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)r^(3)`C. `T^(2)=(1)/(2pir)sqrt((2klambdaq)/(m))`D. `T=(1)/(2pir)sqrt((m)/(2klambdaq))` where `k=(1)/(4piepsilon_(0))`

A particle of charge `-q` and mass m moves in a circle of radius r aro

1.	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)r^(3)`C. `T^(2)=(1)/(2pir)sqrt((2klambdaq)/(m))`D. `T=(1)/(2pir)sqrt((m)/(2klambdaq))` where `k=(1)/(4piepsilon_(0))`
Answer» Correct Answer - A We have centripetal force equation `q((2klambda)/(r))=(mv^(2))/(r) sov=sqrt((2kqlambda)/(m))`Now, `T=(2pir)/(v)=sqrt((m)/(2klambdaq))` where `k=(1)/(4piepsilon_(0))`

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