
A conductor wire, having ${{10}^{29}}$ free electrons per $~{{m}^{3}}$ carries a current of 20A. If the cross-section of the wire is $1m{{m}^{2}}$, then the drift-velocity of electrons will be the order of:
A. ${{10}^{-5~}}m{{s}^{-1}}$
B. ${{10}^{-3}}m{{s}^{-1}}$
C. $~{{10}^{-4}}m{{s}^{-1}}$
D. $10m{{s}^{-1}}$
Answer
581.1k+ views
Hint: Drift velocity is small, constant velocity with which the charge carriers seem to drift slightly in the conductor. It is the average velocity obtained by all the charge carriers to move in random directions. It depends on the no. of free charge carriers and cross-sectional area of the conductor.
Formula used:
Drift velocity is denoted by ${{V}_{d}}$ and its formula is given as:
\[\begin{align}
& ~{{V}_{d}}=I/e\times n\times A \\
& I\text{ }={{V}_{d}}\times e\times n\times A. \\
\end{align}\]
Where, I= current applied, n= no. of electrons, A= area of cross- section of the conductor.
Complete answer:
Drift velocity is the average velocity of all the charge carriers in the conducting materials.
In the conductors, there are charge carriers such as electrons that get randomness. It is also called a free gas model of conductions.
Drift velocity is the constant, small velocity with which electrons drift slightly with the direction opposite to the applied field.
Now, in question the no. of electrons and current is given.
According to the formula of drift velocity we have:
$\begin{align}
& {{V}_{d}}=I/n\times e\times A \\
& \Rightarrow {{V}_{d}}=20/{{10}^{29}}\times (1.6\times {{10}^{-19}})\times {{10}^{-6}} \\
& \Rightarrow {{V}_{d}}=20/1.6\times {{10}^{-4}} \\
& \Rightarrow {{V}_{d}}=1.25\times {{10}^{-3}} \\
& \Rightarrow {{V}_{d}}=1\times {{10}^{-3}} \\
& \Rightarrow {{V}_{d}}={{10}^{-3}}m{{s}^{-1}} \\
\end{align}$
So, option B is the correct option.
Note:
Drift velocity is the average velocity of all the charged carriers in the conductor wire. The charge carriers in the conductors get random motion when an external electric field is applied to it. By remembering the formula for drift velocity with some given terms we can solve ${{V}_{d}}$ easily.
Formula used:
Drift velocity is denoted by ${{V}_{d}}$ and its formula is given as:
\[\begin{align}
& ~{{V}_{d}}=I/e\times n\times A \\
& I\text{ }={{V}_{d}}\times e\times n\times A. \\
\end{align}\]
Where, I= current applied, n= no. of electrons, A= area of cross- section of the conductor.
Complete answer:
Drift velocity is the average velocity of all the charge carriers in the conducting materials.
In the conductors, there are charge carriers such as electrons that get randomness. It is also called a free gas model of conductions.
Drift velocity is the constant, small velocity with which electrons drift slightly with the direction opposite to the applied field.
Now, in question the no. of electrons and current is given.
According to the formula of drift velocity we have:
$\begin{align}
& {{V}_{d}}=I/n\times e\times A \\
& \Rightarrow {{V}_{d}}=20/{{10}^{29}}\times (1.6\times {{10}^{-19}})\times {{10}^{-6}} \\
& \Rightarrow {{V}_{d}}=20/1.6\times {{10}^{-4}} \\
& \Rightarrow {{V}_{d}}=1.25\times {{10}^{-3}} \\
& \Rightarrow {{V}_{d}}=1\times {{10}^{-3}} \\
& \Rightarrow {{V}_{d}}={{10}^{-3}}m{{s}^{-1}} \\
\end{align}$
So, option B is the correct option.
Note:
Drift velocity is the average velocity of all the charged carriers in the conductor wire. The charge carriers in the conductors get random motion when an external electric field is applied to it. By remembering the formula for drift velocity with some given terms we can solve ${{V}_{d}}$ easily.
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