A particle having charge q=1 C and mass m=1kg is released from rest at origin. There are electric and manetic fields given by
`E=(10hati)N//C "for" xle1.8m` and
`B=(-5hatK)T "for" 1.8mlexle2.4m` A screen is placed parallel to y-z palane at x=3.0m. Neglect gravity forces.
y-coordinate of particle where it collides with be screen is .....m.

Options

A. `(0.6(sqrt(3)-1))/(sqrt(3))`
B. `(0.6(sqrt(3)+1))/(sqrt(3))`
C. `1.2(sqrt(3)+1)`
D. `(1.2(sqrt(3)-1))/(sqrt(3))`

Answer

D

Explanation

In magnetic field path of the particle is circle. Radius of circular path is
`r=(mv)/(qB)=((1)(6))/((1)(5))=1.2m`
`d=2.4m-1.8m=0.6m`
Since `d lt r sin theta=d/r=0.6/1/2 = 1/2`
`implies theta=30^(@)`
`AE=AD-DE=r-r cos theta`
`=r(1-cos theta) = 1.2 (1-(sqrt(3))/(2))=0.6(2-sqrt(3))`
`FC=BF tan theta = 0.6/(sqrt(3))`
`:.` y-co-ordinate `=AE+FC`
`=0.6 (2-sqrt(3))+(0.6)/(sqrt(3))`
`0.6[2-sqrt(3)+(1)/(sqer(3))]=(1.2(sqrt(3)-1))/(sqrt(3))m`.
Hence choice (d) is correct and other choices are wrong.