How can we apply Ampere's circuital law in a wire to calculate magnetic field around a straight current carrying wire? The length of the wire is not infinite.

Using the Biot-Savart law we get

$$B = \frac{\mu_0 I}{4\pi r}(\sin\alpha + \sin \beta)$$

Using the Ampère law, $$\oint \boldsymbol B\cdot \mathrm d\boldsymbol l = \mu_0 I,$$ I couldn't get it.

For symmetry let's try it for equal $\alpha = \beta$

Edit: I understood that for applying ampere circuital law, we require symmetry along entire length of wire and not at ant point.

My new doubt is how come Maxwell used this law to find displacement current ? How the situation included symmetry there?

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    $\begingroup$ Ampere's Law is not applicable in the way you have applied it. The system does not have enough symmetry. Alternatively, note that you can't have a current appear out of nothing and the disappear into nothing. The Biot-Savart law doesn't suffer from these problems. $\endgroup$
    – garyp
    Commented Mar 4, 2016 at 18:57
  • $\begingroup$ @garyp Then let's consider at centre of wire where its bilateral symmetry $\endgroup$ Commented Mar 4, 2016 at 19:01
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    $\begingroup$ Amplere's law gives the wrong answer in this situation. See this: physics.stackexchange.com/questions/14078/… $\endgroup$ Commented Mar 4, 2016 at 19:42
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    $\begingroup$ @Anubhav Goel: garyp is right, there is not enough symmetry here. In fact you lack translational symmetry because your wire is not infinite. To understand why Ampère's law doesn't work in this case go through its proof from Maxwell's equations. $\endgroup$
    – Dimitri
    Commented Mar 8, 2016 at 14:12
  • $\begingroup$ @Dimitri How did maxwell proved displacement current? Wasn't their poor symmetry around capacitor $\endgroup$ Commented Mar 9, 2016 at 9:16

2 Answers 2


First of all I would like to give you an answer to what the magnetic field will be for an infinitely long straight current carrying wire.

Magnetic field due to straight current carrying wire (infinite length)

Consider a wire of infinite length, carrying current $I$. The magnetic field strength due to that wire at a some point $P$ situated at distance $r$ from the wire can be calculated as follows:

From Ampere's Law,

$$\oint \boldsymbol B\cdot \mathrm d\boldsymbol l = \mu_0 I,$$

Where $\oint \boldsymbol B\cdot \mathrm d\boldsymbol l$ = Line integral of magnetic field along circular path. As angle between the vector $B$ and $\mathrm d\boldsymbol l$ is $0^0$,

$$\oint \boldsymbol B\cdot \mathrm d\boldsymbol l = \oint \boldsymbol B\cdot \mathrm d\boldsymbol l \cos0 = \boldsymbol B\oint \mathrm d\boldsymbol l$$

But $\oint \mathrm d\boldsymbol l = \boldsymbol {2\pi r}$ (Circumfrence of the circular path of radius $\boldsymbol r$)

$$\oint \boldsymbol B\cdot \mathrm d\boldsymbol l=\boldsymbol B \times \boldsymbol {2\pi r}$$

But $\boldsymbol B \times \boldsymbol {2\pi r}=\boldsymbol \mu_0$ thus,

$$\boldsymbol B = \frac{\mu_0 I}{2\pi r}= \frac {\mu_0}{4\pi} \frac {2I}{r}$$


Explanation (Why can't you use Ampere's law for a finite length wire):

What you have to understand in the above case is that the assumption of an infinite wire means that the Magnetic field will have the same magnitude (at distance $\boldsymbol r$ from the wire) at any point parallel to the axis of the wire.

In case of a finite wire, the Magnetic field will vary in strength depending on how far from the ends of the wire the point in space is, and its direction is no longer exactly parallel with the circle drawn around the wire.

In such circumstances it is more ideal to use the Biot-Savart law instead of Ampere's law

  • $\begingroup$ @AccidentalFourierTransform Sorry i didn't complete my answer, i accidentally posted without completing it :p $\endgroup$
    – hxri
    Commented Mar 10, 2016 at 9:40
  • $\begingroup$ "its direction is no longer exactly parallel with the circle drawn around the wire." I had considered the central point of wire. At this point, the perpendicular components would cancel, so field would be exactly parallel to it. $\endgroup$ Commented Mar 11, 2016 at 6:42
  • $\begingroup$ Please refer my new edit. $\endgroup$ Commented Mar 11, 2016 at 6:48

In fact, the problem in applying ampere's circuital law for a finite wire doesn't lie in complexity raising from absence of symmetry.The problem lies in the fact that the current 'I' appearing in ampere's law is a one forming a complete, closed circuit,not a portion of a circuit ( you can't apply it for a finite wire, a semi-loop ... ).

Ampere's law is generally valid for closed circuits, and the infinite wire is a special case. To understand this better, search for derivation of ampere's law.


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