Why Cant ampere s law be used to calculate magnetic field at a point due to a finite current carrying wire.... (please dun give reason dat magnetic field due to other wires connected to the main wire also are involved and hence d integral in ampere s law can not be calculated ... this is a wrong reason )

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TopNewestIf you are referring to the Ampere's Law given by \(\displaystyle \oint _C \vec{B} . \vec{\text{dl}} = \mu _0 I_{enclosed} \) , then actually the statement is incomplete and a modification has been made to it to rectify the mistake.

Actually, the above statement is interpreted as follows: The line integral of the magnetic field around any arbitrary closed curve \( C \) equals \(\displaystyle \mu _0 \) times the current enclosed by the closed curve. Now how do you define the current enclosed by the closed curve? So, it is interpreted in a very simple manner as the current passing though any arbitrary surface \(S\) which is bound by the closed curve along which the line integral is computed. It is intuitive that if a closed curve bounds a surface then the surface cannot enclose a three dimensional space.

Perhaps a very easy to understand analogy for a surface bound by a closed curve is a pot. Imagine a pot which we use to store water. Basically the exterior surface of the pot ( basically its body) is the surface \(S \) and the circumference of its mouth is the closed curve which bounds the surface. Now it is also easy to realize that the pot does not actually enclosed a three dimensional volume.

Continuing with the analogy of a pot, another very important thing to realize is the current that is involved in the equation i.e., \(\displaystyle I_{enclosed} \). Let us first consider only a single current. By definition, the current which is to be taken should satisfy the fact that it enters through the mouth of the pot (i.e., intercepts the plane of the closed curve ) and pierces the pot

at any point on the surface and comes out. (The notation of entry through the mouth must interpreted as the interception of the plane of the mouth). The piercing and coming out is extremely essential. And the \(\displaystyle I_{enclosed} \) is the algebraic sum of all such currents.So why it fails with a finite wire? I believe the reason is now pretty obvious. Say the pot is held such that the wire terminates inside the pot and since the choice of \(S \) is completely left upon me, I can design a pot of any shape I like such that the wire terminates inside the pot. So now, by definition \(\displaystyle I_{enclosed} = 0 \) and so is the line integral. which means either the magnetic field is always perpendicular to the \(\displaystyle \vec{\text{dl}} \) which is obviously not the case or the field is zero which also is not possible. Implies, a contradiction, a really big one.

Why it fails is because the law when written in such a manner does not incorporate the possibility of existance of magnetic fields due to time varying electric fields but that is a step ahead.

Hope this helps. – Sudeep Salgia · 2 years, 10 months ago

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– Hemant Khatri · 2 years, 10 months ago

How did u get to knw abt such a concept ? I mean such a cncept in not listed in any of d famous books!!! Am nt sure abt ur ans bt I find dis pretty correct...thanksLog in to reply

Ofcourse , it is not . Maxwell proved it. You can use : \(\displaystyle \int \vec{B}.\vec{\text{dl}} = \mu_{0}(I_{encl.} + i_{d})\) Where \(i_{d} \) is the maxwell displacement current. Clearly, the charges are getting accumulated at ends and this results into change of electric flux, and hence, a displacement current. Apply \(\phi_{E} = \dfrac{q_{1}-q_{2}}{2 \epsilon_{0}}(1 - \cos \theta)\) for the loop,(where \(q_{1}\) and \(q_{2}\) are charges at ends) and differentiate and multiply by \(\epsilon_{0}\) to get displacement current. Putting the value, we get exactly the same \(\vec{B}\) as given by Biot-Savart's Law. – Jatin Yadav · 2 years, 9 months ago

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Ampere's law states that, the closed integral of B over the loop enclosed it equals uI, where u = permeability of the material and I = current "passing" through the loop.

I feel confused, because, should the current be extended to infinity? I mean, when we have an infinite wire of current I, everything works very good; on the other hand, when we have a finite wire, this law FAILs(hm...I think it fails, am I right? please correct my viewpoint)

The above example seems to imply that the current should be extended to infinity.

However, I have another example which shows that the current no need to be extended to infinity!

Let' look at a solenoid, we can calculate the magnetic field B along the central of the solenoid by using Ampere's law, yet the wire (current) is not infinitely long! The method we use is: we construct a rectangular loop that one of the sides lies along the central of the solenoid, 1 side outside the solenoid and parallel to the previos one, and the remaining 2 sides connect the previos 2 sides and perpendicular to them. This example shows that, we can use ampere's law here even though the current is not extended to infinity!

So, should the current extended to infinity so that we can use Ampere's law? Notice that for each side (for and against) there is always a contradiction which is highlightened in the above 2 examples! – Kïñshük Sïñgh · 2 years, 10 months ago

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– Hemant Khatri · 2 years, 10 months ago

Kinshuk I asked some well known teachers on this issue and according to them magnetic field should be constant at all points of space if you want to calculate magnetic field from ampere 's law and this follows for an infinite wire and for magnetic field calculation INSIDE A LONG SOLENOID though am not sure of this but I find it pretty good .....Log in to reply

– Kïñshük Sïñgh · 2 years, 10 months ago

Ya but we can calculate the magnetic field B along the central of the solenoid by using Ampere's law, yet the wire (current) is not INFINITELY LONG!!!!! This is something contradicting and I'm confused about thisLog in to reply