A cyclotron is operated at an oscillator frequency of $$24\,MHz$$   and has a dee radius $$R = 60\,cm.$$   What is magnitude of the magnetic field $$B$$ (in Tesla) to accelerate deuterons $$\left( {{\text{mass}} = 3.34 \times {{10}^{ - 27}}} \right)kg$$     ?

Solution :
Given : $$f = 24 \times {10^6}\,Hz,R = 0.60\,m$$
We know that, $$R = \frac{{mv}}{{qB}}$$
$$\eqalign{ & \therefore B = \frac{{mv}}{{qR}}\,......\left( {\text{i}} \right) \cr & {\text{where}}\,\,v = \omega R = 2\pi fR \cr & = 2\pi \times 24 \times {10^6} \times 0.60 \cr & = 9.04 \times {10^7}\,m/s \cr} $$
From (i),
$$B = \frac{{\left( {3.34 \times {{10}^{ - 27}}} \right)\left( {9.04 \times {{10}^7}} \right)}}{{1.6 \times {{10}^{ - 19}} \times 0.60}} = 3.2\,T.$$