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Students often learn the bar magnet and the current loop as two separate topics, with two sets of formulas. They are one idea. A current loop is a magnetic dipole, a bar magnet behaves like a stack of loops, and the same three formulas cover torque, energy and the far field for both. This post sets out those formulas, works through a coil in a field, and shows where the electron's magnetic moment comes from.
What does the syllabus include?
The Syllabus for JEE (Main) 2026 lists "current loop as a magnetic dipole and its magnetic dipole moment, bar magnet as an equivalent solenoid" and "torque on a magnetic dipole in a uniform magnetic field" under Magnetic Effects of Current and Magnetism. The JEE (Advanced) 2026 Information Brochure lists "magnetic moment of a current loop" and the "effect of a uniform magnetic field on a current loop". Check the syllabus for your exam year.
What is the magnetic moment of a loop?
The NCERT Class 12 chapter Moving Charges and Magnetism (section 4.9) defines the magnetic moment of a current loop as m = IA, with direction given by the right-hand thumb rule: curl your fingers along the current and your thumb points along m. For N closely wound turns, m = NIA. The unit is A m².
NCERT then shows that the torque on the loop in a uniform field is
τ = m × B, so |τ| = mB sin θ,
where θ is the angle between m and B. The net force on the loop in a uniform field is zero; only the torque acts. NCERT notes that when m and B are parallel the equilibrium is stable, and when they are antiparallel it is unstable.
What is the energy of a dipole in a field?
In Magnetism and Matter, NCERT integrates the torque to get the potential energy
U = −mB cos θ = −m · B,
with the zero chosen at θ = 90°. It is lowest (−mB) at θ = 0°, the most stable position, and highest (+mB) at θ = 180°. The work needed to turn a dipole from θ₁ to θ₂ is W = mB (cos θ₁ − cos θ₂).
Worked example 1: a coil in a field
A circular coil of 50 turns and radius 4.0 cm carries 2.0 A. It sits in a uniform 0.30 T field with its magnetic moment at 30° to the field.
Area: A = π × 0.040² ≈ 5.03 × 10⁻³ m².
Moment: m = NIA = 50 × 2.0 × 5.03 × 10⁻³ ≈ 0.503 A m².
Torque: τ = mB sin 30° = 0.503 × 0.30 × 0.50 ≈ 0.075 N m.
Energy: U = −mB cos 30° = −0.503 × 0.30 × 0.866 ≈ −0.131 J.
Work to turn it from 30° to 90°: W = mB (cos 30° − cos 90°) ≈ 0.131 J.
Work to flip it from 0° to 180°: W = 2mB ≈ 0.302 J.
The trap version. If the question says the plane of the coil makes 30° with the field, the moment is at 60° to the field, and τ = 0.503 × 0.30 × sin 60° ≈ 0.131 N m. Same numbers, different answer.
What field does a loop make far away?
NCERT's section 4.5 gives the field on the axis of a loop: B = μ₀IR² / [2(x² + R²)^(3/2)]. When x ≫ R, this becomes B ≈ μ₀IR² / (2x³). Since m = IπR², that is
B_axial ≈ (μ₀ / 4π) × 2m / x³.
This has the same form as the axial field of a short bar magnet, which is why NCERT treats a bar magnet as an equivalent solenoid: the magnetic moment of a bar magnet equals the moment of the solenoid that produces the same field.
Worked example 2. For the coil above (m ≈ 0.503 A m²), find B on the axis 0.50 m away. Here x is 12.5 times R, so the approximation is fair. B ≈ 10⁻⁷ × 2 × 0.503 / 0.50³ = 1.006 × 10⁻⁷ / 0.125 ≈ 8.0 × 10⁻⁷ T.
Where does an electron's magnetic moment come from?
Treat an electron in a circular orbit of radius r at speed v as a current loop. It passes a point once per period 2πr/v, so I = ev/(2πr). The moment is m = I × πr² = evr/2. Since the orbital angular momentum is L = mₑvr, this is m = (e / 2mₑ) L.
If L = h/2π, the smallest orbital value in the Bohr model, m becomes eh/(4πmₑ), called the Bohr magneton. NIST lists the Bohr magneton as 9.274 010 0657 × 10⁻²⁴ J/T. NCERT adds that particles such as electrons also carry an intrinsic magnetic moment not accounted for by circulating currents.
Which mistakes are most common?
Using the angle with the plane. Use the normal.
Forgetting N. m = NIA for a coil.
Expecting a net force in a uniform field. There is only a torque.
Getting the sign of U wrong. The stable position has the lowest energy, −mB.
Using the far-field formula close to the loop. Check x ≫ R.
For forces on the individual sides of the loop, see our guide to right-hand-rule mistakes in magnetic force.
A practice routine
Draw m with the right-hand rule for five loops with different current directions.
For each worked example, compute τ and U at 0°, 90° and 180°.
Derive m = (e/2mₑ)L once from scratch.
If dipoles still feel like two topics, the Fermi Concept Library covers every concept in physics with its own lesson, so you can read the current loop and bar magnet ideas side by side.
Pick one coil problem from your book and solve it twice: once with the angle to the normal and once with the angle to the plane.
