The potential difference across a conducting wire of length 20 cm is 30 V . If the electron mobility is 2 × 10^{-6} \, {m}² \, {V}^{-1} \, {s}^{-1}, then the drift velocity of the electrons is [AP EAPCET 24 May 2025 S1]

Correct Answer is : 3×10−4 m s−13 × 10^{-4} \, {m}\,{s}^{-1}3×10−4ms−1

1.5 × 10^{-3} \, {m}\,{s}^{-1}

1.5 × 10^{-4} \, {m}\,{s}^{-1}

Correct Answer is : 3×10−4 m s−13 × 10^{-4} \, {m}\,{s}^{-1}3×10−4ms−1

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Current Electricity

Section: Physics

Question : 2 of 37

Marks: +1, -0

2 \times 10^{-6} \, \mathrm{m}^2 \, \mathrm{V}^{-1} \, \mathrm{s}^{-1}

Solution:

We know the electron mobility

(\mu)

and want to find the drift velocity

(v_d)

The formula is:

\mu = \frac{v_d}{E}

Here,

E

stands for the electric field. For a wire,

E = \frac{V}{l}

, where

V

is the voltage and

l

is the length of the wire.

So, the drift velocity can be found as:

v_d = E \mu = \frac{V}{l} \cdot \mu

Plug in the values:

V = 30\,\mathrm{V},\ l = 20\,\mathrm{cm} = 20 \times 10^{-2}\,\mathrm{m}

, and

\mu = 2 \times 10^{-6} \, \mathrm{m}^2 \, \mathrm{V}^{-1} \, \mathrm{s}^{-1}

v_d = \frac{30 \times 2 \times 10^{-6}}{20 \times 10^{-2}}

Do the calculation step by step:

30 \times 2 = 60,\ 20 \times 10^{-2} = 0.2

v_d = \frac{60 \times 10^{-6}}{0.2} = 300 \times 10^{-6} = 3 \times 10^{-4} \, \mathrm{m}\,\mathrm{s}^{-1}

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