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Question: Find the current which flows through a copper wire of length \[0.2\text{ m}\], area of cross-section...

Find the current which flows through a copper wire of length 0.2 m0.2\text{ m}, area of cross-section 1 mm21\text{ m}{{\text{m}}^{2}}, when connected to a battery of 4 V4\text{ V}. Given that electron mobility is 4.5×10−6 m2V−1s−14.5\times {{10}^{-6}}\text{ }{{\text{m}}^{2}}{{\text{V}}^{-1}}{{\text{s}}^{-1}} and charge of an electron is 1.6×10−19 C1.6\times {{10}^{-19}}\text{ C}. The number density of electrons in copper wire is 8.5×1028 m−38.5\times {{10}^{28}}\text{ }{{\text{m}}^{-3}}.

Explanation

Solution

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 vd{{v}_{d}} of electrons across a wire of length l is given by:
vd=μeE{{v}_{d}}={{\mu }_{e}}E, where μe{{\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=VlE=\dfrac{V}{l}
The current i flowing through the wire is given by:
i=neAvdi=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 vd{{v}_{d}} is the drift velocity of the electrons.

Complete step by step answer:
The length of the copper wire, l=0.2 ml=0.2\text{ m}
The area of cross-section of the wire, A=1 mm2=10−6 m2A=1\text{ m}{{\text{m}}^{2}}={{10}^{-6}}\text{ }{{\text{m}}^{2}}
The potential difference across the wire, V=4 VV=4\text{ V}
The mobility of electrons, μe=4.5×10−6 m2V−1s−1{{\mu }_{e}}=4.5\times {{10}^{-6}}\text{ }{{\text{m}}^{2}}{{\text{V}}^{-1}}{{\text{s}}^{-1}}
The charge on an electron, e=1.6×10−19 Ce=1.6\times {{10}^{-19\text{ }}}\text{C}
The number density of electrons in copper wire, n=8.5×1028 m−3n=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=4 V0.2 m=20 V/mE=\dfrac{4\text{ V}}{0.2\text{ m}}=20\text{ V/m}.
Using E=20 V/mE=20\text{ V/m} and μe=4.5×10−6 m2V−1s−1{{\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:

& {{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{aligned}$$ Substitute the values of n, e, A and $${{v}_{d}}$$ in the current-formula to calculate current: $$\begin{aligned} & 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{aligned}$$ So, the current in the copper wire is $$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}$$