A charged particle of specific charge s passes undeviated through region 1 as shown in Fig.

Options

A. Velocity of paticle in rigion 1 is `v=E/B`
B. Work done to move the charged particle in region 1 and region 2 is zero.
C. The radius of the trajectory of the charged particle in Region 2 is `(E//s BB_(0))`
D. The particle emerges from region 2 with a velocity `vec(v)` where `vec(v')` = -`vec(v)` for `l_(2) gt (E)/(BB_(0))`

Answer

A, B, C, D

Explanation

Net force on charge should be zero in region 1,
`qE=qvB implies v=E//B`
In region 1, net force is zero on particle, hence on work is done on particle in region 1,
In region 2, only magnetic force is there, hence no work is done in this region also.
In region 2: `r=(mv)/(qB_(0))=(E//B)/((q//m)(B_(0)=(E)/(sBB_(0))))=(E)/(sBB_(0))`
If`l_(2)gtr` or` l_(2)gt(E)/(sBB_(0))`, then particle will complete haf revolution in region 2 and come out of it with same velocity in opposite direction.