An electron is shot into one end of a solenoid.As it enters the uniform magnetic field within the solenoid, its speed is `800 m//s` and its velocity vector makes an angle of `30^(@)` with the central axis of the solenoid. The solenoid carries `4.0 A` current and has `8000` turn along its length.Find number of revolutions made by the electron within the solenoid by the time it emerges from the solenoid's opposite end. (Use charge of mass ratio `e/m` for electron `=sqrt3xx10^(11) C//kg`)Fill your answer in multiple of `10^(3)` (neglect end effect)

Answer

`1600`

Explanation

`B_(sol)=mu_(0)N/Li` where `N` is total no. of turns `&` `L` is length of the solenoid.
`T=(2pim)/(qB)` and picth `=V_(||)T`
No. of revolution `=(L qB)/pitch =L/(V_(||).2pim)`
using `B=mu_(0)N/L. irArr(mu_(0).Ni)/(V_(||).2pi).q/m`
using values `rArr(4pixx10^(-7)xx8000xx4xxsqrt3xx10^(11))/(400.sqrt3xx2pi)=16xx10^(5)`