In case of closed surfaces the area vector is directed outwards the surface. But what is the direction of the area vector in case of an open surface e.g. A thin lamina type of surface. Does it depend upon the curvature of the surface i.e. concave or convex side.
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1$\begingroup$ Related: physics.stackexchange.com/q/14165/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jun 13, 2016 at 15:19
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$\begingroup$ It depends on which direction you take it. $\endgroup$– Soubhadra MaitiCommented Jun 13, 2016 at 18:52
3 Answers
For an open orientable surface there are two possible, equivalent normals: $\vec n$ and $-\vec n$.
The usual convention is that you choose a direction in which the perimeter of the surface is traversed and define the positive direction as the direction given by the right hand rule, as shown in the following picture.
This can also be done for non-simply connected surfaces (i.e. surfaces with holes), as it is shown in the following animation:
But there are also open surfaces which are non-orientable. For such surfaces it is impossible to differentiate between a positive and negative orientation (positive/negative normal), as it should be clear from the following picture, showing a famous non-orientable surface, the Moebius strip.
By convention, for a flat lamina or a plane surface, the area vector is a vector whose magnitude is the area of the surface and whose direction points in a direction perpendicular to the surface.
If you have a curved surface, then you have to consider elemental areas, i.e: small patches of area denoted by $dA$ whose direction is perpendicular to the small patch of area.
Mathematically, you can denote the area vector as
$$\vec{A} = |A| \hat{n}$$
where $|A|$ is the magnitude of the area and $\hat{n}$ is the unit vector whose direction is perpendicular to the patch of area.
The differential form of the above equation can be written as,
$$d\vec{A} = \hat{n}dA$$
Using the above differential equation, you can calculate the vector sum of elemental area vectors to get the area vector for the irregularly curved surface.
Your statement "In case of closed surfaces the area vector is directed outwards the surface" is not correct. The area vector of a closed surface is zero always. But the direction of the area vector at a given point (infinitesimally small area element) on the closed surface will be directed perpendicularly outwards.
The sign of the vector does depend on the type of surface, i.e: convex or concave, depending on your conventions.
They are linked by the "law of the right hand": The preferred direction of dℓ⃗ dℓ→ along the loop is that from the palm to fingertips of your right hand when it surrounds the loop. Then, the associated preferred direction of dA⃗ dA→ is indicated by the thumb.
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$\begingroup$ What's dl in the above explanation ? $\endgroup$ Commented Jun 13, 2016 at 13:47
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$\begingroup$ @Prabhdeep Sing is telling you a trick to find the direction of the area vector for a current loop. If you have a simple current loop, folding your palm in the direction of the flow of the current, the direction in which the thumb points gives you the direction of the area vector. $\endgroup$– YashasCommented Jun 13, 2016 at 14:54
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$\begingroup$ IMO, The $dl$ is an elemental part of the current loop. $\endgroup$– YashasCommented Jun 13, 2016 at 14:55
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$\begingroup$ dl as it means is a small part of the entire length of your loop(l) . You can identify if the current is going clockwise or anticlockwise and thus get the direction of your area vector $\endgroup$ Commented Jun 14, 2016 at 13:29