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Magnetic Reluctance

A description of the magnetic reluctance, also discussing a way to calculate it

Overview

Key facts

The magnetic reluctance is defined as:

\mathcal{R} = \frac{\mathcal{F}}{\Phi}

where \inline \mathcal{F} is the magnetomotive force, and \inline \Phi the magnetic flux.

For a magnetic circuit of length \inline l, cross-sectional area \inline A, and relative magnetic permeability \inline \mu_r, the magnetic reluctance can be calculated with:

\mathcal{R} = \frac{l}{\mu_0 \mu_r A}

where \inline \mu_0 is the magnetic permeability of free space.

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Constants

\mu_0 = 4 \pi \cdot 10^{-7} \; \frac{N}{A^2}

The magnetic reluctance \inline \mathcal{R} of a magnetic circuit can be regarded as the formal analog of the resistance in an electrical circuit. The magnetic reluctance can be expressed as:

where \inline \mathcal{F} is the magnetomotive force (mmf), and \inline \Phi is the magnetic flux.

In order to calculate the magnetic reluctance, consider a magnetic circuit of length \inline l and cross-sectional area \inline A, as diagramed in Figure 1.

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Figure 1

We know that the magnetic field strength \inline H can be written as:

where \inline I is the current in the coil, and \inline N is the number of turns (for a more detailed discussion on the magnetic field strength see Field Strength ). Furthermore, \inline H can be related to the magnetic flux density \inline B with the equation:

where \inline \mu_0 is the magnetic permeability of free space, and \inline \mu_r the relative magnetic permeability of the material.

As the magnetic flux \inline \Phi is defined as:

equation (3) can also be written as:

from which the magnetic field strength becomes:

Considering that \inline H is uniform, equation (2) becomes:

Using the expression form of \inline H from (7) in (6), we get that:

which leads to:

As the magnetomotive force \inline \mathcal{F} of a coil is given by:

equation (9) becomes:

or:

Taking into account the definition of the magnetic reluctance from (1), we get that \inline \mathcal{R} can be calculated as:

Example:
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Example - Magnetic flux and flux density of a toroid
Problem
Consider a toroid with the mean length of \inline 20 \; cm, the cross section of \inline 2 \; cm^2, and the relative magnetic permeability of \inline 6700. What is the magnetic flux and the magnetic flux density if the coil has 10 turns and the current is 2 amperes ?
Workings
As the magnetic reluctance \inline \mathcal{R} is given by:

and, in our case, \inline l = 20 \; cm (\inline = 0.2 \; m), and \inline A = 2 \; cm^2 (\inline =2 \cdot 10^{-4} \; m^2), we get that:

from which we obtain:

The magnetic flux \inline \Phi can be written as:

where \inline \mathcal{F}, the magnetomotive force, is given by:

As, in our case, \inline N=10, \inline I=2, and also considering (3), we obtain the magnetic flux:

Taking into account that the cross-sectional area is \inline A = 2 \; cm^2 (\inline =2\cdot 10^{-4} \; m^2), the magnetic flux density becomes:

As a side note, if the toroid has an air gap of length \inline l_g, then its total magnetic reluctance, \inline \mathcal{R}_t, would be the magnetic reluctance of the toroid plus the magnetic reluctance of the air gap:

Thus, in this case, the total magnetic flux, \inline \Phi_t, would be given by:

Solution
\Phi = 1.68 \cdot 10^{-4} \; Wb

B = 0.84 \; Wb/m^2