If $\oint_C [q\vec{V} \times \vec{B}] \cdot d\vec{l}=0$, then why $\vec{F}=q\vec{V} \times \vec{B}$ isn't conservative ?
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1it is velocity-dependent. The usual statement that the vanishing of the line-integral for all closed curves implies the existence of a potential holds for position-dependent vector fields only. See this answer for more. – peek-a-boo Sep 28 '23 at 23:51