Question

A proton is accelerated in a cyclotron where the applied magnetic field is 0.5 T. If the potential gap is 10 kV, then how much revolutions the proton has to make between the dee's to acquire a kinetic energy of 12 MeV?​

Answer 1

Answer:

potential gap= 10 kv= 10× 10^3 V

kinetic energy= 12Mev

[ 1mev= 1.6× 10^-13 joule ]

than

12Mev= 12× 1.6× 10^-13 joule

.

so, number of turns = K.E÷ 2qv.

= 12× 1.6× 10^-13 ÷ 2× 1.6× 10^-19×10× 10^3.

solve it and you will get your answer