Particle accelerators: A cyclotron of radius 0.25 m accelerates nonrelativistic protons in a 1.7-T magnetic field. What is the kinetic energy of the protons when they leave the cyclotron? (1 eV = 1.60 × 10^-19 J, e = 1.60 × 10^-19 C, m_proton = 1.67 × 10^-27 kg) A) 3.7 MeV B) 7.8 keV C) 8.7 keV D) 7.8 MeV E) 8.7 MeV

Question

Quizplus · Accepted Answer

The cyclotron equation is given by:
$$\frac{qB}{m}r=\frac{mv^2}{r}$$
where $q$ is the charge of the particle (in this case, a proton), $B$ is the magnetic field strength, $m$ is the mass of the particle, $r$ is the radius of the cyclotron, and $v$ is the speed of the particle.
Solving for $v$, we get:
$$v=\sqrt{\frac{qBr^2}{m}}$$
The kinetic energy of a particle is given by:
$$K=\frac{1}{2}mv^2$$
Substituting the expression for $v$ and solving for $K$, we get:
$$K=\frac{1}{2}m\left(\frac{qBr^2}{m}ight)=\frac{q^2B^2r^2}{2m}$$
Substituting the given values and noting that $q=e$, we get:
$$K=\frac{(1.60 	imes 10^{-19}	ext{ C})^2 (1.7	ext{ T})^2 (0.25	ext{ m})^2}{2(1.67 	imes 10^{-27}	ext{ kg})}=8.7	ext{ MeV}$$
Therefore, the answer is E: 8.7 MeV.

Particle Accelerators: a Cyclotron of Radius 0