Question

In hydrogen discharge tube, it is observed that through a given cross-section $3.31\times {10}^{15}$ electrons are moving from right to left and $3.12\times {10}^{15}$ protons are moving from left to right. The current in the discharge tube and its direction will be

• A. $2mA$ towards left
• B. $2mA$, towards right
• C. $1mA$, towards right
• D. $2mA$, towards left

Answer 1

Answer: (C)

$1mA$, towards right

Here, number of electrons $n_e=3.31\times {10}^{15}$
and number of protons $n_p=3.12\times {10}^{15}$

As both the charge carriers are moving opposite to each other thus the current due to each carrier will add up.

Current due to both the charge carriers is in right direction.
Current $I=n_e{q}_e+n_p{q}_p$
$=3.31\times {10}^{15}\times 1.6\times {10}^{-19}+3.12\times {10}^{15}\times 1.6\times {10}^{-19}$
$=1.03\times {10}^{-3}\approx 1mA$ towards right.