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

(A) The angular momentum $\vec{l}$ is defined as the cross product of the position vector $\vec{r}$ and the linear momentum vector $\vec{p}$:
$\vec{l} = \vec{r} \times \vec{p} = (x \hat{i} + y \hat{j} + z \hat{k}) \times (p_x \hat{i} + p_y \hat{j} + p_z \hat{k})$
Using the determinant form:
$\vec{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix} = \hat{i}(y p_z - z p_y) - \hat{j}(x p_z - z p_x) + \hat{k}(x p_y - y p_x)$
Comparing components,we get:
$l_x = y p_z - z p_y$
$l_y = z p_x - x p_z$
$l_z = x p_y - y p_x$
If the particle moves only in the $x-y$ plane,then $z = 0$ and $p_z = 0$. Substituting these into the expressions:
$l_x = y(0) - (0)p_y = 0$
$l_y = (0)p_x - x(0) = 0$
$l_z = x p_y - y p_x$
Since $l_x = 0$ and $l_y = 0$,the angular momentum has only a $z$-component.