According to Ampere's Ciruital Law:

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Now consider two straight wires, each carrying current I, one of infinite length and another of finite length l. If you need to find out magnetic field because of each, at a point (X) whose perpendicular distance from wire is d.

You get magnetic field as $\frac{\mu I}{2 \pi d}$. Same for both.


Magnetic field due to infinitely long wire is : $\frac{\mu I}{2 \pi d}$

Magnetic field due to wire of finite length l : $\frac{\mu I (\sin(P)+\sin(Q)) }{2 \pi d}$, where P & Q are the angles subtended at the point by the ends of the wire.

Why are we getting wrong value for using Ampere's circuit law?


1 Answer 1


There are two things to notice here.

  1. You can only make the assignment $\oint \mathbf{B} \cdot dl = 2 \pi d B(d)$ if the situation is radially symmetric.
  2. In the case of a finite wire you either have charge building up at the ends or you have not specified the whole current distribution yet. Question: can you specify a radially symmetric return path and if so do you expected it make up the difference?
  • $\begingroup$ Regarding the finite case is it possible to get the B-field just at the finite end of a semi-infinite wire which extends to infinity only in one direction? I know I should take the charge accumulation at the finite end into account since it will generate a time-dependent E-field, hence a displacement current. Yet I am unaware of the mathematical details since the field of a point charge(assuming there is a point charge generated at the finite end of the wire) is not defined at the origin $r=0$. We need it in the integral to get the displacement current from the current density. $\endgroup$
    – Vesnog
    May 8, 2015 at 10:21
  • $\begingroup$ I suppose that for many cases you could assume that the wire ends on a small but finite-sized spherical capacitor, thus avoiding the need for a singularity. Then choose your surface to not intersect the capacitor. $\endgroup$ May 8, 2015 at 16:28
  • $\begingroup$ Can you check my latest question on this site physics.stackexchange.com/questions/182785/… $\endgroup$
    – Vesnog
    May 8, 2015 at 17:17
  • $\begingroup$ I had recently posted same question but slight variation showing symmetry by considering field at centre, i.e., $\alpha = \beta$ $\endgroup$ Mar 5, 2016 at 9:17

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