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Does a charged rod or a shell have a magnetic potential energy when it is kept in a uniform magnetic field?

Also, can anyone quantitatively explain why a wire loop carrying current would tend to be circular?

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The magnetic moment of a dipole (in this case, the loop) is given by $$ \mu=I A $$ where $I$ is the current carried by the wire, and $A$ its area.

The potential energy of a dipole, when in presence of a magnetic field $B$ is $$ W=-\mu B \cos\theta $$ where $\theta$ is the angle between the direction normal to the wire and the direction of the magnetic field. In this case, we are taking $\theta=0$ (because the wire is perpendicular to the magnetic field), so $W=-IAB$.

With this, you can see that $W$ is minimised when $A$ is maximised.

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  • $\begingroup$ So, does the area of the wire automatically increase? like,on its own? I thought only torques were produced... $\endgroup$
    – Quark
    Mar 2, 2016 at 20:17
  • $\begingroup$ 1) Yes: the loop gets stretched out into a circle. I've been trying to find a video on youtube, but with no luck, sorry 2) It's not just on its own: it's the effect of the magnetic force, which means the loop gets "pulled" by this force. 3) only torques are produced on circular loops. If the wire is not a loop, first it gets stretched out into a circle, and then it feels a torque. Hope this answers your question :) $\endgroup$ Mar 2, 2016 at 20:25
  • $\begingroup$ If I prove this by Lorentz Force on the wire, would it be just? $\endgroup$
    – Quark
    Mar 2, 2016 at 20:29
  • $\begingroup$ yes, but I believe its easiest to prove using energy arguments (i.e., to prove that a circle maximeses area, and therefore mininises potential energy). If you manage to prove this using the Lorentz Force, post it here: I would like to see it! (though I fear its not easy to do so...) $\endgroup$ Mar 2, 2016 at 20:46
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    $\begingroup$ It would be tedious to integrate and find out by equations using Lorentz Force and also, you would have to have the shape or some initial conditions about the shape to start off with, yours is still a better explanation though, Thanks. $\endgroup$
    – Quark
    Mar 2, 2016 at 21:09

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