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A conducting wire of cross-sectional area $1 \mathrm{~cm}^2$ has $3 \times 10^{23}$ charge carriers per $\mathrm{m}^3$. If wire carries a current of $24 \mathrm{~mA}$, then drift velocity of carriers is
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The correct answer is:
$5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
Given, cross-sectional area,
$A=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2$
Charge density, $n=3 \times 10^{23} \mathrm{~m}^{-3}$
Current through wire, $I=24 \times 10^{-3}$ A
We know that, for conductors,
$I=n A e v_d$
$\Rightarrow \quad v_d=\frac{I}{n A e}=\frac{24 \times 10^{-3}}{3 \times 10^{23} \times 10^{-4} \times 1.6 \times 10^{-19}}$
$=\frac{8}{1.6} \times 10^{-3}=5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
$A=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2$
Charge density, $n=3 \times 10^{23} \mathrm{~m}^{-3}$
Current through wire, $I=24 \times 10^{-3}$ A
We know that, for conductors,
$I=n A e v_d$
$\Rightarrow \quad v_d=\frac{I}{n A e}=\frac{24 \times 10^{-3}}{3 \times 10^{23} \times 10^{-4} \times 1.6 \times 10^{-19}}$
$=\frac{8}{1.6} \times 10^{-3}=5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
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