Published by Patrick Mutisya · 14 days ago
Understand that the magnetic field inside a solenoid is increased when a ferrous core is inserted, and be able to calculate the resulting field.
A long solenoid of length \(l\), carrying a current \(I\) through \(N\) turns, produces a magnetic field approximately uniform inside:
where \(\mu_0 = 4\pi\times10^{-7}\,\text{H/m}\) is the permeability of free space.
When a core of relative permeability \(\mu_r\) is inserted, the magnetic field becomes
\[
B{\text{core}} = \mu0 \mu_r n I
\]
The core increases the field by a factor of \(\mur\). Typical ferrous materials have \(\mur\) ranging from 100 to 10,000.
The absolute permeability of the core is
\[
\mu = \mu0 \mur
\]
For a linear material, \(\mur\) is constant. In practice, \(\mur\) decreases with increasing magnetic field strength due to saturation.
Calculate the magnetic field inside a solenoid of length \(0.5\,\text{m}\), with \(2000\) turns, carrying a current of \(2\,\text{A}\). The solenoid is filled with a core of relative permeability \(\mu_r = 500\).
Thus the core increases the field by a factor of 500.
| Parameter | Symbol | Expression | Units |
|---|---|---|---|
| Number of turns per unit length | n | \(\dfrac{N}{l}\) | turns m\(^{-1}\) |
| Magnetic field (air core) | B | \(\mu_0 n I\) | T |
| Magnetic field (ferrous core) | Bcore | \(\mu0 \mur n I\) | T |
| Absolute permeability | \(\mu\) | \(\mu0 \mur\) | H m\(^{-1}\) |