Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

In a cyclotron, centrifugal force needed for rotation is provided by the magnetic force. Thus \frac{m v^{2}}{r}=e v B \Rightarrow \frac{v}{r}=\frac{e B}{m} \Rightarrow f=\frac{1}{T}=\frac{v}{2 \pi r}=\frac{e B}{2 \pi m} (a) Now for a proton m=1.6726 \times 10^{-27} \mathrm{~kg} e=1.6 \times 10^{-19} Here B=1.5 \mathrm{~T}. Thus frequency of oscillation is f=\frac{e B}{2 \pi m}=\frac{1.6 \times 10^{-19} \times 1.5}{2 \pi \times 1.6726 \times 10^{-27}} \mathrm{~Hz}=2.284 \times 10^{7} \mathrm{~Hz} (b) We know that \frac{m v^{2}}{r}=e v B \Rightarrow p=m v=e B r \therefore K E=\frac{p^{2}}{2 m}=\frac{e^{2} B^{2} r^{2}}{2 m} Here r=0.7 \mathrm{~m}. Thus kinetic energy is K E=\frac{e^{2} B^{2} r^{2}}{2 m}=\frac{\left(1.6 \times 10^{-19}\right)^{2} \times 1.5^{2} \times 0.7^{2}}{2 \times 1.6726 \times 10^{-27}} \mathrm{~J}=8.437 \times 10^{-12} \mathrm{~J}   ...