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 ( $\inline \tau$ ) = $\inline I\omega\Omega$ (In the limit).

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

Overview

Key facts

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

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

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

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Constants

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

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

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

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

(2)

where $\inline V$ is the volume, $\inline \mu_0$ the magnetic permeability of free space, and $\inline \mu_r$ 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 $\inline \delta x$ will be:

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

(3)

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

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

(4)

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

$W = F \delta x$

(5)

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

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

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

(6)

from which the magnetic pull force becomes:

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

(7)

Example:

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Example - Magnetic pull force of an electromagnet

Problem

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

Workings

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

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

(1)

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

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

(2)

which gives:

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

(3)

The total magnetic flux is given by:

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

(4)

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

$\mathcal{F} = Ni$

(5)

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

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

(6)

Taking into account that the magnetic flux density $B$ is given by:

$B = \frac{\Phi}{A}$

(7)

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

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

(8)

As the magnetic pull force is given by:

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

(9)

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

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

(10)

Thus, we obtain the total magnetic pull force:

Solution

$F= 2 \cdot 1345.6 = 2691 \; N$

(11)