Ten million electrons pass from point P to point Q in one microsecond. The current and its direction is?
A) $1.6\times {{10}^{-14}}A$ $\text{From point P to point Q}$
B) $3.2\times {1}{{{0}}^{-14}}A$ $\text{From point P to point Q}$
C) $1.6\times {{10}^{-14}}A$ $\text{From point Q to point P}$
D) $3.2\times {1}{{{0}}^{-14}}A$ $\text{From point Q to point P}$
Answer
Verified
116.4k+ views
Hint: Current is the flow of electrons, but current and electrons flow in the opposite direction. Current flows from positive to negative and electrons flow from negative to positive. The current is determined by the number of electrons passing through the cross-section of a conductor in one second.
Complete step by step solution:
A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current I is defined to be:
\[{I = }\dfrac{\Delta {Q}}{\Delta t}\]
where \[\Delta Q\] is the amount of charge passing through a given area in time \[\Delta t=t\]. (As in previous chapters, initial time is often taken to be zero, in which case \[\Delta t=t\]. The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère.
Since \[{I = }\dfrac{\Delta {Q}}{\Delta t}\] we see that an ampere is one coulomb per second: One ampere is equivalent to one coulomb of charge flowing per second,
Here, number of electrons,
\[n=10000000={{10}^{7}}\]
Total charge on ten million electrons is Q = ne
Where
$\Rightarrow e=1.6\times {{10}^{-19}}$
$\Rightarrow {Q = 1}{{{0}}^{7}}\times 1.6\times {{10}^{-19}}$
$\therefore {Q = 1}{.6}\times {1}{{{0}}^{-12}}{ C}$
Time taken by ten million electrons to pass from point P to point Q is
$\Rightarrow t = 1 \mu s$
$\Rightarrow t = 10^{-6}$
$\Rightarrow \text{The current}$ ${I = }\dfrac{\Delta {Q}}{\Delta {t}}$
$\Rightarrow {I = }\dfrac{1.6\times {{10}^{-12}}}{{{10}^{-6}}}$
$\therefore { I = 1}{.6 }\times { 1}{{{0}}^{-6}}{ A}$
Since the direction of the current is always opposite to the direction of flow of electrons. Therefore, due to flow of electrons from point P to point Q the current will flow from point Q to point P.
Note: One electron has a charge of \[1.602\times {{10}^{-19}}{ coulombs}\]. Consequently, you can find the number of electrons in 1 coulomb of charge by taking the reciprocal of this number.
\[{1 coulomb}={ }\dfrac{1}{1.602\times {{10}^{-19}}}{ = 6}{.242}\times {{10}^{18}}{ Electrons}{.}\]An electric current of 1 ampere is equal to 1 coulomb of charge passing a point in a circuit every second:
Therefore, a current of 1 ampere = \[{6}{.242}\times {{10}^{18}}\]electrons moving past any point in a circuit every second.
Complete step by step solution:
A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current I is defined to be:
\[{I = }\dfrac{\Delta {Q}}{\Delta t}\]
where \[\Delta Q\] is the amount of charge passing through a given area in time \[\Delta t=t\]. (As in previous chapters, initial time is often taken to be zero, in which case \[\Delta t=t\]. The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère.
Since \[{I = }\dfrac{\Delta {Q}}{\Delta t}\] we see that an ampere is one coulomb per second: One ampere is equivalent to one coulomb of charge flowing per second,
Here, number of electrons,
\[n=10000000={{10}^{7}}\]
Total charge on ten million electrons is Q = ne
Where
$\Rightarrow e=1.6\times {{10}^{-19}}$
$\Rightarrow {Q = 1}{{{0}}^{7}}\times 1.6\times {{10}^{-19}}$
$\therefore {Q = 1}{.6}\times {1}{{{0}}^{-12}}{ C}$
Time taken by ten million electrons to pass from point P to point Q is
$\Rightarrow t = 1 \mu s$
$\Rightarrow t = 10^{-6}$
$\Rightarrow \text{The current}$ ${I = }\dfrac{\Delta {Q}}{\Delta {t}}$
$\Rightarrow {I = }\dfrac{1.6\times {{10}^{-12}}}{{{10}^{-6}}}$
$\therefore { I = 1}{.6 }\times { 1}{{{0}}^{-6}}{ A}$
Since the direction of the current is always opposite to the direction of flow of electrons. Therefore, due to flow of electrons from point P to point Q the current will flow from point Q to point P.
Note: One electron has a charge of \[1.602\times {{10}^{-19}}{ coulombs}\]. Consequently, you can find the number of electrons in 1 coulomb of charge by taking the reciprocal of this number.
\[{1 coulomb}={ }\dfrac{1}{1.602\times {{10}^{-19}}}{ = 6}{.242}\times {{10}^{18}}{ Electrons}{.}\]An electric current of 1 ampere is equal to 1 coulomb of charge passing a point in a circuit every second:
Therefore, a current of 1 ampere = \[{6}{.242}\times {{10}^{18}}\]electrons moving past any point in a circuit every second.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
Young's Double Slit Experiment Step by Step Derivation
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
Dual Nature of Radiation and Matter Class 12 Notes CBSE Physics Chapter 11 (Free PDF Download)
Charging and Discharging of Capacitor
JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Physics Average Value and RMS Value JEE Main 2025