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:

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

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:

$H\cdot s = I;\quad s = 2\pi r;\quad H = \frac{I}{2\pi r}$

## 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:

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

## 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:

$H = H_0 + H_1$

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:

$H = H_0 + \chi H_0 = (1 + \chi)H_0 = \mu H_0$

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:

$B = \mu_0 \cdot H = \mu_0 \cdot \mu \cdot H_0$

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:

$\Phi = \int_{A}{}B\cdot cos(\alpha)dA$

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:

$\Phi = B\cdot A$

The total magnetic flux linkage around the solenoid is:

$\Psi = N\cdot B\cdot A$

## Inductance

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

$\Psi = L\cdot I$

Then, collecting previous formulas together gives:

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

Where Inductance is:

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

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.