What is the applied electric field that will impose a drift velocity equal to 10% of themean speed, u = 1.2 X 106 m/s of the conduction electrons in copper? The driftmobility of the electron is 32 cm2/Vs.
Question
What is the applied electric field that will impose a drift velocity equal to 10% of themean speed, u = 1.2 X 106 m/s of the conduction electrons in copper? The driftmobility of the electron is 32 cm2/Vs.
Solution
To find the applied electric field that will impose a drift velocity equal to 10% of the mean speed of conduction electrons in copper, we can use the formula:
v_d = μ * E
where v_d is the drift velocity, μ is the drift mobility, and E is the electric field.
Given that the drift mobility of electrons in copper is 32 cm^2/Vs and the mean speed of conduction electrons is 1.2 x 10^6 m/s, we can calculate the drift velocity:
v_d = 0.1 * (1.2 x 10^6 m/s) = 1.2 x 10^5 m/s
Now, we can rearrange the formula to solve for the electric field:
E = v_d / μ
Converting the drift mobility from cm^2/Vs to m^2/Vs:
μ = 32 cm^2/Vs = 32 x 10^-4 m^2/Vs = 3.2 x 10^-3 m^2/Vs
Substituting the values into the formula:
E = (1.2 x 10^5 m/s) / (3.2 x 10^-3 m^2/Vs)
E = 3.75 x 10^7 V/m
Therefore, the applied electric field that will impose a drift velocity equal to 10% of the mean speed of conduction electrons in copper is 3.75 x 10^7 V/m.
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