How this instrument works
A charge moving in a circle is, for every purpose a magnetic field cares about, a small loop of current. Ampère showed a current loop of area A carrying current I behaves like a tiny bar magnet with moment μ = I·A; slice a spinning charged body into countless such loops, one per radius, and sum their contributions, and the total collapses — remarkably — into one clean product: μ = g·(q ⁄ 2m)·L, charge-to-mass ratio times angular momentum times a dimensionless bookkeeping factor g.
That factor g is doing real physical work. Whenever charge density tracks mass density exactly — a uniformly charged sphere, disk, or ring, spinning as one rigid piece — the loop-by-loop sum reduces to exactly g = 1, no matter the shape, because every infinitesimal ring contributes charge and mass in the same fixed proportion. Real particles rarely oblige: the electron's g-factor sits near 2.0023, and the proton's near 5.5857, each number a fingerprint of how charge is actually arranged relative to mass inside something far stranger than a spinning ball.
The formula is linear in charge, so its sign carries through: a negative charge produces a moment pointing opposite its angular momentum, a positive one keeps the two aligned, which is why an electron's magnetic moment points against its spin while a proton's points with it. It is also a single-axis relation, tracking magnitude and sign along one direction; a wobbling or precessing body needs the full vector form. And L must already be in kg·m² ⁄ s — this instrument does not derive it from a spin rate, so a rotation speed has to be converted through moment of inertia first.
- Set g-factor — leave it at 1 for a body whose charge tracks its mass exactly, or use a measured value like the electron's ≈2.0023.
- Enter Particle charge in nC, µC, or C — the amount of charge doing the spinning.
- Enter Particle mass in mg, g, or kg — the mass carrying that charge around.
- Enter Angular momentum, kg·m²/s — compute it first as L = I·ω if you only have a spin rate.
- Read Magnetic moment (from angular momentum), A·m² — the product this instrument returns.
Worked example — a 1-gram charged bead on a spindle
Take a 1-gram plastic bead — mass 0.001 kg — charged by friction to 1 nC (1×10⁻⁹ C) and spun on a low-friction spindle until its angular momentum reads 0.01 kg·m²⁄s, a routine electrostatics-lab setup. Because a uniformly charged sphere carries its charge and mass in the same proportion everywhere, g-factor stays at its baseline value of 1. The product works out to μ = 1 × (1×10⁻⁹ ⁄ (2 × 0.001)) × 0.01 = 1 × (5×10⁻⁷) × 0.01 = 5×10⁻⁹ A·m², or 5 nA·m² — the figure this instrument returns for exactly these four inputs.
Keep the same relation but swap in the electron's own numbers — g-factor near 2.0023 and its intrinsic spin angular momentum of ħ⁄2, about 5.273×10⁻³⁵ kg·m²⁄s — and the identical formula returns roughly 9.28×10⁻²⁴ A·m², close to the Bohr magneton. Nothing about an electron literally spins the way the bead does, yet the same bookkeeping formula still balances once the measured g-factor is supplied, which is precisely why that near-integer number surprised physicists in the first place.
Questions
What does the g-factor actually represent?
It's a dimensionless multiplier describing how a body's charge is arranged relative to its mass. Whenever charge density is everywhere proportional to mass density — a uniformly charged sphere, disk, or ring spinning as a rigid piece — the value works out to exactly 1, independent of the exact shape. Real particles depart from that baseline: the electron's g-factor is close to 2.0023, and the proton's is close to 5.5857, each number reflecting an internal charge structure no rigid classical body can match.
Why is the ratio divided by 2m and not just m?
The factor of two comes from loop geometry, not convention. A ring of charge dq spinning at angular rate ω carries current dq·ω⁄2π; multiplied by its enclosed area πr², the 2π cancels and leaves a magnetic moment contribution of half dq·ω·r². The matching angular momentum contribution from that same ring is plainly dm·ω·r², with no such factor, because moment of inertia counts distance-squared directly. Divide the first by the second and that lone factor of two survives — it sits in front of q ⁄ m, not folded away inside it.
Can the magnetic moment point opposite the angular momentum?
Yes — the formula is linear in charge, so a negative charge flips the sign of the result relative to a positive one. A negatively charged spinning body, or an electron, produces a magnetic moment pointing antiparallel to its angular momentum; a positively charged body, like a spinning proton, keeps the two aligned. This sign is exactly what determines which way spins precess in a magnetic field, in techniques from NMR to electron-spin resonance.
What's the most common mistake when filling in this sheet?
Typing a spin rate straight into Angular momentum. This field wants L in kg·m²⁄s, not rpm or rad/s — if you only know how fast something is spinning, you first need its moment of inertia I and the relation L = I·ω before this instrument's number means anything. Skipping that conversion and entering a raw rotation speed produces a magnetic moment off by whatever factor I actually is.
Does doubling the mass really halve the magnetic moment?
Yes, when charge and angular momentum are held fixed. The formula divides by mass but not by charge, so a body carrying the same charge and the same angular momentum but twice the mass returns exactly half the magnetic moment — 2.5 nA·m² instead of 5 nA·m² if mass rises from 1 gram to 2 grams while charge stays at 1 nC and angular momentum at 0.01 kg·m²⁄s. More inertia dilutes the same charge's contribution to the spin.
Is there a maximum sensible value for the g-factor?
Not mathematically — in this formula g is just a multiplier you supply. Physically, though, values far from 1 point to composite or quantum structure rather than a uniform spinning body: the proton's g-factor of about 5.5857 reflects the motion of quarks and gluons inside it, not a rigid charge distribution, while any classical object built so charge tracks mass exactly will always return g = 1 regardless of its shape or size.