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  4. 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

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Q. 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 through a given cross section. The magnitude and direction of current through the cross section

Current Electricity

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 $i_{e}=\frac{n_{e} q_{e}}{t}==3 \times 10^{18} \times 1.6 \times 10^{-19}=$
$0.48 A$ (Opposite to the motion of electrons, i.e. right to left)
Current due to protons $i_{p}=\frac{n_{p} q_{p}}{t}=2 \times 10^{18} \times 1.6 \times$ $10^{-19}=0.32\, A$ (Right to left) so total $I= i _{ e }+ i _{ p }=0.48+$ $0.32=0.80 \,A$ (Right to left) Hence correct answer is $d$