magnetic field inside a material

jd12345

IF a material say a ferromagnet is placed in a magnetic field Bo then there will be an additional field due to ferromagnet which is equal to uoM (M is magnetisation)

My doubt is how is the additional field equal to uoM. By considering the units of both this expression is indeed equal to the magnetic field but still i'm not convinced

M is magnetisation which is total dipole moment in the material divided by volume. And if we multiply it by uo we get the magnetic field produced by it?? I'm very unconvinced by this. I just cant belive stuff with no proofs
So plz help me - give me an intuitive proof or something so that my mind is convinced

Hassan2

You talk abou the additional field, additional to what?

Well, in a ferro-magnetic material, B=u0(H+M) , where H=Hd+Ha. Hd is called the demagnetizing field and Ha is the applied field. The difference is then u0(Hd+M), which is less than B=u0M because Hd has a negative effect on B. Hd depends on the geometry of the material and on M. For a uniformly magnetized sphere, Hd=-1/3*M

In order to understand Hd, imagine numerous magnet bars ( in fact, the magnetic dipoles) parallel and packed, with the same polarity. The field lines of each bar closes through the fields of other magnets and attenuates them. We can reduce this Hd to zero by making a closed magnetic circuit with no gap.

tiny-tim

hi jd12345!
see http://hyperphysics.phy-astr.gsu.edu.../solenoid.html for a visual demonstration of the connection between magnetic field and magnetic (dipole) moment …

the magnetic field in empty space inside an ordinary electric solenoid (straight and long, or circular) with current I, and n turns per metre, is found experimentally to be B = µ_onI

but the magnetic moment density is, by definition, M_o = nI


this shows the relation between magnetic fields measured in tesla (T) and measured in ampere-metre (A-m)

(and if you put a magnetic core inside the solenoid, it will have induced magnetic moment density M, and the field will now be B = µ_o(M_o + M))

Hassan2

Sorry , I believe your last formula is incorrect except for an infinitely long core.

If you puts a cylinder core with length L inside an infinitely long solenoid, the B-field in the core is less than if you put a a cylinder with with length 2L. Your equation is not compatible with this fact.

tiny-tim

yes, i said straight and long

tiny-tim

further thoughts …

µ_o is a constant which should be "1" (see bottom)

the magnetic field is measured by its effect (via the Lorentz force), the magnetisation field is measured by its cause (induction by an actual or a theoretical system of currents in loops) …

if µ_o was "1", that link between cause and effect wouldn't be surprising

let's see a "visual" proof of that link:

any magnetic field can be replaced by an identical solenoidal field, as follows:

a solenoidal field is a region R of space with a "honeycomb" of thin hexagonal solenoids (they needn't be hexagonal: but that makes them fit nicely ), each with a (different) current I_i, and a (different) pitch, n_i (pitch is turns per length)

the solenoids aren't straight, they can be curved into any shape

that causes ("induces") the whole region R to be filled with a magnetic field, of "Lorentzian" strength µ_on_iI_i inside each solenoid

this is of course the same as µ_oM_i, where M_i is the magnetic moment density of each solenoid, measured in amp-turns per metre

(magnetic moment per turn = I_iA, where A is the cross-section area of that turn; so magnetic moment per volume = I_iA times turns per volume = I_i times turns per length = I_in_i = M_i)

now consider any B field in any region R

we can fill R with an imaginary honeycomb of solenoids whose sides follow the B field lines (ie lines of constant |B|, and whose tangent at each point is parallel to the B field at that point), and whose current or pitch (or both) are adjusted so that the solenoidal field equals the B field along the centre line of each solenoid …

and by making the number of solenoids large enough (ie, the diameters small enough), we can make the solenoidal field match the whole B field to any required degree of accuracy

in other words: in the limit, any actual B field can be replaced by a purely solenoidal field

B fields are naturally measured by their effect, in units of force per charge per speed (N.s/C.m = N/A.m = tesla)

(we could call these either "Lorentzian units" (named after the Lorentz force q(E + v x B)), or "Laplacian units" (named after the Laplace force qv x B))

solenoidal fields are naturally measured by their cause, in units of magnetic moment density, measured in amp-turns per metre, or A-turn/m (or, in SI units, simply but confusingly A/m)

the conversion ratio between these cause and effect measurements is a universal constant, µ_o, which should be "1", in units of N/A² (newtons per amp per amp-turn)

why isn't this unit "1" ?

well, it would be , buuuut …

i] in SI units, a factor of 4π keeps cropping up! … so we multiply by 4π

ii] that would make the amp that current which in a pair of wires a metre apart would produce a force between them of 2 N/m …

which would make most electrical appliances run on micro-amps! …

so, for practical convenience only, we make µ_o 10⁷ smaller, and the amp 10⁷ larger!

(so the amp is that current which in a pair of wires a metre apart would produce a force between them of 2 10^-7 N/m, and µ_o is 4π 10^-7 N/A² (= 4π 10^-7 H/m))

(for historical detaills, see http://en.wikipedia.org/wiki/Magnetic_constant)

jd12345

fair enough - thank you

Hassan2

Thanks tiny-tim. Now I understand why a divergence-free field is called a solenoidal field.