Question 332: Current Electricity

Question

When $$5V$$ potential difference is applied across a wire of length $$0.1$$ $$m,$$ the drift speed of electrons is $$2.5 	imes {10^{ - 4}}\,\,m{s^{ - 1}}.$$ If the electron density in the wire is $$8 	imes {10^{28}}\,\,{m^{ - 3}},$$ the resistivity of the material is close to :

Accepted Answer

$$V = IR = \left( {neA{v_d}} ight)ho {\ell  \over A}$$
$$	herefore$$ $$ho  = {V \over {{V_d}\ln e}}$$
Here $$V=$$ potential difference
$$l = $$ length of wire
$$n=$$ no. of electrons per unit volume of conductor.
$$e=$$ no. of electrons
Placing the value of above parameters we get resistivity
$$ho  = {5 \over {8 	imes {{10}^{28}} 	imes 1.6 	imes {{10}^{ - 19}} 	imes 2.5 	imes {{10}^{ - 4}} 	imes 0.1}}$$
$$ = 1.6 	imes {10^{ - 5}}\Omega m$$

Answer

$$1.6 	imes {10^{ - 6}}\Omega m$$

Answer

$$1.6 	imes {10^{ - 5}}\Omega m$$

Answer

$$1.6 	imes {10^{ - 8}}\Omega m$$

Answer

$$1.6 	imes {10^{ - 7}}\Omega m$$

Current Electricity

When $5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}. If the electron density in the wire is 8 \times {10^{28}}\,\,{m^{ - 3}},$ the resistivity of the material is close to :

Choose the correct answer: