Magnetic Pull Force — Magnetism

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

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

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

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

(3)

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

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

(4)

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

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

(5)

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

$$W = F \delta x$$

(6)

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

Taking into account equations (5) and (6), we get that:

\calc{A*(B^2/(8*π*10^(-7)*mu_r))} "Instant calculator eq(6)"

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

(7)

from which the magnetic pull force becomes:

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

(8)

Example 1 [metric]

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:

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

(9)

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:

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

(10)

which gives:

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

(11)

The total magnetic flux is given by:

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

(12)

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

$$\mathcal{F} = Ni$$

(13)

As, in our case, $N=200$, $i=1\; A$, and $\mathcal{R}=10.88 \cdot 10^4 \; At/Wb$ (from equation 11), we obtain from (12) and (13) that the total magnetic flux is:

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

(14)

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

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

(15)

and also considering (14) and that $A=10\; cm^2$ ($=10\cdot 10^{-4} \; m^2$), we obtain the magnetic flux density in the air gap:

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

(16)

As the magnetic pull force is given by:

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

(17)

and also considering (16), 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:

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

(18)

Thus, we obtain the total magnetic pull force:

Solution

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

(19)