Drift speed of electrons,when 1.5 , A of curr | vedclass

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(A) The relationship between current $I$ and drift velocity $v_d$ is given by the formula: $I = n e A v_d$.Here,$I = 1.5 \, A$,$n = 9 	imes 10^{28} \

Vedclass · Accepted Answer

$0.02$

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(A) The relationship between current $I$ and drift velocity $v_d$ is given by the formula: $I = n e A v_d$.Here,$I = 1.5 \, A$,$n = 9 	imes 10^{28} \, m^{-3}$,$e = 1.6 	imes 10^{-19} \, C$,and $A = 5 \, mm^2 = 5 	imes 10^{-6} \, m^2$.Substituting the values into the formula:$1.5 = (9 	imes 10^{28}) 	imes (1.6 	imes 10^{-19}) 	imes (5 	imes 10^{-6}) 	imes v_d$.$1.5 = (9 	imes 1.6 	imes 5) 	imes 10^{28-19-6} 	imes v_d$.$1.5 = 72 	imes 10^3 	imes v_d$.$v_d = \frac{1.5}{72 	imes 10^3} = \frac{1.5}{72} 	imes 10^{-3} \, m/s$.$v_d \approx 0.0208 	imes 10^{-3} \, m/s = 0.0208 \, mm/s$.Thus,the value of $v$ is close to $0.02 \, mm/s$.

Drift speed of electrons,when $1.5 \, A$ of current flows in a copper wire of cross-section $5 \, mm^2$,is $v$. If the electron density in copper is $9 \times 10^{28} \, m^{-3}$,the value of $v$ in $mm/s$ is close to (Take charge of electron to be $1.6 \times 10^{-19} \, C$).

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