3
$\begingroup$

I have just learnt Ampere's Law, useful for calculating the magnetic field in situations having a high degree of symmetry. However, I have some conceptual doubts regarding it:

Before I begin, I would like to point out an important fact regarding Gauss Law, which is usually overlooked. Gauss Law states that the integral of E.dS over a closed surface equals the (charge enclosed)/€0. It is mentioned everywhere that 'E' appearing on the left side should be the NET field due to all charges(inside or outside) the surface. However, even if 'E' only includes the field created by those charges enclosed by the surface, the equation still holds true, though it is not particularly useful in this form. The simple reason behind this is that the total electric flux due to any external charge through the closed surface turns out to be zero. Thus it doesn't make a difference whether we consider their contribution or not. And if we can ignore the flux, we can ignore the field as well IN THE EQUATION. Even then, to make best use of Gauss Law, we include the field contribution by ALL charges.

Having established this, I have a doubt about Ampere's Law. Ampere's Law states that the integral of B.dl along a closed loop equals u0 times the net current that penetrates any surface for which the chosen loop acts as a periphery.

(I) Just like in Gauss Law, is the integral B.dl due to any current NOT piercing the surface always zero?

(II) If this is the case, I have another doubt. Consider a wire of finite length carrying a current I. Now, consider any point P which lies on the perpendicular bisector of this wire, at a distance R from it. The field at this point can be easily calculated using Biot Savart law. But, using Ampere's Law and symmetry arguments, we get an incorrect answer, u0I/2*pi*r, which is what we should get for an infinitely long wire. Fortunately, my book(Halliday and Resnick) has an answer for this: The wire mentioned has to be part of some closed circuit. In presence of this circuit, the integral of B.dl will turn out to be u0I, but B cannot be factored out of the integral. So here is my question. When we say that the integral for a current NOT PIERCING the surface is zero, what kind of currents do we mean? Technically speaking, can't the rest of the circuit be regarded as a current external to the loop? How will the inclusion of the entire circuit make a difference to the value of the integral?

In other words, where should a particular wire or a section of a wire be "positioned" in order to have a non zero contribution to integral of B.dl?

$\endgroup$

closed as too broad by ACuriousMind, user36790, Gert, knzhou, honeste_vivere Jun 27 '16 at 17:34

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Please do not ask unrelated questions in the same post. Ask one focused question per post. Also, use MathJax for formulae and *text* for emphasis instead of capitals. $\endgroup$ – ACuriousMind Jun 26 '16 at 19:44
  • $\begingroup$ This is what you are looking for: en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential. These integrals can be used to describe the electromagnetic fields of a general configuration of moving charges. In the static current case they can be reduced to what you already know. $\endgroup$ – CuriousOne Jun 27 '16 at 3:31
1
$\begingroup$

Just like in Gauss Law, is the integral B.dl due to any current NOT piercing the surface always zero?

Correct

enter image description here

The problem which you face using Ampere's law with a wire of finite length is not to do with the current going through the surface it is to do with the magnetic fields produced at point $X$ on the loop along which the integration is to be done produced by wires $BC$, $CD$ and $AD$ as well as the magnetic field produced by the wire under consideration $AB$.

The finite length wire $AB$ carrying the current is the only one which penetrates the surface but it is part of a complete circuit as shown in the diagram and each part of that circuit produces a magnetic field which must be added at point $X$ to perform the integral.

In other words, where should a particular wire or a section of a wire be "positioned" in order to have a non zero contribution to integral of B.dl?

As far as I know it cannot be done other than by making the lengths of each of the wires very, very long and then you have the magnetic field due to a wire of infinite length.


Please ask your completely unrelated question in another post.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.