
#1
Feb2312, 11:37 PM

P: 260

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 



#2
Feb2412, 09:56 AM

P: 404

You talk abou the additional field, additional to what?
Well, in a ferromagnetic 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. 



#3
Feb2512, 02:02 AM

Sci Advisor
HW Helper
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P: 26,167

hi jd12345!
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_{o}nI but the magnetic moment density is, by definition, M_{o} = nI so B = µ_{o}M_{o} (and H = M_{o}) this shows the relation between magnetic fields measured in tesla (T) and measured in amperemetre (Am) (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)) 



#4
Feb2512, 02:29 AM

P: 404

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



#6
Feb2712, 07:25 AM

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P: 26,167

further thoughts …
µ_{o} is a constant which should be "1" (see bottom)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_{o}n_{i}I_{i} inside each solenoid this is of course the same as µ_{o}M_{i}, where M_{i} is the magnetic moment density of each solenoid, measured in ampturns per metre (magnetic moment per turn = I_{i}A, where A is the crosssection area of that turn; so magnetic moment per volume = I_{i}A times turns per volume = I_{i} times turns per length = I_{i}n_{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 fieldB 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 ampturns per metre, or Aturn/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/A^{2} (newtons per amp per ampturn) why isn't this unit "1" ? 



#7
Feb2712, 10:27 AM

P: 260

fair enough  thank you




#8
Feb2712, 11:31 AM

P: 404

Thanks tinytim. Now I understand why a divergencefree field is called a solenoidal field.



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