A metal wire of cross-sectional area 0.5 mm ^2 and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1\ . The density, atomic mass and electrical conductivity of the metal are 6.35 × 10^3

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 0.208

The number density of conduction electrons

n

is given by the number of atoms per unit volume, since there is one conduction electron per atom:

n = \dfrac{d \times N_A}{M}

Substituting the given values:

n = \dfrac{6.35 \times 10^3 \times 6 \times 10^{23}}{63.5 \times 10^{-3}} = 6 \times 10^{28} \text{ m}^{-3}

The resistance of the wire

R

is:

R = \dfrac{L}{\sigma A}

R = \dfrac{100}{2 \times 10^8 \times 0.5 \times 10^{-6}} = \dfrac{100}{10^2} = 1\ \Omega

The total resistance of the circuit is

R_{total} = R + r = 1 + 1 = 2\ \Omega

.

The current in the circuit is:

I = \dfrac{E}{R_{total}} = \dfrac{2}{2} = 1 \text{ A}

The drift velocity

v_d

is given by the relation

I = n e A v_d

:

v_d = \dfrac{I}{n e A}

v_d = \dfrac{1}{6 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.5 \times 10^{-6}}

v_d = \dfrac{1}{4.8 \times 10^3} \text{ m s}^{-1}

Converting to mm s

^{-1}

:

v_d = \dfrac{1000}{4800} \text{ mm s}^{-1} = \dfrac{5}{24} \text{ mm s}^{-1} \approx 0.208 \text{ mm s}^{-1}

Answer:

0.208

A metal wire of cross-sectional area 0.5 mm ^2 and length 100 m is...

Options

Correct answer

Step-by-step solution

More from Current Electricity