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 $p_{ r }, p_{ y }$ and $p_{z} .$ Show that if the particle moves only in the $x -y$ plane the angular momentum has only a $z-$component.

$l_{ x }=y p_{ z }-z p_{ y } l_{ y }$

$=z p_{ x }- x p_{ z } l_{ z }$

$=x p_{y}-y p_{x}$

Linear momentum of the particle, $\vec{p}=p_{ x } \hat{ i }+p_{y} \hat{ j }+p_{2} \hat{ k }$

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

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

$=(x \hat{ i }+y \hat{ j }+z \hat{ k }) \times\left(p_{x} \hat{ i }+p_{y} \hat{ j }+p_{z} \hat{ k }\right)$

$=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ x & y & z \\ p_{x} & p_{y} & p_{z}\end{array}\right|$

$l_{ x } \hat{ i }+l_{ y } \hat{ j }+l_{ z } \hat{ k }$$=\hat{ i }\left(y p_{z}-z p_{ y }\right)-\hat{ j }\left(x p_{z}-z p_{ x }\right)+\hat{ k }\left(x p_{y}-z p_{ x }\right)$

Comparing the coefficients of $\hat{ i }, \hat{ j },$ and $\hat{ k },$ we get:

$\left.\begin{array}{l}l_{ x }=y p_{ z }-z p_{ y } \\ l_{ y }=x p_{ z }-z p_{ x } \\ l_{z}=x p_{y}-y p_{ x }\end{array}\right\}$ $\dots(i)$

The particle moves in the $x$ $-y$ plane. Hence, the $z$ -component of the position vector

and linear momentum vector becomes zero, i.e., $z=p_{z}=0$

Thus, equation ( $i$ ) reduces to:

$l_{x}=0$

$l_{y}=0$

$l_{z}=x p_{y}-y p_{x}$

Therefore, when the particle is confined to move in the $x-y$ plane, the direction of angular momentum is along the $z$ -direction.