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inductance

Straight Wire Inductor

Calculate the inductance of a straight wire conductor.

A straight wire inductor is a type of conductor frequently used in electrical circuits to produce inductance and magnetic fields. It can be as simple as a single straight wire or multiple wires stacked in parallel, functioning as a basic type of inductor in a variety of electrical circuits.

When current flows through a straight wire inductor, its main function is to store and release energy in the form of a magnetic field. In circuits, it produces inductance — resisting variations in current flow and aiding in controlling voltage levels, removing noise, and stabilizing electrical signals.

Understanding Straight Wire Inductors

A straight wire possesses inductive properties, albeit with limited efficiency compared to a coiled wire. This is referred to as self-inductance — the wire's capacity to store energy within its magnetic field.

Factors Affecting Inductance

FactorEffect
Wire LengthLonger wires generally exhibit greater inductance
Wire DiameterThicker diameters contribute to higher inductance compared to thinner ones
Surrounding MediumThe surrounding medium influences inductance — variations are observed between air and ferromagnetic surroundings

Straight Wire vs. Coiled Wire

The inductance of a straight wire is notably lower than that of a coiled wire due to the less concentrated magnetic field. As a result:

  • Wire coils are preferred in applications requiring higher inductance
  • Straight wires suffice for applications where lower inductance is adequate

It is important to recognize that while inductance is commonly associated with wire coils, straight wires also display inductive characteristics — typically at lower levels.

Applications

  • Filter Circuits
  • Power Supplies
  • RF Circuits
  • Electronic Components
  • Structural Integrity

About This Calculator

This online calculator determines the inductance of a straight wire conductor using only the wire's length and diameter.

Formula

L=0.00508×a×(log(2ad)0.75)L = 0.00508 \times a \times \left(\log\left(\frac{2a}{d}\right) - 0.75\right)

where:

  • LL = Inductance (μH)
  • aa = Length of the Wire (in)
  • dd = Diameter of the Wire (in)

Inputs

Length of the wire in inches

Diameter of the wire in inches

Results

Length must be greater than zero
InductanceµHSelf-inductance of the straight wire in microhenries