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04:54
JR0006 : HCV-34-O1-10
Question:
Let $\vec{E}$ and $\vec{B}$ denote electric and magnetic fields in a frame $S$ and $\vec{E'}$ and $\vec{B'}$ in another frame $S'$ moving with respect to $S$ at a velocity $\vec{v}$. Two of the following equations are wrong. Identify them.
(a) $B'_y=B_y+\frac{vE_z}{c^2}$
(b) $E'_y=E_y-\frac{vB_z}{c^2}$
(c) $B'_y=B_y+vE_z$
(d) $E'_y=E_y+vB_z$
I had some doubts with the above problem and looked the solution. But this was solved based on dimensional analysis. Is there any way of solving this question without using shortcuts like dimensional analysis, sir?
 
2 hours later…
06:51
@JohnRennie: Hi sir. If you're free could you see the question above?
@GuruVishnu I'm dealing with some problems at work. Hopefully I'll be done in 15 minutes or so.
@JohnRennie No problem sir. I'm sorry for disturbing. Please take your time.
07:42
@GuruVishnu Yes, the way you do this is to Lorentz transform the fields into the new frame.
Lorentz transformations mix up magnetic and electric fields i.e. when you Lorentz transform a pure magnetic or pure electric field the result is a mixture of magnetic and electric fields.
This is the well known result that electric and magnetic fields look different to observers moving at different velocities.
@JohnRennie I haven't learnt about "Lorentz transform" yet. This is what I know: Electric and magnetic fields are two aspects of the same type of field - the electromagnetic field. Electric field in one frame can be magnetic field or a mixture of two or vice versa. When we are in a fixed frame we can treat these two as independent quantities, sir.
@JohnRennie Yes sir. I understood that point.
Yes, that's basically correct. The EM field requires an object called a tensor to describe it. A tensor is kind of a combination of two vectors, which is why it can describe the electric and magnetic fields at the same time.
In electromagnetism, the electromagnetic tensor or electromagnetic field tensor (sometimes called the field strength tensor, Faraday tensor or Maxwell bivector) is a mathematical object that describes the electromagnetic field in spacetime. The field tensor was first used after the four-dimensional tensor formulation of special relativity was introduced by Hermann Minkowski. The tensor allows related physical laws to be written very concisely. == Definition == The electromagnetic tensor, conventionally labelled F, is defined as the exterior derivative of the electromagnetic four-potential, A,...
@JohnRennie In short, is tensor something like a vector sum of two different physical quantities?
The EM field is not affected by a coordinate change, but the way it splits onto magnetic and electric fields is affected.
@GuruVishnu no. You need to learn about tensors properly to do justice to them. But that's degree level stuff.
@JohnRennie Ok sir. So in advanced physics, we would convert magnetic and electric fields to EM fields in a different frame and then again decompose them into magnetic and electric components. Is this what Lorentz transformation, mean in simple terms sir?
@JohnRennie :-) Ok sir.
07:51
@GuruVishnu it's ... complicated
In physics, the Lorentz transformations are a one-parameter family of linear transformations from a coordinate frame in spacetime to another frame that moves at a constant velocity (the parameter) relative to the former. The respective inverse transformation is then parametrized by the negative of this velocity. The transformations are named after the Dutch physicist Hendrik Lorentz. The most common form of the transformation, parametrized by the real constant v , {\displaystyle v,} representing a velocity confined to the x-direction, is expressed...
@JohnRennie Ok sir. For the given question, is dimensional analysis the only easy way to solve or are there any other fundamental ways that could be understood by me?
Basically pre-degree level dimensional analysis is the only option.
And it's by far the simplest option.
Ok sir. Thank you for your time and help. I thought of doing justice to the question by using some other method other than dimensional analysis. It seems every other method is beyond my current scope :-)
08:33
@JohnRennie: Hi sir :-)
Are you free now?
@GuruVishnu yes, I've finished work now so I'm free for the rest of the morning :-)
Hmm... Was that a server fault?
And how is the weather there?
I totally forgot. I remember you mentioned it was stormy some days ago.
The storm blew over last Sunday. The weather since has been cold and wet. Typical UK weather :-)
@JohnRennie Have you seen solid ice or snow there? Here it's totally impossible.
In Chester we get snow maybe twice during the winter, though so far this winter there has been no snow. In the north of the UK and Scotland snow is routine in winter.
08:40
@JohnRennie Cool!
May I ask an academic question?
Ok sir. Thank you.
How to prove that force acting on a current carrying wire joining two fixed points $a$ and $b$ in a uniform magnetic field, is independent of the shape of the wire?
Particularly is there any intuitive way of explaining it as I did in the following answer, today? :
0
A: How to prove that the net force on an irregular current carrying loop in a uniform magnetic field is zero?

Guru VishnuThe mathematical proof is provided by the accepted answer. This will be an intuitive way to explain why the net force on an arbitrarily shaped closed loop, exerted by a uniform magnetic field, is zero. I hope the reader must have come across the following formula: $$d\vec{F}=i(d\vec l\times \ve...

Join the points a and b with a straight line to make a loop.
We know the net force on the loop is zero.
@JohnRennie But wouldn't that affect the geometry?
aka the given situation, sir?
The force on our straight line doesn't change as we change the wiggly path from a to b, and since the total force (zero) is constant that must mean the force on the wiggly segment of wire is also constant.
08:48
@JohnRennie Ok sir. Understood this point.
@GuruVishnu your answer assumes we can decompose the loop into pairs of equal and opposite $d\ell$s, and I'm not sure that's true.
Consider a semicircle with a straight line joining the two ends. There is only one $d\ell$ on the curved region parallel to the straight region.
@JohnRennie Oops. Could you explain why is it incorrect? I hope so, because it was my own theory which I felt was reasonable.
So you can't argue that you can pair up parallel elements.
Or consider a triangle. There are no pairs of parallel elements there.
@JohnRennie There is a mistake here. What about the vertical components which add up, sir?
And also a mistake in my assumption.
Round a loop the vertical and horizontal components obviously cancel.
The proof would be to expand the cross product and show that cancellation of the components makes the total result zero.
08:54
@JohnRennie Ok sir. Understood. The way I explain doesn't hold true in all cases. Deleted the answer :-)
@JohnRennie And that's what the first answer did. I tried a non-mathematical way and failed.
Chris Drost's answer seems pretty intuitive to me ...
@JohnRennie I just read the first subdivision completely. I need to give the second part a bit more attention. Currently I just skipped it.
I don't think the second part is necessary. The first part is a short and elegant proof.
@JohnRennie Then that's fine sir. I don't find the first part (at least) difficult. So this must be some kind of fun - "How to do it the hard way"
09:18
A cool image I found on MSE ^
@JohnRennie: How did you think the following fact? It was really a great idea.
33 mins ago, by John Rennie
The force on our straight line doesn't change as we change the wiggly path from a to b, and since the total force (zero) is constant that must mean the force on the wiggly segment of wire is also constant.
I have probably read it somewhere over the years and it was lurking somewhere in my subconscious memories :-)
You accumulate a lot of cunning tricks in forty years.
:-)
"cunning tricks"
Sir, what does kick-mute this user mean? What will it do? I can understand mute but not kick in this context.
@JohnRennie: Hi sir. Are you busy?
09:33
If you are a room owner, and a user in your room is behaving badly, the you can kick mute them and this ejects them from the room and prevents them re-entering the room for 30 seconds.
If you kick mute a second time it ejects then for 5 minutes, and a third time for 30 minutes.
I think after three kick mutes the SE moderators are alerted.
@JohnRennie Ok sir. Thank you for letting me know. I was about to ask whether you could kick me out for fun :-)
But now it seems it alerts mods unnecessarily.
83
Q: Impose a re-entry delay on users kicked out of a chat room

user102937Currently, moderators can kick users out of a room. I've done this to two users now, and they just come right back into the room. I conclude that it's not effective for its presumed purpose (slowing down trolls or other troublemakers, on a room-specific basis). I propose that a delay be impose...

That has all the details.
@JohnRennie Ok sir. I'll read it after my dinner. Now may I ask a doubt from a problem?
Two metal strips each of length $l$ are clamped parallel to each other on a horizontal floor with a separation $b$ between them. A wire of mass $m$ lies on them perpendicularly. A vertically upward magnetic field of strength $B$ exists. The metal strips are smooth but the coefficient of friction between the wire and the floor is $\mu$. A current $i$ us established when the switch is closed at the instant $t=0$. Discuss the motion of the wire after the switch is closed.
--------
I know to solve each and every part. But I don't understand whether we should count the normal force exerted by the metal rails. If so, is it possible to determine that force as I think it would have an indirect impact on the frictional force on the wire in motion.
09:42
You mean an additional normal force due to the (vertical) magnetic field i.e. a component of the normal force that is in addition to the weight of the moving wire?
@JohnRennie Not due to the magnetic field. I was referring to the normal force (mechanical force) exerted by the metal rails on the vertical wire, sir.
But that is the frictional force i.e. the frictional force is $\mu m g$ where $m$ is the mass of the moving wire.
@JohnRennie I just wanted to know whether the frictional force $\mu$ X normal force is altered due to the normal force exerted by the metal rails.
Because whenever two things are in contact, there exists contact/normal force.
What normal force exerted by the rails? The wire is resting on the rails so the only (vertical) force present is the weight of the wire.
I think I didn't explain the question properly. The system looks something like the following image:
Where the blue lines are the metal rails and the red line is the moving wire, sir.
09:48
Yes. That's the view from above.
So there are two forces acting. The Lorentz force to the right and the weight of the red wire that acts downwards into the page.
> The wire is resting on the rails so the only (vertical) force present is the weight of the wire.
That's where I'm not sure whether to include the normal force exerted by the blue line on the red line as they are in contact.
@JohnRennie And the force exerted by the rail in the vertically upward direction (?)
Let me redraw the diagram
Ok sir.
@JohnRennie That looks great sir.
09:57
There, that's a perspective view showing all the forces.
The frictional force at each contact with the rails is $mg/2$.
@JohnRennie Yes sir. Understood. But it seems, according to the problem the floor is also in contact with the moving wire.
@JohnRennie There is no friction between the rails and the wire according to the following statement sir:
> The metal strips are smooth but the coefficient of friction between the wire and the floor is $\mu$
Ah, OK, yes. That makes the question impossible to answer since we don't know what proportion of the weight is being carried by the rails and what proportion is carried by the floor, so we cannot calcuate the frictional force.
@JohnRennie This is what I was looking for. If you see, this is totally unrelated to the main question which is from Magnetic field. However, shall we assume that weight is uniformly distributed?
That doesn't help unless you know the size of the contact patch with the rails compared to the size of the contact patch with the floor.
@JohnRennie Ah. Yes sir. But suppose we have some values about it. Shall we assume uniform weight distribution or does it depend on the pair of materials in contact?
10:03
But I suspect it doesn't matter. There is some frictional force, even though we can't calculate exactly what it is, and the wire accelerates only when the Lorentz force is larger than the frictional force.
It's impossible to say without more details.
@JohnRennie Ok sir. I was thinking of non-uniform distribution due to difference in the material pairs.
Thank you for your help and time sir :-)
Especially the 3d diagram was great.
I also tried some in the past like:
@JohnRennie: What is meant by "force of compression" on a circular wire if each element is under a magnetic force which it towards the centre?
Is that tension?
@GuruVishnu yes
Ok sir. Thank you.
But from the above diagram, I could only infer that the wire breaks. Or in other words, I think there could be no equilibrium as net force is not equal to zero, @JohnRennie sir.
@GuruVishnu I would do it using virtual work ...
10:18
@JohnRennie Cool! "virtual work" = I haven't learnt about that so far. Is that "real" work we're used to?
Could you explain how to use virtual work? It seems like a good tool.
Suppose the radius of the wire decreases by $dr$.
Ok sir.
Then the work done is $dr$ times the total inwards force on the wire $BI\ell$. OK so far?
@JohnRennie Yes sir.
I see, now you'd say the work done by the compression force is opposite and finally I get $F=iBl$
Now suppose the tension in the wire is $T$. As the wire moves inwards its length changes by $dL = 2\pi dr$ i.e. it decreases from $2\pi r$ to $2\pi (r + dr)$.
10:27
@JohnRennie Ok sir.
So the work done is $dW = T2\pi dr$
And the two works must be the same.
So equate the two works and solve for $T$.
@JohnRennie: Yes sir. This method is really great and I obtained $F=iBl$ when you were explaining itself (but using some hypothetical assumptions :)). And I got the correct answer. But could you tell whether there is any mistake in the above force diagram? Am I missing any other forces?
Your diagram looks fine.
@JohnRennie But it seems every single particle will not be in equilibrium as net force is not zero, sir.
Each particle in the wire exerts an equal and opposite force on every other particle. The forces cancel so the ring doesn't move.
10:32
I'm wrong the current answer is $T=iBr$
@JohnRennie Still I don't understand sir. I think we need to include only one of the action-reaction pair of forces to find the equilibrium.
This is how the free body diagram looks like.
The virtual work method gives me the correct answer. Do you want to go through the virtual work calculation?
@JohnRennie If it's fine for you, I'd say "yes" sir.
OK. Step 1: the length of the wire is $\ell = 2\pi r$ so the inwards force is $F = B i 2\pi r$.
@JohnRennie Yes sir,
So the work done by the inwards force as the radius decreases by $dr$ is $dW = Bi2\pi r dr$.
OK so far?
10:41
@JohnRennie Yes sir.
As the wire moves in its circumference dereases from $2\pi r$ to $2\pi (r - dr)$ so the circumference changes by $2\pi dr$.
If the tension in the wire is $T$ then the work done by the tension as the length of the wire changes is $dW = T 2\pi dr$.
@GuruVishnu OK so far?
@JohnRennie Does the tension as seen in the previous diagram have a direction opposite to that in the diagram?
The tension is the force in the wire i.e. it is tangential to the wire.
@JohnRennie Fine then we shall consider the tension vectors to be nearly vertical but not tangential. If they were exactly tangential, then the net force will be towards the centre and the tension can't counteract the magnetic force, sir.
Let's get this virtual work calculation done and worry about the stability in a moment.
10:47
@JohnRennie Ok sir.
8 mins ago, by John Rennie
So the work done by the inwards force as the radius decreases by $dr$ is $dW = Bi2\pi r dr$.
6 mins ago, by John Rennie
If the tension in the wire is $T$ then the work done by the tension as the length of the wire changes is $dW = T 2\pi dr$.
Do you agree with my working so far?
@JohnRennie Understood this but
@JohnRennie not completely this one sir.
Because of the direction argument. So I think it's safe to proceed.
OK. It's the same work we are calculating, just calculated in two different ways, and that means the two $dW$s must be the same.
So we equate the two expressions, and solve for $T$.
@JohnRennie Ok sir. And on equating we get $T=iBr$
And it gives the correct answer. I understood this point. If possible, shall we discuss about that direction, sir?:
Let me draw a diagram
10:54
Ok sir. I arrived at the following after some tangential modifications:
Here are some metal atoms in a circle.
When we try and shrink the circle the atoms exert forces on each other and don't move.
Ok sir.
The forces are indeed not tangential, but metal atoms are very small so if we use a macroscopic radius, i.e. mm or cm, the forces are so nearly tangential that in practice the deviation from the tangential direction is undetectably small.
@JohnRennie Ok sir.
More precisely, if we consider an element of the circle $d\ell$ then the force becomes tangential in the limit of $d\ell \to 0$.
Which is fine for us, because we are doing the calculation of the virtual work in the limit $dr \to 0$.
11:01
@JohnRennie Ok sir. Then how does it stay in equilibrium. There is a force exerted by the magnetic field towards the centre.
Because In the real world $d\ell$ never goes to zero since atoms have a finite size.
@JohnRennie Ok sir. If so is the following nearly accurate as it answers the equilibrium:
The tension is massive enough so that even it small horizontal component matches the inwards magnetic force.
@JohnRennie Thank you sir :-)
 
2 hours later…
12:43
[JR0006 : Virtual work ] conversation ended 2 hours ago.

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