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Magnetic Reluctance

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

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Overview

Key facts

The magnetic reluctance is defined as:

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

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

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

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

where $\mu_0$ 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 $\mathcal{R}$ 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:

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

(1)

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

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

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

$\int{H dl} = NI$

(2)

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

$B = \mu_0 \mu_r H$

(3)

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

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

$\Phi = B \cdot A$

(4)

equation (3) can also be written as:

$\frac{\Phi}{A} = \mu_0 \mu_r H$

(5)

from which the magnetic field strength becomes:

$H = \frac{\Phi}{\mu_0 \mu_r A}$

(6)

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

$H l = NI$

(7)

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

$NI = \frac{\Phi l}{\mu_0 \mu_r A}$

(8)

which leads to:

$\Phi = NI \div \frac{l}{\mu_0 \mu_r A}$

(9)

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

$\mathcal{F} = NI$

(10)

equation (9) becomes:

$\Phi = \mathcal{F} \div \frac{l}{\mu_0 \mu_r A}$

(11)

or:

$\frac{\mathcal{F}}{\Phi} = \frac{l}{\mu_0 \mu_r A}$

(12)

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

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

(13)

Example:

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Example - Magnetic flux and flux density of a toroid

Problem

Consider a toroid with the mean length of $20 \; cm$ , the cross section of $2 \; cm^2$ , and the relative magnetic permeability of $6700$ . 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 $\mathcal{R}$ is given by:

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

(1)

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

$\mathcal{R} = \frac{0.2}{6700 \cdot 4 \pi \cdot 10^{-7} \cdot 2 \cdot 10^{-4}}$

(2)

from which we obtain:

$\mathcal{R} = 1.19 \cdot 10^5 \; At/Wb$

(3)

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

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

(4)

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

$\mathcal{F} = NI$

(5)

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

$\Phi = \frac{20}{1.19 \cdot 10^5} = 1.68 \cdot 10^{-4} \; Wb$

(6)

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

$B = \frac{\Phi}{A} = \frac{1.68 \cdot 10^{-4}}{2 \cdot 10^{-4}} = 0.84 \; Wb/m^2$

(7)

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

$\mathcal{R}_t = \frac{l}{\mu_0 \mu_r A} + \frac{l_g}{\mu_0 A}$

(8)

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

$\Phi_t = NI \div \frac{l}{\mu_0 \mu_r A} + \frac{l_g}{\mu_0 A}$

(9)

Solution

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

$B = 0.84 \; Wb/m^2$