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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 ($\inline&space;\tau$) = $\inline&space;I\omega\Omega$ (In the limit).
• $\inline&space;I$ = Moment of Inertia.
• $\inline&space;\omega$ = Angular velocity
• $\inline&space;\Omega$ = Angular velocity of precession.

## Overview

Key facts

The magnetic reluctance is defined as:

$\mathcal{R}&space;=&space;\frac{\mathcal{F}}{\Phi}$

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

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

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

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

<br/>

Constants

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

The magnetic reluctance $\inline&space;\mathcal{R}$ 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:

$\mathcal{R}&space;=&space;\frac{\mathcal{F}}{\Phi}$

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

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

##### MISSING IMAGE!

We know that the magnetic field strength $\inline&space;H$ can be written as:

$\int{H&space;dl}&space;=&space;NI$

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

$B&space;=&space;\mu_0&space;\mu_r&space;H$

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

As the magnetic flux $\inline&space;\Phi$ is defined as:

$\Phi&space;=&space;B&space;\cdot&space;A$

equation (4) can also be written as:

$\frac{\Phi}{A}&space;=&space;\mu_0&space;\mu_r&space;H$

from which the magnetic field strength becomes:

$H&space;=&space;\frac{\Phi}{\mu_0&space;\mu_r&space;A}$

Considering that $\inline&space;H$ is uniform, equation (3) becomes:

$H&space;l&space;=&space;NI$

Using the expression form of $\inline&space;H$ from (8) in (7), we get that:

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

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

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

$\mathcal{F}&space;=&space;NI$

equation (10) becomes:

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

or:

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

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

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

Example:
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##### Example - Magnetic flux and flux density of a toroid
Problem
Consider a toroid with the mean length of $\inline&space;20&space;\;&space;cm$, the cross section of $\inline&space;2&space;\;&space;cm^2$, and the relative magnetic permeability of $\inline&space;6700$. 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 $\inline&space;\mathcal{R}$ is given by:

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

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

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

from which we obtain:

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

The magnetic flux $\inline&space;\Phi$ can be written as:

$\Phi&space;=&space;\frac{\mathcal{F}}{\mathcal{R}}$

where $\inline&space;\mathcal{F}$, the magnetomotive force, is given by:

$\mathcal{F}&space;=&space;NI$

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

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

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

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

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

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

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

$\Phi_t&space;=&space;NI&space;\div&space;\frac{l}{\mu_0&space;\mu_r&space;A}&space;+&space;\frac{l_g}{\mu_0&space;A}$
Solution
$\Phi&space;=&space;1.68&space;\cdot&space;10^{-4}&space;\;&space;Wb$

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