Lorentz Force on a Current Carrying Wire Does the Lorentz Force on a Current Carrying wire given by the equation
$$\mathbf{F} = I \int \text{d}\ell \times \mathbf{B}$$
constitute an action reaction pair? That is, if i have two arbitrarily shaped current carrying wires, is it true that force on any one of them due to the magnetic field of the other is equal to the force on the other due to the magnetic field of the first?
 A: Yes this is true, for example Lets take 2  parallel infintite wires carrying a current $I_1$ and $I_2$ that located distance $d$ from each other
We know from biot savart law that magnetic field of wire  at distance $d$ is $$\frac{\mu_0 I}{2\pi d}\hat \phi$$ so now we can calculate to Force of wire 1 on wire 2
by your formula $$\vec{F} = I \int \vec{d}\ell \times \vec{B}$$ and getting the result$$F=I_{2}\left(\frac{\mu_{0} I_{1}}{2 \pi d}\right) \int d \ell_{2}$$
and the force per unit lengh will be $$f=\frac{\mu_{0} I_{1} I_{2}}{2 \pi d}$$
Now you can see that if $I_1$ and $I_2$ are in the same direction we getting repeling force, and if $I_1$ and $I_2$ are in opposite directions we getting attractive force.
Of course we could do the same by calculate the force of wire 2 on wire 1 and getting the same result.
A: edited
No, it seems that the law of action is true for the cherry-picked case of parallel wires as shown by @Sagigever's response. For instance, this is certainly not true for the case of magnetic Lorentz force exerted by two charges (yellow in figure) moving perpendicular to each other (along the blue and red lines). In this case, the $\vec{F_{12}}$ is perpendicular to $\vec{F_{21}}$ (the dotted lines represent the magnetic field due to the charges).

A: Newton's third law in modern terms states conservation of momentum. Electrostatic forced conserve $P_{kin}=\sum_i m_i p_i$ but magnetic forces do not , as argued by @Rohit. In the presence of electromagnetic fields the conserved momentum is $P = P_{kin} + P_{pot}$, where $P_{pot} = - \sum_i q_i \vec A_i$.
