Question

Find the current which flows through a copper wire of length \[0.2\text{ m}\], area of cross-section \[1\text{ m}{{\text{m}}^{2}}\], when connected to a battery of \[4\text{ V}\]. Given that electron mobility is \[4.5\times {{10}^{-6}}\text{ }{{\text{m}}^{2}}{{\text{V}}^{-1}}{{\text{s}}^{-1}}\] and charge of an electron is \[1.6\times {{10}^{-19}}\text{ C}\]. The number density of electrons in copper wire is \[8.5\times {{10}^{28}}\text{ }{{\text{m}}^{-3}}\].

Answer 1

Hint: The strength of electric current in a conductor is measured by the magnitude of electric charge flowing per second though a cross-section of the conductor, and is directly proportional to the drift velocity of the electrons.

Formula used: The drift velocity \[{{v}_{d}}\] of electrons across a wire of length l is given by:
\[{{v}_{d}}={{\mu }_{e}}E\], where \[{{\mu }_{e}}\]is the electron mobility and E is the intensity of electric field at every point of the wire.
If the potential difference across a wire of length l is V , then the intensity of electric field E at every point of the wire is given by:
\[E=\dfrac{V}{l}\]
The current i flowing through the wire is given by:
\[i=neA{{v}_{d}}\]
Where, n implies the number density of electron in the wire; e implies the charge on an electron; A implies the area of cross-section of the wire and \[{{v}_{d}}\] is the drift velocity of the electrons.

Complete step by step answer:
The length of the copper wire, \[l=0.2\text{ m}\]
The area of cross-section of the wire, \[A=1\text{ m}{{\text{m}}^{2}}={{10}^{-6}}\text{ }{{\text{m}}^{2}}\]
The potential difference across the wire, \[V=4\text{ V}\]
The mobility of electrons, \[{{\mu }_{e}}=4.5\times {{10}^{-6}}\text{ }{{\text{m}}^{2}}{{\text{V}}^{-1}}{{\text{s}}^{-1}}\]
The charge on an electron, \[e=1.6\times {{10}^{-19\text{ }}}\text{C}\]
The number density of electrons in copper wire, \[n=8.5\times {{10}^{28}}\text{ }{{\text{m}}^{-3}}\]
Now, substitute the values of V and l in the electric field intensity formula to calculate the intensity of electric field at every point of the wire:
\[E=\dfrac{4\text{ V}}{0.2\text{ m}}=20\text{ V/m}\].
Using \[E=20\text{ V/m}\] and \[{{\mu }_{e}}=4.5\times {{10}^{-6}}\text{ }{{\text{m}}^{2}}{{\text{V}}^{-1}}{{\text{s}}^{-1}}\], calculate the drift velocity of the electrons by the formula:
\[\begin{align}
  & {{v}_{d}}={{\mu }_{e}}E \\
 & {{v}_{d}}=(4.5\times {{10}^{-6}}\text{ }{{\text{m}}^{2}}{{\text{V}}^{-1}}{{\text{s}}^{-1}})(20\text{ V/m)} \\
 & {{v}_{d}}=9\times {{10}^{-5}}\text{ m/s} \\
\end{align}\]
Substitute the values of n, e, A and \[{{v}_{d}}\] in the current-formula to calculate current:
\[\begin{align}
  & i=(8.5\times {{10}^{28}}\text{ }{{\text{m}}^{-3}})(1.6\times {{10}^{-19\text{ }}}\text{C )(1}{{\text{0}}^{-6}}\text{ }{{\text{m}}^{2}})(9\times {{10}^{-5}}\text{ m/s)} \\
 & i=1.224\text{ A} \\
\end{align}\]
So, the current in the copper wire is \[1.224\text{ A}\].

Note:
Make sure the physical quantities are in the same unit system, preferably in the S.I. system
If the potential difference across a wire of length l is V , then the intensity of electric field E at every point of the wire is given by:
\[E=\dfrac{V}{l}\]