Question

The $z$ component of the angular momentum of a particle whose position vector is $\vec{r}$ with components $x,y$ and $z$ and linear momentum is $\vec{p}$ with components $p_x,p_y$ and $p_z$ is

• A. $x{p}_y-y{p}_x$
• B. $y{p}_z-z{p}_y$
• C. $z{p}_x-x{p}_z$
• D. $x{p}_y+y{p}_x$

Answer 1

Answer: (A)

Here, $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$
$\vec{p}=p_x\hat{i}+p_y\hat{j}+p_z\hat{k}$
Let $\vec{L}=L_x\hat{i}+L_y\hat{j}+L_z\hat{k}...\left(i\right)$
As $\vec{L}=\vec{r}\times \vec{p}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x& y& z\\ p_x& p_y& p_z\end{array}\right|$
$=\hat{i}\left(y{p}_z-z{p}_y\right)+\hat{j}\left(z{p}_x-x{p}_z\right)+\hat{k}\left(x{p}_y-y{p}_x\right)...\left(ii\right)$
Comparing the coefficients of $\hat{i},\hat{j}$ and $\hat{k}$ in eqs. $\left(i\right)$ and $\left(ii\right)$. we get
$L_x=y{p}_z-z{p}_y$
$L_y=z{P}_x-x{p}_z$
$L_z=x{p}_y-y{p}_x$