# How do we convert between the maximum value of the adhesive force (in kilograms or in Newton) of a magnet and its magnetic field strength (in tesla)?

On the following website they indicate how powerful is a magnet by giving the theoretical maximum value of its adhesive force in kilograms or in Newton (its adhesive force at a distance of zero).

http://www.supermagnete.fr/eng/faq/How-strong-is-this-magnet

I would like to know how to convert this in tesla or in gauss.

More precisely, I would like to know how many tesla does a NdFeB magnet with a maximum adhesive force of 200 kg = 2000 N, a size of 10x10x1.5 cm and a mass of 1.3 kg generates at a distance of 50 cm ?

(I don't need a very accurate result. So if you can give me a rough approximation/estimation it will be good enough.)

You will need to know these things:

• The coenergy in the system, actually the air between magnet and object

$$W_m^*=\int_V\int_H BdHdV=\frac{B^2 Ax}{\mu}$$

• The circulation of the magnetic field according to Ampere

$$\int_\ell Hd\ell=\sum\theta$$

• The magnetomotive force

$$\theta_{PM}=\frac{B_Rh_{PM}}{\mu_0\mu_{PM}}$$

So that you can finally calculate the force:

$$F=-\frac{dW_m^*}{dx}$$ You need to calculate the magnetic coenergy in the air between the magnet and the object. The thing is that you won't be able to get a lot of the field 50cm away. The reason is that the field lines have a path of comparable resistance (directly between the poles of the NdFeB magnet), that they will take for a large part.

The advertised case is the scenario when the object is directly stuck to the magnet. That way almost all of the field lines go through it and bond it strongly.

I haven't calculated it but my guess would be that you actually won't get any usable attractive force at this distance.

I have found the answer to my question thanks to this online calculator: http://www.kjmagnetics.com/calculator.asp

At 50 cm (along the center axis of the magnet) my magnet generates a magnetic field strength of 3 Gauss.

I even got a more accurate result taking into account the inclination to the center axis of the magnet! With an inclination of 30° I got a magnetic field strength of 1.8 Gauss.

Not much, but still it's 4 times the strength of Earth's magnetic field at the surface. I will probably have to reduce the distance to at least 30 cm.