Physics · Moving Charges And Magnetism · NEET
Far away, yes. The magnetic field pattern on the axis of a current loop at large distances matches the field of a bar magnet: B_axial = (μ₀/4π)·(2m)/x³, the same 1/x³ shape as a magnetic dipole. So a loop is treated as a magnetic dipole with moment m = NIA. Up close inside the loop the field looks different, but for NEET 'dipole' problems we use the far-field/dipole model.
Use the right-hand thumb rule. Curl the fingers of your right hand along the direction of conventional current in the loop; your thumb points along the axis in the direction of m. One face of the loop then acts as the North pole (where m comes out) and the other as South. m is a vector along the axis, not in the plane of the loop.
They are two different things. Field at the centre B = μ₀NI/(2r) is measured in tesla and tells you how strong the field is at one point. Magnetic moment m = NIA = N·I·πr² is measured in A·m² and describes the loop as a whole magnet (how strongly it responds to an external field and how strong its far field is). Do not mix the two formulas.
On area. m = I·A = I·πr² for a single circular turn, so m is proportional to r² (not r). If you double the radius with the same current, the moment becomes 4 times larger. This is a very common NEET trap—students write m ∝ r instead of m ∝ r².
N is the number of turns. Each turn carries the same current I around the same area A, and their moments all point the same way along the axis, so they simply add: m = N·(IA) = NIA. For a single loop N = 1, giving m = IA.
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 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 and the magnitude of the magnetic moment of this coil are, respectively (μ₀ = 4π × 10⁻⁷ T m/A):
A uniform conducting wire of length 12a and resistance R is wound as a coil in the shape of (i) an equilateral triangle of side a, (ii) a square of side a. The magnetic dipole moments in each case respectively are:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
m = NIA = N·I·(πr²), where N is the number of turns, I is the current, and r is the radius. Its SI unit is ampere-metre² (A·m²), and its direction is along the axis given by the right-hand thumb rule.
For a point on the axis at distance x much greater than the radius, B = (μ₀/4π)·(2m)/x³, where m = NIA. This is the same form as a bar magnet's axial field, which is why the loop is called a magnetic dipole.
The face out of which the magnetic moment m points is the North pole. Curl your right-hand fingers along the current; the thumb points to the North face. The opposite face is South.
Ampere-metre squared (A·m²). It can also be written as joule per tesla (J/T), since torque τ = mB has units of joule.
It links three high-weightage ideas: the centre field of a loop, the magnetic moment m = NIA, and the torque τ = m × B on a loop. Nearly every year NEET asks a numerical using m = NIA, so mastering m ∝ r² and the right-hand direction rule scores easy marks.
