There exist uniform magnetic field `vec(B)=5T(-hat(j))` from `x=0` to `x=2m`. From `x=2m` to `x=4m` a uniform electric field `vec(E)=50N//C(-hat(k))`. A point chage `(q=2C,m=1kg)` is projected from origin with velocity `vec(V)=20m//s(hat(k))`. Speed with which the charge particle will leave the electric fields is `10sqrt(n)m//s`. Find the value of `n`

Answer

5

Explanation

Path of the particle is circle in xz plane in magnetic field centre at `(2,0,0)` and it enter in electic field at `(2,0,2)` parallel to `x`-axis so
From O to A `=T/4=(pim)/(2qB)=(pi)/20s`
A to P `=1/10s`
Total `=((pi+2)/20)sv_(z)=u_(z)+a_(z)t=10m//s`
`( :' a_(z)=100m//s^(2))`
`v=10sqrt(5)`