A particle of mass `1xx 10^(-26) kg` and charge `+1.6xx 10^(-19) C` travelling with a velocity `1.28xx 10^6 ms^-1` in the `+x` direction enters a region in which uniform electric field E and a uniform magnetic field of induction B are present such that `E_x = E_y = 0, E_z= -102.4 kV m^-1, and B_x = B_z =0, B_y = 8xx 10^-2.` The particle enters this region at time `t=0.` Determine the location (x,y,z coordinates) of the particle at `t= 5xx10^-6 s.` If the electric field is switched off at this instant (with the magnetic field present), what will be the position of the particle at `t= 7.45xx10^-6 s` ?

Explanation

The Lorentz force on the charged particle is
`F=q(vecE+vecvxxvecB)`.
The electric force on the charged particle, `F_E=qE_z` which acts
towards negative z direction.

The magnetic force on the charged particle, `F_B=qv_xB_y`.
As velocity of charge is in +x direction and magnetic field is
along +y direction, from right hand rule the magnetic force acts
along +z direction. The resultant force,
`F=F_E+F_B=q(E_z+v_xB_y)`
`=q[-102.4xx10^3+1.28xx10^6xx8xx10^-2]=0`
During time `t=0` to `t_1=5xx10^-6`, the resultant force on the
particle is zero, it moves with uniform velocity `v_x`. The position
of the particle `(X_1,Y_1,Z_1)` after time `t_1` is
`X_1=v_xt_1=(1.28xx10^6)xx(5xx10^-6)=6.4m`
When electric field is switched off, the particle circulates in xz-
plane under the influence of magnitude field.
Radius R of circulation is
`R=(mv_x)/(qB_y)=(10^-26xx1.28xx10^6)/(1.6xx10^-19xx8xx10^-2)=1m`
`theta=(P_1P_2)/R=(v_x(t_2-t_1))/R=(1.28xx10^6)xx(2.45xx10^-6)/1`
`=3.136=piradian`
The coordinates of the particle are
`X_2=X_1+R sin theta=X_1+R sin pi=X_(1)=16.4m`
and `Z_2=R-Rcos theta=R-Rcos pi=2R=2m`
Note that `theta=pi` implies that `t_2=T//2`, where T is time period of
circulation. We could have written the result directly.