Why force on small wire is not same as force on infinite wire due to magnetic field?

Question

Question:- Find magnetic force between wires as shown in figure. The infinte wire with current 'I' and the small finite wire of length 'l' with current 'i' at a distance of x from infinte wire (direction shown in figure).

I got force on small wire in negative x-axis direction. But as Newton's third law says, for every action there is an equal and opposite reaction. So according to this there must be same magnitude of force applying on infinite wire in positive x-axis (like that in second figure) But how is this possible, because of magnetic field of finite wire, we have simple cross product for force so force must be perpendicular to both L and B but here it is parallel to L. (L is length vector and B is magnetic field vector) Now if we go particularly for small wire we get something like this(in figure 3) We get torque and F net on infine wire equals to zero. How is this even possible. Neither we are getting force in opposite direction nor of same magnitude. Does this question violates Newton's third law? But this can't. I am sure something I am missing. Please help me through this! Are there more general cases like this where problem like this arises?

Edit: I was asked in comments that how I solved for force, so here it is

Answer 1

I think @jensen paull is a bit dismissive of Newton 3. It is the case that the force between short current elements is not symmetric: $$K\mathbf{dl_2}\times(\mathbf{dl_1}\times\mathbf{r_{12})} =K[(\mathbf{dl_2}.\mathbf{dl_1})\mathbf{r_{12}}-(\mathbf{dl_2}.\mathbf{r_{12}})\mathbf{dl_1}]$$ is the force on element 2 ($K=I_1I_2\mu_0/4\pi r_{12}^3$ is a constant). The first term is equal and opposite if we interchange 1 and 2, but the second is not. So at the level of current elements Newton 3 is not true. But current elements are unphysical: they require creation of charge at one end and destruction at the other. If we integrate around a circuit the second term vanishes, so at the level of whole circuits in magnetostatics (ie constant current) Newton 3 is valid.

In a time-dependent situation we can recover conservation of momentum (which is the big prediction of Newton 3) only by introducing the momentum of the EM field.

Answer 2

Summary: Apparent violations of Newton's Third Law can occur in the presence of magnetic forces due to the fact that electric & magnetic fields can carry momentum. Newton's Third Law is, after all, a consequence of the conservation of momentum.

In this case, the field momentum arises because the upper wire is not a closed loop. Because charge must be conserved, this means there must be a time-varying point charge of magnitude (plus a constant) at its tip. This means that there is also a time-varying electric field $\vec{E}$ present, of the form $$ \vec{E} = \frac{Q_0 - I t}{r^2} \hat{r}. $$ (A time-varying magnetic field may also be present, though I suspect that this is not the case due to the symmetry of the configuration.)

But it is also known that electromagnetic fields have a momentum density of $\vec{\mathscr{p}} = \epsilon_0 \vec{E} \times \vec{B}$. This quantity must be integrated over all of space and added to the momentum of the charges to find the total momentum of the system. While the momentum of the charges in this system is constant, the momentum of the fields is (probably) not; I have not done the explicit calculation, but it seems likely that the field momentum is increasing linearly with time, since the electric field will also be increasing linearly with time.

Now, if wire 1 and wire 2 obeyed Newton's Third Law ($\vec{F}_{12} = -\vec{F}_{21}$), then we would expect the net rate of change of the momentum of the system to be zero. But it's not; as shown above, the system's momentum is changing. This means that we should not necessarily expect to have these two wires exert equal & opposite forces.

For an explicit demonstration of the fact that $\vec{F}_{12} \neq -\vec{F}_{21}$ when calculating forces from the Biot-Savart Law and the Lorentz force law, see my answer to this question.

Answer 3

$$F_{1}=-F_{2}$$

$$\frac{dp_{1}}{dt}=-\frac{dp_{2}}{dt}$$

$$\int_{t_{0}}^{t_{1}}\frac{dp_{1}}{dt}dt=-\int_{t_{0}}^{t_{1}}\frac{dp_{2}}{dt}dt$$

$$p_{1}(t_{1})-p_{1}(t_{0}) = p_{2}(t_{0})-p_{2}(t_{1})$$

$$= p_{1}(t_{0}) + p_{2}(t_{0})= p_{1}(t_{1}) + p_{2}(t_{1}) $$

Newtons third law implies conservation of mechanical momentum, the momentum of 1 and 2 at time $t_{0}$ is the same as the total momentum of 1 and 2 at time $t_{1}$

From electromagnetic theory we also know that the electromagnetic field stores momentum

$$\vec{p}_{EM} = \epsilon_{0} \vec{E} × \vec{B}$$

We can also show that [given the fields vanish at infinity] the total momentum conservation from electromagnetic theory in of space is

$$- \frac{\partial\vec{p}_{mech}}{\partial t} = \frac{\partial \vec{p}_{EM}}{\partial t} $$

When our expression for the EM field momentum increases, the total mechanical momentum decreases. Ie - Newtons third law isn't correct in general as this predicts mechanical momentum is NOT conserved.

Regarding your comments on other answers:

I haven't actually done the math, but I think @CWPP is saying that the total force on the wire is inline with Newtons thirdlaw, and you are not integrating, only considering a single element. But then again I haven't looked at the problem.

P.s

Single current elements like

$$\vec{J} = \delta^3(r) \hat k$$

Can be described, however would require infinite charge density, with charge moving from one point to another [Although moving to the same point, yes its weird]

$$-\frac{\partial \rho}{\partial t} = \nabla \cdot \vec{J} = \nabla \cdot [ \delta^3(r) \hat k]$$

$$\rho = -t \hat k \cdot \nabla \delta^3(r)$$

The charge density increases, but its integral is zero. So Net zero charge [Yes very strange]

However this is NOT what is going on with analysing a single element of a continuous distribution. That is just false.