Question

A cyclotron is opened at an oscillator frequency of $12MHz$ and has a dee radius $R=50cm$. What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron?

Answer 1

Step 1: Given data

Frequency of the cyclotron, $f=12MHz=12\times {10}^{6}Hz$

Dee radius of the cyclotron, $R=50cm$

Magnitude of the magnetic field needed to accelerate the proton, $B=?$

Step 2: Assumptions

Mass of proton, $m=1.67\times {10}^{-27}Kg$

Charge on electron, $q=1.6\times {10}^{-19}C$

Step 3: Formula used

$B=\frac{2\pi fm}{q}$………………………….(a)

Step 4: Calculation of the magnitude of the magnetic field needed to accelerate the proton

Substituting the given values in equation (a), we get

$$\begin{gathered} B=\frac{2\times 3.14\times 12\times {10}^{6}Hz\times 1.67\times {10}^{-27}Kg}{1.6\times {10}^{-19}C} \\ B=0.78T \end{gathered}$$

Hence, the magnitude of the magnetic field needed to accelerate the proton is equal to $0.78T$.