Permeability of Free Space Dimensional Formula: [MLT⁻²I⁻²]

The permeability of free space dimensional formula, written with every power shown, is [M¹ L¹ T⁻² I⁻²], or [MLT⁻²A⁻²] with A standing for the ampere. In words, μ₀ has the dimensions of a force divided by the square of a current.

That neat reading comes from the force between two long parallel wires, which is proportional to μ₀ times the product of the two currents. Rearrange it for μ₀ and the currents land in the denominator as I². The numerical value used in exams is μ₀ = 4π × 10⁻⁷ T m/A, about 1.257 × 10⁻⁶ H/m.

Quick facts

Dimensional formula
[M¹ L¹ T⁻² I⁻²]
Defining relation
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$
SI unit
H/m, equal to N/A² and T m/A
Value
4π × 10⁻⁷ H/m ≈ 1.257 × 10⁻⁶ H/m
Type
Scalar constant
Link to light
c = 1/√(μ₀ε₀)

Deriving the permeability of free space dimensional formula

Two long straight wires a distance d apart carry currents I₁ and I₂. The force on a length l of either wire is

$F = \frac{\mu_0 I_1 I_2\, l}{2\pi d}$

Rearrange: $\mu_0 = \frac{2\pi F d}{I_1 I_2\, l}$ .
2π has no dimensions, and d over l is length over length, so it also has none.
That leaves $[\mu_0] = \frac{[F]}{[I]^{2}} = \frac{[MLT^{-2}]}{[I^{2}]}$ .
Final bracket: $[MLT^{-2}I^{-2}]$ .

A second route starts from $B = \frac{\mu_0 I}{2\pi r}$ , the field beside a straight conductor. Solving gives $\mu_0 = \frac{2\pi r B}{I}$ , and with B as [MT⁻²I⁻¹] you get $\frac{[L][MT^{-2}I^{-1}]}{[I]}$ , the same result. The magnetic field dimensions are worked out separately if you need that middle step.

Units of μ₀ and why three names are correct

Swap the base dimensions for kg, m, s and A and the bracket gives kg m s⁻² A⁻². Group the first three as a newton and you get N/A². Group them using the henry, the unit of inductance, and you get H/m, which is how tables of constants usually print it. From B = μ₀I/(2πr), the unit T m/A also works.

Equivalent units for μ₀
Form	Where it comes from
N A⁻²	Force between currents
T m A⁻¹	Field of a straight wire
H m⁻¹	Inductance per length
kg m s⁻² A⁻²	Pure base units

Any of these earns full marks, provided you can show they reduce to the same base units.

Dimensional proof that 1/√(μ₀ε₀) is a speed

Example: show that $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ has the dimensions of speed, then find its value.

Permittivity: ε₀ carries [M⁻¹L⁻³T⁴I²].

Product: add the powers, $[M^{1-1}L^{1-3}T^{-2+4}I^{-2+2}] = [L^{-2}T^{2}]$ . Mass and current cancel.

Square root and invert: $[L^{-2}T^{2}]^{-1/2} = [LT^{-1}]$ , a speed.

Numbers: $\mu_0\varepsilon_0 = (1.257 \times 10^{-6})(8.854 \times 10^{-12}) = 1.113 \times 10^{-17}\ \text{s}^2/\text{m}^2$ . Then $c = 1/\sqrt{1.113 \times 10^{-17}} = 3.00 \times 10^{8}\ \text{m/s}$ .

This is one of the most popular dimension questions in electromagnetism, and it links straight to the speed of light dimensions.

μ₀ beside ε₀ and relative permeability

Students often swap the two vacuum constants. Permeability μ₀ sets the strength of magnetic effects and has M and L to positive powers. Permittivity ε₀ sets the strength of electric effects and has M⁻¹ and L⁻³. Their product, as the example shows, is the inverse square of a speed.

The permeability of a material is written $\mu = \mu_r \mu_0$ . Relative permeability μᵣ is a ratio of two permeabilities, a pure number carrying neither dimensions nor a unit. Iron can have μᵣ in the thousands while air has μᵣ very close to 1, yet every μ in the problem keeps the bracket of μ₀.

Slips with the dimensional formula of permeability

Writing I⁻¹. Two currents multiply in the force law, so the power is minus 2.
Giving ε₀'s bracket. Check the sign of the M power: positive for μ₀, negative for ε₀.
Treating 4π as dimensional. 4π is a pure number and never enters a bracket.
Giving a unit to μᵣ. Relative permeability has none.

The vacuum constants are listed together under dimensional formulas, where μ₀ sits next to G, h and k.

Frequently asked questions

What is the dimensional formula of permeability of free space?

It is [M¹ L¹ T⁻² I⁻²]. You get it by dividing the dimensions of force, [MLT⁻²], by current squared, from the force between two parallel wires.

What is the dimensional formula of permeability of a material?

Absolute permeability μ has the same bracket as μ₀, [MLT⁻²I⁻²]. Relative permeability μᵣ = μ/μ₀ is dimensionless.

What is the SI unit of μ₀?

It is usually written H/m. N/A² and T m/A are equal forms; each one reduces to kg m s⁻² A⁻² in base units.

Which bracket does the product μ₀ε₀ have?

The product has dimensions [L⁻² T²], the inverse square of a speed. That is why 1/√(μ₀ε₀) equals the speed of light.

What is the value of μ₀?

The value used in school and entrance exam work is 4π × 10⁻⁷ H/m, which is about 1.257 × 10⁻⁶ H/m.