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This was a question I was wondering about, when I solved the similar problem of a square loop near a long wire.

Q A long wire carries current $I_0$ and lies in the plane of a circular current-carrying loop of radius $R$. The loop carries current $i$ and its centre is at a distance of $a$ ($a>R$) from the wire. What is the magnitude of magnetic force exerted on the loop by the wire?

My attempt

I worked on the question on my own, and this is what I got.

  • If the magnetic field were uniform, there would be no force on the loop, as it is a closed figure.

  • However, that isn't the case here. So, I proceeded via integration, but the solution was nowhere in sight.

  • Since $\vec{dF}=i(\vec{dl}×\vec{B})$, and based on the conditions given, I figured out that the magnetic force on any current element must be perpendicular to it, and in the plane of the coil.

  • Also, the net magnetic force on two symmetrically placed current elements must be perpendicular to the long wire, either directed towards it or away, depending on the direction of current, still in the plane of the coil.

  • However, the integration proved to be too problematic, and I can't include it here without a figure. Anyway, I wasn't able to obtain the final answer.

Could someone please help?

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Did you get$$F=\frac{\mu_0 I_1 I_2}{2 \pi} \int_{0}^{2 \pi} \frac{sin \theta\ d \theta}{\frac{a}{R} + sin \theta}$$in which $\theta$ is the angle around the coil measured from a radius parallel to the straight wire?

The wonderful Wolfram integration site gives the indefinite integral - quite a formidable looking expression. If I didn't want to cheat I'd expand $(\frac{a}{R}+sin\ \theta)^{-1}\ $ binomially for the case when $\frac{a}{R}>>1$.

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  • $\begingroup$ I need to work on it, and your answer definitely serves as a useful reference point for where to go. Thanks, I've accepted and upvoted your answer! $\endgroup$ – Harry Weasley Feb 28 '18 at 10:36
  • $\begingroup$ By the way, did you mean chords parallel to the wire? There can only be two radii parallel to the wire. $\endgroup$ – Harry Weasley Feb 28 '18 at 10:37
  • $\begingroup$ Thank you. You have checked my formula, haven't you? And I meant one of those two radii ! That's alright, isn't it? $\theta$ is measured around the circle, so it would't make sense to start from a chord. $\endgroup$ – Philip Wood Feb 28 '18 at 11:26
  • $\begingroup$ Got it now! :) I was thinking of angles subtended by chords parallel to the wire, at the centre, but now I understand. $\endgroup$ – Harry Weasley Feb 28 '18 at 12:33

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