Do magnetic forces obey Newton’s third law. Verify for two current elements $\overrightarrow {d{l_1}}  = dl\left( {\hat i} \right)$ located at the origin and $\overrightarrow {d{l_2}}  = dl\left( {\hat j} \right)$ located at $ (0, R, 0)$. Both carry current $\mathrm{I}$.

According to Biot-Savart law, direction of magnetic field is in the direction of $\mathrm{I} \overrightarrow{d l}$.

Magnetic field for $d l_{2}$ element at $(0, \mathrm{R}, 0)$ is given by,

$\overrightarrow{\mathrm{B}}=\mathrm{I} \overrightarrow{d l}_{2} \times \vec{r}$ $=\mathrm{I} d l(\hat{i}) \times r \hat{j}$

$=\mathrm{I} d l r(\hat{i} \times \hat{j})$ $\therefore\overrightarrow{\mathrm{B}}=\mathrm{I} d l r(\hat{k})$

That means, it is in $z$-direction.

Force on this element,

$\overrightarrow{\mathrm{F}}_{2}=\mathrm{I} \overrightarrow{d l}_{2} \times \overrightarrow{\mathrm{B}}$ $=\mathrm{I} d l(\hat{i}) \times \mathrm{B}(\hat{k})$ $=\mathrm{I} d l \mathrm{~B}(\hat{i} \times \hat{k})$ $=\mathrm{I} d l \mathrm{~B}(-\hat{j})$

This force is in $y$-direction.

Magnetic force on $d l_{1}$ element at origin,

$\overrightarrow{\mathrm{I}} \overrightarrow{d l}_{1} \times \vec{r}=\mathrm{I} d l \hat{j} \times r(-\hat{j})$

$\quad=0$

$\vec{r}=r(-\hat{j})$ because first element is at $(0, \mathrm{R}, 0)$ with respect to that this element is in

$y$-direction. So, magnetic field is zero at this point. So, magnetic force on $\overrightarrow{d l}_{1}$ due to $\overrightarrow{d l}_{2}$ be zero.