Magnetic Dipole Moment of a Current Loop (m = NIA)

Physics · Moving Charges And Magnetism · NEET

A current loop of N turns, each carrying current I and enclosing area A, behaves like a tiny bar magnet with magnetic dipole moment m = NIA. Its direction is along the axis of the loop, given by the right-hand curl rule (curl fingers along current, thumb points to m). Memory hook: "N I A = Number, current, Area" - just multiply the three, and the answer is in ampere per metre squared (A·m²).
Magnetic Dipole Moment of a Current Loop: m = N I AI (current)mArea AN turns, current I, area AmRight-hand rule:curl fingers along I,thumb gives mIm is along the loop axis, unit A·m²
A current loop acts like a bar magnet: the magnetic moment m = NIA points along the loop's axis, found by curling the right hand's fingers along the current so the thumb gives m. Its unit is A·m².

Your doubts, answered

Is the magnetic moment m = IA or m = NIA?

For a single loop (one turn) m = IA. When the same wire is wound into N turns, each turn adds its own IA and they all point the same way, so m = NIA. In NEET numericals, always check the number of turns N. If a coil has 100 turns, m is 100 times larger than a single loop with the same current and area.

What decides the direction of the magnetic moment?

Use the right-hand curl rule. Curl the fingers of your right hand along the direction of current flow around the loop; your thumb points along the magnetic moment m. This is the same axis as the loop's magnetic field at the centre (the North-pole side). m is perpendicular to the plane of the loop, never in the plane.

What is the SI unit of magnetic dipole moment?

The unit of m = NIA is ampere times metre squared, written A·m². You can check it: current (A) multiplied by area (m²). An equivalent unit is joule per tesla (J/T), which is useful because energy U = -m·B and torque tau = m×B.

Does the magnetic moment change if I move the loop into a stronger magnetic field?

No. m = NIA depends only on the loop itself (turns, current, area), not on any external field B. What changes in a stronger field is the torque (tau = mB sin theta) and the potential energy (U = -mB cos theta), not m. Students often confuse m with torque because both appear together in tau = m×B.

A wire of fixed length is bent into a coil - how do I find m?

First find how many turns N the wire makes: N = total length / perimeter of one turn. Then find the area A of one turn, and use m = NIA. Fewer, larger turns can give more or less m than many small turns, so you must compute N and A carefully - this is a very common NEET trap.

⚠️ The NEET trap
Forgetting the number of turns N and writing m = IA for a multi-turn coil, or thinking m increases when the coil is placed in a stronger field B.
Always use m = NIA with the correct N. The magnetic moment depends only on the loop (N, I, A) and is independent of the external field B; B only affects torque and energy.
🧠 m belongs to the loop, not to the field. Count the turns first, then multiply.

Real NEET questions

2026

A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively: (Take μ₀ = 4π × 10⁻⁷ T m/A)

A · 2 A, 10 A·m²
B · 2.5 A, 20 A·m²
C · 2 A, 4 A·m²
D · 2.5 A, 2 A·m²
Solution: Step 1 - find current from the field at the centre. B = μ₀NI/(2R). So I = 2RB/(μ₀N) = (2 × 0.05 × 3.14 × 10⁻³) / (4π × 10⁻⁷ × 100) = (3.14 × 10⁻⁴)/(1.256 × 10⁻⁴) = 2.5 A. Step 2 - apply m = NIA with A = πR². m = N I (πR²) = 100 × 2.5 × 3.14 × (0.05)² = 100 × 2.5 × 3.14 × 2.5 × 10⁻³ ≈ 2 A·m². So I = 2.5 A and m = 2 A·m² → option D.
2025

A 2 A current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be:

A · 2 : 1
B · 4 : 1
C · 1 : 4
D · 1 : 2
Solution: Magnetic moment of a single-turn coil: m = IA = I(πr²). The current is the same (2 A) for both, so m ∝ r². Ratio = r₁² : r₂² = 1² : 2² = 1 : 4 → option C. Note how m depends on area, which grows as radius squared, not as radius.

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Frequently asked

What is the formula for the magnetic dipole moment of a current loop?

m = NIA, where N is the number of turns, I is the current, and A is the area enclosed by one turn. Its direction is along the loop axis by the right-hand curl rule, and its unit is A·m² (also J/T).

Why does a current loop act like a magnetic dipole?

A current loop produces a magnetic field with a North and South pole just like a small bar magnet. Its field pattern at large distances matches that of a magnetic dipole, so we describe it by a single vector m = NIA pointing from the South to the North face.

Is magnetic moment a vector or a scalar?

It is a vector. Its magnitude is NIA and its direction is perpendicular to the plane of the loop, along the axis given by the right-hand curl rule. This direction matters in tau = m×B and U = -m·B.

How is magnetic moment related to torque?

When the loop is placed in a magnetic field B, the field exerts a torque tau = m×B, magnitude tau = mB sin theta = NIAB sin theta. The moment m is a property of the loop; the torque only appears when an external field is present.

Does the shape of the loop matter for m?

Only through the area A. For the same current and turns, a loop enclosing more area has a larger moment. When a fixed length of wire is bent into different shapes, both N and A change, so you must recompute m = NIA for each shape.