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Relation Between Current and Drift Velocity

Relation Between Current and Drift Velocity :- Let a battery of e.m.f. V is applied across a conductor of length L. An electric field E (= V/L) is set up inside the conductor and the electrons drift towards the positive terminal of the battery with drift velocity v_d.

Let

L = length of the conductor

A = area of cross-section of the conductor

n = number of free electrons per unit volume of the conductor

V = potential difference applied across the conductor

v_d = drift velocity of the electrons

∴ Total number of free electrons in the conductor = nAL

Total free charge in the conductor (Q) = nALe

Time taken by an electron to cross the conductor, t = L/v_d

Current flowing in the conductor, I = Q/t = nALe/(L/v_d)

I = neAv_d

This is the required relation between current and drift velocity.

Example 1.
Find free electrons per unit volume in a metallic wire of density 10⁴ kg/m³, atomic mass number 100 and number of free electron per atom is one.

Solution

Number of free electrons per unit volume (n) = total free charge particles/total volume

∴ Number of free electron per unit volume (n) = total number of atoms per unit volume (as given in question that there is one free electron per atom)

Total number of atoms(N), is given by

$\displaystyle N=\frac{m}{{{M}_{W}}}\times {{N}_{A}}$

∴ Number of free electron per unit volume (n) = N/V

$\displaystyle n=\frac{\frac{m}{{{M}_{W}}}\times {{N}_{A}}}{V}=\frac{m}{V}\times \frac{{{N}_{A}}}{{{M}_{W}}}=\rho \times \frac{{{N}_{A}}}{{{M}_{W}}}={{10}^{4}}\times \frac{6.023\times {{10}^{23}}}{100\times {{10}^{-3}}}$

n = 6.023 × 10²⁸

Example 2.

A current of 1.34 A exists in a copper wire of cross–section 1.0 mm². Assuming each copper atom contributes one free electron, calculate the drift speed of the free electrons in the wire. The density of copper is 8990 kg/m³ and atomic mass = 63.50.

Solution :

As I = neAv_d

$\displaystyle {{v}_{d}}=\frac{I}{neA}$

Here n = number of free electrons per unit volume

⇒ n = number of copper atoms per unit volume of the wire (as given in the question)

$\displaystyle n=\frac{\text{Total number of atoms}}{\text{Volume of wire}}=\frac{{{N}_{a}}\times \frac{m}{M}}{V}=\frac{{{N}_{a}}}{M}\times \frac{m}{V}=\frac{{{N}_{a}}}{M}\times \rho$

$\displaystyle n=\frac{6\times {{10}^{23}}}{63.50}\times (8990\times {{10}^{3}})=8.4\times {{10}^{28}}$

$\displaystyle \therefore {{v}_{d}}=\frac{I}{neA}=\frac{1.34}{8.4\times {{10}^{28}}\times 1.6\times {{10}^{-19}}\times {{10}^{-6}}}={{10}^{-4}}m/s$

Next Topic :- Ohm’s Law

Previous Topic :- Drift Velocity

Relation Between Current and Drift Velocity

Relation Between Current and Drift Velocity

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