# 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?)

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: $$w_{cycle} =\int_{C} H\ dB$$
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.