KCET · Physics · Current Electricity
A current of \(5 \mathrm{~A}\) is passing through a metallic wire of cross-sectional area \(4 \times 10^{-6} \mathrm{~m}^{2}\) If the density of charge carriers of the wire is \(5 \times 10^{26} \mathrm{~m}^{-3}\) the drift velocity of the electronswill be
- A \(1 \times 10^{2} \mathrm{~ms}^{-1}\)
- B \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\)
- C \(1.56 \times 10^{-3} \mathrm{~ms}^{-1}\)
- D \(1 \times 10^{-2} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Drift velocity, \(v_{d}=\frac{1}{n e A}\)
\(\begin{aligned}\Rightarrow v_{d} &=\frac{5}{\left(5 \times 10^{26}\right) \times\left(1.6 \times 10^{-19}\right) \times\left(4 \times 10^{-6}\right)} \\
&=\frac{1}{64}=1.56 \times 10^{-2} \mathrm{~ms}^{-1}\end{aligned}\)
\(\begin{aligned}\Rightarrow v_{d} &=\frac{5}{\left(5 \times 10^{26}\right) \times\left(1.6 \times 10^{-19}\right) \times\left(4 \times 10^{-6}\right)} \\
&=\frac{1}{64}=1.56 \times 10^{-2} \mathrm{~ms}^{-1}\end{aligned}\)
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