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In a gas discharge tube, if 3 times 10^1...

In a gas discharge tube, if `3 times 10^18` electrons are flowing per sec from left to right and `2 times 10^18` protons are flowing per second from right to left of a given cross-section, the magnitude and direction of current through the cross-section :

A

0.48 A, left to right

B

0.48 A, right to left

C

0.80 A, left to right

D

0.80 A, right to left

Text Solution

Verified by Experts

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=(n_eq_e)/t=3 times 10^18 times 1.6 times 10^-19=0.48A`
(Opposite to the motion of electrons, i.e., right to left) Current due to protons
)`i_p=(n_qq_p)/t=2 times10^18 times 1.6 times 10^-19=0.32 A`
So total `I=j_e+i_p, 0.48+0.32=0.80` RIght to left
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