A magnetic field `vec(B)=-B_(0)hati` exists within a sphere of radius `R=v_(0)T sqrt(3)` where `T` is the time period of one revolution of a charged particle starting its motion from origin and moving with a velocity `vec(v)_(0)=(v_(0))/2sqrt(3)hati-(v_(0))/2 hatj`. Find the number of turns that the particle will take to come out of the magnetic field.

Options

A. `5`
B. `1`
C. `4`
D. `2`

Answer

D

Explanation

`v_(bot er)=-(v_(0))/2` will contribute to circular motion.
`v_(||)=(v_(0)sqrt(3))/2` will contribute to helical path.
`T=2pim/(B_(0)q)`
Pitch, `P=v_(||)T=(v_(0)sqrt(3))/2 T`
Numbers of turns `=R/P=(v_(0)Tsqrt(3)(2))/(v_(0)sqrt(3)T) =2`