Permeability of free space in AP Physics 2

The permeability of free space (μ₀ = 4π × 10⁻⁷ T·m/A) is the fundamental constant that converts electric current into magnetic field strength in a vacuum, appearing in the AP Physics 2 equation B = μ₀I/(2πr) for the field around a long, straight, current-carrying wire.

Verified for the 2027 AP Physics 2 examLast updated June 2026

What is the permeability of free space?

The permeability of free space, written μ₀ ("mu naught"), is a fundamental physical constant equal to 4π × 10⁻⁷ T·m/A. It answers a simple question. If you push a certain amount of current through a wire, how strong is the magnetic field it creates? μ₀ is the conversion factor. Think of it as nature's exchange rate between amps and teslas.

In AP Physics 2, μ₀ lives inside the equation B = μ₀/(2π) (I/r), which gives the magnetic field magnitude a perpendicular distance r from a long, straight wire carrying current I. The field wraps around the wire in concentric circles (right-hand rule), gets stronger with more current, and weaker as you move away. μ₀ is what sets the overall scale. The "free space" part means vacuum, though air is so close to vacuum magnetically that you treat them the same on the exam.

Why the permeability of free space matters in AP® Physics 2

μ₀ sits at the heart of Topic 12.3 (Magnetism and Current-Carrying Wires) in Unit 12. Learning objective 12.3.A asks you to describe the magnetic field produced by a current-carrying wire, and the essential knowledge hands you the equation B = μ₀/(2π) (I/r) directly. Without μ₀, you can describe the field's shape and direction, but you can't calculate its actual magnitude. The constant is also the bridge in two-wire problems. Wire 1 creates a field using μ₀ (LO 12.3.A), and that field then pushes on wire 2 through F_B = IℓB sin θ (LO 12.3.B). Conceptually, μ₀ is also evidence for one of the biggest ideas in the course, that electricity and magnetism are two faces of the same interaction. A single constant ties moving charge to magnetic field.

How the permeability of free space connects across the course

B = μ₀/(2π) (I/r) (Unit 12)

This is μ₀'s home equation. The constant sets the scale, the current I cranks the field up, and the distance r dilutes it. Everything else in the formula is geometry; μ₀ is the physics.

F_B = IℓB sin θ (Unit 12)

Chain these two equations together for parallel-wire problems. One wire's current makes a field (that's where μ₀ enters), and that field exerts a force on the second wire. Parallel currents attract, antiparallel currents repel.

Permittivity of free space, ε₀ (Unit 10)

ε₀ is μ₀'s electric twin. It sets the strength of the electric force between charges in Coulomb's law, the same way μ₀ sets the strength of the magnetic field from a current. The deep payoff is that the two constants together determine the speed of light, since c = 1/√(μ₀ε₀).

Is the permeability of free space on the AP® Physics 2 exam?

μ₀ shows up almost entirely through B = μ₀/(2π) (I/r). Multiple-choice questions ask you to identify what μ₀ stands for in the formula, calculate B given I and r, solve backward for μ₀ from measured field data, or pick out which factors do and don't affect field strength (hint: μ₀ is a constant, so it's never the variable changing the field). The good news is you don't memorize the value; μ₀ is on the AP Physics 2 constants sheet. In free-response settings, μ₀ powers any quantitative claim about field strength near a wire, and proportional reasoning is the favorite move. Double the current and B doubles. Double the distance and B halves. No released FRQ hinges on the constant by name, but any wire-field calculation runs through it.

The permeability of free space vs Permittivity of free space (ε₀)

Permeability (μ₀) is magnetic; permittivity (ε₀) is electric. μ₀ relates current to the magnetic field it produces and equals 4π × 10⁻⁷ T·m/A. ε₀ relates charge to the electric force and field it produces, hiding inside Coulomb's constant k = 1/(4πε₀). A quick memory hook is that 'permeability' and 'magnetic' both describe how field lines pass through (permeate) space around a current. If the equation has I, you want μ₀; if it has q, you want ε₀.

Key things to remember about the permeability of free space

  • The permeability of free space, μ₀ = 4π × 10⁻⁷ T·m/A, is the constant that converts current into magnetic field strength in a vacuum or air.

  • μ₀ appears in B = μ₀/(2π) (I/r), the equation for the magnetic field a distance r from a long, straight wire carrying current I.

  • Because μ₀ is a constant, only current and distance actually change the field strength around a wire; B is proportional to I and inversely proportional to r.

  • You don't need to memorize μ₀'s value since it's provided on the AP Physics 2 reference sheet, but you do need to recognize what it means in an equation.

  • Don't mix up μ₀ (permeability, magnetic, goes with current) and ε₀ (permittivity, electric, goes with charge).

  • In two-wire problems, μ₀ enters when one wire's current creates a field, and F_B = IℓB sin θ then gives the force that field exerts on the other wire.

Frequently asked questions about the permeability of free space

What is the permeability of free space in AP Physics 2?

It's the constant μ₀ = 4π × 10⁻⁷ T·m/A that relates the magnetic field around a current-carrying wire to the current producing it. It appears in the Topic 12.3 equation B = μ₀/(2π) (I/r).

Do I have to memorize the value of μ₀ for the AP exam?

No. μ₀ is listed on the AP Physics 2 constants and equations sheet you get during the exam. What you actually need is to recognize it in B = μ₀/(2π) (I/r) and use it correctly in calculations.

What's the difference between permeability (μ₀) and permittivity (ε₀)?

μ₀ is the magnetic constant, linking current to magnetic field, while ε₀ is the electric constant, linking charge to electric force in Coulomb's law. If the formula involves current I, you're using μ₀; if it involves charge q, you're using ε₀.

Does μ₀ change the magnetic field strength around a wire?

No, μ₀ is a fixed constant, so it never changes. In B = μ₀/(2π) (I/r), only the current I and the perpendicular distance r actually vary the field strength, which is exactly the kind of distinction multiple-choice questions test.

Why is μ₀ written as 4π × 10⁻⁷ instead of a decimal?

Because the 4π form cancels neatly with the 2π in B = μ₀/(2π) (I/r), leaving 2 × 10⁻⁷ T·m/A out front. That makes mental math on the exam much faster. As a decimal it's about 1.26 × 10⁻⁶ T·m/A.