Question

A particle of charge `-q` and mass m moves in a circle of radius r around an infinitely long line charge of linear charge density `+lamda`. Then, time period will be
where , `k=1/4piepsilon_0)`
A. `T=2pir sqrt((m)/(2klambdaq))`
B. `T^(2)=(4pi^(2)m)/(2klambdaq)`
C. `T= 1/(2pir)sqrt((2klambdaq)/(m)`
D. `T T= 1/(2pir)sqrt((m)/(2klambdaq)`

Answer 1

Correct Answer - a
`(mv^(2))/(R )=q(lambda)/(2pi epsilon_(0)r)implies v=sqrt((q lambda)/(2piepsilon_(0)m))= sqrt((2kqlambda)/(m))`
`T = (2pir)/(v)= 2pirsqrt((m)/(2kqlambda))`