Bohr Magneton — Physica

Equivalent forms

\mu = g\,\mu_B\,\frac{J}{\hbar}

Combine the electron's charge, spin unit ℏ, and mass and out drops the exact size of an atom's magnetism.

Symbol	Name	SI unit	Dimension	Range
$\mu_B$	Bohr magnetonoutput Fundamental unit of magnetic moment	J/T	L^2*I	—
$e$	Elementary charge Electron charge magnitude	C	I*T	—
$\hbar$	Reduced Planck constant h/2π	J*s	ML^2T^-1	—
$m_e$	Electron mass Mass of the electron	kg	M	—
$B$	Magnetic field Applied field (for Zeeman energy μ_B B)	T	MT^-2I^-1	0 – 5

Where it holds

Sets the scale for electronic magnetism. Nuclear moments

use the \sim 2000\times

smaller nuclear magneton

(m_e \to m_p)

. The electron's actual moment is

g/2 \approx 1.00116 \times \mu _B

due to QED.

Dimensional analysis

\mu _B \to [L^2*I] (J/T = A\cdot m^{2})

e\hbar /2m_e \to [I\cdot T][M L^2\,\mathrm{T}^-1]/[M] = [I*L^2]

Charge

\times

action over mass gives current

\times

area.

\checkmark

Discovery

Niels Bohr / Wolfgang Pauli · 1920

The unit emerged from the Bohr–Sommerfeld model: an electron in the first orbit has angular momentum ℏ and thus magnetic moment eℏ/2m_e. Pauli and others formalized it. The Stern–Gerlach experiment (1922) then showed atomic magnetic moments are quantized in these units, and the electron's spin moment is very nearly one μ_B (g≈2).

Research status: stable

Real-world applications

MRI and NMR energy scales
Electron spin resonance (ESR/EPR) spectroscopy
Quantifying magnetic moments of ions in magnetic materials

Common misconceptions

The electron spin moment $is \sim 1 \mu _B$ , not 1/2, because $g\approx 2$ compensates the spin-1/2
The nuclear magneton is far smaller (uses proton mass), so nuclear magnetism is weak
$\mu _B$ is a unit/scale, not a maximum — atoms can carry several $\mu _B$

Experimental verification

Stern–Gerlach beam splitting, Zeeman line splitting

\Delta E = \mu _B B

, and electron-spin-resonance frequencies all quantify magnetism in units

of \mu _B

to high precision.

Derivation

A charge

-e

moving in a circle of radius r at speed v is a current

I = -e/T = -ev/2\pi r

enclosing area

A = \pi r^{2}

.

Its magnetic moment

\mu = IA = -evr/2 = -(e/2m_e)L

, where

L = m_e vr

is the angular momentum.

Setting

L = \hbar

(one quantum) gives the magnitude

\mu _B = e\hbar /2m_e

.

The proportionality constant e/2m_e is the classical gyromagnetic ratio.

What if…

What if we used the proton instead of the electron?

Replace m_e by $m_p (1836\times$ heavier): the nuclear magneton $is \sim 1836\times$ smaller, which is why NMR needs strong fields.

What if g were exactly 2?

The spin moment would be exactly $\mu _B$ ; QED corrections make it $1.00116 \mu _B$ , one of physics' best-tested predictions.

1

Zeeman splitting in a 1 T field

Given ·

B:: 1 T
μ B:: 9.274e-24 J/T

Find · energy splitting

\Delta E

Steps

Each spin state shifts $by \pm \mu _B B$
$\Delta E = 2 \times 9.274e-24 \times 1 = 1.85e-23\,\mathrm{J}$
In frequency, $\Delta E/h \approx 28\,\mathrm{GHz}$

Result ·

\Delta E = 2\mu _B B \approx 1.85e-23\,\mathrm{J} \approx 116

µ

eV (\approx 28\,\mathrm{GHz})

stern gerlachzeeman effectspin magnetic moment