Answer
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Hint:Before going through the question let us first know about current. A movement of particles (electrons) passing through wires and components is referred to as an electric current. It is the rate of charge movement. We say there is an electric current in a conductor when an electric charge passes through it. Electrons make up a charge flow in circuits that use metallic wires.
Complete step by step answer:
As current rises, the potential gap across rises as well. Since the overall resistance decreases, the potential difference will grow as the total current in the circuit grows. As current rises, the potential gap across rises as well. We know that,
\[I = \dfrac{{\Delta \theta }}{{\Delta t}}\]
In this case, $\Delta \theta $ = (number of electrons striking in the time interval $\Delta t$ ) (electronic charge)
\[{\mathbf{I}} = {\mathbf{3}}.{\mathbf{2mA}},\]
$\Delta t=1\,s$ and
Electronic charge = \[{\mathbf{1}}.{\mathbf{0}} \times {\mathbf{10}} - {\mathbf{19c}}\;,\] so,
\[{\mathbf{3}}.{\mathbf{2}} \times {\mathbf{1}}{0^{ - 3}}\]=\[\dfrac{{\Delta n}}{{{{(1.6 \times {{10}^{ - 19}})}^{ - 1}}}}(1)\]
\[\dfrac{{3.2 \times {{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 19}}}} = \Delta n\]
where \[\Delta {\mathbf{n}}\] = No. of electrons striking per sec.
$\therefore \Delta n = 2 \times {10^{16}}$ electrons striking the target per second
Hence, the correct option is A.
Note:The battery in a torch provides a charge flow, which means the torch's bulb glows due to the electric current. What is the purpose of a switch? The switch connects the cell to the bulb by forming a conducting bond. If the circuit is broken, the current stops flowing and the lamp does not light up.
Complete step by step answer:
As current rises, the potential gap across rises as well. Since the overall resistance decreases, the potential difference will grow as the total current in the circuit grows. As current rises, the potential gap across rises as well. We know that,
\[I = \dfrac{{\Delta \theta }}{{\Delta t}}\]
In this case, $\Delta \theta $ = (number of electrons striking in the time interval $\Delta t$ ) (electronic charge)
\[{\mathbf{I}} = {\mathbf{3}}.{\mathbf{2mA}},\]
$\Delta t=1\,s$ and
Electronic charge = \[{\mathbf{1}}.{\mathbf{0}} \times {\mathbf{10}} - {\mathbf{19c}}\;,\] so,
\[{\mathbf{3}}.{\mathbf{2}} \times {\mathbf{1}}{0^{ - 3}}\]=\[\dfrac{{\Delta n}}{{{{(1.6 \times {{10}^{ - 19}})}^{ - 1}}}}(1)\]
\[\dfrac{{3.2 \times {{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 19}}}} = \Delta n\]
where \[\Delta {\mathbf{n}}\] = No. of electrons striking per sec.
$\therefore \Delta n = 2 \times {10^{16}}$ electrons striking the target per second
Hence, the correct option is A.
Note:The battery in a torch provides a charge flow, which means the torch's bulb glows due to the electric current. What is the purpose of a switch? The switch connects the cell to the bulb by forming a conducting bond. If the circuit is broken, the current stops flowing and the lamp does not light up.
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