A/m Unit Explained: Magnetic Fields

[ctech4285]

hello,
I can't seem to make sense out of amper/meter...one amp flowing throw a wire with a length of 0.5m would be 2A/m ? how does relate to the magnetic field?

 

[mfb]

In a coil, you can find something like "ampere per meter": With a current of 1 Ampere and 100 windings per meter, you get "100 ampere per meter". In a similar way, a cylinder with constant current density (around the interior), can have the property "ampere per meter". And the magnetic field strength inside just depends on this quantity.

 

[ctech4285]

thank you that made it a little bit clearer, but a solenoid has length width height and copper winding length. does m in (A/m) refer to the length of the copper wire?

 

[mfb]

No, it does not. The dimensions are not relevant, as long as the coil is long compared to its other two dimensions (and even if not, it is just a dimensionless prefactor for the geometry). The length of the copper wire is mainly geometry-related. The relevant quantity is the winding density (with unit 1/m), multiplied by the current.

 

[ctech4285]

density can only exist in two or three dimensional space no?!?

we can have 100 ampere turns in one cubic meter, but in a dimensionless meter??

you say dimensions are irrelevant, but i can not see how you can have density without dimensions...

 

[mfb]

A 1-dimensional density is nothing mysterious. On a highway, you can have 10 cars per kilometer, for example. "Cars per cubic kilometer" is not a useful quantity.

If your coil has a length of 2 meters and 100 windings, you have 50 windings per meter.

 

[ctech4285]

ah ok this makes sense now, magnetic field is proportional to the distance and quantity of charge travel per unit time
thank you for your help

 

[tiny-tim]

hello ctech4285!

when i see "A/m", i always read it as "amp-turns per metre" …

the magnetic field is the amps times the number of turns divided by the length of the solenoid

(or the amps times the pitch)