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We say that the magnetic field produced by current carrying conductor produces no force on itself but I am not able to understand this simple argument. Consider a long current carrying conductor.We observe that any small part of the wire has electrons in it which produce magnetic field independently.Now my question is at any other place in the conductor,electrons lie and therefore they must experience a force due to magnetic field produced by electrons in that small part of the wire which leads us to conclude that the conductor as a whole experiences force?Where lies the fallacy in my reasoning?

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  • $\begingroup$ Consider an infinite long conducting wire carrying current $i$. Imagine, a small element; and the force on it from all the parts of the same wire. Wouldn't the top and the bottom half average out to give net force zero? $\endgroup$ – sbp May 16 '18 at 18:09
  • $\begingroup$ Your argument is for long wire only.Consider a finite wire.The Element near the ends will experience unsymmetric forces? $\endgroup$ – Panda May 16 '18 at 18:11
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The magnetic field due to a small element of current is given by the Biot-Savart Law: $$ B(z) = \frac{\mu_0}{4\pi} I \int \frac{d\textbf{l'}\times \textbf{r}}{r^2} dr= \frac{\mu_0}{4\pi} I \int \frac{dl'}{r^2} \sin\theta dr $$ Which is basically the found using the cross product of the direction to which you’ll measure the field and the direction of the current. Notice that along the wire the cross product will be of two parallel vectors are direction you’re measuring along and current are parallel as $\theta$ is 0 so $sin(\theta)$ is 0. Hence there is no magnetic field in any part of the wire and hence the electrons in the wire don’t experience any force and the wire can be at equilibrium.

Note: This wire is a thin wire and it can work for both finite and infinite lengths of the wire. I am ignoring any drastic effects of Joule-Heating due to power dissipation of the resistance of the wire as I assume the current is not as high. But nevertheless even with a high current, it would not experience at least any magnetic forces as there is no field inside the wire.

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  • $\begingroup$ What if the wire is not straight? What if it is parabolic? $\endgroup$ – Panda May 16 '18 at 20:51
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We say that the magnetic field produced by current carrying conductor produces no force on itself

We most certainly not NOT say that. Here is the image of a copper conductor that had a strong current in it. This was part of testing that followed the realization that lightning rods were crushed in this fashion.

Someone will complain that this is a hollow conductor and claim it does not apply. However, I know of this image because I used it as an illustration of the Lorentz force in a plasma, which was used as a fusion confinement system in the 1950s. The same technique, with limits, remains in use in the tokamak designs like ITER. The plasma columns are continuous, not hollow.

Now someone will complain that it is a fluid, and so on.

Pollock and Barraclough, 1905

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  • $\begingroup$ The fact that this wire is hollow is why it has experienced a force. But for a straight thin wire it will indeed not experience at least a magnetic force. It may ofcourse experience Joule-Heating by dissipating electrical power and cause a plethora of faults. $\endgroup$ – Tausif Hossain May 16 '18 at 19:15
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    $\begingroup$ It will see all sorts of forces, depending on details of the current. AC, for instance, will be subject to the skin effect and thus show the same end result. DC pulses will do the same. Which is exactly why I stated that in my post. $\endgroup$ – Maury Markowitz May 16 '18 at 19:35
  • $\begingroup$ It’s true but it’s mostly in high current cases but the forces are not due to magnetic effects. $\endgroup$ – Tausif Hossain May 16 '18 at 19:37
  • $\begingroup$ Sure they are. I think you are confusing the net effect part. See the answer directly below. $\endgroup$ – Maury Markowitz May 16 '18 at 19:53
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A conductor with a current travelling through it experiences no net magnetic force. Individual pieces can indeed experience a force, it's just the sum of all these forces that adds up to zero.

This is obvious if you think about conservation of momentum, since if an isolated conductor with a current through it feels a net force its momentum will change.

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  • $\begingroup$ That is why I am asking.Net force is zero means net momentum change is zero but if force is acting this means torque may act which means if we induce current in a lopsided wire it may not remain stationary.I am just asking as while solving question in books there is no assumption stated while solving for several things like force etc. $\endgroup$ – Panda May 16 '18 at 20:57
  • $\begingroup$ @yashasvigrover Torque may act, but again, the net torque is still zero, by conservation of angular momentum. The wire need not remain stationary, though: it may move in other ways. A small loop of wire with slack in it will tend to expand when current is driven through it, for instance. $\endgroup$ – Chris May 16 '18 at 21:08
  • $\begingroup$ As far as I know,It will tend to expand when placed in EXTERNAL magnetic field(in fact I have even solved for tension in the wire as an excercise).But if there is no external field,will it still expand? $\endgroup$ – Panda May 16 '18 at 21:23
  • $\begingroup$ @yashasvigrover Yes, it will still expand with no external magnetic field. $\endgroup$ – Chris May 17 '18 at 4:45
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Before reading: this is just a classic explanation, it is not used q.m.
Think to this experiment: an electron is moving through the empty space. There is an external generic electromagnetic field which produces a force on it. You are an observer who measures electron's speed $\mathbf{v}(t)$. It means that you always measure an electric and a magnetic field respectively due to the charge of the electron and to its speed: $\mathbf{E}=\mathbf{E}(x,y,z,t)$; $\mathbf{B}=\mathbf{B}(x,y,z,t)$. If you maesure these two fields on the point in which instantly is situated the electron you find $\mathbf{E}=\mathbf{B}=\mathbf{0}$, it means that the electron doesn't auto-influence his motion, because it can't see the fields it generates. Now if you put another electron, the previous considerations stay true, but now the two electrons see one the fields that the other one generates, so they reciprocally influence their motion. You can solve some equations and get the two trajectories. You can consider a system with n electrons. If you want to know the total force acting on the system of the n electrons at a certain time, you have to sum all the forces. But the sum of the internal forces of the system is zero (third principle). So the only force that acts on the system of n electrons is the sum of all the forces due to the external electromagnetic field, on each electrons. This shows that the total force acting on the system just depends on the the external electromagnetic field, and it is zero if the external field is zero. If you have a current carrying conductor the electrons which make the current don't generate a force on the conductor, for this you need an external field.
Hope it's clear.

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  • $\begingroup$ If you have system of n electrons net force is zero on the system but indivually they experience force and they DO move and gain kinetic energy. So will an unsymmetric wire Move around due to torque even if NET force is zero as at microscopic level each electron experiences force? $\endgroup$ – Panda May 16 '18 at 21:28
  • $\begingroup$ If n is very very big, and the initial conditions of the electrons are random, also the total torque must be zero. Two electrons have opposite torque if on them acts the same force and if they are at the same distance from the centre $0$. If n is big you can always find couple of electrons in this situation and by summing you get zero. $\endgroup$ – Landau May 16 '18 at 21:51

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