A potential difference of 5 ,V is applied acr | vedclass

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(B) The mobility $\mu$ of electrons is defined as the ratio of drift velocity $v_d$ to the electric field $E$.Given:Potential difference $V = 5 \,V$Le

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$5 	imes 10^{-6}$

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(B) The mobility $\mu$ of electrons is defined as the ratio of drift velocity $v_d$ to the electric field $E$.Given:Potential difference $V = 5 \,V$Length $l = 10 \,cm = 0.1 \,m$Drift velocity $v_d = 2.5 	imes 10^{-4} \,m/s$The electric field $E$ is given by $E = \frac{V}{l} = \frac{5}{0.1} = 50 \,V/m$.Now,calculate mobility $\mu$:$\mu = \frac{v_d}{E} = \frac{2.5 	imes 10^{-4}}{50}$$\mu = \frac{2.5}{50} 	imes 10^{-4} = 0.05 	imes 10^{-4} = 5 	imes 10^{-6} \,m^2 V^{-1} s^{-1}$.

$A$ potential difference of $5 \,V$ is applied across a conductor of length $10 \,cm$. If the drift velocity of electrons is $2.5 \times 10^{-4} \,m/s$,then the electron mobility will be ............ $m^2 V^{-1} s^{-1}$.

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