Question

In a hydrogen discharge tube, it is observed that through a given cross-section $3.13\times {10}^{15}$ electrons are moving per second from right to left and $3.12\times {10}^{15}$ protons are moving from left to right per second. The electric current in the discharge tube and its direction will be

• A. $1.6\mu A$ towards left
• B. $1.6\mu A$ towards right
• C. $1mA$ towards left
• D. $1mA$ towards right

Answer 1

The correct option is D

$1mA$ towards right

As electrons and protons are moving in opposite directions, the current due to them will add up.
$I=(n_1+n_2)\frac{q}{t}=(3.13+3.12)\times {10}^{15}\times 1.6\times {10}^{-19}=10\times {10}^{-4}A=1mA$
The current will be in direction of motion of protons, i.e., towards right.