KCET · Physics · Current Electricity
A charged particle is moving in an electric field of \(3 \times 10^{-10} \mathrm{Vm}^{-1}\) with mobility \(2.5 \times 10^6 \mathrm{~m}^2 / \mathrm{V}\)-s, its drift velocity is
- A \(8.33 \times 10^{-4} \mathrm{~m} / \mathrm{s}\)
- B \(25 \times 10^4 \mathrm{~m} / \mathrm{s}\)
- C \(1.2 \times 10^{-4} \mathrm{~m} / \mathrm{s}\)
- D \(7.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(D) \(7.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Given, electric field, \(E=3 \times 10^{-10} \mathrm{Vm}^{-1}\)
Mobility, \(\mu=25 \times 10^6 \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}\)
If drift velocity is \(v_d\), then we know that
\(\begin{aligned}\mu & =\frac{v_d}{E} \Rightarrow v_d=\mu E \\& =25 \times 10^6 \times 3 \times 10^{-10}=7.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\end{aligned}\)
Mobility, \(\mu=25 \times 10^6 \mathrm{~m}^2 / \mathrm{V}-\mathrm{s}\)
If drift velocity is \(v_d\), then we know that
\(\begin{aligned}\mu & =\frac{v_d}{E} \Rightarrow v_d=\mu E \\& =25 \times 10^6 \times 3 \times 10^{-10}=7.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\end{aligned}\)
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