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In the Feynman lectures he derives the (mechanical) energy of a dipole

$$ U = -\boldsymbol{\mu}\cdot \boldsymbol{B} + \mathrm{constant}$$

by considering only the torque on it in a uniform field. He then says that by the principle of virtual work one can also use this to evaluate the force on the dipole in a non-uniform field.

This argument seems a bit dodgy to me; the above expression was calculated assuming the position $\boldsymbol{r}$ of the dipole is fixed and only its angle with respect to the field rotates. Thus what Feynman calls $\mathrm{constant}$ could possibly depend on $\boldsymbol{r}$, and it is only by 'luck' that $U$ as given above in fact captures the $\boldsymbol{r}$ dependence too.

Is there some trick I am missing to explain why what Feynman does is allowed?

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  • $\begingroup$ "Is there some trick I am missing to explain..." Isn't this explained in the paragraphs following his statement about virtual work and forces? I.e., the paragraphs starting off with: "We can show for our rectangular loop that $U_{mech}$..." $\endgroup$
    – hft
    Commented Apr 3 at 18:35
  • $\begingroup$ But in other words he had to check explicitly that $U_\mathrm{mech}$ was also the work done in bringing in the dipole? It wouldn't be safe to just assume this to be so? $\endgroup$
    – Ghorbalchov
    Commented Apr 3 at 18:37
  • $\begingroup$ I am asking because my lecturer did just that; derived the force from the torque without further elaboration like Feynman. $\endgroup$
    – Ghorbalchov
    Commented Apr 3 at 18:40
  • $\begingroup$ If you already know the answer then it is safe to assume so... but I think he is trying to explain why this is safe... This seems to be what he is saying in the two sentences surrounding his Eq. 15.11. $\endgroup$
    – hft
    Commented Apr 3 at 18:40
  • $\begingroup$ Hard to say why your lecturer did what they did. $\endgroup$
    – hft
    Commented Apr 3 at 18:42

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Feynman's treatment does seem a bit convoluted. His derivation of equation 15.4 seems to assume that B is constant. Then he derives equation 15.11 by using the spatial variation of B. All of this for a square loop to simplify things. It would be better, but not Feynman's style, to derive the energy directly from energy considerations. Also, many magnetic moments are intrinsic magnetic moments that cannot be derived in classical electromagnetism, but are given by quantum mechanics, starting with the equation, $U=-\mu\cdot{\bf B}$.

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  • $\begingroup$ I also agree it would be better to derive it directly from energy - like what is done for an electric dipole. $\endgroup$
    – Ghorbalchov
    Commented Apr 6 at 17:25

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