For cyclotron
\(\frac{m v^{2}}{r}\) = qvB
Here m is mass of charge particle, v is velocity of particle, Bis magnetic field.
∴ Radius of the path
r = \(\frac{m v}{q B}\)
Periodic time, T = \(\frac{2 \pi r}{v}=\frac{2 \pi m}{q B}\)
∴ Time spent in semicircle
\(\frac{T}{2}=\frac{\pi m}{q B}\)