In the exchange Bias phenomenon, the AFM spins, which are ferro-magnetically aligned to the FM Spins, will exert microscopic torque opposing the change in the direction of magnetism. The same effect will favour the rotation of the spins in switching the direction of magnetism in other directions. So my question is if the torque is of same magnitude how will the increase in coercivity occur? Of-course the shift in hysteresis loop will be there but the how is the coercivity increased??
There are two "types of spins" at the interface which constitute the antiferromagnetic interface magnetization. One species is virtually stationary when the reversal of the ferromagnet takes place and gives rise to the pinning of the ferromagnet resulting in the exchange bias field aka shifting of the FM hysteresis loop. There are also loose spins at the interface, which cannot pin the ferromagnet. They will rotate together with the ferromagnet on reversal. They provide a channel where energy is dissipated (kind of like friction). More dissipated energy means more area under the hysteresis loop (Integral MdH is propo to dissipated energy). Increasing the coercive fields means increasing the area under the loop, means more energy is dissipated on reversal in comparison to a ferromagnetic film not in proximity with the antiferromagnet. There is a nice mean field theory which qualitatively explains the coercivity to be dependent on the susceptibility of the antiferromagnetic interface magnetization. In an ideal Meiklejohn Bean scenario, where the antiferromagnetic spins are rigid, there is no susceptibility and thus no coercivity increase. Only if the antiferromagnetic interface spins (at least a fraction of them) shows a reaction on the rotation of the ferromagnet there is susceptibility and there is coercivity increase.
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