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. $xp_y - yp_x$
• B. $yp_z - zp_y$
• C. $zp_x - xp_z$
• D. $xp_y + yp_x$

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} \quad...\left(i\right) $
As $\vec{L} = \vec{r} \times\vec{p} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ x&y&z\\ p_{x}&p_{y}&p_{z}\end{vmatrix}$
$ =\hat{i}\left(yp_{z} -zp_{y}\right) +\hat{j} \left(zp_{x} -xp_{z}\right) +\hat{k}\left(xp_{y}-yp_{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} = yp_{z} -zp_{y} $
$ L_{y} = zP_{x} -xp_{z} $
$ L_{z} = xp_{y} -yp_{x}$