Magnetic constant (μ₀)

μ0
Electromagnetic constant
$1.256\,637\,061\,27 \times 10^{-6}$ N A−2
Dimensions: M L Q−2  ·  Relative uncertainty: $1.6 \times 10^{-10}$
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At a glance
The magnetic constant, also called the vacuum permeability, can be expressed as $\mu_0 = \frac{4\pi\,m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$ — four $\pi$ times the ratio of Planck mass times Planck length to Planck charge squared. In physical formulas, $\mu_0$ mediates forces between electric currents; in Planck units, it is a fixed ratio of mechanical to electromagnetic Planck quantities. The ratio $\frac{\mu_0}{4\pi} = \frac{F_{\mathrm{P}}}{I_{\mathrm{P}}^2}$ (Planck force divided by Planck current squared) remains invariant across scales, from the Planck current through the ampere scale of everyday experiments. Prior to the 2019 SI redefinition, $\mu_0$ was defined exactly as $4\pi \times 10^{-7}$ N A$^{-2}$; it is now a measured quantity, constrained by the fine-structure constant $\alpha$.

Universal form

The Planck-unit decomposition of $\mu_0$ is (derived in Understanding the natural units, Eur. J. Phys. 45, 055802):

Magnetic constant, Planck-unit form
$$\mu_0 \;=\; 4\pi F_{\mathrm{P}} \left(\frac{t_{\mathrm{P}}}{q_{\mathrm{P}}}\right)^2$$

Since the Planck force $F_{\mathrm{P}} = m_{\mathrm{P}} l_{\mathrm{P}}/t_{\mathrm{P}}^2$, this simplifies to:

Equivalent compact form
$$\mu_0 \;=\; \frac{4\pi\,m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$$

The Planck charge squared $q_{\mathrm{P}}^2$ appears in the denominator, and the product $m_{\mathrm{P}} l_{\mathrm{P}}$ (a mass times a length, or equivalently a momentum times a time, carrying dimensions M·L) appears in the numerator. The factor $4\pi$ is a geometric coefficient that arises from the spherical symmetry of the Coulomb field.

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Dimensions
$\mu_0$ contributes M L Q−2. In the decomposition $\frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$: the numerator $m_{\mathrm{P}} l_{\mathrm{P}}$ contributes M·L; the denominator $q_{\mathrm{P}}^2$ contributes Q2. Ratio: M L Q−2. ✓

The same expression can be written using the Planck current $I_{\mathrm{P}} = q_{\mathrm{P}}/t_{\mathrm{P}}$ and Planck force:

Scale-invariant ratio
$$\frac{\mu_0}{4\pi} \;=\; \frac{F_{\mathrm{P}}}{I_{\mathrm{P}}^2}$$

This ratio of Planck force to Planck current squared is dimensionally consistent ($[\mathrm{F}/I^2] =$ N A$^{-2}$ = H m$^{-1}$) and is the scale-invariant quantity that $\mu_0$ encodes: $\mu_0/(4\pi) = F_{\mathrm{P}}/I_{\mathrm{P}}^2 = F/(2I^2)$ — the ratio of magnetic force to current squared is the same at the Planck scale and at the ampere scale used to define the pre-2019 SI.

Equivalent expressions

Several equivalent groupings illuminate different physical roles of $\mu_0$.

From Planck force and current

$$\mu_0 \;=\; \frac{4\pi F_{\mathrm{P}}}{I_{\mathrm{P}}^2} \;=\; \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$$

The magnetic constant as a Planck force per Planck current squared, scaled by $4\pi$. This form appears directly in the force between parallel wires, where $F/L = \mu_0 I_1 I_2/(2\pi r)$.

From the Coulomb constant

$$\mu_0 \;=\; \frac{1}{\varepsilon_0 c^2}$$

In Planck units: $\mu_0\,\varepsilon_0 = 1/c^2 = t_{\mathrm{P}}^2/l_{\mathrm{P}}^2$. This product relation is the electromagnetic constraint that ensures the wave speed of electromagnetic radiation equals the speed of light — a consequence of Maxwell’s equations.

From the vacuum impedance

$$\mu_0 \;=\; \frac{Z_0}{c} \;=\; \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}^2}{t_{\mathrm{P}} q_{\mathrm{P}}^2} \cdot \frac{t_{\mathrm{P}}}{l_{\mathrm{P}}} \;=\; \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$$

The vacuum impedance $Z_0 = \mu_0 c = 376.73$ Ω; dividing by $c = l_{\mathrm{P}}/t_{\mathrm{P}}$ recovers $\mu_0$. This relation connects the vacuum’s resistive and inductive properties.

From the fine-structure constant

$$\mu_0 \;=\; \frac{2\alpha}{c} \cdot \frac{h}{e^2} \;=\; \frac{2\alpha R_K}{c}$$

Since $Z_0 = 2\alpha R_K$ (where $R_K = h/e^2$ is the von Klitzing constant) and $\mu_0 = Z_0/c$, the magnetic constant is related to the fine-structure constant and the quantized Hall resistance. After the 2019 SI redefinition, both $h$ and $e$ are exact, so $\mu_0$ can in principle be determined from $\alpha$ alone. It is now a measured quantity, not a defined one.

The scale-invariant ratio

$$\frac{\mu_0}{4\pi} \;=\; \frac{m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2} \;=\; \frac{F_{\mathrm{P}}}{I_{\mathrm{P}}^2} \;=\; 10^{-7}\;\mathrm{N\,A}^{-2}$$

This ratio of force to current squared is invariant across scales — the same number whether evaluated at the Planck scale or in the CGS system. This is why the pre-2019 SI defined $\mu_0/(4\pi) = 10^{-7}$ N A$^{-2}$ exactly: it was fixing one representative value of this scale-invariant ratio.

Dimensional verification

Multiple independent formulas for $\mu_0$ reduce to the same Planck-unit expression.

From Ampère’s force law

$$\frac{F}{L} \;=\; \frac{\mu_0 I_1 I_2}{2\pi r}$$
Planck-unit reduction
Substitute $\mu_0 = \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$, $I_1 = I_2 = I_{\mathrm{P}}$, $r = l_{\mathrm{P}}$$F/L = \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2} \cdot \frac{q_{\mathrm{P}}^2/t_{\mathrm{P}}^2}{2\pi l_{\mathrm{P}}}$
Simplification$F/L = \frac{2 m_{\mathrm{P}}}{t_{\mathrm{P}}^2} = \frac{2 F_{\mathrm{P}}}{l_{\mathrm{P}}}$

Force per unit length between two Planck currents at Planck separation is $2F_{\mathrm{P}}/l_{\mathrm{P}}$ — twice the Planck force per Planck length. All charge factors cancel; the ratio $m_{\mathrm{P}} l_{\mathrm{P}}/q_{\mathrm{P}}^2$ combined with $q_{\mathrm{P}}^2/t_{\mathrm{P}}^2$ gives a pure mechanical quantity.

Numerical verification

$\mu_0 = \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$
$m_{\mathrm{P}}$$2.176\,434 \times 10^{-8}$ kg
$l_{\mathrm{P}}$$1.616\,255 \times 10^{-35}$ m
$q_{\mathrm{P}}$$1.875\,546 \times 10^{-18}$ C
Numerator $4\pi\, m_{\mathrm{P}} l_{\mathrm{P}}$$4.420 \times 10^{-42}$
Denominator $q_{\mathrm{P}}^2$$3.518 \times 10^{-36}$
Ratio$1.2566 \times 10^{-6}$ N A$^{-2}$ ✓

Physical characterization

The magnetic constant characterizes how strongly the vacuum responds to electric currents — how much magnetic field a given current produces, and how strongly two current-carrying wires attract or repel. In Planck units, it is the conversion factor between mechanical force and the square of electric current. Its structure $\mu_0 = \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$ can be read as placing magnetic forces, alongside gravitational and electrostatic forces, on a common Planck-scale footing scaled by dimensionless ratios of the system’s properties to the Planck scale.

Magnetic force between parallel wires

Two long parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ attract (or repel) with force per unit length:

$$\frac{F}{L} \;=\; \frac{\mu_0 I_1 I_2}{2\pi r} \;=\; 2F_{\mathrm{P}} \cdot \frac{I_1}{I_{\mathrm{P}}} \cdot \frac{I_2}{I_{\mathrm{P}}} \cdot \frac{l_{\mathrm{P}}}{r}$$

The Planck decomposition shows three dimensionless ratios: the two current ratios $I_{1,2}/I_{\mathrm{P}}$ and the inverse distance ratio $l_{\mathrm{P}}/r$. The Planck current $I_{\mathrm{P}} = q_{\mathrm{P}}/t_{\mathrm{P}} \approx 3.479 \times 10^{25}$ A is enormous. No practical current comes close to it. A typical laboratory current of $1$ A represents $I/I_{\mathrm{P}} \approx 2.9 \times 10^{-26}$, which is why magnetic forces at the laboratory scale are small compared to the Planck force.

Numerical verification — force between $1$ A wires at $1$ m
$F/L = \mu_0 I^2/(2\pi r)$$1.2566 \times 10^{-6} \times 1^2 / (2\pi \times 1)$
Result$2.000 \times 10^{-7}$ N m$^{-1}$ ✓

This was the historic definition of the ampere (before 2019): one ampere was the current that produces a force of $2 \times 10^{-7}$ N per meter between parallel wires $1$ m apart.

Why the magnetic and electric constants are complementary

The electric constant (vacuum permittivity) $\varepsilon_0$ and the magnetic constant $\mu_0$ are not independent: they are related by $\mu_0 \varepsilon_0 = 1/c^2$. In Planck units, the comparison is instructive:

$$\varepsilon_0 \;=\; \frac{1}{4\pi F_{\mathrm{P}}} \left(\frac{q_{\mathrm{P}}}{l_{\mathrm{P}}}\right)^2 \qquad \mu_0 \;=\; 4\pi F_{\mathrm{P}} \left(\frac{t_{\mathrm{P}}}{q_{\mathrm{P}}}\right)^2$$

Their product is:

$$\mu_0 \varepsilon_0 \;=\; \left[4\pi F_{\mathrm{P}} \frac{t_{\mathrm{P}}^2}{q_{\mathrm{P}}^2}\right]\left[\frac{1}{4\pi F_{\mathrm{P}}} \frac{q_{\mathrm{P}}^2}{l_{\mathrm{P}}^2}\right] \;=\; \frac{t_{\mathrm{P}}^2}{l_{\mathrm{P}}^2} \;=\; \frac{1}{c^2}$$

The $4\pi F_{\mathrm{P}}$ and $q_{\mathrm{P}}^2$ factors cancel exactly, leaving only the Planck velocity ratio. This cancellation is why Maxwell’s equations predict that electromagnetic waves travel at $c$, and why the electric and magnetic constants are structurally complementary rather than independent. The difference between them is a factor of $l_{\mathrm{P}}^2/t_{\mathrm{P}}^2 = c^2$ — they carry opposite powers of Planck length and Planck time.

The CGS connection and electromagnetic unit systems

The scale-invariant ratio $\frac{\mu_0}{4\pi} = \frac{m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2} = 10^{-7}$ N A$^{-2}$ played a central role in the history of electromagnetic unit systems. The CGS electromagnetic unit system (emu) defined the abcoulomb by setting this ratio to $1$, which gives $q_{\mathrm{P}}^2 = l_{\mathrm{P}} m_{\mathrm{P}}$, yielding $q_{\mathrm{P}} = \sqrt{l_{\mathrm{P}} m_{\mathrm{P}}}$ — charge in units of M$^{1/2}$L$^{1/2}$. The pre-2019 SI instead fixed $\mu_0/(4\pi) = 10^{-7}$ N A$^{-2}$ exactly, choosing a specific scale-invariant value of this ratio consistent with the ampere as defined. After 2019, $\mu_0$ is determined by measurement through $\alpha$: the current best value is $\mu_0 = 2\alpha h/(c e^2)$. The Planck-unit decomposition $\mu_0 = \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$ clarifies why different electromagnetic unit systems disagree on the “dimensions” of charge — each system treats one of these ratios as dimensionless in a different way.

Planck unit calculations from the magnetic constant

Because $\mu_0$ contains Planck-unit quantities in its dimensions, it can be combined with $G$, $c$, and the elementary charge $e$ to calculate the Planck-unit values directly — without any reference to $\hbar$. This route runs through the magnetic side of electromagnetism, paralleling the electric-side route through $\varepsilon_0$. The fine-structure constant $\alpha$ appears as the bridge between the electromagnetic and quantum regimes.

Solving for the Planck constant in terms of the magnetic constant

Starting from $\alpha = e^2/(4\pi\varepsilon_0\hbar c)$ and using the Maxwell relation $\varepsilon_0 = 1/(\mu_0 c^2)$, we solve for $\hbar$ in terms of the magnetic constant:

$$\hbar = \frac{e^2 \mu_0 c}{4\pi \alpha}$$

Substituting this expression for $\hbar$ into the standard Planck-unit formulas eliminates $\hbar$ entirely:

$$l_{\mathrm{P}} = \sqrt{\frac{\hbar G}{c^3}} = \frac{e}{c} \sqrt{\frac{G \mu_0}{4\pi \alpha}}$$ $$m_{\mathrm{P}} = \sqrt{\frac{\hbar c}{G}} = ec \sqrt{\frac{\mu_0}{4\pi \alpha G}}$$ $$t_{\mathrm{P}} = \sqrt{\frac{\hbar G}{c^5}} = \frac{e}{c^2} \sqrt{\frac{G \mu_0}{4\pi \alpha}}$$

Each expression recovers the same Planck-unit value as the standard $\hbar$-based formula. The electromagnetic constants $\mu_0$ and $e$, together with $G$, $c$, and $\alpha$, contain sufficient Planck-unit content to pin down the Planck length, Planck mass, and Planck time without invoking the Planck constant.

Numerical verification — Planck length from $\mu_0$
Using $e = 1.602\,176\,634 \times 10^{-19}$ C, $\mu_0 = 1.256\,637 \times 10^{-6}$ N A$^{-2}$, $G = 6.674\,30 \times 10^{-11}$ m³ kg⁻¹ s⁻², $c = 2.998 \times 10^{8}$ m/s, $\alpha = 7.297 \times 10^{-3}$:
$G\mu_0/(4\pi\alpha)$$9.146 \times 10^{-16}$ m³ kg⁻¹ N A$^{-2}$
$\sqrt{G\mu_0/(4\pi\alpha)}$$3.024 \times 10^{-8}$
$e/c$$5.344 \times 10^{-28}$ C s/m
$l_{\mathrm{P}} = (e/c)\sqrt{G\mu_0/(4\pi\alpha)}$$1.616 \times 10^{-35}$ m ✓

Without the Planck constant at all

If $\hbar$ is dropped from the input set entirely, what length can be assembled from $\{e, \mu_0, c, G\}$ alone? Dimensional analysis forces a unique answer:

$$\ell = \frac{e}{c} \sqrt{G \mu_0} = \sqrt{4\pi \alpha} \cdot l_{\mathrm{P}}$$

The four constants $\{e, \mu_0, c, G\}$ determine a length — but it is $\sqrt{4\pi\alpha} \approx 0.3028$ times the Planck length, not the Planck length itself. This is the same observation as for the $\varepsilon_0$ route: $\alpha$ is what distinguishes the elementary charge from the Planck charge, and the purely classical-magnetic set $\{e, \mu_0, c, G\}$ does not encode that distinction.

Why the fine-structure constant has to appear

From the five dimensional constants $\{\mu_0, c, G, \hbar, e\}$, exactly one dimensionless combination can be formed, and that combination is $\alpha$ itself. Any expression that substitutes the magnetic constant in place of $\hbar$ must therefore carry $\alpha$ as its trade-in factor. The two electromagnetic routes, electric through $\varepsilon_0$ and magnetic through $\mu_0$, are equivalent expressions of the same fact: both constants contain Planck-unit quantities in their dimensions, and either provides a path to the Planck-unit values when combined with $G$, $c$, and $e$.

The constant in context

Ampère’s force law

$$\frac{F}{L} \;=\; \frac{\mu_0 I_1 I_2}{2\pi r} \;=\; 2F_{\mathrm{P}} \cdot \frac{I_1 I_2}{I_{\mathrm{P}}^2} \cdot \frac{l_{\mathrm{P}}}{r}$$

Force per unit length between parallel wires. In Planck units, the force is the Planck force scaled by the product of current ratios and the inverse distance ratio.

Biot–Savart law

$$B \;=\; \frac{\mu_0 I}{2\pi r} \;=\; \frac{2 m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2} \cdot \frac{I}{r}$$

The magnetic field of a long straight wire. The field strength scales as the current-to-distance ratio, with $\mu_0/(2\pi)$ as the proportionality constant. In Planck units this is $\frac{2m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2} = 2 \times 10^{-7}$ T m A$^{-1}$.

Vacuum impedance

$$Z_0 \;=\; \mu_0 c \;=\; \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}^2}{t_{\mathrm{P}} q_{\mathrm{P}}^2} \;\approx\; 376.73\;\Omega$$

The characteristic impedance of free space. This determines the ratio of electric to magnetic field in an electromagnetic wave. In Planck units it is $4\pi$ times the Planck impedance $\frac{m_{\mathrm{P}} l_{\mathrm{P}}^2}{t_{\mathrm{P}} q_{\mathrm{P}}^2}$.

Energy stored in a magnetic field

$$U \;=\; \frac{B^2}{2\mu_0} \cdot V \;=\; \frac{q_{\mathrm{P}}^2 B^2 V}{8\pi m_{\mathrm{P}} l_{\mathrm{P}}}$$

The energy density of a magnetic field is $B^2/(2\mu_0)$. In Planck units, the natural magnetic energy density is $F_{\mathrm{P}}/l_{\mathrm{P}}^2 = E_{\mathrm{P}}/l_{\mathrm{P}}^3$; actual magnetic energy densities are this quantity scaled by $(B/B_{\mathrm{P}})^2$ where $B_{\mathrm{P}} = F_{\mathrm{P}}/(I_{\mathrm{P}} l_{\mathrm{P}})$ is the Planck magnetic field.

Inductance

$$L \;=\; \frac{\mu_0 N^2 A}{\ell} \;=\; \frac{4\pi N^2 m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2} \cdot \frac{A}{\ell}$$

The inductance of a solenoid with $N$ turns, cross-sectional area $A$, and length $\ell$. In Planck units, the natural inductance is $\frac{m_{\mathrm{P}} l_{\mathrm{P}}^2}{q_{\mathrm{P}}^2}$ — the Planck inductance.

Why this value?

Before the 2019 SI redefinition, $\mu_0$ was exactly $4\pi \times 10^{-7}$ N A$^{-2}$ by definition. This was because the ampere was defined by the force between parallel wires, and the pre-2019 ampere was set so that $\mu_0/(4\pi) = 10^{-7}$ N A$^{-2}$ exactly. This was not a measurement but a convention that defined the unit of current.

After the 2019 SI redefinition, the ampere is defined by fixing the elementary charge $e = 1.602\,176\,634 \times 10^{-19}$ C exactly. As a result, $\mu_0$ is no longer defined. It is now a measured quantity, with a value determined by the fine-structure constant $\alpha$. The relation $\mu_0 = 2\alpha h/(ce^2)$, combined with the exact values of $h$ and $e$, gives $\mu_0$ a relative uncertainty of about $1.6 \times 10^{-10}$ — the same as $\alpha$.

In the Planck decomposition, $\mu_0 = \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2}$, the specific numerical value depends on the values of the Planck units, which are customarily calculated from $\hbar$, $G$, and $c$. In particular, $m_{\mathrm{P}} l_{\mathrm{P}} = \hbar/c$ and $q_{\mathrm{P}}^2 = 4\pi\varepsilon_0\hbar c$, so $\mu_0 = \frac{4\pi \hbar}{c \cdot 4\pi\varepsilon_0\hbar c} = \frac{1}{\varepsilon_0 c^2}$ — recovering the expected relation.

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The scale-invariant reading
The ratio $\frac{\mu_0}{4\pi} = \frac{m_{\mathrm{P}} l_{\mathrm{P}}}{q_{\mathrm{P}}^2} = 10^{-7}$ N A$^{-2}$ takes the same numerical value at the Planck scale and at every scale accessible to experiment. This scale-invariance is the signature that the magnetic constant can be read as a structural ratio — the relationship between the vacuum’s mechanical and electromagnetic properties — rather than as an arbitrary convention.

Connections

Vacuum permittivity: $\varepsilon_0 = \frac{q_{\mathrm{P}}^2}{4\pi F_{\mathrm{P}} l_{\mathrm{P}}^2} = \frac{t_{\mathrm{P}}^2 q_{\mathrm{P}}^2}{4\pi m_{\mathrm{P}} l_{\mathrm{P}}^3}$ — the complementary electromagnetic constant. The product $\mu_0\varepsilon_0 = 1/c^2$; the ratio $\mu_0/\varepsilon_0 = Z_0^2$ defines the vacuum impedance squared.

Speed of light: $c = l_{\mathrm{P}}/t_{\mathrm{P}} = 1/\sqrt{\mu_0\varepsilon_0}$ — the vacuum permeability and permittivity together determine the wave speed of electromagnetic radiation. The Planck-unit form makes the cancellation transparent: $\mu_0\varepsilon_0$ carries $t_{\mathrm{P}}^2/l_{\mathrm{P}}^2$ exactly.

Vacuum impedance: $Z_0 = \mu_0 c = \frac{4\pi m_{\mathrm{P}} l_{\mathrm{P}}^2}{t_{\mathrm{P}} q_{\mathrm{P}}^2} \approx 376.73$ Ω — the ratio of electric to magnetic field amplitude in a plane electromagnetic wave. Also $Z_0 = 2\alpha R_K = 2\alpha h/e^2$.

Fine-structure constant: $\alpha = e^2/(4\pi\varepsilon_0\hbar c) = (e/q_{\mathrm{P}})^2$ — since 2019, $\mu_0 = 2\alpha h/(ce^2)$ with $h$ and $e$ exact, making $\mu_0$ proportional to $\alpha$. A better measurement of $\alpha$ is now a better measurement of $\mu_0$.

Planck charge: $q_{\mathrm{P}} = \sqrt{4\pi\varepsilon_0\hbar c} \approx 1.876 \times 10^{-18}$ C — appears squared in the denominator of $\mu_0$. The magnetic constant weighs the vacuum’s mechanical content ($m_{\mathrm{P}} l_{\mathrm{P}}$) against its electromagnetic content ($q_{\mathrm{P}}^2$).

Planck force: $F_{\mathrm{P}} = m_{\mathrm{P}} l_{\mathrm{P}}/t_{\mathrm{P}}^2 = 1.210 \times 10^{44}$ N — the magnetic constant is $\mu_0 = 4\pi F_{\mathrm{P}}/I_{\mathrm{P}}^2$. This encodes the magnetic constant as a ratio of the Planck force to the square of the Planck current.