Chapter 20, Problem 22P

To determine

To show: A charged particle moving in a circular field has an angular momentum of $qB{r}^2$ .

Answer to Problem 22P

The angular momentum of a charged particle is equal to $qB{r}^2$ .

Explanation of Solution

Introduction:

According to the conservation of energy principle, the electrical potential energy is equal to the kinetic energy as it starts accelerating.

Since charged particle moving in a circular path, thus magnetic force applied on it will be perpendicular to it. Now, according to the energy conservation principle,

  $\begin{array}{l}{F}_{\max }=qvB=\frac{m{v}^2}{r}\\ v=\frac{qBr}{m}\end{array}$

Since angular momentum is given as $L=mvr$ . Inserting the value of velocity in the formula of angular momentum, we get

  $\begin{array}{l}L=m\frac{qBr}{m}r\\ L=qB{r}^2\end{array}$

Conclusion:

Thus, the angular momentum $L$ of a charged particle is equal to $qB{r}^2$ .