0
$\begingroup$

enter image description here

Let a hysteresis loop be as shown in the figure, with $B_0$ the applied external field due to a current $I$ in a toroid, and $B_M$ the additional field due to the ferromagnetic material. Hence we have that \begin{align*} \mathbf{B} = \mathbf{B}_0 + \mathbf{B}_M, \end{align*} where $B$ is the magnitude of the total field.

I got a couple questions now. My textbook says that in one cycle, a lot of energy is transformed to thermal energy (friction) due to the realigning of the magnetic domains. Do they mean magnetic potential energy due to the extern field ($B_0$)? Also, if some energy is transformed to thermal energy, where does the rest go?

Suppose I want to calculate the work done on one cycle, which is (I figure) just the area under the loop. How should I go about doing this? I would just do \begin{align*} W_h = V \oint B_0 \ dB_M, \end{align*} where $V$ is the volume of the ferromagnetic material enclosed. I'm not sure if this is correct though, nor what my integration bounds should be.

Any clarification would be helpful.

$\endgroup$
  • $\begingroup$ The current source maintaining the field $B_0$ works on the magnetic sample. The work is the sum of two terms: thermal energy as the result of the friction between the domain walls and magnetic potential energy that is the realignment of the magnetic dipoles in the field. The enclosed area of the hysteresis curve is the dissipated thermal energy $\endgroup$ – hyportnex Mar 15 '15 at 18:03
1
$\begingroup$

The integral is correct, and gives the area of the hysteresis loop. Integrate from d to b along the right-hand curve. Then subtract the integral from d to b along the left-hand curve.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.