The electric field is swiched off at time `t=0.2xx10^(-1) sec`. What will be the postion of the particle at time `t_(0)=0.6xx10^(-1) sec`, if the ration of initial velocity, `v` to the radius of the curve traced by the particle is `(100 pi)/4`?

Options

A. `(v/50,0,(mv)/(2qB))`
B. `(v/25,0,(mv)/(qB))`
C. `(v/25,0,(2mv)/(qB))`
D. `(v/50,0,(2mv)/(qB))`

Answer

D

Explanation

As electric field is switched off, particle will describe a circle.

Radius of circle, `r=(mv)/(qB)`
`x=x_(0)+r sin theta, y=0`
`z=r(1-cos theta)`
But `S=r_(theta)=v(0.6-0.2)xx10^(-1)`
So, `theta=v/r(0.4xx10^(-1))=((100 pi)/4)xx0.04=3.14`
`=pi radian`
So, `x=v/50+0=v/50`
`z=(mv)/(qB)(1-cos pi)=(2mv)/(qB)`