Electricity & Magnetism
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Field Inside a Solenoid

Magnetic field inside an ideal solenoid.

Formula

B=μ0nIB = \mu_0 n I

Variables

\mu_0Permeability
nTurns per length
ICurrent

How it works

For an ideal solenoid that is long compared with its diameter, the magnetic field inside is nearly uniform and parallel to the axis. The field strength is proportional to how many turns are packed into each metre and to the current in the coil, with the constant of proportionality set by the magnetic permeability of free space. This comes from Ampère’s law: when you choose a loop running partly inside the solenoid, the enclosed current is the number of turns per unit length times the current, which gives B = μ0 n I for the interior field.

When to use it

Use this relation when analysing long air-core solenoids, electromagnets, inductors, or any problem where the field inside the coil is assumed to be uniform and edge effects are negligible. It is especially useful for finding the field from the coil geometry and current, or for solving for the current needed to produce a target field.

Worked example, step by step

A long solenoid has 1,200 turns spread uniformly over a length of 0.40 m and carries a current of 2.5 A. Step 1: Find the turns per unit length. n = 1200 / 0.40 = 3000 m^-1 Step 2: Use B = μ0 n I. B = (4π × 10^-7 T·m/A)(3000 m^-1)(2.5 A) Step 3: Multiply the values. B = 4π × 10^-7 × 7500 T B = 3π × 10^-3 T Step 4: Evaluate numerically. B ≈ 9.42 × 10^-3 T Final answer: B ≈ 9.4 mT

Common mistakes to avoid

  • Using the total number of turns N instead of the turn density n = N/L
  • Forgetting that the formula assumes a long ideal solenoid and breaks down near the ends
  • Mixing up units and using length in cm instead of metres when calculating n

Did You Know?

A coiled solenoid concentrates its magnetic field inside like a bar magnet you can switch on and off.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026