In a gas discharge tube if 3 × 1018 electrons are flowing per sec from left to right and 2 × 1018 protons are flowing per second from right to left through a given cross section. The magnitude and direction of current through the cross section

Question

In a gas discharge tube if 3 × 1018 electrons are flowing per sec from left to right and 2 × 1018 protons are flowing per second from right to left through a given cross section. The magnitude and direction of current through the cross section (A) 0.48 , left to right (B) 0.48 A, right to left (C) 0.80A, left to right (D) 0.80 A, right to left

Tardigrade Admin · Accepted Answer

As current is rate of flow of charge in the direction in which positive charge will move, the current due to electron will be

i_{e} = \frac{n _{e} q _{e}}{t} == 3 \times 1 0^{18} \times 1.6 \times 1 0^{- 19} =

0.48 A

(Opposite to the motion of electrons, i.e. right to left)
Current due to protons

i_{p} = \frac{n _{p} q _{p}}{t} = 2 \times 1 0^{18} \times 1.6 \times

1 0^{- 19} = 0.32 A

(Right to left) so total

I = i_{e} + i_{p} = 0.48 +

0.32 = 0.80 A

(Right to left) Hence correct answer is

d

Q. In a gas discharge tube if 3×1018 electrons are flowing per sec from left to right and 2×1018 protons are flowing per second from right to left through a given cross section. The magnitude and direction of current through the cross section

0.48 , left to right

0.48 A, right to left

0.80A, left to right

0.80 A, right to left

Q. In a gas discharge tube if $3 \times 1 0^{18}$ electrons are flowing per sec from left to right and $2 \times 1 0^{18}$ protons are flowing per second from right to left through a given cross section. The magnitude and direction of current through the cross section