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

Answer 2

Explanation:

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Given, Magnetic field B=2T

Potential gap ΔV=100kV

We know that, r= qB/mv

⇒V=100×10 3

=10 5volts

So, KE gained in each revolution =e(V)+e(V)=2e(V)=2e×10 5

Thus, number of revolution N= 2×10 5/20×10 6

=100revolution

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