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# Magnetic Pull Force

An analysis of the magnetic pull force which arises between the poles of an electromagnet
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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

For an electromagnet characterized by the area $\inline&space;A$, the magnetic flux density $\inline&space;B$, and the relative magnetic permeability $\inline&space;\mu_r$, the magnetic pull force is:

$F&space;=&space;A&space;\frac{B^2}{2&space;\mu_0&space;\mu_r}$

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

<br/>

Constants

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

Consider an electromagnet of area $\inline&space;A$ and magnetic flux density $\inline&space;B$, and also imagine a displacement of $\inline&space;\delta&space;x$ as highlighted in Figure 1.

##### MISSING IMAGE!

We know that the energy stored in a magnetic field of no magnetic saturation is given by:

$E_{stored}&space;=&space;V&space;\frac{B^2}{2&space;\mu_0&space;\mu_r}$

where $\inline&space;V$ is the volume, $\inline&space;\mu_0$ the magnetic permeability of free space, and $\inline&space;\mu_r$ the relative magnetic permeability (for a more detailed discussion on the energy stored in a magnetic field see Stored Energy ).

Thus, the change in energy stored following the displacement $\inline&space;\delta&space;x$ will be:

$\delta&space;E_{stored}&space;=&space;\delta&space;V&space;\frac{B^2}{2&space;\mu_0&space;\mu_r}$

where $\inline&space;\delta&space;V$ ($\inline&space;=A&space;\delta&space;x$) is the change in volume. This leads to:

$\delta&space;E_{stored}&space;=&space;A&space;\delta&space;x&space;\frac{B^2}{2&space;\mu_0&space;\mu_r}&space;(=W)$

where $\inline&space;W$ refers to the work done. However, we also know that work can also be defined as:

$W&space;=&space;F&space;\delta&space;x$

where $\inline&space;F$ is the force ($\inline&space;Newtons$).

Taking into account equations (4) and (5), we get that:

$F&space;\delta&space;x&space;=&space;A&space;\delta&space;x&space;\frac{B^2}{2&space;\mu_0&space;\mu_r}$

from which the magnetic pull force becomes:

$F&space;=&space;A&space;\frac{B^2}{2&space;\mu_0&space;\mu_r}$

Example:
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##### Example - Magnetic pull force of an electromagnet
Problem
Consider the electromagnet diagramed in Figure E1, characterised by the lengths $\inline&space;l_1&space;=&space;20&space;\;&space;cm$, $\inline&space;l_2&space;=&space;15&space;\;&space;cm$, and $\inline&space;l_g&space;=&space;0.001&space;\;&space;cm$, and the area $\inline&space;A&space;=&space;10&space;\;&space;cm^2$. Given that a current of $\inline&space;i&space;=&space;1&space;&space;A$ passes through a coil with $\inline&space;N&space;=&space;200$ turns and relative magnetic permeability of $\inline&space;\mu_r&space;=&space;3000$, find the total magnetic pull force.

Workings
We know that the total magnetic reluctance of a magnetic circuit of length $\inline&space;l$, cross-sectional area $\inline&space;A$, and relative magnetic permeability $\inline&space;\mu_r$, with an air gap of length $\inline&space;l_g$, is given by:

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

As, in our case, $\inline&space;l=l_1+l_2=20+15&space;\;&space;cm$ ($\inline&space;=(20+15)\cdot&space;10^{-2}&space;\;&space;m$), $\inline&space;A=10\;&space;cm^2$ ($\inline&space;=10\cdot&space;10^{-4}&space;\;&space;m^2$), $\inline&space;\mu_r&space;=&space;3000$, and $\inline&space;l_g=0.001&space;\;&space;cm$ ($\inline&space;=0.001&space;\cdot&space;10^{-2}&space;\;&space;m$), we obtain the total magnetic reluctance:

$\mathcal{R}&space;=&space;\frac{(20+15)\cdot&space;10^{-2}}{4\pi&space;\cdot&space;10^{-7}\cdot&space;3000&space;\cdot&space;10&space;\cdot&space;10^{-4}}&space;+&space;2\cdot&space;\frac{0.001&space;\cdot&space;10^{-2}}{4\pi&space;\cdot&space;10^{-7}&space;\cdot&space;10&space;\cdot&space;10^{-4}}$

which gives:

$\mathcal{R}&space;=&space;10.88&space;\cdot&space;10^4&space;\;&space;\frac{At}{Wb}$

The total magnetic flux is given by:

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

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

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

As, in our case, $\inline&space;N=200$, $\inline&space;i=1\;&space;A$, and $\inline&space;\mathcal{R}=10.88&space;\cdot&space;10^4&space;\;&space;At/Wb$ (from equation 3), we obtain from (4) and (5) that the total magnetic flux is:

$\Phi&space;=&space;\frac{200\cdot&space;1}{10.88\cdot&space;10^4}&space;=&space;18.38&space;\cdot&space;10^{-4}&space;\;&space;Wb$

Taking into account that the magnetic flux density $\inline&space;B$ is given by:

$B&space;=&space;\frac{\Phi}{A}$

and also considering (6) and that $\inline&space;A=10\;&space;cm^2$ ($\inline&space;=10\cdot&space;10^{-4}&space;\;&space;m^2$), we obtain the magnetic flux density in the air gap:

$B_g&space;=&space;\frac{18.38&space;\cdot&space;10^{-4}}{10\cdot&space;10^{-4}}&space;=&space;1.838&space;\;&space;\frac{Wb}{m^2}$

As the magnetic pull force is given by:

$F&space;=&space;A&space;\frac{B^2}{2&space;\mu_0&space;\mu_r}$

and also considering (8), and that $\inline&space;A=10\;&space;cm^2$ ($\inline&space;=10\cdot&space;10^{-4}&space;\;&space;m^2$), and the relative magnetic permeability of air is $\inline&space;\mu_r&space;=&space;1$, the magnetic pull per pole becomes:

$A&space;\frac{B_g^2}{2&space;\mu_0&space;\mu_r}&space;=&space;10&space;\cdot&space;10^{-4}&space;\cdot&space;\frac{1.838^2}{2&space;\cdot&space;4\pi&space;\cdot&space;10^{-7}&space;\cdot&space;1}&space;=&space;1345.6&space;\;&space;N$

Thus, we obtain the total magnetic pull force:
Solution
$F=&space;2&space;\cdot&space;1345.6&space;=&space;2691&space;\;&space;N$