If there were two charges placed at a large distance, won't their (say) magnetic fields interact? What if that large distance is something like close to infinity? Even if they experience a feeble force from each other, doesn't that imply that the magnetic field lines are not closed, as the particle could have been placed anywhere around the source charge at any distance?

Would not this generally say that every electrically charged particle in the universe is part of an infinitely large magnetic field?

  • 2
    $\begingroup$ Are you aware that there are no magnetic monopoles? So there are no magnetic charges and field lines can only be produced by magnetic dipoles... $\endgroup$
    – Fabian
    Jan 2, 2017 at 17:25
  • $\begingroup$ That's how magnetic fields are observed and detected experimentally. Its their property. There's no why in that. $\endgroup$
    – UKH
    Jan 2, 2017 at 18:03

1 Answer 1


Since monopoles don't exist. What it means is that a north pole can never be extracted independent of a south pole. So independent sources and sinks of lines of force don't exist unlike electrostatics, where a positive charge can be isolated from a negative charge (but a north pole can not be isolated from a south pole). Hence because sources and sinks don't exist, there is no concept of beginning/termination of lines of force (unlike electrostatics wherein lines of force begin in the +ve charge and terminate in the negative charge). The only way a curve can have no beginning or ending is if the curve loops into itself. Hence magnetic lines of forces create closed loops. Hope this helps.

  • $\begingroup$ It is possible for magnetic field lines to neither close nor extend to infinity. See this post and this paper. $\endgroup$ Oct 4, 2021 at 6:35
  • $\begingroup$ @VincentThacker: Absolutely, but what needs to happen is that, it needs to curl such that for any path, $\int_C \vec{F} \cdot \mathrm{d}\vec{r} \neq 0$, else $\vec{F} = \nabla \, f$ and the magnetic vector potential $\vec{F} : \vec{B} = \nabla \times \vec{F}$ will be a conservative field with an embedded scalar potential. Any helical shape of arrow quivers which would revolve its life over to infinity with $\nabla \times \vec{F} \neq \vec{0}$ should work. This closed lines pedagogy is a simplistic one where we can avoid talking about curls and divergences. $\endgroup$ Oct 10, 2021 at 13:33
  • $\begingroup$ I added the comment in response to a suggested edit on your post as I felt that it would be better left as a comment. $\endgroup$ Oct 10, 2021 at 16:00

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