On this page
A question gives you a charged ring, or a disc, or a rod, spinning about an axis, and asks for its magnetic moment. There is no wire and no battery, so many students do not see the current. This post shows how a spinning charge becomes a current loop, works the numbers for a ring and a disc, explains why the ratio of magnetic moment to angular momentum comes out as q/2m again and again, and shows the case where that shortcut gives the wrong answer.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "current loop as a magnetic dipole and its magnetic dipole moment" in Unit 13, and the NEET (UG) 2026 syllabus lists the same line. The JEE (Advanced) 2026 brochure lists "magnetic moment of a current loop". The NCERT Class 12 chapter Moving Charges and Magnetism (2026-27 reprint) defines the moment as m = IA (Eq. 4.22), with unit A m², and gives the torque as τ = m × B (Eq. 4.23). NCERT does not work through spinning bodies, so treat them as an application of that definition. Check your own exam year's syllabus.
How does a spinning ring carry a current?
Take a ring of radius R with charge q spread along it, spinning at angular speed ω about its axis. Stand at one point on the ring's path and count: the whole charge q passes you once every period T = 2π/ω. Current is charge per unit time, so
I = q/T = qω/2π
The ring encloses area πR², so
μ = IA = (qω/2π) × πR² = qωR²/2
Its angular momentum about the axis is L = mR²ω, so μ/L = q/2m. For positive charge, μ points along ω; for negative charge it points the other way.
Worked example 1: a spinning ring
These are sample numbers. A ring of radius 10 cm and mass 20 g carries 5.0 µC and spins at 600 revolutions per minute.
Frequency f = 600/60 = 10 rev/s, so ω = 2π × 10 = 62.8 rad/s.
Current I = qf = 5.0 × 10⁻⁶ × 10 = 5.0 × 10⁻⁵ A.
Moment μ = I × πR² = 5.0 × 10⁻⁵ × π × 0.010 = 1.57 × 10⁻⁶ A m².
Angular momentum L = mR²ω = 0.020 × 0.010 × 62.8 = 1.26 × 10⁻² kg m²/s.
Ratio μ/L = 1.25 × 10⁻⁴ C/kg. Check: q/2m = 5.0 × 10⁻⁶ / 0.040 = 1.25 × 10⁻⁴ C/kg.
In a 0.50 T field at right angles to the axis, the torque is μB = 7.9 × 10⁻⁷ N m.
Worked example 2: a disc, built from rings
Now spread the same charge evenly over a disc of the same radius and mass. The surface charge density is σ = q/πR².
Pick a thin ring of radius r and width dr. Its charge is dq = σ × 2πr dr.
It is a current loop: dI = dq × ω/2π = σωr dr.
Its moment is dμ = dI × πr² = πσωr³ dr.
Add the rings from 0 to R: μ = πσωR⁴/4 = qωR²/4.
With the numbers above, μ = 7.85 × 10⁻⁷ A m², half the ring's value, because most of the charge now sits closer to the axis. The angular momentum also halves, to L = ½mR²ω = 6.28 × 10⁻³ kg m²/s, so μ/L is again 1.25 × 10⁻⁴ C/kg.
The same four steps give a rod of length ℓ spinning about one end: μ = qωℓ²/6, with L = mℓ²ω/3. A uniform solid sphere gives μ = qωR²/5, with L = (2/5)mR²ω. If the moment of inertia part feels shaky, our guide to finding moment of inertia by integration uses the same slicing idea.
Why does q/2m keep appearing?
Each small piece of the body, of mass dm and charge dq, moves in its own circle. A single charge in a circle has μ = (charge/2 × mass) × L, the same result our post on the magnetic moment of an orbiting electron derives for the Bohr atom. If every piece carries charge in the same proportion, dq/dm = q/m, then every piece has the same ratio q/2m, and adding them up keeps it. Shape does not matter. The proportion does.
When does the shortcut fail?
Take a disc of uniform mass m whose charge q sits only on its rim. The moment is the ring's, qωR²/2, but the angular momentum is the disc's, ½mR²ω. The ratio is q/m, twice the shortcut.
The electron's spin is another exception. NIST lists the electron g-factor as about −2.0023, which means its spin magnetic moment is roughly twice what the q/2m rule predicts for its spin angular momentum. A spinning charged ball is not a good model of an electron. Spin magnetism is not in the JEE Main physics syllabus, so treat this as background.
Which mistakes cost marks?
Writing I = qω. The current is qω/2π, or simply q × f.
Using rpm as if it were rad/s. Convert to revolutions per second, then multiply by 2π.
Giving every ring in a disc the area πR². Each ring has its own area πr².
Assuming q/2m without checking where the charge and mass sit.
Dropping the direction. For negative charge, μ is opposite to ω.
A practice routine
Day 1: derive the ring and disc results without notes.
Day 2: do a rod spinning about its centre. You should get μ = qωℓ²/24 and, again, q/2m.
Day 3: redo the rim-charged disc and a hollow sphere.
Day 4: add a field. Find the torque at 30° and the work to turn the ring from 0° to 90°.
Day 5: a closed-book mixed set. Dunlosky and colleagues (2013) rated practice testing and spaced practice as high-utility techniques.
If the picture of a spinning charge as a current is hard to hold in your head, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than to read.
Tonight, redo example 2 for a flat washer with inner radius R/2 and check whether μ/L is still q/2m.
