JR0008 : Ampere's Law

Guru Vishnu

Feb 22, 2020 09:57

@JohnRennie Sir, in Ampere's law circulation of the dot product of magnetic field and length vector of the current element equals permeability times the total current passing through the loop. I don't understand why we consider the magnetic field due to all current sources irrespective of whether they are inside or outside the loop but only consider the total current of current sources inside the loop on the RHS. Is there any reason behind this, or is this just because Ampere's law is a "law"?

John Rennie

Magnetic field lines are always loops. They cannot begin or end because field lines can only begin or end on a charge and there are no magnetic monopoles.

Since the field lines go round the wire through the loop formed by the wire that means the flux through the inside of the loop has to be the same as the flux that flows through the outside of the loop.

Guru Vishnu

@JohnRennie Is this explanation similar to that of Gauss's law where the flux passing through the Gaussian surface is non-zero for an internal charge but zero for an external charge, sir?

John Rennie

Basically yes. If there are no charges inside the surface all the field lines entering the surface must exit it again because they cannot end inside the surface.

Guru Vishnu

@JohnRennie But it seems, it's different for magnetic field lines due to their difference compared to electric field lines. I think I'm unable to understand your explanation properly, sir.

First thing, the LHS is not a magnetic flux term, instead it's a line intergral.

John Rennie

No, it's the same for both electric and magnetic field lines. It's just that since no magnetic charges exist magnetic field lines cannot begin or end anywhere.

So a magnetic field line can only exist if it forms a loop i.e. has no beginning or end.

This can happen with electric field lines as well. Even when no charge is present a changing magnetic creates an electric field and since no charge is present the electric field lines form loops.

Guru Vishnu

Feb 22, 2020 10:06

@JohnRennie I understand the reason behind Gauss law for magnetism which is similar to your explanation. But is it the same for Ampere's circuital law, sir?

John Rennie

Well no. The surface in Ampere's law isn't closed for one thing.

Guru Vishnu

Or in the first place, did I explain the question properly? Or am I misunderstanding it?

John Rennie

I'm not sure what you are asking.

Guru Vishnu

@JohnRennie Ok sir. I'll explain it in a bit different way. $\int B.dl=\mu_0 I_{int}$.

John Rennie

Yes

Guru Vishnu

Feb 22, 2020 10:09

@JohnRennie The above equation is the Ampere's circuital law. and the integral is circulation. I have to learn how to type it in TeX.

On the LHS, the $B$ term includes magnetic field due to all current carrying wires, whether they are inside the closed loop or outside the loop. Fine sir?

John Rennie

Yes

Guru Vishnu

@JohnRennie: And the $I$ term on the RHS included only the current passing though the inside of the closed loop. It's the total of all currents passing thought the area with proper signs which depend on the direction of flow - whether it's into the plane or outside it? Ok so far, sir?

John Rennie

Yes

Guru Vishnu

@JohnRennie: I'm just asking the reason behind: including all sources of magnetic field for $B$ on the LHS and only the inner sources for the $I$ term on the RHS.

Is this just because Ampere verified it experimentally or has a good explanation for net zero flux as explained above for a different situation?

John Rennie

Let me draw a diagram ...

Guru Vishnu

Feb 22, 2020 10:15

Ok sir.

John Rennie

Guru Vishnu

@JohnRennie Ok sir. Red dot is the wire. Red circles are the corresponding field lines and black circle is the line integral's path.

John Rennie

Hmm, hang on, maybe this won't work ...

I need to draw more field lines ...

I'll have to think about how to make an intuitive argument for this.

The approach I was going to use isn't very convincing.

Guru Vishnu

Ok sir. Then, could you ping me afterwards?

@JohnRennie Or do you know of any Q/A on the main site regarding this. I used the following query before I asked you:

physics.stackexchange.com/…

As you could see it was extremely inefficient and I don't have enough time and patience to read 458 tuples.

John Rennie

4

Q: Ampere's law and external currents

In Ampere's law, the current outside the curve taken is not included in the expression. Does this mean that only the currents crossing the area bounded by the curve taken contribute to the magnetic field one calculates?

Guru Vishnu

Feb 22, 2020 10:32

@JohnRennie Aha! It seems a user named Guru Vishnu edited the question 2 hours ago. Yes sir. I read that and its duplicate. Still I don't get it.

Especially this answer there was useful but didn't solve my question.

John Rennie

I have to go now. I'll have a think about if there is a nice intiuive way to understand this.

Guru Vishnu

Ok sir. Thanks. Bye.

Guru Vishnu

Feb 22, 2020 11:55

Sir, this question occupied a lot of processing power, so I asked it on the main site. If you wish you may also read it to ensure I've explained it properly in the chat above:

0

Q: Reason behind the inclusion and exclusion of current sources while using Ampere's law for the total magnetic field and total current

The following definition of Ampere's law is from Concepts of Physics by Dr. H.C.Verma, from chapter 35, "Magnetic Field due to a Current", page 241: The circulation $\oint\vec B.d\vec l$ of the resultant magnetic field along a closed, plane curve is equal to $\mu_0$ times the total current cr...

electromagnetism magnetic-fields laws-of-physics

Further, if you invent your intuitive solution please share it by any means? Thank you sir.

I've also added a diagram to make the question clear.

Guru Vishnu

Feb 22, 2020 13:51

Analysing two perpendicular wires carrying equal current:
When one wire is kept fixed and the other one is free to move, initially, only a torque acts on the free wire which tends to make the wires parallel in a way the two wires carry current in the same direction.
As it starts to slightly rotate, attractive forces between the wires start to appear and as the angle between these two wires decrease, this force increases and becomes maximum when both wires become coplanar.

@JohnRennie: When you find time, kindly tell whether my conclusion on the interaction between skew current carrying wires is correct or not. Thank you.

Guru Vishnu

Feb 23, 2020 04:53

Deleted

physics.stackexchange.com/q/532593/238167

Guru Vishnu

Feb 23, 2020 11:04

@JohnRennie: Hi sir :-) Good morning.

John Rennie

@GuruVishnu hi :-)

Afternoon in India, morning in the UK :-)

Guru Vishnu

@JohnRennie Aha! :-) It seems you were quick enough to notice it.

I thought I would make that change before you'd see it.

John Rennie

@GuruVishnu yes, your summary of the forces between the wires is correct.

Guru Vishnu

@JohnRennie Thank you sir :-)

Did you invent any intuitive way for the choice in Ampere's law, sir?

John Rennie

Kind of.

Guru Vishnu

Feb 23, 2020 11:08

:-)

John Rennie

I can post some diagrams I drew that give a feel for it if not a complete proof.

Guru Vishnu

Ok sir. If you wish you may do so. I'd also try if I could figure any way out. Thank you.

John Rennie

Guru Vishnu

For an internal current all field lines pass in and pass out equally unlike electric charges.

John Rennie

This is a current inside the circle. If you look at the two field lines that are tangents to the circle you'll see that they point the same direction round the circle i.e. when we sum $\mathbf B \cdot d\ell$ the two are going to add up.

Guru Vishnu

Feb 23, 2020 11:13

Ok sir.

And for the rest, will they just add to zero? I think yes, due to symmetry.

John Rennie

And this is a current outside the circle. Again I've shown field lines that are tangent to the circle.

Guru Vishnu

Yes sir. Now I think I can understand it. They all add to zero.

John Rennie

But now the two elements at the tangent points are going to have opposite signs so they will tend to cancel each other out.

This isn't a proof because the field strength is different at the two tangent points so you can't just claim they are equal and opposite.

But it gives you a feel for why currents outside give $\int \mathbf B \cdot d\ell = 0$

Guru Vishnu

@JohnRennie Yes sir. Understood your point.

@JohnRennie Sure. I find this method to be interesting. Thank you sir :-)

But it gives even more to think about...

I got two answers for the same question on the main site. But they deal with the mathematical explanation for this one. I'm not sure whether they answer the question properly however, I'm sure both of them are good.

I had some issues in understanding the second answer as it involved something called "Stokes theorem". So I've to learn about it before I read it further and completely.

John Rennie

You don't need to know how Stokes' theorem is derived, just that it links a volume integral to a surface integral.

It's used all over physics but I bet most physicists have forgotten the derivation :-)

Guru Vishnu

Feb 23, 2020 11:25

@JohnRennie Hmm... Does that include even you :-) ?

John Rennie

Absolutely :-)

Guru Vishnu

:-)

John Rennie

I remember only stuff I need. My brain is too cluttered already without adding more junk. If I ever need the derivation I can Google it.

Are you done for now? I need to go.

Guru Vishnu

I consider it to be a better way. I myself doubt what is the point in remembering the ore of copper (say) when I can easily Google it?

@JohnRennie Ok sir. Then let's see tomorrow. Good bye :-)

Thank you for your time and the intuitive way.

John Rennie

Bye :-)

Guru Vishnu

Feb 24, 2020 08:34

@JohnRennie: Please reply when you find time:
What does $\nabla \times \vec F=\vec0$ mean? Especially I've seen $\nabla$ in place of $d/dx$ (gradient) but here it appears as an independent quantity.

John Rennie

$\nabla\times$ is the curl of a vector.

Curl (mathematics)

In vector calculus, the curl is a vector operator that describes the infinitesimal rotation of a vector field in three-dimensional Euclidean space. At every point in the field, the curl of that point is represented by a vector. The attributes of this vector (length and direction) characterize the rotation at that point. The direction of the curl is the axis of rotation, as determined by the right-hand rule, and the magnitude of the curl is the magnitude of rotation. If the vector field represents the flow velocity of a moving fluid, then the curl is the circulation density of the fluid. A vector...

Guru Vishnu

Thank you sir. Isn't it also used for $d/dx$?

At least that's what I understood from Wiki's article on \nabla (also MathWorld)

John Rennie

$\nabla$ is the vector $(d/dx, d/dy, d/dz)$.

So for example $\nabla\cdot\mathbf a$ is the dot product $(d/dx, d/dy, d/dz) \cdot (a_x, a_y, a_z) = da_x/dx + da_y/dy + da_z/dz$

Guru Vishnu

Feb 24, 2020 09:59

Ok sir. Thank you very much for the clarification :)

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