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Question: Every atom has one free electron in copper. If 2.2 ampere current is flowing in the wire of copper h...

Every atom has one free electron in copper. If 2.2 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper = 9×10310^3 kgm−3^{-3} and atomic weight = 63)

A

0.3 mm/sec

B

0.1 mm/sec

C

0.2 mm/sec

D

0.2 cm/sec

Answer

0.2 mm/sec

Explanation

Solution

The drift velocity vdv_d of electrons in a conductor is given by the formula:

vd=IneAv_d = \frac{I}{neA}

where:

II is the current flowing through the wire.

nn is the number density of free electrons (number of free electrons per unit volume).

ee is the elementary charge (1.6×10−191.6 \times 10^{-19} C).

AA is the cross-sectional area of the wire.

First, we need to calculate the number density of free electrons, nn.

The density of copper is ρ=9×103\rho = 9 \times 10^3 kg/m3^3.

The atomic weight of copper is 63. This means the molar mass of copper is M=63×10−3M = 63 \times 10^{-3} kg/mol.

Avogadro's number is NA=6.022×1023N_A = 6.022 \times 10^{23} atoms/mol.

The number density of copper atoms is given by the number of atoms per mole divided by the volume per mole. The volume per mole is the molar mass divided by the density.

Volume per mole =Mρ=63×10−3 kg/mol9×103 kg/m3=7×10−6 m3/mol= \frac{M}{\rho} = \frac{63 \times 10^{-3} \text{ kg/mol}}{9 \times 10^3 \text{ kg/m}^3} = 7 \times 10^{-6} \text{ m}^3\text{/mol}.

The number density of copper atoms is NAVolume per mole=6.022×1023 atoms/mol7×10−6 m3/mol=6.0227×1029 atoms/m3\frac{N_A}{\text{Volume per mole}} = \frac{6.022 \times 10^{23} \text{ atoms/mol}}{7 \times 10^{-6} \text{ m}^3\text{/mol}} = \frac{6.022}{7} \times 10^{29} \text{ atoms/m}^3.

Since every atom of copper has one free electron, the number density of free electrons is n=6.0227×1029 m−3≈0.8603×1029 m−3n = \frac{6.022}{7} \times 10^{29} \text{ m}^{-3} \approx 0.8603 \times 10^{29} \text{ m}^{-3}.

Next, we calculate the cross-sectional area of the wire, AA.

The diameter of the wire is d=1d = 1 mm =1×10−3= 1 \times 10^{-3} m.

The radius is r=d/2=0.5r = d/2 = 0.5 mm =0.5×10−3= 0.5 \times 10^{-3} m.

The area is A=πr2=π(0.5×10−3)2=π(0.25×10−6) m2A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 = \pi (0.25 \times 10^{-6}) \text{ m}^2.

Using π≈3.14159\pi \approx 3.14159, A≈0.7854×10−6 m2A \approx 0.7854 \times 10^{-6} \text{ m}^2.

The current flowing is I=2.2I = 2.2 A.

The charge of an electron is e=1.6×10−19e = 1.6 \times 10^{-19} C.

Now, we can calculate the drift velocity:

vd=IneA=2.2(0.8603×1029 m−3)×(1.6×10−19 C)×(0.7854×10−6 m2)v_d = \frac{I}{neA} = \frac{2.2}{(0.8603 \times 10^{29} \text{ m}^{-3}) \times (1.6 \times 10^{-19} \text{ C}) \times (0.7854 \times 10^{-6} \text{ m}^2)}

vd=2.2(0.8603×1.6×0.7854)×1029−19−6 m/sv_d = \frac{2.2}{(0.8603 \times 1.6 \times 0.7854) \times 10^{29 - 19 - 6}} \text{ m/s}

vd=2.2(0.8603×1.6×0.7854)×104 m/sv_d = \frac{2.2}{(0.8603 \times 1.6 \times 0.7854) \times 10^4} \text{ m/s}

Calculate the product in the denominator: 0.8603×1.6×0.7854≈1.0800.8603 \times 1.6 \times 0.7854 \approx 1.080.

vd≈2.21.080×104 m/s≈2.037×10−4 m/sv_d \approx \frac{2.2}{1.080 \times 10^4} \text{ m/s} \approx 2.037 \times 10^{-4} \text{ m/s}.

We need to convert the drift velocity to mm/sec.

1 m=1000 mm1 \text{ m} = 1000 \text{ mm}.

vd≈2.037×10−4 m/s×1000 mm1 m=2.037×10−4×103 mm/sec=2.037×10−1 mm/sec=0.2037 mm/secv_d \approx 2.037 \times 10^{-4} \text{ m/s} \times \frac{1000 \text{ mm}}{1 \text{ m}} = 2.037 \times 10^{-4} \times 10^3 \text{ mm/sec} = 2.037 \times 10^{-1} \text{ mm/sec} = 0.2037 \text{ mm/sec}.

Comparing this value with the given options, the calculated value 0.2037 mm/sec is closest to 0.2 mm/sec.