Uniform electric and magnetic fields with strength E and induction B, respectively, are along y-axis as shown in Fig. A particle with specific charge `q//m` leaves the origin O in the direcction of x-axis with an initial non-relativistic velocity `v_0`

The angle `alpha` between the particle's velocity vector and y-axis at and moment is

Options

A. `tan^-1((3v_0B)/(2pinE))`
B. `tan^-1((v_0B)/(pinE))`
C. `tan^-1((v_0B)/(sqrt2pinE))`
D. `tan^-1((v_0B)/(2pinE))`

Answer

d

Explanation

Particle's acceleration in in y direction.
`(dv_y)/(dt)=(qE)/m=` constant
The motion of the particle is equivalent to circular motion in xz
plane with uniform acceleration in y-direction.
Hence, `v_0^2=v_x^2+v_z^2= constant`
The magnetic force cannot change the magnitude of `v_0`.
The y-component of velocity,
`v_y=(qE)/mt`
The y-coordinate at time t, `y=1/2 (qE)/mt^2`
The time period of circular motion in xz plane.
`T=(2pim)/(Bq)`
Let the particle cross y-axis after n rotations, then
`t=nT=(2pimn)/(qB)`
Thus, `y_n=(qE)/(2m)xx((2pimn)/(qB))^2=(2pi^2mn^2E)/(qB^2)`
As `v_y=a_yt=((qE)/m)((2pimn)/(qB))=(2pinE)/B`
Thus, `tan alpha=(v_0)/(v_y)=(v_0B)/(2pinE) implies alpha=tan^-1((v_0B)/(2pinE))`