When I was studying about Ampere's circuital law. Then there comes a question in my mind that "whether this law is applicable for finite current carrying wire or not"
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4Just for thoroughness, note that "a finite current carrying wire" s not a complete description of a situations, because either charge is accumulating (meaning an electric field is developing) or there is a return mechanism that you have no mentioned. Both possibilities affect the outcome and should be included in a full calculation of the situation. – dmckee --- ex-moderator kitten Aug 18 '14 at 12:54
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I presume by this you mean the form of the law where
$$\oint {\bf B} \cdot d{\bf l} = \mu_0 I\, ,$$
where the LHS is the path integral around a closed loop and the current $I$ is the current passing through that loop.
This law works whether the wire you are considering is finite or not, BUT, there is a subtlety. The assumption of an infinite wire means that the B-field will have the same magnitude (at some distance $r$ from the wire) at any coordinate parallel to the axis of the wire. It will also always be directed in a way that is parallel to a circle drawn around the wire. These simplifications make the LHS easy to evaluate.
$$\oint {\bf B} \cdot d{\bf l} = |{\bf B}| 2 \pi r \, .$$
If you relax the assumption of an infinite wire then neither simplification is true. The B-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 (except at the centre of the stretch of wire) with a circle drawn around the wire.
In these circumstances it is more usual to use the Biot-Savart law. e.g. see http://www.phys.uri.edu/~gerhard/PHY204/tsl216.pdf
ProfRob
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