Particle Accelerator

Guru Vishnu

06:57

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

@JohnRennie Are you free now sir? I need some hints to solve a problem which I analysed it only qualitatively.

John Rennie

@GuruVishnu I'm working, but only for a few minutes. Post the question and I'll have a look as soon as I'm free.

Guru Vishnu

Ok sir. This is the question:
A long straight wire carries a current $i$. A particle having a positive charge $q$ and mass $m$, kept at a distance $x_0$ from the wire is projected towards it with a speed $v$. Find the minimum separation between the wire and the particle.

Thank you.

John Rennie

07:16

@GuruVishnu that looks hard to me ...

Guru Vishnu

@JohnRennie Hmm... The situation of non-uniform magnetic field with varying distance makes the situation hard. Is there anyway I can determine its path? Anything as simple as virtual work?

One thing I can say for sure is: The speed remains constant. I know how the magnitude of force varies. But I'm wondering on how to find its direction at different points of time.

John Rennie

The motion would look like this

Guru Vishnu

@JohnRennie Is that symmetrical about the horizontal axis? I think yes.

John Rennie

Yes, it would be.

Guru Vishnu

@JohnRennie So, will it be an elliptical path?

John Rennie

07:26

But I can't see how to calculate the trajectory. I doubt it would be elliptical.

Guru Vishnu

If we consider the motion beyond the right edge of the image?

John Rennie

This is surely not a JEE question?

Guru Vishnu

@JohnRennie Ok sir. But I suspect it would be much different than planetary motion as speed here remains constant, whereas in planetary motion speed is maximum at perigee and minimum at apogee.

@JohnRennie I think this is not a JEE question. But it seems I could answer it based on my understanding, although I couldn't figure a way out.

John Rennie

The differential equation is easy to write down. Solving it is the problem!

Guru Vishnu

@JohnRennie Fine. Can you just give a brief idea on how to approach it?

At least I'd like to have a good idea of it rather than solving it.

John Rennie

07:37

Take $x$ to be the distance from the wire and $y$ the vertical position. Then we can write the velocity of the particle as the vector $\mathbf v = (v_x, v_y)$

Guru Vishnu

Ok sir. And $v^2_x+v^2_y=v^2=\text{constant}$

John Rennie

The force on the charge is the Lorentz force $F = q \mathbf v \times \mathbf B$ where $\mathbf B = (0, 0, \mu_0 I/2\pi x)$

Guru Vishnu

Ok sir.

John Rennie

If you take the cross product you'll fund the force is always in the plane of the page so you can write it as $F = (f_x, f_y)$. Then the acceleration is $\mathbf a = (a_x, a_y)$

And you'll end up with a differential equation $\mathbf a = d\mathbf v/dt = f(\mathbf v, x)$

Guru Vishnu

Ok sir. It seems we're just decomposing two dimensional motion into two single dimensional motion along orthogonal directions to make the situation easier like we used to do for projectile motion.

John Rennie

07:41

Good luck solving that!

Guru Vishnu

@JohnRennie :-)

@JohnRennie: Is there anyway I could find "Magnetic potential energy" at some distance from a current carrying wire? Is it similar to other such forces like gravity and electrostatic?

I think we may be able to apply law of conservation of energy?

John Rennie

I don't know of any such method.

Guru Vishnu

Ok sir. Thank you.

¯\_(ツ)_/¯

I think it's a challenging task.

John Rennie

$$ \frac{dv_x}{dt} = \frac{\mu_o I q}{2\pi m} v_y / x$$

$$ \frac{dv_y}{dt} = -\frac{\mu_o I q}{2\pi m} v_x / x$$

That's what I get for the equations of motion.

Guru Vishnu

I got $v/x$ instead of $vx$ sir.

For both $x$ and $y$ components.

John Rennie

07:53

Oops, yes

Guru Vishnu

I'm planning to do that after some time.

So I might get some better ideas.

John Rennie

08:57

@GuruVishnu what did you want to ask?

Guru Vishnu

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

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

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

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

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

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

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.

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♦ Particle Accelerator