Question

In a certain cyclotron a proton moves in a circle of radius 0.500 m. The magnitude of the magnetic field is 1.20T.

(a)What is the oscillator frequency?

(b)What is the kinetic energy of the proton, in electron-volts?

Short answer

1. The oscillator frequency is 1.83 x 10⁷ Hz.
2. The kinetic energy of the proton is 1.72 x 10⁷ Hz.

Step-by-step solution

Listing the given quantities

• The radius of a circular orbit, r = 0.500m
• Magnetic field, B = 1.20 T
• Charge of the proton, $q=1.60\times {10}^{-19}C$
• Mass of proton,$m_p=1.67\times {10}^{-27}kg$

Understanding the concept of the frequency of the oscillator, kinetic energy

We can find the frequency of the oscillator by using the value of the magnetic field. We can find the kinetic energy by using the formula for the radius of the circle in which the proton moves and the kinetic energy of the proton by substituting the given values.

Formula:

Oscillator frequency, $f_{osc}=\frac{qB}{2{\mathrm{\pi m}}_{\mathrm{p}}}$

The radius of the circle in which the proton moves, $r=\frac{mv}{\left|q\right|B}$

The kinetic energy of the proton,$K=\frac{1}{2}{m}_p{V}^2$

(a) Calculation of the oscillating frequency

The oscillating frequency is given by:

$\begin{array}{rcl}{f}_{osc}& =& \frac{qB}{2{\mathrm{\pi m}}_{\mathrm{p}}}\\ & =& \frac{\left(1.60\times {10}^{-19}\right)\left(1.20\right)}{2\mathrm{\pi }\left(1.67\times {10}^{-27}\right)}\\ & =& 1.83\times {10}^{7}Hz\end{array}$

The oscillator frequency is $1.83\times {10}^{7}Hz$.

(b) Calculation of the kinetic energy

The radius oftheorbit in whichtheproton moves in a circular orbit is:

$r=\frac{m_{p}v}{\left|q\right|B}$

And the kinetic energy of the proton is:

$K=\frac{1}{2}{m}_p{v}^2$

Rearranging the equation for v :

$v=\sqrt{2K\overline{)m_p}}$

The radius becomes,

$$\begin{gathered} r=\frac{m_p\sqrt{2K\overline{)m_p}}}{\left|q\right|B} \\ r=\frac{\sqrt{2m_{p}K}}{\left|q\right|B} \end{gathered}$$

We can rearrange this equation to get kinetic energy as:

$\begin{array}{rcl}K& =& \frac{{\left(rqB\right)}^2}{2m_p}\\ & =& \frac{{\left[\left(0.500\right)\left(1.60\times {10}^{-19}\right)\left(1.20\right)\right]}^2}{\left[2\left(1.67\times {10}^{-27}\right)\left(1.60\times {10}^{-19}\right)\right]}\\ & =& 1.72\times {10}^{7}eV\end{array}$

The kinetic energy of the proton is $\begin{array}{rcl}& & 1.72\times {10}^{7}eV\end{array}$.