Australian Magnetic Solutions
IMS Home Page
Search
Force Calculation

Force Calculation

The following example illustrates how a simple change in the magnetic circuit increased the lifting force of a magnetic device by a factor of 15.

Let us consider the situation where a lifting device is made using the same magnet in two different arrangements. The magnet selected for the example has dimensions of 25.4 mm x 25.4 mm x 5 mm.

Force calculation example - Figure A Single magnet with one pole face secured to a lifting attachment

The first example shown in figure A utilises a single magnet with one pole face secured to a lifting attachment which is connected to a spring scale to measure the force on the magnet. The other face of the magnet is placed on a solid steel plate which is secured to a bench. The magnetic field holds the magnet to the steel plate.

The magnetic force holding the steel plate to the magnet can be measured on the scale. The approximate force can also be calculated using the following formula:

F ˜ Bm2Am
8π x 10-7

where Bm is the magnetic flux density and Am is the area of the magnet contacting the steel plate (6.45 x 1O-4 m2). In this example the force of attraction is 41 Newtons. Bm would be about one third of the magnet's maximum flux density of 1.2 Tesla because of the large air gaps in the circuit, say 0.4 Tesla.

Force calculation example - Figure B Single magnet with square steel section completing the magnetic circuit giving 15 times the force

Now in the second example as shown in figure B square steel section is used to complete the magnetic circuit between the upper side of the magnet and the steel plate significantly reducing the air gap. The steel section is secured to the upper face of the magnet and the scale. This results in an effective connection of both pole faces of the magnet to the steel plate.

Now the pulling force can be calculated using both surfaces contacting the steel plate with the following formula:

F ˜ Bm2Am + Bs2As
8π x 10-7

where Bs is the flux density entering the steel plate from the square steel section and AS is the surface area of the square steel section making contact with the steel plate. In this case Bm will approach the maximum flux density of the magnet say 1.1 Tesla.

If we assume that Bm is equal to Bs and that As is equal to Am then the force will be 621 Newtons, which is approximately 15 times that of the original example.

So, 15 times the force without increasing the size of magnet.