A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T magnetic field perpendicular to the proton orbits. When the protons have acheived a kinetic energy of 2.7 MeV, what is the radius of their circular orbit and what is their angular speed?

Answer;

Radius = 0.0818 m

Angular velocity = 2.775 * 10^7 rad/sec

Explanation;

The mass of proton m=1.6748 * 10^-27 kg;

Charge of electron e = 1.602 * 10^-19 C;

kinetic energy E = 2.7 MeV

= 2.7 * 10^6 * 1.602 * 10^-19 J;

= 4.32 * 10^-13 Joules

But; K. E = 0.5m*v^2,

Hence v=√ (2K. E/m)

Velocity = 2.27 * 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √ (2K. E/m) = ωr

r = √ (2E/m) / w = √E*√ (2*m) / (eB)

= √E * √ (2*1.6748*10^-27) / (1.602*10^-19 * 2.9)

but E = 4.32 * 10^-13 Joules

r = 0.0818 m

Angular speed

Angular velocity, ω = v/r, where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 * 10^7 / 0.0818 m

= 2.775 * 10^7 rad / sec