Area in hysteresis loop = energy lost? Experimenting with ferromagnetics in college, the following formula and line appeared in the lab manual:
"The energy loss per unit volume of material per cycle round the loop is $\int HdB $, i.e. the area enclosed by the B vs H plot."
Where does this come from? I know that the magnetic energy density is defined as $\rho = \frac{1}{2} H \cdot B $, so why isn't there a factor of 1/2 missing in the integral? (Also, isn't the integral strictly taken zero, considering that the loop is symmetric about O?)
 A: Consider the simple ferromagnetic system shown below. The instantaneous power consumed by the system at any time $t$ is simply:
$$p(t) = v(t)i(t)$$
Where $v(t)$ and $i(t)$ are simply the voltage and current in the terminal of our coil. Now the total energy dissipated from time $t_A$ to $t_B$ in the system is:
$$W_{A-B} = \int_{t_A}^{t_B} p(t)dt =\int_{t_A}^{t_B} v(t)i(t)dt$$
But from Faraday's law, we know that $v(t) = N  \frac{d\phi(t)}{dt}$ where $\phi(t)$ is the magnetic flux through the cross section of the system. Thus:
$$W_{A-B} =N \int_{t_A}^{t_B} \frac{d\phi(t)}{dt}i(t)dt=N \int_{A}^{B} id\phi$$
On the other hand, $\phi =\int_s\mathbf B.d\mathbf a\simeq BA$ (the magnetic field is taken to be uniform inside the ferromagnetic material because of its high permeability). Also, from Ampere's law $Ni=\oint_c\mathbf H.d\mathbf x \simeq H l$. Plugging   These into the expression for the energy, we get:
$$W_{A-B} =N \int_{A}^{B} \frac {l}{N} H A \ dB= lA \int_{A}^{B}  H\ dB = V\int_{A}^{B}  H\ dB$$
Thus, the energy consumption per unit volume, when the system travels from point A to point B in its hysteresis curve is just the following integral, which is just the area under the H-B curve from point A to point B.
$$w_{A-B} = \frac {W_{A-B}}V=\int_{A}^{B}  H\ dB$$
Now when the system undergoes a full cycle of its hysteresis loop, the negative parts of the integral cancel the positive ones and this just becomes an integral describing the area inside the hysteresis curve:
$$w_{cycle} =\int_{C}  H\ dB$$


Edit: Generalization
As a response to Prasanna's comment, you can easily generalize this relation for a non-uniform $\mathbf H$ and $\mathbf B$, although this is quite uncommon. The above approximations are standard in analyzing magnetic circuits because of the very high permeability ($\mu_r \sim 10^5$) of the used ferromagnetic materials (see for example these notes from MIT).
The generalization goes as follows:
$$W_{A-B} = \int_{t_A}^{t_B} p(t)dt =\int_{t_A}^{t_B} v(t)i(t)dt$$
$$        = \int_{t_A}^{t_B} dt N \frac{d\phi(t)}{dt}i(t)$$
Now consider some curve $C$, defined to be an arbitrary $\mathbf B$-field line inside the material. Ampere's law along this curve tells us that $Ni = \oint_C d\mathbf x \cdot \mathbf H = \oint_C dl H$, where the scalar product is simply the product of magnitudes because $C$ is a field line (I'm assuming that the material is isotropic, such that $\mathbf H$ and $\mathbf B$ have the same field lines). On the other hand, consider a cross section of the material $S$, such that $S$ is normal to the field lines at every point (i.e. $C\perp S$). The flux is simply $\phi=\int_S d\mathbf a \cdot \mathbf B = \int_S da B$. Plugging these into our relation for the energy gives:
$$W_{A-B}  = \int_{t_A}^{t_B} dt  \frac{d}{dt}\Big[\int_S da B(\mathbf x,t)\Big]\Big[\oint_C dl H(\mathbf x,t)\Big]$$
Since $\frac{d}{dt}\Big[\int_S da B(\mathbf x,t)\Big]=\int_S da \frac{\partial}{\partial t}B(\mathbf x,t)$ (the Leibniz integral rule), we get:
$$W_{A-B}  =\int_S \oint_C  da dl  \Big[\int_{t_A}^{t_B} dt  \frac{\partial B(\mathbf x,t)}{\partial t} H(\mathbf x,t) \Big]$$
$$=\int_S \oint_C  da dl  \Big[\int_{A}^{B} dB(\mathbf x,t) \ H(\mathbf x,t) \Big]$$
where $dB(\mathbf x,t)$ is understood as the variation of $B$ at constant $\mathbf x$, due to $t$, i.e. $dB=dt \frac{\partial B}{\partial t}$. Finally, since $S \perp C$, $da \times dl$ is simply a volume element $dV$, resulting in:
$$W_{A-B}  = \int_{\Gamma}  dV \ w_{A-B}(\mathbf x)$$
where $\Gamma$ is the volume of the whole ferromagnetic system, and the energy density $w_{A-B}(\mathbf x)$ is:
$$w_{A-B}(\mathbf x) \equiv \int_{A}^{B} dB(\mathbf x,t) \ H(\mathbf x,t)$$
which means we are done.
