Particle Accelerator

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Guru Vishnu

04:53

Deleted

physics.stackexchange.com/q/532593/238167

6 hours later…

Guru Vishnu

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

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

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

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.