Mu Metal wire and Tube
What is field attenuation?
This is also known as the shielding factor (S) and is a ratio of the magnetic field strength outside of the magnetic shield (Ha) and the resultant field on the inside of the shield ie Ha/Hi (no units) or S = 20 x log(Ha/Hi) (Db). There are various formula based on the permeability of the material, the shape and size of the shield and the material thickness. In most cases these formulae are only approximate.
For a closed shielding can :
S = 4/3 X (Mu x d/D) where Mu : The permeability(relative)
d : material thickness
D : Shielding Diameter
For a long hollow cylinder in a magnetic transverse field :
S = Mu x d/D
For a cubic shielding box :
S = 4/5 X (Mu x d/a)
a : box side length.
In the case of multiple layer shields with air gaps provided by insulating spacers the shielding factors of the individual shields are multiplied together resulting in excellent shielding factors.
For a double layer shield :
S= S1 x ((S2 x (2 x change in diameter /diameter) )
How would these formulas work if I wanted to shield a Magnet inside a box?
E.g. the field strenght inside is large and I want the field strenght on the outside to be as small as possible.
The shielding formulas work in exactly the same way whether the source of interference is on the inside of the shield or on the outside – this is assuming that the magnet to be shielded is within the saturation induction range of a single or multi layer shield (see properties of mu-metal, supra 50 and multi layer shielding)
It is also worth noting that whilst a shield housing a magnet will negate the effect of the magnetic field produced, the magnet will still be attracted to the shield as magnetic shielding alloys are ferromagnetic. It would therefore need housing in a non-magnetic, isolative structure (such as polyurethane), within the shield to make this attraction impossible through means of restriction of movement for optimal shielding.
What is the shielding factor S for a double layer cubic shielding box?
Dear Laurent,
Thanks for the mail. Our shielding formula for shielding boxes can be found on our website : http://www.magneticshields.co.uk under frequently asked questions.
For the second layer please use the box side length for all instead of using diameter measurements.
Best Regards
Steve Locker
If I want to shield a thick wire with maybe 200 Amps in it so that it’s field (about a few mT, I think, in air) will not (to 99%) penetrate outside the ellipsoidal mu-metal shield, how thick must it be (d) if the ellipsoid’s minor axis D is about 3 cm and major D is 10 cm (wire length). If too thick, then dor 90% attenuation what is d?
Thanks,
Hugh
Dear Hugh,
At present there is no known formula for ellipsoidal mumetal shields so we need to base the calculations on a closed shielding cylinder and a cubic box.
For the minor axis we would assume a diameter of 30mm, therefore to reduce the field by 99% a layer of 0.2mm mumetal would be required
For the major axis you cannot use the formula for a closed shielding cylinder, you would need to use the formula for a cubic box with a side length of 100mm. On this basis, to give an attenuation of 99% the mumetal thickness would need to be 0.8mm thick.