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The area of cross section of a current carrying conductor is A₀ and $\frac{A_{0}}{4}$ at section (1) and (2) respectively. If $v_{d_{1}}, v_{d_{2}}$ and $E_{1}, E_{2}$ be the drift velocity and electric field at sections 1 and 2 respectively then :

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$v_{d_{1}} : v_{d_{2}} = 4 : 1$

$E_{1} : E_{2} = 1 : 4$

$v_{d_{1}} : v_{d_{2}} = 1 : 4$

$E_{1} : E_{2} = 4 : 1$

answer is A, D.

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Detailed Solution

$V = \frac{i}{neA}; (a) i_{1} = i_{2}$

$\Rightarrow ne A_{0} V_{d_{1}} = ne \frac{A_{0}}{4} V_{d_{2}} \Rightarrow \frac{V_{d_{1}}}{V_{d_{2}}} = \frac{1}{4}$

$And also V_{d} = \frac{eET}{me}$

$\Rightarrow \frac{V_{d_{1}}}{V_{d_{2}}} = \frac{E_{1}}{E_{2}} = \frac{1}{4}$

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