A cyclotron is designed to accelerate protons (mass $1.67 \times 10^{-27} \mathrm{~kg}$ ) up to a kinetic energy of $2.5 \times 10^{-13} \mathrm{~J}$. If the radius of the dipole magnets is $96 \mathrm{~cm}$, what is the magnetic field used in the cyclotron?
A) $0.13 \mathrm{~T}$
B) $0.21 \mathrm{~T}$
C) $35 \mathrm{mT}$
D) $0.19 \mathrm{~T}$

Question

Quizplus · Accepted Answer

The kinetic energy of a proton is given by:$$K = \frac{1}{2}mv^2$$where $m$ is the mass of the proton and $v$ is its speed. The magnetic force on a proton is given by:$$F = qvB$$where $q$ is the charge of the proton, $v$ is its speed, and $B$ is the magnetic field strength. The centripetal force on a proton is given by:$$F_c = \frac{mv^2}{r}$$where $r$ is the radius of the circular path. In a cyclotron, the magnetic force and the centripetal force are equal, so we can set these two equations equal to each other and solve for $B$:$$qvB = \frac{mv^2}{r}$$$$B = \frac{mv}{qr}$$Plugging in the given values, we get:$$B = \frac{(1.67 \times 10^{-27} \mathrm{~kg})(2.5 \times 10^{-13} \mathrm{~J})}{(1.6 \times 10^{-19} \mathrm{C})(0.96 \mathrm{~m})}$$$$B = 0.19 \mathrm{~T}$$Therefore, the correct answer is D) $0.19 \mathrm{~T}$.

A Cyclotron Is Designed to Accelerate Protons (Mass 1.67×10−27 kg1.67 \times 10^{-27} \mathrm{~kg}1.67×10−27 kg

A Cyclotron Is Designed to Accelerate Protons (Mass $1.67 \times 10^{-27} \mathrm{~kg}$