Qstnid -Iv-11-a Copper Wire Of Radius 0.1 Mm And Resistance Kω Is Connected Across A Power Supply Of 20 V. (A) How Many Electrons Are Transferred Per Second Between The Supply And The Wire At One End? (B) Write Down The Current Density In The Wire. (A) How Many Electrons Are Transferred Per Second Between The Supply And The Wire At One End? (B) Write Down The Current Density In The Wire. - 32: Electric Current In Conductors

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NEET-XII-Physics

32: Electric Current in Conductors

with Solutions - page 4

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#11
A copper wire of radius 0.1 mm and resistance kΩ is connected across a power supply of 20 V. (a) How many electrons are transferred per second between the supply and the wire at one end? (b) Write down the current density in the wire. (a) How many electrons are transferred per second between the supply and the wire at one end? (b) Write down the current density in the wire.
Ans : Given:
Radius of the wire, r = 0.1 mm = 10^-4 m
Resistance, R = 1 kΩ = 10³ Ω
Voltage across the ends of the wire, V = 20 V (a) Let q be the charge transferred per second and n be the number of electrons transferred per second.
We know:
`` i=\frac{V}{R}``
`` \Rightarrow i=\frac{20\,\mathrm{\,V\,}}{{10}^{3}\,\mathrm{\,\Omega \,}}``
`` \Rightarrow i=20\times {10}^{-3}=2\times {10}^{-2}\,\mathrm{\,A\,}``
`` q\mathit{=}it``
`` \Rightarrow q=2\times {10}^{-2}\times 1``
`` \Rightarrow q=2\times {10}^{-2}\,\mathrm{\,C\,}``
Also,
q = ne
`` \Rightarrow n=\frac{q}{e}=\frac{2\times {10}^{-2}}{1.6\times {10}^{-19}}``
`` \Rightarrow n=1.25\times {10}^{17}`` (b) Current density of a wire,
`` j=\frac{i}{A}``
`` \Rightarrow j=\frac{2\times {10}^{-2}}{3.14\times {10}^{-8}}``
`` \Rightarrow j=6.37\times {10}^{5}\,\mathrm{\,A\,}/{\,\mathrm{\,m\,}}^{2}``
Page No 198: (a) Let q be the charge transferred per second and n be the number of electrons transferred per second.
We know:
`` i=\frac{V}{R}``
`` \Rightarrow i=\frac{20\,\mathrm{\,V\,}}{{10}^{3}\,\mathrm{\,\Omega \,}}``
`` \Rightarrow i=20\times {10}^{-3}=2\times {10}^{-2}\,\mathrm{\,A\,}``
`` q\mathit{=}it``
`` \Rightarrow q=2\times {10}^{-2}\times 1``
`` \Rightarrow q=2\times {10}^{-2}\,\mathrm{\,C\,}``
Also,
q = ne
`` \Rightarrow n=\frac{q}{e}=\frac{2\times {10}^{-2}}{1.6\times {10}^{-19}}``
`` \Rightarrow n=1.25\times {10}^{17}`` (b) Current density of a wire,
`` j=\frac{i}{A}``
`` \Rightarrow j=\frac{2\times {10}^{-2}}{3.14\times {10}^{-8}}``
`` \Rightarrow j=6.37\times {10}^{5}\,\mathrm{\,A\,}/{\,\mathrm{\,m\,}}^{2}``
Page No 198:

#11-a
How many electrons are transferred per second between the supply and the wire at one end?
Ans : Let q be the charge transferred per second and n be the number of electrons transferred per second.
We know:
`` i=\frac{V}{R}``
`` \Rightarrow i=\frac{20\,\mathrm{\,V\,}}{{10}^{3}\,\mathrm{\,\Omega \,}}``
`` \Rightarrow i=20\times {10}^{-3}=2\times {10}^{-2}\,\mathrm{\,A\,}``
`` q\mathit{=}it``
`` \Rightarrow q=2\times {10}^{-2}\times 1``
`` \Rightarrow q=2\times {10}^{-2}\,\mathrm{\,C\,}``
Also,
q = ne
`` \Rightarrow n=\frac{q}{e}=\frac{2\times {10}^{-2}}{1.6\times {10}^{-19}}``
`` \Rightarrow n=1.25\times {10}^{17}``

#11-b
Write down the current density in the wire.
Ans : Current density of a wire,
`` j=\frac{i}{A}``
`` \Rightarrow j=\frac{2\times {10}^{-2}}{3.14\times {10}^{-8}}``
`` \Rightarrow j=6.37\times {10}^{5}\,\mathrm{\,A\,}/{\,\mathrm{\,m\,}}^{2}``
Page No 198: