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Why do we consider a circle as a closed loop while deriving the magnetic field of a infinitely long wire. How does it affect if I take another shape.

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  • $\begingroup$ Are you asking why we choose a circular (constant radius) path vs an ellipse, or why we choose a closed path? $\endgroup$ – Bill N Mar 29 '18 at 12:48
  • $\begingroup$ @BillN this is not the main question but I have problem in which you mentioned also $\endgroup$ – user190600 Mar 29 '18 at 15:03
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Because of the symmetry of the problem, we know that the magnitude of the magnetic field is dependent only on the distance from the wire (because the wire looks the same from any angle and after any displacement parallel to the wire). Therefore, the magnitude of the magnetic field should be constant along a circle with the wire at its center. This makes the line integral present in Ampere's Law very simple; for a circle of radius $r$,

$$\oint \vec{B}\cdot d\vec{\ell}=2\pi rB$$

Circles are considered closed loops because we are doing a line integral over a one-dimensional path (as opposed to in Gauss's law, in which we do a surface integral over a two-dimensional boundary). Any other shape would still be a perfectly valid choice for Ampere's Law calculations (so long as it formed a closed loop), but in almost every other case, doing the line integral is very difficult without basically knowing the magnetic field already, so we use circles because they take advantage of the symmetry of the problem.

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  • $\begingroup$ Thanks for helping. I have a request for you. Please answer my another question "Why do we consider a rectangle as a closed loop while deriving the magnetic field of a solenoid. I think I could have any other shape. How does it affect if I take another shape?" $\endgroup$ – user190600 Mar 28 '18 at 15:27
  • $\begingroup$ I have posted the above question. $\endgroup$ – user190600 Mar 28 '18 at 15:29
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    $\begingroup$ The answer is basically the same as the answer here. The rectangle is the shape that best takes advantage of the symmetry of the problem (because the line integral vanishes along three of the four sides of the rectangle, and the field is constant along the remaining side). Any other shape will also work, assuming you can do the corresponding line integral (but you almost never can). $\endgroup$ – probably_someone Mar 28 '18 at 15:29