I want to know how much the saturation magnetization is in Ni.Co.Fe2O4 nanoferrite (with superparamagnetic property)?? and How about Ni.Co.Fe2O4 in bulk form??
the magnetic moment of Ni(II) , Co(II) and Fe(III) are usually stated as 3.2 , 5.2 and 5.8 Borh magnetons. SInce Ni and Co prefer the tetrahedral A sites in the spinel structure, and Fe(III) go to octahedral (B) sites, the net magnetic moment per formula unit is the vectorial sum of all (eight) atoms at B sites and those (four) atoms at A sites.
Thing is that, depending on your synthesis method, the inversion degree can be different for the same stoichimetry, yielding different saturation magnetization values.
By the way, the fact that the material is superparamagnetic does not affect the saturation magnetization, since SPM is a state for single-domain particles. Both bulk and SPM nanoparticles will saturate at the same Ms values, the difference is only that SPM will usually require a larger applied field to get saturation, due to the thermal energy effects. Hope this helps.
The value of Ms depends upon cation distribution of contribution ions and the value of Ms in ferrites arises is given by relation Ms= MB -MA. Site preference energy decides the cation distribution in particular ferrite system. The synthesis route as well as sintering temperature affect the cation distribution and hence difference in Ms values is observed in bulk and the nano ferrites. For NiFe2O4 Ms= 56 emu, due to introduction of Co it will change accordingly and the spin glass behavior in nano ferrite will also affect the coupling between A and B site.