0
$\begingroup$

I am reading Maxwell's treatise vol 2 page 157 and I have a doubt to clear in the following page:

enter image description here

My analysis of this derivation:

$\vec{ds}$ and $\vec{ds'}$ are the elements of circuits; $\cos{\epsilon}$ is the angle between them; $r$ the distance between them and $\rho$ and $Q$ are unknown functions of $r$.

The only difference between my analysis and Maxwell's derivation is that I use: $\vec{r}=(x-x')\hat{i}+(y-y')\hat{j}+(z-z')\hat{k}=\xi\hat{i}+\eta\hat{j}+\zeta\hat{k}$ while Maxwell uses: $\vec{r}=(x'-x)\hat{i}+(y'-y)\hat{j}+(z'-z)\hat{k}=\xi\hat{i}+\eta\hat{j}+\zeta\hat{k}$

where $(x,y,z)$ are the coordinates of field circuit $s$ while $(x',y',z')$ are the coordinates of source circuit $s'$

enter image description here enter image description here enter image description here enter image description here

In this derivation, as highlighted in yellow, $\dfrac{d\rho}{dx}\cos{\epsilon}=\dfrac{d(\rho\cos{\epsilon})}{dx}$ as if $\dfrac{d(\cos{\epsilon})}{dx}=0$. It means when an element $\vec{ds}$ changes its position, its orientation remains unaltered.

Why is it necessary for the orientation of circuit $s$ to be unaltered when it moves in a path while doing work? Guys please help. I have been toiling over this problem for a long time.

This book also says the circuit should be moved as a whole without rotation:

enter image description here

On the other hand, magnetic shells which are equivalent to closed circuits, doesn't have this limitation. The work done by a magnetic shell in the preesence of another magnetic shell only depends on the initial and final positions of shell whether or not the shell rotates in the path.

So what is going on here. Do we really need to make the orientation of circuit $s$ unaltered while doing work? Why?

Edit

Thanks to the answer, I understand that even if we rotate the circuit, the work done in rotation is determined by the variation of $M$. But how can it be proved using the above equations.

$\endgroup$
  • 2
    $\begingroup$ While it's not a bad source - after all Maxwell wrote it - you may want to look elsewhere to learn electrodynamics, it is incredibly dated. $\endgroup$ – JamalS Oct 26 '17 at 13:15
  • $\begingroup$ Please....... I need help badly. There is nowhere else I can find these old and original derivations of electrodynamics. I have deciphered this whole chapter except this doubt in it. I can give you if you want. It is very lengthy. $\endgroup$ – user173411 Oct 26 '17 at 14:07
  • $\begingroup$ Hi user173411 and welcome. Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems to verify that your question is probably out of scope of Physics SE. You might therefore, in case you don't get an answer here, want to consider posting elsewhere, e.g., at the PhysicsForums. $\endgroup$ – stafusa Oct 26 '17 at 14:10
  • $\begingroup$ Why do you visit the site as a new today member while you are the user @faheemahmed400 therein :(1) Confusion in Maxwell's derivation of Ampere's Force Law - Part III. (2) Confusion in reaction force of Ampere's Force Law. (3) etc $\endgroup$ – Frobenius Oct 26 '17 at 14:40
  • $\begingroup$ ...your "quiet some time" is 2,5 months ago (Aug 10). $\endgroup$ – Frobenius Oct 26 '17 at 14:54
1
+50
$\begingroup$

My simple answer to the question would be: The author doesn't consider rotations of the closed loop, because this makes calculations easier. The author wants to calculate a force on the whole loop, and it's an easy thing for him to calculate energy differences between displaced loops, if they haven't been tilted. This doesn't mean that the orientation musn't be altered in general (and as you have stated yourself, they often are altered, for example in a solenoid within an a magnetic field). It just means that if you only allow displacements without rotation, you will see a conservative force acting on the whole loop.

What would happen if you would allow rotations? In that case the step you have problems with can't be performed like that, you would have to make additional calculations concerning the other term that won't vannish. The end result would be not only a force acting on the position, but only a torque acting on the orientation of the loop.

$\endgroup$
  • $\begingroup$ Thanks. Can you please show the "additional calculations" we have to make (concerning the other term that won't vanish) if we allow rotations? $\endgroup$ – Joe Nov 2 '17 at 5:14
  • $\begingroup$ Can you please show the calculations for torque a bit more elaborately. $\endgroup$ – Joe Nov 3 '17 at 9:19
  • $\begingroup$ The work done on rotations indeed the change in $M$. Apart from the magnetic shell example this book page 107 also conforms it. But I do not have any idea of how to show it from the above equations. Please give a hint if you do have an idea. $\endgroup$ – Joe Nov 3 '17 at 9:33
  • $\begingroup$ To be honest, I don't have a good approach to it. I just think that it must be possible in principle, but I don't have a clue on how to perform the calculations. It is understandable to me if this is not satisfying to you. $\endgroup$ – Quantumwhisp Nov 3 '17 at 16:05
  • $\begingroup$ From the equation $\vec{F}=-\vec{\nabla}M$, we can only say force is conservative if we keep the orientation unaltered. By altering the orientations, force may or may not be conservative. However from the magnetic shell theory, we can say even by altering the orientations force is conservative. Anyway thanks for the answer. $\endgroup$ – Joe Nov 3 '17 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.