Do magnetic forces obey Newton's third law? Verify for two current elements $\overrightarrow{dl_1} = dl(\hat{i})$ located at the origin and $\overrightarrow{dl_2} = dl(\hat{j})$ located at $(0, R, 0)$. Both carry current $I$.

(N/A) According to the Biot-Savart law,the magnetic field $d\vec{B}$ produced by a current element $I\overrightarrow{dl}$ at a position vector $\vec{r}$ is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\overrightarrow{dl} \times \vec{r}}{r^3}$.
$1$. Magnetic field at the position of element $1$ due to element $2$:
Element $2$ is at $(0, R, 0)$ with $\overrightarrow{dl_2} = dl\hat{j}$. The position vector of element $1$ (at origin) relative to element $2$ is $\vec{r}_{12} = -R\hat{j}$.
Since $\overrightarrow{dl_2} \times \vec{r}_{12} = (dl\hat{j}) \times (-R\hat{j}) = 0$,the magnetic field $\vec{B}_2$ at the origin is $0$. Thus,the force $\vec{F}_{12} = I\overrightarrow{dl_1} \times \vec{B}_2 = 0$.
$2$. Magnetic field at the position of element $2$ due to element $1$:
Element $1$ is at $(0, 0, 0)$ with $\overrightarrow{dl_1} = dl\hat{i}$. The position vector of element $2$ relative to element $1$ is $\vec{r}_{21} = R\hat{j}$.
The magnetic field $\vec{B}_1$ at $(0, R, 0)$ is $\frac{\mu_0}{4\pi} \frac{I(dl\hat{i}) \times (R\hat{j})}{R^3} = \frac{\mu_0 I dl}{4\pi R^2} \hat{k}$.
The force on element $2$ is $\vec{F}_{21} = I\overrightarrow{dl_2} \times \vec{B}_1 = I(dl\hat{j}) \times (B_1\hat{k}) = I dl B_1 \hat{i}$.
Since $\vec{F}_{12} = 0$ but $\vec{F}_{21} \neq 0$,magnetic forces between current elements do not obey Newton's third law in the strong sense (action-reaction pairs are not equal and opposite).