Why Transformers Hum

Equivalent forms

\Delta\ell/\ell \propto B^2

B(t) = B_0\sin(2\pi f_{line} t)

A squared dependence on a sinusoid silently doubles the frequency — you literally hear the math of B².

Symbol	Name	SI unit	Dimension	Range
$f_{hum}$	Hum frequencyoutput Acoustic frequency of the core	Hz	T⁻¹	2 – 2000
$f_{line}$	Line frequency Grid AC frequency	Hz	T⁻¹	1 – 1000

Unit systems

Symbol	SI
$f_{hum}$	Hz
$f_{line}$	Hz

Where it holds

The

2\times

fundamental is exact for the

B^{2}

-driven magnetostrictive strain. Harmonics (4f, 6f …) appear from core saturation and loose laminations; mechanical resonances of the tank can amplify particular harmonics.

Dimensional analysis

[2][f_line $] =$ (dimensionless $)\cdot Hz = Hz \checkmark$

Discovery

James Prescott Joule · 1842

Joule discovered magnetostriction — that iron changes length when magnetized. A century later it became the textbook explanation for the ubiquitous 100/120 Hz drone of the power grid.

Try this

Stand near a power transformer and you hear a steady drone — and it's an octave above the grid frequency, not at it. Why double?

On a 50 Hz grid, at what frequency does a transformer's core hum, and why is it twice the line frequency?

Research status: stable

Real-world applications

Transformer noise specification and abatement
Diagnosing loose laminations by harmonic content
Designing quiet audio-equipment transformers
Grid-frequency forensics from recordings

Common misconceptions

The hum is at the line frequency — it's at double, because strain depends on $B^{2}$ , not B
The sound is from current in the windings — it's mainly the core's magnetostrictive flexing and laminations

Experimental verification

A microphone and FFT on any mains transformer shows a dominant 100 Hz (or 120 Hz) line with weaker even harmonics; the fundamental sits at exactly twice the supply frequency.

Derivation

Magnetostrictive strain

\Delta

ℓ/ℓ

\approx \lambda _s\cdot (B/B_s)^{2}

depends on

B^{2}

.

With

B(t) = B_{0} \sin (\omega t)

,

B^{2} = B_{0}^{2} \sin ^{2}(\omega t) = \tfrac{1}{2}B_{0}^{2}(1 - \cos 2\omega t)

.

The oscillating part is at

2\omega

, so the core flexes — and radiates sound — at twice the line frequency, independent of B's sign.

Limiting cases

f_line

= 50\,\mathrm{Hz}

(Europe)⟶

f_hum = 100\,\mathrm{Hz}

The familiar deep European mains drone.

f_line

= 60\,\mathrm{Hz}

(North America)⟶

f_hum = 120\,\mathrm{Hz}

A slightly higher pitch — you can hear which grid you're on.

What if…

What if you're in North America?

60 Hz mains gives a 120 Hz hum — a noticeably higher pitch than Europe's 100 Hz.

What if the laminations come loose with age?

Saturation and rattling add strong even harmonics (200, 240 Hz…), and the transformer gets buzzier — a maintenance warning sign.

1

European 50 Hz grid

Given ·

f {line}:: 50

Find · hum frequency

Steps

Strain $\propto B^{2} and B^{2}$ peaks twice per cycle
So mechanical vibration is at $2\times f$ _line
$f_hum = 2\times 50 = 100\,\mathrm{Hz} (G_{2}-ish$ pitch)

Result · 100 Hz — an octave above the 50 Hz supply.

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