If n, e, and $τ$ m have their usual meanings, then the resistance of a wire of length l and cross-sectional area A is given by –

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$\frac{ml}{{ne}^{2} τA}$

$\frac{{ne}^{2} τA}{2 ml}$

$\frac{mτA}{{ne}^{2} l}$

$\frac{{ne}^{2} A}{2 mτl}$

answer is B.

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Detailed Solution

From Ohms law, V=I R
$R = \frac{V}{I} = \frac{E \cdot l}{n.e.A. v_{d}} = \frac{E \cdot l}{A.n.e. (\frac{eE}{m} τ)}$
[Where E is electric field and v_d is drift velocity]
$R = (\frac{m}{{ne}^{2} τ}) \frac{l}{A}$

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