Question
Question: In a gas discharge tube if 3 × 10<sup>18</sup> electrons are flowing per sec from left to right and ...
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
C
0.80A, left to right
D
0.80 A, right to left
Answer
0.80 A, right to left
Explanation
Solution
As current is rate of flow of charge in the direction in which positive charge will move, the current due to electron will be ie = 
= 3 × 1018 × 1.6 × 10−19 = 0.48 A (Opposite to the motion of electrons, i.e. right to left)
Current due to protons, ip =
= 2 × 1018 × 1.6 × 10−19
= 0.32 A (Right to left) so total I = ie + ip
⇒ = 0.48 + 0.32 = 0.80 A (Right to left)
Hence correct answer is