Timeline for Magnetic field appearing in equation for free energy
Current License: CC BY-SA 4.0
7 events
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| Nov 5, 2023 at 12:41 | comment | added | hyportnex | You have misunderstood the meaning of this integral. Here the function is between the two fields as $H=H(B)$, yes they are both space dependent, say $B=B(r)$ and then there is the indirect function as $h(r)=H =H(B(r))$ but there is a direct constitutive relationship between the two fields. Changing one will change the other at the SMAE place by the material relationship between the two. The integral $\int HdB$ is at one fixed location, and the integral $\int dv\int HdB$ is all of those locations together, and when is space homogeneous then it is the latter integral in your question. | |
| Nov 5, 2023 at 12:37 | comment | added | Riemann | I think my confusion comes from the space dependence. So I thought that it implicitly meant $\int_V dr \int_0^{B(r)} H(r) dB$. But that looks strange and simplifies to $\int_Vdr H(r)\cdot B(r)$. However if $H$ and $B$ are considered to be space-independent, then it would be $V\int_0^B H\cdot dB$. Which of the two is true? | |
| Nov 5, 2023 at 12:32 | comment | added | hyportnex | If H were independent of B then the integral would be HB but that cannot happen. What can happen is that they are proportional, $B=\mu H$ and $\mu=const$, as in vacuum where $\mu = \mu_0$, or in a non-magnetic (dielectric) material and then the integral is HB/2. In a paramagnetic material you can have $\mu \approx const$ for sufficiently low H intensity as long as $T=const$. | |
| Nov 5, 2023 at 12:12 | comment | added | Riemann | Is $\int_0^B H\cdot dB$ equal to $H\cdot B$? If not, why? | |
| S Nov 28, 2022 at 21:56 | history | suggested | Schiele | CC BY-SA 4.0 |
fixed LateX typos
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| Nov 28, 2022 at 21:28 | review | Suggested edits | |||
| S Nov 28, 2022 at 21:56 | |||||
| Apr 13, 2017 at 0:00 | history | answered | hyportnex | CC BY-SA 3.0 |