Question 4: Current Electricity

Question

A current of $$10 \mathrm{~A}$$ is passing through a metallic wire of cross-sectional area $$4 	imes 10^{-6} \mathrm{~m}^2$$. If the density of the aluminium conductor is $$2.7 \mathrm{~gm} / \mathrm{cc}$$ considering aluminium gives 1 electron per atom for conduction, then find the drift velocity of the electrons if molecular weight of aluminium is $$27 \mathrm{~gm}$$.

Accepted Answer

Given, current, $$I=10 \mathrm{~A}$$
Area of cross-section, $$A=4 	imes 10^{-6} \mathrm{~m}^2$$
Density of conductors, $$ho=2.7 \mathrm{~gm} / \mathrm{cc}$$
$$=2.7 	imes 10^3 \mathrm{~kg} / \mathrm{m}^3$$
Molecular weight of aluminium,
$$M_w=27 \mathrm{~gm}=27 	imes 10^{-3} \mathrm{~kg}$$
If $$n$$ be the total number of electrons in the conductor per unit volume, then
$$\begin{aligned}
n & =\frac{	ext { Total number of electrons }}{	ext { Volume of conductor }(V)} \
& =\frac{	ext { Number of atoms per mole } 	imes 	ext { Number of moles }}{V} \
& =\frac{	ext { Avogadro Number }}{V} 	imes\left(\frac{M}{M_w}ight) \
& =6 	imes 10^{23} 	imes \frac{ho}{M_w}=6 	imes 10^{23} 	imes \frac{2.7 	imes 10^3}{27 	imes 10^{-3}} \
	herefore \quad & n=6 	imes 10^{28} \quad 	ext{.... (i)}
\end{aligned}$$
We know that, drift velocity
$$\begin{aligned}
v_d=\frac{I}{n e A} & =\frac{10}{6 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 4 	imes 10^{-6}} \
& =2.6 	imes 10^{-4} \mathrm{~m} / \mathrm{s}
\end{aligned}$$

Answer

$$1.6 	imes 10^{-4} \mathrm{~m} / \mathrm{s}$$

Answer

$$3.6 	imes 10^{-4} \mathrm{~m} / \mathrm{s}$$

Answer

$$2.6 	imes 10^{-4} \mathrm{~m} / \mathrm{s}$$

Answer

$$1.5 	imes 10^{-4} \mathrm{~m} / \mathrm{s}$$

Current Electricity

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