Question

A particle of mass $1\times {10}^{-26}kg$ and charge $+1.6\times {10}^{-19}C$ travelling with a velocity $1.28\times {10}^{6}m{s}^{-1}$ 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 $E_x=E_y=0,E_z=-102.4kV{m}^{-1}$, and $B_x=B_z=0,B_y=8\times {10}^{-2}$. The particle enters this region at time $t=0$. Determine the location (x, y, z coordinates) of the particle as $t=5\times {10}^{-6}s$. If the electric field is switched off at this instant (with the magnetic field present), what will be the position of the particle at $t=7.45\times {10}^{-6}s$?

Answer 1

The Lorentz force on the charged particle is: $F=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$.
The electric force on the charged particle, $F_E=q{E}_z$ which acts toward negative z direction.
The magnetic force on the charged particle, $F_B=q{v}_x{B}_y$.
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_x{B}_y)$$=q[-102.4\times {10}^3+1.28\times {10}^6\times 8\times {10}^{-2}]=0$

During time $t=0$ to $t_1=5\times {10}^{-6}$, 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: $X_1=v_x{t}_1=(1.28\times {10}^6)\times (5\times {10}^{-6})=6.4m$
When electric field is switched off, the particle circulates in xz-plane under the influence of magnetic field.
Radius R of circulation is:

$R=\frac{m{v}_x}{q{B}_y}=\frac{{10}^{-26}\times 1.28\times {10}^6}{1.6\times {10}^{-19}\times 8\times {10}^{-2}}=1m$

$\theta =\frac{P_1{P}_2}{R}=\frac{v_x(t_2-t_1)}{R}=\frac{(1.28\times {10}^6)\times (2.45\times {10}^{-6}}{1}$$=3.136=\pi radian$

The coordinates of the particle are:
$X_2=X_1+Rsin\theta =X_1+Rsin\pi =X_1=6.4m$
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.