When 5 V potential difference is applied acro | vedclass

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(B) The current $I$ in a conductor is given by $I = n e A v_d$,where $n$ is the electron density,$e$ is the elementary charge,$A$ is the cross-section

Vedclass · Accepted Answer

$1.6 	imes 10^{-5} \ \Omega m$

Vedclass · Answer

(B) The current $I$ in a conductor is given by $I = n e A v_d$,where $n$ is the electron density,$e$ is the elementary charge,$A$ is the cross-sectional area,and $v_d$ is the drift speed.From Ohm's law,$V = I R$,where $R = ho \frac{l}{A}$.Substituting $I$ and $R$ into the equation: $V = (n e A v_d) 	imes (ho \frac{l}{A}) = n e v_d ho l$.Rearranging for resistivity $ho$: $ho = \frac{V}{n e v_d l}$.Given values: $V = 5 \ V$,$l = 0.1 \ m$,$n = 8 	imes 10^{28} \ m^{-3}$,$v_d = 2.5 	imes 10^{-4} \ m/s$,and $e = 1.6 	imes 10^{-19} \ C$.Substituting these values: $ho = \frac{5}{8 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 2.5 	imes 10^{-4} 	imes 0.1}$.$ho = \frac{5}{3.2 	imes 10^5} = 1.5625 	imes 10^{-5} \ \Omega m \approx 1.6 	imes 10^{-5} \ \Omega m$.

When $5\ V$ potential difference is applied across a wire of length $0.1\ m$,the drift speed of electrons is $2.5 \times 10^{-4} \ m/s$. If the electron density in the wire is $8 \times 10^{28} \ m^{-3}$,the resistivity of the material is close to:

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