Respuesta :
Answer:
The drift velocity is [tex]v_d=2.9\times 10^{-4}\ m/s.[/tex]
Explanation:
Given :
Area of metallic wire, [tex]A = 3\times 10^{-6}\ m^2[/tex].
Current through wire , [tex]I=6 \ A.[/tex]
Mobile charge density , [tex]n=4.24\times 10^{28} \ carriers/m^3.[/tex]
Charge value , [tex]e=1.6\times 10^{-19}\ C.[/tex]
We need to find drift velocity , [tex]v_d.[/tex]
Now, we know :
[tex]I=neAv_d[/tex]
Therefore, [tex]v_d=\dfrac{I}{neA}[/tex]
Putting all given values in above equation we get,
[tex]v_d=\dfrac{6}{4.24\times 10^{28}\times 1.6\times 10^{-19} \times 3 \times 10^{-6}}[/tex]
[tex]v_d=2.9\times 10^{-4}\ m/s.[/tex]
Hence, this is the required solution.
The drift velocity of the mobile charge carriers is [tex]2.9*10^{-4}m/s[/tex]
Drift velocity :
It is a velocity gained by the free electrons of a conductor, with which the electrons get drifted under the influence of an electric field.
Drift velocity is the average velocity attained by charged particles in a material due to an electric field.
It is given by formula,
[tex]v_{d}=\frac{I}{neA}[/tex]
Where,
- [tex]I[/tex] is current
- [tex]A[/tex] is cross section area
- [tex]n[/tex] is mobile charge density
- [tex]e[/tex] is charge on electron
Given that, [tex]I=6A,A=3*10^{-6}m^{2} ,n=4.24*10^{28} ,e=1.6*10^{-19}C[/tex]
Substitute all values in above formula.
[tex]v_{d}=\frac{6}{4.24*10^{28}*1.6*10^{-19} *3*10^{-6} } \\\\v_{d}=2.9*10^{-4}m/s[/tex]
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