Question 381: Current Electricity

Question

A cylindrical conductor of length 2 m and area of cross-section $0.2 \mathrm{~mm}^2$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha 	imes 10^{-3} \mathrm{~m}^2 / \mathrm{V} . \mathrm{s}$. The value of $\alpha$ is :
(electron concentration $=5 	imes 10^{28} / \mathrm{m}^3$ and electron charge $=1.6 	imes 10^{-19} \mathrm{C}$ )

Accepted Answer

The mobility $(\mu)$ of free electrons in a conductor is defined as the magnitude of their drift velocity $\left(\mathrm{v}_{\mathrm{d}}ight)$ per unit electric field (E).
$$ \mu=\frac{v_d}{E} $$
The electric field ( E ) established across a conductor of length L connected to a potential difference V is given by:
$$ \mathrm{E}=\frac{\mathrm{V}}{\mathrm{~L}} $$
The electric current (I) flowing through a conductor with cross-sectional area A and free electron number density n is:
$$ \mathrm{I}=\mathrm{neA} \mathrm{v}_{\mathrm{d}} $$
$$ \Rightarrow \mathrm{v}_{\mathrm{d}}=\frac{1}{\mathrm{ne} \mathrm{~A}} $$
Using mobility formula,
$\mu=\frac{\left(\frac{\mathrm{I}}{\mathrm{ne} \mathrm{A}}ight)}{\left(\frac{\mathrm{V}}{\mathrm{L}}ight)}$
$$\Rightarrow $$ $$ \mu=\frac{\mathrm{IL}}{\mathrm{ne} \mathrm{AV}} $$
We have given,
Length, $\mathrm{L}=2 \mathrm{~m}$
Current, $\mathrm{I}=1.6 \mathrm{~A}$
Voltage, $\mathrm{V}=2 \mathrm{~V}$
Electron concentration, $\mathrm{n}=5 	imes 10^{28} \mathrm{~m}^{-3}$
Electron charge, $\mathrm{e}=1.6 	imes 10^{-19} \mathrm{C}$
Area of cross-section, $\mathrm{A}=0.2 \mathrm{~mm}^2=0.2 	imes 10^{-6} \mathrm{~m}^2=2 	imes 10^{-7} \mathrm{~m}^2$
Substituting the values,
$$ \mu=\frac{1.6 	imes 2}{\left(5 	imes 10^{28}ight) 	imes\left(1.6 	imes 10^{-19}ight) 	imes\left(2 	imes 10^{-7}ight) 	imes 2} $$
$$\Rightarrow $$ $\mu=\frac{3.2}{32 	imes 10^2}$
$$\Rightarrow $$ $\mu=0.001 \mathrm{~m}^2 / \mathrm{V} \cdot \mathrm{s}$
$$\Rightarrow $$ $\mu=1 	imes 10^{-3} \mathrm{~m}^2 / \mathrm{V} \cdot \mathrm{s}$
$\Rightarrow 1 	imes 10^{-3}=\alpha 	imes 10^{-3} \Rightarrow \alpha=1$

Current Electricity

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