CBSE 2024 · Region 3 · Set 3 · Q22 · 3 marks
A current of $\displaystyle 1.6$ A flows through a wire when a potential difference of $\displaystyle 1.0$ V is applied across it. The length and cross-sectional area of the wire are $\displaystyle 1.0$ m and $\displaystyle 1.0 \times 10^{-7} \mathrm{~m}^{2}$ respectively. Calculate :(a)Electric field across the wire(b)Current density(c)Average relaxation time ( $\displaystyle \tau$ ) (Number density of free electrons in the wire is $\displaystyle 9.0 \times 10^{28} \mathrm{~m}^{-3}$ )
A current of $\displaystyle 1.6$ A flows through a wire when a potential difference of $\displaystyle 1.0$ V is applied across it. The length and cross-sectional area of the wire are $\displaystyle 1.0$ m and $\displaystyle 1.0 \times 10^{-7} \mathrm{~m}^{2}$ respectively. Calculate :
(a)
Electric field across the wire
(b)
Current density
(c)
Average relaxation time ( $\displaystyle \tau$ ) (Number density of free electrons in the wire is $\displaystyle 9.0 \times 10^{28} \mathrm{~m}^{-3}$ )
Marking-scheme solution
(a)
$\displaystyle E = \dfrac{V}{l}$
$\displaystyle = \dfrac{1.0\,V}{1.0\,m} = 1.0$ V/m
(b)
$\displaystyle J = I/A$
$\displaystyle J = \dfrac{1.6\,A}{1.0 \times 10^{-7}\,m^{2}} = 1.6 \times 10^{7}$ A/m$\displaystyle ^{2}$
(c)
$\displaystyle \tau = \dfrac{m}{ne^{2}}\dfrac{J}{E}$
$\displaystyle = \dfrac{9.1 \times 10^{-31} \times 1 \times 1.6}{9 \times 10^{28} \times \left(1.6 \times 10^{-19}\right)^{2}}$
$\displaystyle = 6.31 \times 10^{-14}\,\text{s}$
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