Formula
Torque on current loop
τ = N B I A sin θ, where N is number of turns, A is loop area, θ is angle between B and normal to loop. Magnetic dipole moment m = NIA; τ = m × B.
-- NCERT Class 12 Physics, Ch. 4, p. 126Current Loop Dipole Moment
8 MCQs1 revision card9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Current Loop Dipole Moment, explained for NEET
A current loop has no magnet in it, no poles, no iron. Yet at distances large compared with its size, its field is indistinguishable from that of a tiny bar magnet. That equivalence is what "magnetic dipole moment" encodes, and the quantity that carries it is
m = NIA
for a plane coil of N turns, each of area A, carrying steady current I. NCERT Class 12 Physics, Chapter 4, page 128 introduces it in the torque expression τ = m × B, where the whole geometry of the loop collapses into the single vector m.
Three things about that definition earn marks.
It is a vector, and its direction is not in the plane of the loop. m points along the normal to the loop's plane, with the sense fixed by the right-hand rule: curl the fingers along the current, and the thumb gives m. A loop lying flat in the horizontal plane has a vertical moment.
N multiplies, A multiplies, but the shape of A does not matter. Square, circular, triangular — only the enclosed area enters. This is why rewinding a given length of wire into fewer, larger turns changes m even though the wire and current are unchanged.
Its unit is A·m², not tesla and not weber. A dimensional check settles most option lists instantly: current × area.
For NEET, m is rarely asked alone. It appears as the first step of a torque or dipole-field problem, and the marks are lost there — an N dropped, a diameter used where a radius was needed (A = πR², so using the diameter inflates m fourfold), or the moment reported along the plane instead of the normal.
Watch-out: "area" in m = NIA means the area of one turn, not N turns' worth. The N is already counted separately.
Can you answer these Current Loop Dipole Moment 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
The SI unit of magnetic dipole moment is
Show answer and why every option is right or wrong
Answer: C. C is correct. Magnetic moment is defined as m = NIA — current multiplied by area — so its unit is ampere × metre² = A·m², as listed with the torque relation in NCERT Class 12 Physics, Chapter 4, page 128.
Why A is wrong: A is wrong because the tesla is the unit of magnetic field B, the quantity m is multiplied by to give torque — not of m itself.
Why B is wrong: B is wrong because the weber is the unit of magnetic flux (B × area), not of current × area; the two differ by a factor with the dimensions of B/I.
Why D is wrong: D is wrong because T·m has no standing as a unit of moment; it arises from mixing the field unit with a length rather than applying m = NIA.
MCQ 2Easy RecallPractice
For a plane coil of N turns, each of area A, carrying a steady current I, the magnitude of the magnetic dipole moment is
Show answer and why every option is right or wrong
Answer: B. B is correct. Each turn contributes a moment IA and the N turns are coplanar and carry the same current, so the contributions add directly to give m = NIA — the form used inside τ = NBIA sin θ in NCERT Class 12 Physics, Chapter 4, page 128.
Why A is wrong: A is wrong because it divides by N; additional turns increase the moment, they do not dilute it.
Why C is wrong: C is wrong because it divides by area; a larger loop at the same current encloses more, and the moment grows with A rather than shrinking.
Why D is wrong: D is wrong because the turns add linearly, not quadratically; N² would double-count the turns, which enter once each.
MCQ 3Easy RecallPractice
The direction of the magnetic dipole moment vector of a plane current loop is
Show answer and why every option is right or wrong
Answer: A. A is correct. The moment is an axial vector directed along the loop's normal, its sense fixed by curling the right hand's fingers along the current so the thumb gives m — the convention that makes τ = m × B in NCERT Class 12 Physics, Chapter 4, page 128 a well-defined cross product.
Why B is wrong: B is wrong because a tangential direction changes from point to point around the loop and so cannot define a single vector for the loop as a whole.
Why C is wrong: C is wrong because a radial in-plane direction is likewise not unique around the loop, and it would make the moment perpendicular to the loop's axis rather than along it.
Why D is wrong: D is wrong because m is fixed to the loop and tilts with it; a loop turned on its side has a horizontal moment.
MCQ 4Direct ApplicationPractice
A flat circular coil of 200 turns (exact) and radius 7.00 cm carries a steady current of 0.500 A. The magnitude of its magnetic dipole moment is closest to
Show answer and why every option is right or wrong
Answer: A. A is correct. A = πR² = π(7.00 × 10⁻² m)² = 1.539 × 10⁻² m², so m = NIA = 200 × 0.500 × 1.539 × 10⁻² = 1.54 A·m², applying m = NIA from NCERT Class 12 Physics, Chapter 4, page 128.
Why B is wrong: B is wrong because it omits the turn count, giving the moment of a single turn (IA) rather than of the 200-turn coil.
Why C is wrong: C is wrong because it uses the radius in centimetres without converting to metres in A = πR², inflating the area, and the moment, by a factor of 10⁴.
Why D is wrong: D is wrong because it doubles the correct value, as happens when the loop area is taken as 2πR² — mixing the circumference formula into the area.
MCQ 5Direct ApplicationPractice
A single circular loop of diameter 12.0 cm carries a current of 2.00 A. Its magnetic dipole moment is
Show answer and why every option is right or wrong
Answer: B. B is correct. The diameter is 12.0 cm, so R = 6.00 × 10⁻² m and A = πR² = 1.131 × 10⁻² m²; with N = 1, m = IA = 2.00 × 1.131 × 10⁻² = 2.26 × 10⁻² A·m², following m = NIA in NCERT Class 12 Physics, Chapter 4, page 128.
Why A is wrong: A is wrong because it uses the diameter in place of the radius in A = πR², which multiplies the area — and hence m — by four.
Why C is wrong: C is wrong because it uses the circumference 2πR as though it were the area; the moment requires an enclosed area, in m², not a length.
Why D is wrong: D is wrong because it reports the area πR² alone and drops the factor I = 2.00 A.
MCQ 6Concept TrapPractice
Two plane loops carry the same steady current I and have the same number of turns. Loop P is a square of side L; loop Q is a circle of circumference 4L. Which statement about their magnetic dipole moments is correct?
Show answer and why every option is right or wrong
Answer: C. C is correct. In m = NIA only the enclosed area enters — shape is irrelevant — and for a fixed perimeter 4L the circle encloses 4L²/π ≈ 1.27L² against the square's L², so Q's moment is larger. The area-based definition is the one used in NCERT Class 12 Physics, Chapter 4, page 128.
Why A is wrong: A is wrong because m depends on enclosed area, not on the length of wire used; equal perimeters do not give equal areas.
Why B is wrong: B is wrong because it reverses the isoperimetric comparison — for a fixed perimeter the circle, not the square, encloses the greater area.
Why D is wrong: D is wrong because m = NIA holds for any plane loop; the formula refers to the enclosed area and makes no assumption about the boundary's shape.
MCQ 7Direct ApplicationPractice
A plane coil has a fixed number of turns and a fixed area. Its magnetic dipole moment is 4.80 × 10⁻² A·m² when it carries a steady current of 0.600 A. If the current is raised to 1.50 A while the coil itself is unchanged, its new magnetic dipole moment is
Show answer and why every option is right or wrong
Answer: A. A is correct. With N and A unchanged, m = NIA is directly proportional to I, so m₂ = m₁ × (I₂/I₁) = 4.80 × 10⁻² × (1.50/0.600) = 4.80 × 10⁻² × 2.50 = 1.20 × 10⁻¹ A·m², applying m = NIA from NCERT Class 12 Physics, Chapter 4, page 128.
Why B is wrong: B is wrong because it treats m as a fixed property of the coil's geometry alone, ignoring that m = NIA also scales directly with the current flowing through it.
Why C is wrong: C is wrong because it inverts the current ratio, scaling by (I₁/I₂) = 0.600/1.50 = 0.400 instead of (I₂/I₁), which shrinks the moment rather than growing it.
Why D is wrong: D is wrong because it scales by the square of the current ratio, (2.50)² = 6.25, as if m depended on I² rather than on I to the first power.
MCQ 8CalculationPractice
A rectangular coil of 50 turns (exact) has sides 8.00 cm and 5.00 cm and carries a current of 1.20 A. It is then replaced by a square coil wound from the same number of turns of the same total wire length per turn, carrying the same current. The magnetic moment of the square coil exceeds that of the rectangular coil by approximately
Show answer and why every option is right or wrong
Answer: D. D is correct. The rectangle's perimeter is 2(8.00 + 5.00) = 26.0 cm, so the square has side 6.50 cm; areas are 4.00 × 10⁻³ m² and 4.225 × 10⁻³ m², and Δm = NIΔA = 50 × 1.20 × 2.25 × 10⁻⁴ = 1.35 × 10⁻² A·m², using m = NIA from NCERT Class 12 Physics, Chapter 4, page 128.
Why A is wrong: A is wrong because it takes the square's side as the full 13.0 cm half-perimeter instead of a quarter of the perimeter, overstating the area gain.
Why B is wrong: B is wrong because equal perimeter does not mean equal area; the square encloses more than any non-square rectangle of the same perimeter, so the moments differ.
Why C is wrong: C is wrong because it reports the rectangular coil's own moment (50 × 1.20 × 4.00 × 10⁻³ = 0.240 A·m²) rather than the difference the question asks for.
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Current Loop Dipole Moment: quick recall before you leave
How do you solve a Current Loop Dipole Moment question? A worked example
- 1
Given.
• Number of turns, N = 120 (exact — a count)• Radius of each turn, R = 4.50 cm = 4.50 × 10⁻² m• Current, I = 0.250 A• Coil in the horizontal plane; current clockwise viewed from above
- 2
Required.
The magnitude |m| of the magnetic dipole moment, and its direction.
- 3
Concept.
A plane current loop behaves as a magnetic dipole. Every turn contributes a moment equal to (current × enclosed area), directed along the loop's normal; identical coplanar turns carrying the same current add directly. The direction follows the right-hand rule applied to the current sense.
- 4
Formula.
m = NIA, with A = πR² for a circular turn.
- 5
Substitution.
A = π(4.50 × 10⁻² m)²
m = 120 × 0.250 A × π(4.50 × 10⁻² m)² - 6
Calculation.
(4.50 × 10⁻²)² = 2.025 × 10⁻³
A = π × 2.025 × 10⁻³ = 6.3617 × 10⁻³ m²
m = 120 × 0.250 × 6.3617 × 10⁻³ = 30.0 × 6.3617 × 10⁻³ = 1.9085 × 10⁻¹ A·m²
The turn count N = 120 is an exact counting number and π is a mathematical constant; neither limits the significant figures. The measured inputs are R = 4.50 cm (3 s.f.) and I = 0.250 A (3 s.f.), so the answer carries 3 significant figures. - 7
Final answer.
|m| = 1.91 × 10⁻¹ A·m².
Direction: curling the right hand's fingers clockwise as seen from above sends the thumb downward, so m points vertically downward — along the normal to the coil's plane, not anywhere within it. - 8
Common trap.
Two failure modes account for most lost marks here. First, dropping N and reporting the single-turn moment 1.59 × 10⁻³ A·m², which sits plausibly among the options. Second, being handed a diameter instead of a radius: because A = πR², using 4.50 cm as a diameter rather than a radius changes the answer by a factor of four, and the wrong value is almost always on the option list. Read the geometry word before touching the calculator. A third, quieter trap: reporting the direction as "in the plane of the coil." The moment is always along the normal.
- 9
Similar NEET-style question.
A square coil of 75 turns and side 6.00 cm carries a current of 0.400 A. Compute its magnetic dipole moment, then state by what factor the moment changes if the same wire is rewound into 150 turns of half the side length at the same current. *(Answer: 1.08 × 10⁻¹ A·m²; the moment halves.)*
What to remember before solving Current Loop Dipole Moment questions
Which Current Loop Dipole Moment formulas do you need for NEET?
1 formula — click to collapse
Torque on current loop
Torque on N-turn loop of area A carrying I in field B; magnetic moment m = NIA.
| Symbol | Quantity | SI Unit |
|---|---|---|
| N | turns | - |
| A | loop area | m^2 |
| B | field | T |
| m | magnetic moment | A*m^2 |
| theta | angle between m and B | rad |
Valid when
- Uniform B
- Loop in plane
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 10 formulas · 5 question patterns from its other lessons.
Current Loop Dipole Moment questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
Sources
NCERT refs: Class 12 Physics Chapter 4, p.128
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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