Question

A cyclotron is used to accelerate protons to a kinetic energy of $5\text{MeV}$. If the strength of the applied magnetic field in the cyclotron is $2\text{T}$, the radius of the cyclotron will be
$($Velocity of proton is $3\times {10}^7\text{m/s})$

• A. $0.241\text{m}$
• B. $0.167\text{m}$
• C. $0.423\text{m}$
• D. $0.765\text{m}$

Answer 1

The correct option is B $0.167\text{m}$

Given,
The kinetic energy $\left(K\right)$ of proton,

$K=5\text{MeV}=5\times {10}^6\times (1.6\times {10}^{-19})\text{J}$

$\therefore K=8.0\times {10}^{-13}\text{J}$

The magnetic force acting on proton will act as source of centripetal force, so

$qvB=\frac{m{v}^2}{r}\Rightarrow \frac{qvB}{2}=\frac{1}{2}\frac{m{v}^2}{r}=\frac{K}{r}$

$\therefore r=\frac{2K}{qvB}$

Substituting the values, we get

$r=\frac{2\times 8.0\times {10}^{-13}}{1.6\times {10}^{-19}\times 3\times {10}^7\times 2}=\frac{10}{6}\times {10}^{-1}$

$\therefore r=0.167\text{m}$

Hence, option $\left(b\right)$ is correct answer.