Question
potential diff b/w dees of cyclotron is V , if charge q comes out of it after completing n revolution then gain in KE of charge is 2nqV HOW?
Asked By: SARIKA 1 Month ago
is this question helpfull: 1 0 submit your answer
Joshi sir comment
when charge will move from first dee to second dee it will gain qV and on returning to first dee it will gain qV again so in one revolution it will gain 2qV so in n revolutions it will gain 2qVn
Read 1 Solution.
U=qV
loss of P.E. = gain in K.E.
on passing from one dee to another , PE lost = qv
in one round , it passes 2 times, so total PE lost = 2qV
therefore, KE = 2qV
in n rounds, KE gained = 2nqV 
SANKALP SINGH DEV 10 year ago
is this solution helpfull: 0 0