# Shield Magnetic Field

by strokebow
Tags: field, magnetic, shield
 P: 101 Hi, If I run a current carrying wire through a tube of iron will it shield the magnetic field from outside the tube? That being true... what is the best material to use for the tubing (and why?)? and also, how thick would it need to be? Since thickness will probably be related to the field strength, lets say the current was ~10A. Any Ideas/suggestions? cheers
 PF Gold P: 436 Well..10 A of current is sufficient. but to test your experiment, i think you have used a coil of wire inside the tube. That's better. As for the information on shielding you can visit this website-http://www.lessemf.com/mag-shld.html
 P: 101 Thanks. Why a coil of wire? I understand that using a coil of wire will give me a B field in a given direction but if I just use the wire in one direction I will get a 'circular' B field. But why should that matter when I am attempting to shield the field?
 P: 660 Shield Magnetic Field The best way to shield magnetic fields is Mu-metal, but it is expensive and a ***** to work with. http://en.wikipedia.org/wiki/Mu-metal The best way to shield the magnetic field from a wire is to use a coaxial wire and have the return current in the shield. A single wire will give magnetic field lines around the wire. A coil will give magnetic field lines along the axis of the coil. This is not a good way to test the shielding properties.
 P: 101 Thanks for the reply. Is there an equation for determining how thick your shield should be?
P: 409
 Quote by strokebow Thanks for the reply. Is there an equation for determining how thick your shield should be?
Unlike conductors which shield static electric field quite perfectly, magnetic shielding rooms just weaken the a magnitudes by few orders of magnitudes at best. This is because the relative permeability of magnetic materials is finite and usually less than 10,000, unlike conductors with relative conductivity of infinity.

Also due to non-linearity of magnetic materials, the permeability decreases as the applied filed increase. In strong fields, Magnetic Shielding Rooms are much less effective. The effectiveness depends on the geometry and the configuration too. In general the calculations requires a software which solves Maxwell equation numerically. I have heard of multi-layer shielding rooms giving high shielding factors.
P: 409
 Quote by strokebow Hi, If I run a current carrying wire through a tube of iron will it shield the magnetic field from outside the tube? cheers
No it doesn't. According to Amper's law, the magnitude of the magnetic flux density at a pint in the air with distance r from the pile ( wire) is still equal to: $\frac{μ_{0}I}{2\pi r}$
P: 101
 Quote by Hassan2 No it doesn't. According to Amper's law, the magnitude of the magnetic flux density at a pint in the air with distance r from the pile ( wire) is still equal to: $\frac{μ_{0}I}{2\pi r}$
So clearly the strength of the field reduces with distance from the wire (r). But are you saying the the iron tubing has no effect at all?

The tubing has no influence in the equation. But you said before that the shielding can reduce the field strength by some orders of magnitude.
P: 409
 Quote by strokebow thanks for reply. So clearly the strength of the field reduces with distance from the wire (r). But are you saying the the iron tubing has no effect at all? The tubing has no influence in the equation. But you said before that the shielding can reduce the field strength by some orders of magnitude.
The field strength decreases with distance with or without the tube. This is due not the effect of the tube.

As I said it depends on the geometry and configuration. In case of a straight current-carrying cylinder, the field lines are like circles around the cylinder. They are closed on themselves in the same material. They can't choose between tow materials so that they select the one with higher permeability. In case of an iron hollow cylinder ( with a finite wall thickness), with a limited length inside a solenoid, the situation is different. The field lines come from air, face two choices; the wall with high permeability and the air inside the cylinder and perhaps outside the box, they prefer the wall, so the concentration in the wall becomes higher than in the air. assuming a constant flux, the concentration in the air becomes lower.
 P: 101 thanks But why is it that they prefer the higher permeability material? What property of magnetic fields makes them prefer the higher permeability?
P: 409
 Quote by strokebow thanks But why is it that they prefer the higher permeability material? What property of magnetic fields makes them prefer the higher permeability?

You're welcome.

That was just the rule of thumb to have an idea how the field is effected when a ferromagnetic material like iron is placed in it. Of course it can be explained by mathematics and physics.

Physical description: a ferromagnetic ( also called magnetic) material when magnetized acts like a piece permanent magnet, or equivalently like a coil. This is due to alignment of atomic magnetic dipoles of the material. The polarity is roughly in the direction of the field. The field inside the material is increased then (B=μ0H+M) . On the other hand, the field lines of this piece of magnet leave one end and return back to the other end in the surrounding air as well as in the cavities if any, to close on themselves. This way, they cancel the nearby field greatly. The results is that the field lines concentrate in the material.

In case of the iron tube, The field lines of the wire are circles and the tube is magnetized along the field lines. It's like a magnet with its north pole touching its south pole. The field lines just close there and don't enter the air. They don't affect the surrounding fields then.

I hope that helps.
 P: 101 Thanks hassan. I appreciate you taking time out to answer me but I didnt quite get what you meant. I get you on the molecular level. What is 'M'? I didn't get the rest of what you said. Sorry. Is there anyway you could explain it differently? thanks
 Sci Advisor PF Gold P: 2,063 Place the return wire next to the supply and twist them. The twist depends on the wire size, but for 10A (14 AWG), around 1-2 turns per inch will suffice to dramatically reduce the magnetic fields, and magnetic coupling to other circuits.
P: 409
 Quote by strokebow Thanks hassan. I appreciate you taking time out to answer me but I didnt quite get what you meant. I get you on the molecular level. What is 'M'? I didn't get the rest of what you said. Sorry. Is there anyway you could explain it differently? thanks

M is the magnetization density ( also called magnetization). It's in fact the volume-averaged of electron magnetic dipoles moments ;

An electrons have two kind of magnetic dipole moments ,spin moment and orbital moment. In iron, and other ferromagnetic moments, spin moments matter the most.

M=$\frac{\sum\mu_{k}}{\Delta V}$

$\mu_{k}$ : electron magnetic moment ( it is a vector and has direction)
$\Delta V$ : volume

For iron, In a zero applied field, the average of these atomic magnetic dipoles ( like tiny permanent magnets) is near zero. Under and applied field, this M is is almost proportional to the applied field. A larger M means more tiny magnets are parallel, so they create a stronger field. This field is added to the applied field. A permanent magnet is nothing but a ferromagnetic material with almost all its spin moments parallel, and they remain parallel even without any applied field.

Back to the shielding phenomena,

Suppose we have a uniform magnetic field in the room, in z- direction. first, If we put a solid cylinder made of Iron in the center of the room, with its with its axis in z-direction. What would happens to the uniform field?

Answer: The field wont be uniform anymore, specially iron the iron cylinder. The field lines are more concentrated in the iron.

Why?...

In order to understand this, replace the iron with a permanent magnet of the same shape and magnetization. The magnet has its own field and the field is superposed with the applied. The field inside the magnet is the magnet's own field plus the applied field. Around the magnet , since the magnet's field has different direction with the applied field, tit attenuates the applied field. This is equal to saying the applied field lines have accumulated in the cylinder as if their way has become narrower to the cylinder. An of course have the to leave the cylinder from the other end and spread again at farther distances.

If the cylinder has a void inside, the field inside the void is substantially weakened by the magnet's field for the same reason.

Sorry for the long explanation. I hope things are more clear now.
 P: 2,464 I think this is similar to what marcusl is saying, Could we just run a wire carrying equal current in the opposite direction so the current enclosed is zero.
PF Gold
P: 436
 Quote by strokebow Thanks. Why a coil of wire? I understand that using a coil of wire will give me a B field in a given direction but if I just use the wire in one direction I will get a 'circular' B field. But why should that matter when I am attempting to shield the field?
That's because if u get a good field u can test the shielding even better..so that it is detectable.

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