Tags:
1. Aug 24, 2015

Abid Mir

Imagine an infinitely long wire carrying current i1 in west to east direction. At a small distance 'd' above the wire there is another small current carrying wire of length 'l' carrying current i2 from east to west direction (opposite to the direction of current in the below placed wire). Obviously they are magnetically repelling.And if wire 2(carrying current from east to west) is at rest the magnetic force must be equals to mg. The magnetic force is upward and mg is down.The magnetic force can be written as Uoi1i2/(2pi×d) × l which is equal to mg(wire at rest).Now if magnetic force is greater than mg, the wire moves up. Now magnetic force is up and displacement is up too which means that work done by magnetic force should be positive. HOW IS IT POSSIBLE WHEN WE KNOW THAT WORK DONE BY MAGNETIC FORCE IS ALWAYS ZERO. IS IT THAT AN EMF IS INDUCED WHICH OPPOSES THE CHANGE ? SOMEONE PLEASE HELP ME WITH THIS QUESTION.

2. Aug 24, 2015

Hesch

It's not always zero. When a magnet raises an iron nail from a table, the magnetic field is doing work to the nail ( adding the nail kinetic and potential energy )

Thereby the magnetic field is loosing magnetic energy, because some volume of airgap ( with high energy density ) is substituted by iron ( with low energy density ).

3. Aug 24, 2015

Philip Wood

It is indeed because of the induced emf. In order to maintain the current in the wire when the wire moves, work has to be done on the free electrons (in addition to that needed to overcome the usual collision forces) by the electric field supplied by the power supply. It's the power supply that does the work. That's it in a nutshell. Maybe too small a nutshell.

4. Aug 24, 2015

Abid Mir

F = q(v×b) ?

5. Aug 24, 2015

Abid Mir

''In order to maintain the current in the wire when the wire moves, work has to be done on the free electrons "

How does movement of the wire effect the current?

6. Aug 24, 2015

BvU

F = q(v×b) !

7. Aug 24, 2015

Hesch

In post #2 I just contradict that magnetic force never do work. Regarding the (permanent) magnet and the nail, the force could be expressed as:

F = d( magnetic energy)/ds , where s is the width of the airgap.

8. Aug 24, 2015

Philip Wood

The current drops while the wire is moving because the net emf in the circuit containing the wire is reduced. The reduction is because of the back-emf (due to the cutting of magnetic flux produced by the other wire). This emf opposes the power supply's emf.

9. Aug 24, 2015

Abid Mir

My professor actually put forward this question to us and gave it as a homework to us . What he said is that in this case the work is done but not by magnetic force. He said he will solve the paradox but forgot it then after.

10. Aug 24, 2015

Abid Mir

I have posted
I have posted this question on multiple websites but i feel that your answer is the most appropriote as most of the answers are varying to a large extend. I would like you to discuss such more situations where work done looks positive but it isnt

11. Aug 24, 2015

Abid Mir

But if the two wires are supported by batteries so a dc current flows through them would the emf still will be induced?

12. Aug 24, 2015

Philip Wood

One gains a lot of insight by going back to first principles and looking at the origin of the back-emf in terms of forces.

For simplicity let the charge carriers in the wire be positive. It's easy to modify what follows for negative carriers if you wish. The conclusion will be unaffected.

Let the wires run in the ±x direction, and the current in the lower wire in the -x direction and the current in the top wire in the +x direction. So if the top wire were stationary, the drift velocity of carriers in it would be $\mathbf{v}_{drift}\ =\ v_{drift} \mathbf{i}$. But suppose the top wire is moving upwards (in the +y direction) with velocity $\mathbf{v}_{wire}\ =\ v_{drift} \mathbf{j}$. Thus the resultant velocity of charge carriers in the top wire, in the laboratory frame of reference will be $$\mathbf{v}_{drift}\ +\ \mathbf{v}_{wire} \ = \ v_{drift} \mathbf{i}\ +\ v_{wire} \mathbf{j}.$$
The bottom wire will be producing magnetic flux density in the –z direction, in the vicinity of the top wire. So the top wire is sitting in a field $\mathbf{B} \ = -B \mathbf {k}$.

So the force on a charge carrier in the top wire will be
$$q (\mathbf{v}_{drift}\ +\ \mathbf{v}_{wire})\ \times \mathbf{B} = (v_{drift} \mathbf{i}\ +\ v_{wire} \mathbf{j}) \times B (\mathbf{-k}) = q B( v_{drift} \mathbf{j}\ +\ v_{wire} \mathbf{(-i)}).$$

The second term represents a force on the carrier opposing its motion along the wire. Note that this term is zero when the wire is held stationary! For the wire to move upwards, and the current in it to stay the same, this force has to be opposed by a force from the electric field pushing the carriers through the wire. It is this electric field force which does the work lifting the top wire. The magnetic field acts rather as a pulley does on a rope, altering the direction of the force exerted.

13. Aug 24, 2015

Abid Mir

'The second term represents a force on the carrier opposing its motion. Note that this term is zero when the wire is held stationary' this statement is confusing me. I dont understand fully what you are main point is in that math.

14. Aug 24, 2015

Philip Wood

The main point in the maths is to show that when the wire moves (in the magnetic field from the bottom wire) there is a magnetic force on the charge carriers in the top wire, in the opposite direction to their motion through the top wire. It is work done against this force by the electric field on the charge carriers in the top wire, that supplies the work done to lift the top wire. The magnetic force acts a bit like an anchored pulley, not doing work itself, but changing directions of forces.

If the wire is stationary, so the second term in the final expression is zero, there will be no force parallel to the wire on the charge carriers, so no back-emf, which accords with Faraday's law: if no flux cutting, then no back-emf.

That's the best I can do, at the moment.

Last edited: Aug 24, 2015
15. Aug 24, 2015

Philip Wood

Yes. There are now two emfs acting in the top wire's circuit: that from its battery, opposed by the induced back-emf due to the flux cutting. The only way to maintain the original current (i.e. the current when the wire was stationary) is to increase the battery's emf (add more cells?).

16. Aug 24, 2015

Abid Mir

""If the wire is stationary, so the second term in the final expression is zero, there will be no force parallel to the wire on the charge carriers''

This part is problematic as if the wire is stationary it does not mean that electrons are not flowing in it so the should feel the force. But is it that the force experienced by the electrons is equal to the force experienced by the protons in the wire?

'there is a magnetic force on the charge carriers in the top wire, in the opposite direction to their motion through the top wire.'

Isnt the motion created by the magnetic field of the the below wire.how then the charge carriers feel a new magnetic force in the opposite direction of motion and where from.

OUR PROFESSOR WHEN HE WAS TEACHING THIS TO US, HAD NOT TAUGHT US EMI AT THAT TIME AND THAT IS WHY IT IS HARD TO BELIEVE FOR ME THAT THERE IS ANY CONCEPT OF EMI USED IN THIS QUESTION. MOREOVER THE ANSWERS TO THIS QUESTIONS ARE VERY MUCH CONTRADICTING WHICH MAKES IT HARD TO PICK ONRlE AMONG MANY.

17. Aug 25, 2015

BvU

Well, you're in trouble now! Welcome to the real world of physics and the internet:
1. You have a perfectly reasonable question/problem
2. There are a lot of candidate answers, but your own knowledge isn't sufficient to distinguish which is which, whether they are contradictory or not, etcetera

My advice is to exercise patience: store this one in a precious little box until you have learned a little more and come back to it much later.

I'm sorry my post #6 was so extremely terse. But it was the whole answer.

18. Aug 25, 2015

Philip Wood

The electrons are indeed flowing, and this constitutes an electric current. Therefore they experience a 'magnetic force', a force due to their being in a magnetic field (due to the bottom wire. But, when the top wire is stationary, this force is in a vertical direction; there is no force component along the wire. Only when the wire moves vertically (and work is done by the magnetic force on the wire) is there a magnetic force component along the wire on the charge carriers.
No. The protons are stationary and so don't experience magnetic forces.
That's what my analysis in post 12 showed. Did you follow it? If not, I think you'd be well advised to heed the wise and kindly words of BvU in post 17.

Good luck!

Last edited: Aug 25, 2015
19. Aug 25, 2015

Abid Mir

Yes indeed the answers vary to a large extend. I like the points which philip woods mentioned but on stack exchange they have a completely different answers!

20. Aug 25, 2015

Abid Mir

O

Last edited: Aug 25, 2015
21. Aug 25, 2015

Philip Wood

The thread for which you've just given a link is long, and in my opinion, confusing. The question you posed is entirely answerable in terms of the Lorentz force (fancy name for the electric and magnetic forces experienced by a charge). Bringing magnets into the picture won't help. Magnets only add complications; you have to know a bit about what's going on inside them.
Could you give me a link to what you've been reading on Stackexchange?

22. Aug 25, 2015

BvU

Hello Abid, Philip,

Now that I stuck my curious nose in this thread, I can't leave it at that. Just like you folks, I suspect. Must be one of the better chracteristics of a physicist .

I found a Griffiths 3rd ed (1999 !) and the framed
with a lot of text following, e.g.:

there are many situations in which it appears so manifestly false that one's confidence is bound to waver. ...(magnetic crane) .. something is obviously doing work and it seems perverse to deny that the magnetic force is responsible. Well, perverse or not, deny it we must, and it can be a very subtle matter to figure out what agency does deserve the credit in such circumstances. ...​

In other words: the magnetic component of the Lorentz force does no work.

Two pages further he works out pretty exhaustively the example that features in this thread (the fact that the magnetic field in his example is brought about by unspecified means does not constitute a different situation). Over two full pages all the issues that also apppear in the form thread (and also in the other one) are being treated in a didactically sensible way. Even these non-playing protons appear just above where the example 5.3 starts.

How come these threads rant on and on from one misunderstanding to the next misinterpretation, while it's so easy to read on a bit further and get it all handed on a silver platter ? And if at that point something is still unclear, it's easier to post very specific questions that allow much more compact answers !

23. Aug 25, 2015

Abid Mir

continue reading and they do talk about the same case of the wires which is creating the confusion. page 2-3 doc al(mentor) gets involved and discussion becomes interesting. Go through that.

24. Aug 25, 2015

Abid Mir

Yeah i heard a lot of griffith talking about the paradox.i will loo forward to his text.

25. Aug 25, 2015

Philip Wood

BvU: I've never read (or owned) Griffiths. Judging by the bit you've quoted, he writes most engagingly. Incidentally, 1999 seems very recent to me – despite your "!".

AM: Thanks for the Stackexchange link. I didn't find what was said very clear. It didn't seem to address your specific query - which was very well posed.

Last edited: Aug 25, 2015