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
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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.

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 yes

SANKALP SINGH DEV 6 year ago is this solution helpfull: 0 0

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