0
$\begingroup$

I am just starting to learn Electromagnetism and I am a bit confused about the idea that we need to attach a surface when evaluating the equation for Ampère's Law.

I am not talking about the 'why' in Math. I am talking in principal (Physics), why do we need that surface at all and what does it represent?

$\endgroup$
1
  • $\begingroup$ What surface? ${\rm curl}{\bf H}= {\bf J}$ requires no surface. Perhaps you are thinking of the integral form using Stokes' theorem? $\endgroup$
    – mike stone
    Commented Jun 19, 2020 at 19:24

2 Answers 2

1
$\begingroup$

We most certainly do not 'attach' a surface when evaluating a magnetic field using Ampère's Law.

I think what you are referring to is the imaginary surface, which is called an 'Ampèrian Loop'.

An Ampèrian Loop is an imaginary, closed surface that is assumed to exist, through which the electric current passes. It is usually selected so as to make the calculation of the magnetic field 'simple', so it has a certain element of symmetry to it and thus the choice of an Ampèrian Loop is crucial in order to find the magnetic field. It is defined by:

In Integral Form:

$$\oint_C \mathbf{B}.d\mathbf{l}=\mu_o\iint_S \mathbf {J}.d\mathbf{S}=\mu_oI_{enc.}$$

In Differential Form:

$$\nabla \times \mathbf{B}=\mu_o \mathbf{J}$$

where, $J$ is the current density

$\int_C$ is the closed line integral over a curve C

$\iint_S$ is surface integral over a surface S enclosed by C

Cheers!

$\endgroup$
4
  • $\begingroup$ Forgive me for any dump questions I am just starting but I need more explanation on how it make the calculation easier $\endgroup$ Commented Jun 19, 2020 at 21:32
  • $\begingroup$ What I mean by 'calculation easier' is that, if we make use of the symmetry of the magnetic field lines of the object, then an Ampèrian loop (which is selected by keeping in mind that symmetry) will be much more suitable than a random Ampèrian loop with no correlation to the field lines. For example, if there is an infinite length of wire, then we know that the field lines form concentric circles, so an appropriate choice for an Ampèrian loop will be a circle and not a prism. $\endgroup$ Commented Jun 19, 2020 at 21:40
  • $\begingroup$ I get that. but one more thing the imaginary open surface of that Loop sometimes takes weird shapes like the one we use in proving that there must be a displacement current between plates of a capacitor it looks like that our Amperian loop surface has somewhat stretched how is that not affecting it's symmetry ? google.com/… $\endgroup$ Commented Jun 21, 2020 at 18:43
  • $\begingroup$ Well, the reason for that is the Stokes Theorem. The Stokes Theorem is basically valid for any random surface. So, the line integral can be calculated for any path. So, the symmetry issue will not arise. The integral form of the Ampère's Law will be easy to apply for simple geometries. In more complicated cases, it's always a better idea to apply the differential form of the Ampère's Law. $\endgroup$ Commented Jun 22, 2020 at 7:12
1
$\begingroup$

To apply the integral form of Ampere's law, you need to choose an oriented surface (i.e. a surface with a choice of normal vector direction), simply because the law is stated in terms of this surface $S$ and its boundary $\partial S$ (which is a closed curve). At steady state: $$\oint\limits_{\partial S} \vec{H}·d\vec{\ell} = \iint\limits_S\vec{J}·d\vec{S} .$$

The right hand side is the current through the surface $S$. When we speak of current, we always speak (sometimes implicitly) of the current flowing through some oriented surface.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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