Name: Meter Bridge
Uploaded: 2019-07-19
Duration: 7 min 2 s
Description: Measurement, by comparison, is always accurate. The video talks about a metre bridge device that helps us find the unknown resistance by comparing it with a...

Question

Compare the average drift speed of conduction electrons in a copper wire of cross-sectional area

3.0 \times 10^{- 7} m^{2}

carrying a current of

5 A

with the speed of propagation of electric field along the conductor which causes the drift motion. Assume that each copper atom contributes roughly one conduction electron. The density of copper is

9.0 \times 10^{3} kg m^{- 3}

and its atomic mass is

63.5 u

.

Accepted Answer

Given that,

The cross-sectional area of the copper wire is $A = 3.0 \times 10^{- 7} m^{2}$ .

The current through the wire is $I = 5 A$ .

Charge on an electron is $e = 1.6 \times 10^{- 19} C$

The density of copper is $ρ = 9.0 \times 10^{3} kg m^{- 3} = 9.0 g {cm}^{- 3}$ .

Since $6.0 \times 10^{23}$ copper atoms have a mass of $63.5 g$ and there is one conduction electron per atom of copper, the number density of electrons is $n = \frac{6 \times 10^{23}}{63.5} \times 9.0 = 8.5 \times 10^{22} {cm}^{- 3} = 8.5 \times 10^{28} m^{- 3}$ .

The average drift speed of conduction electrons is $v_{d} = \frac{I}{A n e} = \frac{5 C s^{- 1}}{3.0 \times 10^{- 7} m^{2} \times 8.5 \times 10^{28} m^{- 3} \times 1.6 \times 10^{- 19} C}$

$\Rightarrow v_{d} = 0.0012 m s^{- 1} = 1.2 \times 10^{- 3} m s^{- 1}$ .

The speed of propagation of the electric field is equal to the speed of an electromagnetic wave, $c = 3 \times 10^{8} m s^{- 1}$ .

Thus, $\frac{v_{d}}{c} = \frac{1.2 \times 10^{- 3}}{3 \times 10^{8}} = 0.4 \times 10^{- 11} = 4 \times 10^{- 12}$ .

Important Questions on Current Electricity

Learn and Prepare for any exam you want !