipole field is inverse-cube, so the multiplier is 2³ and not 2².
Why D is wrong: D is wrong because 16 corresponds to r⁻⁴; no factor of 2 from the axial/equatorial distinction applies here, since both points are equatorial.
MCQ 7CalculationPractice
A short bar magnet produces a field of magnitude B₁ at an axial point a distance r from its centre, and a field of magnitude B₂ at an equatorial point a distance 2r from its centre. The ratio B₁/B₂ is
Show answer and why every option is right or wrong
Answer: C. C is correct. B₁ = (μ₀/4π)(2m/r³) and B₂ = (μ₀/4π)(m/8r³), so B₁/B₂ = 2 × 8 = 16 — combining the axial factor with the r⁻³ scaling from NCERT Class 12 Physics, Chapter 5, page 139.
Why A is wrong: A is wrong because it applies an inverse-square distance factor (2² = 4) and then drops the axial factor of 2 entirely.
Why B is wrong: B is wrong because it captures the distance factor 2³ = 8 but omits the extra factor 2 that distinguishes the axial numerator from the equatorial one.
Why D is wrong: D is wrong because it applies the axial factor twice (2 × 2 × 8); only one factor of 2 separates the axial and equatorial expressions.
MCQ 8Concept TrapPractice
A short bar magnet of moment m lies along the x-axis with its north pole pointing towards +x. Point P is on the +x-axis at distance r, and point Q is on the +y-axis at the same distance r, both measured from the magnet's centre. Which statement describes the fields at P and Q?
Show answer and why every option is right or wrong
Answer: B. B is correct. P is axial, so its field is parallel to m (+x) with magnitude (μ₀/4π)(2m/r³); Q is equatorial, so its field is antiparallel to m (−x) with magnitude (μ₀/4π)(m/r³) — the direction and magnitude pairing given in NCERT Class 12 Physics, Chapter 5, page 139.
Why A is wrong: A is wrong on direction: it gets the 2:1 magnitude ratio right but has the equatorial field parallel to m, when it is antiparallel.
Why C is wrong: C is wrong on both counts: the equatorial field lies along the magnet's axis direction (here −x), not along the line joining the magnet to Q, and the magnitudes differ by a factor 2.
Why D is wrong: D is wrong because it reverses both assignments — it gives the axial point the equatorial direction and the equatorial point the larger magnitude.
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Magnetic Field of Dipole: quick recall before you leave
How do you solve a Magnetic Field of Dipole question? A worked example
- 1
Given
• Magnetic dipole moment of a short bar magnet: m = 1.2 A·m²• Distance of point P from the centre, on the perpendicular bisector: r = 1.0 × 10⁻¹ m• Medium: vacuum, μ₀/4π = 1.0 × 10⁻⁷ T·m/A (exact by definition of μ₀)
- 2
Required
The magnitude of the magnetic field at P, and the magnitude at a point on the axis at the same distance.
- 3
Concept
P lies on the perpendicular bisector of the magnet — the equatorial line. The equatorial field of a short dipole is weaker than the axial field at the same distance by exactly a factor of 2, and both fall as r⁻³. The phrase "perpendicular bisector" is the whole signal here; it is what selects the m/r³ expression rather than 2m/r³.
- 4
Formula
B_eq = (μ₀/4π)(m/r³), and for the comparison, B_ax = (μ₀/4π)(2m/r³).
- 5
Substitution
B_eq = (1.0 × 10⁻⁷ T·m/A) × (1.2 A·m²) / (1.0 × 10⁻¹ m)³
- 6
Calculation
r³ = (1.0 × 10⁻¹)³ = 1.0 × 10⁻³ m³
B_eq = (1.0 × 10⁻⁷ × 1.2) / (1.0 × 10⁻³) = 1.2 × 10⁻⁷ ⁺ ³ = 1.2 × 10⁻⁴ T
B_ax = 2 × B_eq = 2.4 × 10⁻⁴ T
The factor 2 in the axial expression and the exponent 3 on r are exact — they come from the structure of the dipole field, not from measurement — so neither limits the significant-figure count. The constant μ₀/4π = 1.0 × 10⁻⁷ T·m/A is exact by the historical definition of μ₀ and likewise imposes no limit. The measured inputs m and r each carry 2 significant figures, so the answers are quoted to 2. - 7
Final answer
Equatorial field: B_eq = 1.2 × 10⁻⁴ T, directed antiparallel to m.
Axial field at the same distance: B_ax = 2.4 × 10⁻⁴ T, directed parallel to m. - 8
Common trap
The stem says "perpendicular bisector," not "equatorial line," and a student scanning quickly sees only "bisector … 1.0 × 10⁻¹ m" and reaches for the 2m/r³ expression out of habit — the axial formula is the one most aspirants rehearse first. The answer 2.4 × 10⁻⁴ T will be sitting in the options. Read for the geometry word before selecting the numerator: on the axis → 2m; on the bisector, broadside, or equatorial → m. A second version of the same trap gets the magnitude right and the direction wrong, marking the equatorial field as parallel to m.
- 9
Similar NEET-style question
A short bar magnet of moment 2.5 A·m² is placed in vacuum. Point A lies on its axis at 2.0 × 10⁻¹ m from the centre and point B lies on its equatorial line at 1.0 × 10⁻¹ m from the centre. Find the ratio of the field magnitudes at A and B, and state the direction of each relative to m. (Take μ₀/4π = 1.0 × 10⁻⁷ T·m/A.)
What to remember before solving Magnetic Field of Dipole questions
Which Magnetic Field of Dipole formulas do you need for NEET?
1 formula — click to collapse
Magnetic dipole field (axial/equatorial)
Magnetic field from dipole moment m at distance r along axis or equatorial.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m | dipole moment | A*m^2 |
| r | distance | m |
Valid when
- r >> dipole size
- Far-field approximation
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 10 formulas · 4 question patterns from its other lessons.
Magnetic Field of Dipole questions from past NEET papers
2 questions from NEET 2023, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
15π2
2128π2
350π2
41280π2
NTA Answer: Option 4(revised_final)
NEET 2023
The net magnetic flux through any closed surface is
1Positive
2Infinity
3Negative
4Zero
NTA Answer: Option 4(final)
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
How does NEET ask about Magnetic Field of Dipole?
1 recurring pattern from past papers — click to collapse
Magnetic dipole field at axial / equatorial point. Axial 2m/r^3, equatorial m/r^3.
InterpretationMedium
Common distractors
uses axial formula equatorial
Mixes axial 2m vs equatorial m factor
Sources
NCERT refs: Class 12 Physics Chapter 5, p.139
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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