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)`

Options

A. `T = 2 pi r sqrt((m)/(2k lambda q))`
B. `T^(2) = (4 pi^(2) m)/(2 k lambdaq)r^(2)`
C. `T = (1)/(2 pi r) sqrt((2k lambda q)/(m))`
D. `T = (1)/(2 pi r ) sqrt(((m)/(2 k lambda q))`

Answer

A

Explanation

For circular motion of particle we have
`qE=(mv^2)/r implies q(lamda/(2pi e_0 r)) = (mv^2)/r implies V=sqrt((q^lamda)/(2 pi e_0))=sqrt((2 k q lamda)/m)`
Time period of circular motion is given as , `T=(2 pi r)/v implies T=2pi rsqrt(m/(2 k lamda q))`