There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is 1 mm2. If the number of free electron

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There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is 1 mm2. If the number of free electrons per cm3 is 8.4 × 1022, then the drift velocity would be:

1. 1.0 mm per sec
2. 1.0 metre per sec
3. 0.1 mm per sec
4. 0.01 mm per sec

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Correct Answer - Option 4 : 0.01 mm per sec

Concept:

• Drift Velocity: The average speed of electrons by which they slowly move inside a conductor under influence of an applied electric field is called drift velocity.
• Drift velocity of the electrons is calculated by:

$v_d=\frac{I}{neA}$

I is current in the conductor, e is the electronic charge, n is the number of free electrons per unit volume A is cross-sectional Area.

Calculation:

Given,

The number of free electrons per unit volume n = 8.4 × 10 22 / cm 3

Current I = 1.344 A

Cross-sectional Area A = 1 mm2 = 10 -2 cm 2

Electronic charge e = 1.602 × 10 -19 C

Drift velocity from the above expression will be

$v_d = \frac{1.344}{(8.4 \times 10^{22})(1.6 \times 10^{-19})(10^{-2})}$

⇒ vd = 10 -2 cm / sec

1 cm = 10 mm

So drift velocity (mm /sec) is

vd = 10 -1 mm / sec

So, correct option is 0.1 mm per sec