Question In a certain cyclotron a proton moves in a circle of radius 0.700 m. The magnitude of the magnetic field is 1.50 T. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton? (a) Number i Units (b) Number i Units

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}   ...