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In Ampere's law: $$ \oint \vec B\cdot d\vec \ell = \mu I $$ the current in the equation is the current which passes through the open surface. If a surface is chosen such that on the outer region of surface too current flows. That current too contributes to magnetic field at a point on that surface. But why we consider only the inner current? Or is it that Ampere's law find magnetic field only due to inner current?enter image description here

BioPhysicist
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GRAVITON PI
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    Do you understand the analogous case with Gauss' law (say, for finding the field inside a uniformly charged sphere at some distance $r < R$?) – jacob1729 Dec 04 '20 at 14:42

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(a) You are right about the field at P being due to currents both inside your surface and outside it.

(b) But Ampère's law isn't about the field at P but about the line integral of the field around the edge of your open surface. It is this line integral that is independent of currents that don't pass through your surface. Here are two arguments that might help you see why..

• Imagine a straight current-carrying wire running past your open surface, but outside it. Around different arcs of the line bounding your surface this wire's field will make contributions to the line integral that turn out to be equal and opposite.

• There is a good analogy with Gauss's law in electrostatics: The surface integral of electric field strength, that is the electric flux, leaving a closed surface is proportional to the charge inside the surface and independent of charge outside it.

(c) But you might say: Surely we do use Ampère's law to find the magnetic field strength at points like P on the edge of an open surface. Yes, if the current density inside the surface has enough symmetry, we do get information about the field at P. But we only get the field due to currents passing through the surface. We don't get information about the contribution to the field at P made by any currents that don't pass through the surface.

Philip Wood
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  • What I don't quite understand is, that why, in the straight wire example, is $\oint \vec{B}\cdot \mathrm d l=0$? That is, how do I prove that? The angles between the entering $\vec{B}$ and $\mathrm dl$ and outgoing $\vec{B} $ and $\mathrm d l$ aren't equal. although $|\vec{B}|$ is definitely equal as it lies on the circular path. – V.G Mar 31 '21 at 02:49
  • @Light Yagami (a) Does "in the straight wire example" refer to my first bullet [•]? (b) What starting point would you accept on which to build a proof? A formula for $\vec B$ due to a long straight wire? – Philip Wood Mar 31 '21 at 12:20
  • Yes. Assuming we know $\vec{B}=\dfrac{\mu_0 i}{2\pi r}$ at a distance $r$ from the axis of the wire. Now, we draw any random path outside it. We see that number of field lines entering it is equal to the number leaving it. We note $\vec{B}~ \cdot ~ \mathrm d l$ for each point, but the angles between $\mathrm d l $ and $\vec{B}$ are changing randomly, so I am not able to show that the line integral would be $0$. – V.G Mar 31 '21 at 13:41
  • Suppose that the long wire is at right angles to the plane of the loop, and passes through part of the plane outside the loop. Locate points on the loop using polar co-ordinates centred on the point on the wire that passes through the plane. Any small displacement $\vec{ds}$ is $\vec{rd\theta} + \vec{dr}$. The wire's field is $\frac{\mu_0 I}{2\pi r} \frac {\vec{rd\theta}}{rd\theta}$, so the dot product $\vec B.\vec{ds} = \frac{\mu_0 I\ d\theta}{2\pi}$. Integratng round the loop, the change in $\theta$ is zero, so the integral is zero. – Philip Wood Mar 31 '21 at 20:01
  • @Light Yagami I replied to your last comment (see above). – Philip Wood Mar 31 '21 at 20:43
  • Thank you. Now I got it. One thing, is change in $\theta$ $0$ because we end up at the same point? And instead if, wire was inside the loop, then change in $\theta$ would be $2\pi$? – V.G Apr 01 '21 at 03:20
  • @Light Yagami That's exactly right. Glad that you're convinced. – Philip Wood Apr 01 '21 at 07:09