Magnetic force on a charged particle is given by `vec F_(m) = q(vec(v) xx vec(B))` and electrostatic force `vec F_(e) = q vec (E)`. A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by `vec(E) = (10 hat(i)) N//C for x = 1.8 m` and `vec(B) = -(5 hat(k)) T` for `1.8 m le x le 2.4 m`
A screen is placed parallel to y-z plane at `x = 3 m`. Neglect gravity forces.
y-coordinate of particle where it collides with screen (in meters) is

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.