A copper wire of radius \(0.1 \mathrm{~mm}\) and resistance \(2 \mathrm{k} \Omega\) is connected across a…

A copper wire of radius \(0.1 \mathrm{~mm}\) and resistance \(2 \mathrm{k} \Omega\) is connected across a power supply of \(40 \mathrm{~V}\). The number of electrons transferred per second between the supply and the wire at one end is
  1. \(2.00 \times 10^{16}\)
  2. \(1.25 \times 10^{17}\)
  3. \(2.85 \times 10^{17}\)
  4. \(3.25 \times 10^{16}\)

Solution

Radius of copper wire, \(r=0.1 \mathrm{~mm}\) \(=1 \times 10^{-4} \mathrm{~m}\) Resistance, \(R=2 \mathrm{k} \Omega=2 \times 10^3 \Omega\) Power supply, \(V=40 \mathrm{~V}\) Current flowing through the wire, \(I=\frac{V}{R}=\frac{40}{2 \times 10^3}=2 \times 10^{-2} \mathrm{~A}\) \(\therefore\) Charge flowing per second \(q=I t=2 \times 10^{-2} \times 1=2 \times 10^{-2} \mathrm{C}\) \(\therefore\) Number of electrons transferred, \(\begin{aligned} n & =\frac{q}{e}=\frac{2 \times 10^{-2}}{1.6 \times 10^{-19}} \\ & =1.25 \times 10^{17} \text { electrons } \end{aligned}\)

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

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