created at November 28, 2022

Solenoid Inductance

Let's introduce the general terms to find out the Inductance formula later.

Magnetic Field Strength

The current I flowing in the wire (conductor) generates a magnetic field around, which has a strength H [A/m]. Magnetic Field Strength is a vector, and Magnetic field is a vector field. There is a formula:

CHds=AjdA=I(1)\oint_{C}^{}Hds=\int_{A}^{}jdA=I\quad (1)

According to which the H field circulation along a closed curve C, which is enclosing an area A, is equals to a surface integral of the current density, flowing through that area. The surface integral of the current density gives the total current I, flowing through the surface.

Magnetic Field Strength around the Wire

Consider a straight wire and current I flowing in it. Current will produce a magnetic field around the wire. Field lines will look as a concentric circles around the wire.

Let's take a closer look at some specific circle having radius r. In every circle point the H vector is tangent to the circle and has constant H magnitude. Also, wire current I is crossing the area A enclosed in this circle. So, for this specific case integral (1) can be reduced to a simple formula:

Hs=I;s=2πr;H=I2πrH\cdot s = I;\quad s = 2\pi r;\quad H = \frac{I}{2\pi r}

Magnetic Field Strength around the Solenoid Inductor

Now consider we have winded the wire around the calendrical surface. The length l of that cylinder is much greater than its diameter. Here we have a solenoid. Current flowing in every single coil is the same and is producing a magnetic field around. The total magnetic field strength H will be nearly equal at every point inside the solenoid and will be very small outside the solenoid. Magnetic field lines will look like this:

We can neglect the outer magnetic field and apply the integral (1) to the solenoid. Consider a loop C, enclosing the half of the solenoid. Magnetic field H inside the solenoid is contributing the field circulation. Since H is nearly constant, we just multiply it by solenoid length l. On the other side, coil current I is crossing the area A enclosed by the loop N times, i.e. the total current is N·I. Then again the integral (1) is reduced to a simple formula:

H=NIlH = \frac{N\cdot I}{l}

Magnetic properties of materials

If we place a material in the outer magnetic field H0, material particles can become oriented so that they produce their own field H1. Then the total field H inside material will be:

H=H0+H1H = H_0 + H_1

Depending on how the materials behave in the outer magnetic field they can be divided into groups:

  • Diamagnetic (weakening the outer field)
  • Paramagnetic (slightly strengthening the outer field)
  • Ferromagnetic (significantly strengthening the outer field)

If the field H1 has a linear dependence on H0 (for dia- and paramegnetics), one can write:

H=H0+χH0=(1+χ)H0=μH0H = H_0 + \chi H_0 = (1 + \chi)H_0 = \mu H_0

were χ is a magnetic susceptibility of the material, and μ = 1 + χ is a magnetic permeability (sometimes called the relative magnetic permeability), which shows how much the material is increasing the magnetic field inside.

Magnetic flux density

Magnetic flux density B [Tesla] is another magnetic field characteristic, which relates to a field strength H through μ coefficient:

B=μ0H=μ0μH0B = \mu_0 \cdot H = \mu_0 \cdot \mu \cdot H_0

where μ0 = 4π10-7 is a magnetic constant (magnetic permeability of the vacuum).

Magnetic flux

The flux of B vector field through the area A is:

Φ=ABcos(α)dA\Phi = \int_{A}{}B\cdot cos(\alpha)dA

We consider the flux of a single coil. Since the H (and also B) values at every coil section point are nearly equal, the integral gives:

Φ=BA\Phi = B\cdot A

The total magnetic flux linkage around the solenoid is:

Ψ=NBA\Psi = N\cdot B\cdot A


Inductance of the solenoid shows its ability to produce magnetic field, when current I flows through it:

Ψ=LI\Psi = L\cdot I

Then, collecting previous formulas together gives:

Ψ=μ0μN2AIl\Psi = \frac{\mu_0\cdot \mu\cdot N^2\cdot A\cdot I}{l}

Where Inductance is:

L=μ0μN2AlL = \frac{\mu_0\cdot \mu\cdot N^2\cdot A}{l}

If the solenoid is empty inside (there are only coils present), μ has value of 1 (magnetic permeability of air). If we place the core made of some ferromagnetic material inside the coil (μ >> 1) - we will increase the Inductance significantly without changing the inductors dimensions.