A charge particle of charge q and mass m is moving with velocity v as shown in fig In a uniform magnetic field B along -ve z-direction.Select the correct alternative (s).

Options

A. Velocity of the particle when it comes out form the magnetic field is `vecv-vcos 60^@hati+vsin 60^@hatj`
B. Time for which the particle was in magnetic field is `(pim)/(3qB)`
C. Distance travelled in magnetic field is `(pimv)/(3qB)`
D. None of these

Answer

(a,b,c)

Explanation

Arc length: `AB=pi/3r=(pimv)/(3qB)`

Time: `t=theta/omega=(pi//3)/(2pi//T)=T/6=(pim)/(3qB)`
`Distance =vt=(pimv)/(3qB)`