I have made a Co/CoO sample of 9 nm in thickness using the in-situ oxidation method by Gruyters et al. After magnetization measurement using SQUID Field Cooled measurement at 10 K gave Ms = 1.48x10-4 emu. The sample size was 0.2x0.3 cm2 and FM Co is 7 nm supposing CoO ~ 2 nm. At 10K exchange bias field = 1990 Oe. When I calculated Ms in emu/cm3 I got 3523 emu/cm3. This is more than the saturation magnetization of colbat which is 1440 emu/cm3. So I got the wrong interficial energy when I use Eex = Ms . Hex . t = 4.8 erg/cm2. Please can anyone help because I don't seem to figure out what I am doing wrongly.
The high values of the interface exchange energy Eex which you obtained mean that the above formula is not correct for very thin films (because it does not account for the AFM layer thickness t_A). The large value of Ms is a result of the exchange bias (EB) shifting of the hysteresis loop (towards negative coercive field Hc). In the presence of the EB with coupling energy J (which, unlike exchange energy Eex, is sensitive to t_A), the coercive fields are assymetric. When J>>K_At_A (where K_A is the anisotropy of the AFM layer), energetically becomes more favorable a situation when AFM and FM spins rotate together. This leads to the shifting of the loop to the left, resulting in a double effect: decrease of the positive coercive field by Hex=J/t_FMs and the corresponding increase of the saturation by Mex=J/t_FHex, where t_F is the thickness of the FM layer. As a result, from the used by you formula, we obtain Eex=Ms Hex t_F=J(t_A) for the overestimated value of the EB interface energy, keeping in mind that J strongly depends on the thickness of the AFM layer. Here is a paper which addresses many practical issues on the subject.
http://www.sciencedirect.com/science/article/pii/S0304885398002662#
Here is a follow-up to my previous answer. First of all, let me clarify the situation about the importance of the AFM layer thickness t_A in the exchange energy Eex. Like I said, only for J>>K_At_A we have a hysteresis loop shift to the left with the resulting increase of the saturation. In the opposite limit, J
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