Question

$n_1$ electrons/sec passes through a given cross-section towards right with a velocity $v_1$ and $n_2$ protons/sec passes through the same cross-section with velocity $v_2$ in the same direction, the current through a given cross-section will be :
$($Given: $n_1=1.5\times {10}^{10},n_2={10}^{10}\text{and charge on electron}=1.6\times {10}^{-19}\text{C})$

• A. ${10}^{-9}\text{A}$
• B. $8\times {10}^{-10}\text{A}$
• C. $0.5\times {10}^{-11}\text{A}$
• D. $4\times {10}^{-9}\text{A}$

Answer 1

The correct option is B $8\times {10}^{-10}\text{A}$

Motion of electron and protons leads to flow of charge through a given cross-section.


Let $i_1$ be the flow of current due to flow of electrons and $i_2$ be the current due to flow of protons.

$i_1=\frac{\Delta q}{\Delta t}=q_1{\left(\frac{dN}{dt}\right)}_e$

And similarly,

$i_2=q_2{\left(\frac{dN}{dt}\right)}_p$

Here, $q_1=-e,q_2=e$ and $\frac{dN}{dt}$ represents the rate of particles passing through the cross-section.

$\left|q_1\right|=\left|q_2\right|=e$

So, net current flow through cross-section :

$I=i_1-i_2$

$\Rightarrow I=e{\left(\frac{dN}{dt}\right)}_e-e{\left(\frac{dN}{dt}\right)}_p$

$\Rightarrow I=e(n_1-n_2)=e(1.5-1.0)\times {10}^{10}$

$\Rightarrow I=1.6\times {10}^{-19}\times 0.5\times {10}^{10}=0.8\times {10}^{-9}$

$\therefore I=8\times {10}^{-10}\text{A}$

Hence, option $\left(b\right)$ is correct.

Why this question ? Tip : The direction of current will be in the same direction due to flow of protons & it will be in the opposite direction due to flow of electrons $i_{net}=i_e+i_p$