Derive the time period of the particle motion in a cyclotron.

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force required for circular motion.
The magnetic force is given by $F_B = qvB$.
The centripetal force is given by $F_c = \frac{mv^2}{r}$.
Equating these forces:

$$qvB = \frac{mv^2}{r}$$

Solving for the radius $r$:

$$r = \frac{mv^2}{qvB}$$ $$r = \frac{mv}{Bq}$$

The angular frequency $\omega$ is related to the linear velocity $v$ and the radius $r$ by the formula $v = \omega r$.
Substituting the expression for $r$ from Step 1 into this relationship:

$$v = \omega \left(\frac{mv}{Bq}\right)$$

Solving for $\omega$:

$$\omega = \frac{v}{\left(\frac{mv}{Bq}\right)}$$ $$\omega = \frac{Bq}{m}$$

The time period $T$ is the time taken for one complete revolution, which corresponds to an angular displacement of $2\pi$ radians.
The relationship between angular displacement $\Delta\theta$, angular frequency $\omega$, and time $t$ is $\Delta\theta = \omega t$.
For a full revolution, $\Delta\theta = 2\pi$ and $t = T$.

$$2\pi = \omega T$$

Substituting the expression for $\omega$ from Step 2:

$$2\pi = \left(\frac{Bq}{m}\right) T$$

Solving for $T$:

$$T = \frac{2\pi m}{Bq}$$

The image also shows a general formula for time $t$ for an angular displacement of $(2\pi - 2\theta)$:

$$t = \frac{(2\pi - 2\theta)m}{Bq}$$

This formula is derived from $\Delta\theta = \omega t$, where $\Delta\theta = (2\pi - 2\theta)$.
If the particle completes a full circle, then $\theta = 0^\circ$ (or the angular displacement is $2\pi$). In this case, the formula simplifies to:

$$t = \frac{(2\pi - 0)m}{Bq}$$ $$t = \frac{2\pi m}{Bq}$$

This confirms that the time for a full revolution is the time period $T$.