A particle of mass 1 x 10 -26 kg and charge + 1.6 x 10 -19 coulomb travelling with a velocity 1.28 x 10 6 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 E x = E y = 0, E z = - 102.4 kV / m and B x = B z = 0, B y = 8 x 10 -2 water/ m 2 . 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 ?

Question

A particle of mass 1 x 10 -26 kg and charge + 1.6 x 10 -19 coulomb travelling with a velocity 1.28 x 10 6 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 E x = E y = 0, E z = - 102.4 kV / m and B x ­ = B z = 0, B y = 8 x 10 -2 water/ m 2 . 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 · Answer

Hello Student,

Please find the answer to your question

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^-26x 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
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