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Magnetic Reluctance

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

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Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\tau$ ) = $I\omega\Omega$ (In the limit).

= Moment of Inertia.
$\omega$ = Angular velocity
$\Omega$ = Angular velocity of precession.

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

, cross-sectional area

, 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:

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

(2)

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

and cross-sectional area

, as diagramed in Figure 1.

We know that the magnetic field strength

can be written as:

$\int{H dl} = NI$

(3)

where

is the current in the coil, and

is the number of turns (for a more detailed discussion on the magnetic field strength see Field Strength ). Furthermore,

can be related to the magnetic flux density

with the equation:

$B = \mu_0 \mu_r H$

(4)

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:

$\Phi = B \cdot A$

(5)

equation (4) can also be written as:

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

(6)

from which the magnetic field strength becomes:

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

(7)

Considering that

is uniform, equation (3) becomes:

(8)

Using the expression form of

from (8) in (7), we get that:

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

(9)

which leads to:

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

(10)

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

$\mathcal{F} = NI$

(11)

equation (10) becomes:

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

(12)

or:

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

(13)

Taking into account the definition of the magnetic reluctance from (2), we get that $\mathcal{R}$ can be calculated as:

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

(14)

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

. 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,

, 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

, 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

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

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