# A particle of mass 1 x 10-26 kg and charge + 1.6 x 10-19 coulomb travelling with a velocity 1.28 x 106 m/s in the +X direction enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present such that Ex  = Ey  = 0, Ez = - 102.4 kV / m  and Bx­ = Bz = 0, By = 8 x 10-2 water/ m2.  The particle enters this region at the origin at time t = 0. Determine that location (x, y and x coordinates ) of the particle at t = 5 x 10-6 s. If the electric field is switched off at this instant (with the magnetic field still present ), what will be the position of the particle at t = 7. 45 x 10-6 s ?

Navjyot Kalra
10 years ago
Hello Student,
m = 10-26 kg, q = + 1.6 x 10-19 C
$\underset{v}{\rightarrow}$= (1.28 x 106 m/s) , = - 102.4 x 103 m/s
$\underset{B}{\rightarrow}$= 8 x 10-2
The force on the charged particle due to electric field.
F1 = qE in the – z direction
= 1.6 x 10-19 x 102.4 x 103 = 163. 84 x 10-16 N
The force on the charged particle due to magnetic field.
F2 = qvB in the + z direction
= 1.6 x 10-19 x 1.28 x 106 x 8 x 10-2 = 163.84 x 10-16 N
∴ The net force in z-direction is zero and thus the net force on the charged particle is zero and hence no acceleration.
Displacement of particle = v x t
= 1.28 x 106x 5 x 10-6 = 6.4 m
If the electric field is switched off at this instant then there will be a force due to magnetic field which is in the + ve z – direction.
This force is acting perpendicular to the velocity throughout and thus makes the particle to move in a circular path of radius
∴ r = mv / qB = 10-26 x 1.28 x 106 / 1.6 x 10-19 x 8 x 10-2 = 1m
The time period of revolution is
T = 2πr / v = 2π x 1 / 1.28 x 106 = 44.9 x 10-6 sec ….(i)
We have to find the position of the particle after time
(7.45 x 10-6 s – 5.0 x 10-6 s ) = 2.45 x 10-6 sec …… (ii)
On comparing (i) and (ii), we find that particle makes half the revolution.
Therefore, its co-ordinates after time 7.45 x 10-6 sec will be (6.4 m, 0,2 m)
Thanks
Navjot Kalra