Find the components along the x, y, z axes of the angular momentum $l$ of a particle, whose position vector is $r$ with components x, y, z and momentum is $p$ with components $p_x,p_y$ and $p_z$. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Linear momentum of particle, $\vec{p}=p_x\hat{i}+p_y\hat{j}+p_z\hat{k}$

Position vector of the particle, $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Angular momentum, $\vec{l}=\vec{r}\times \vec{p}$

$⟹l_x\hat{i}+l_y\hat{j}+l_z\hat{k}=\hat{i}(y{p}_z-z{p}_y)-\hat{j}(x{p}_z-z{p}_x)+\hat{k}(x{p}_y-y{p}_x)$

Therefore on comparison of coefficients,

$l_x=y{p}_z-z{p}_y$

$l_y=z{p}_x-x{p}_z$

$l_z=x{p}_y-y{p}_x$

The particle moves in the x-y plane. Hence the z component of the position vector and linear momentum vector becomes zero.

$z=p_z=0$

Thus $l_x=0$

$l_y=0$

$l_z=x{p}_y-y{p}_x$
Thus when particle is confined to move in the x-y plane, the angular momentum of particle is along the z-direction.