Respuesta :
Answer:
a) [tex] I =\frac{Q}{t}= \frac{420 C}{4800 s}= 0.0875 A[/tex]
b) [tex] I = n |q| v_d A[/tex]
And solving for vd we got:
[tex] v_d = \frac{I}{n |q| A}[/tex]
We know that [tex] A = \pi r^2[/tex] and replacing we have:
[tex] v_d = \frac{I}{n |q| \pi r^2}[/tex]
And replacing we got:
[tex] v_d = \frac{0.0875 A}{\pi (5.8x10^{28}) (1.6x10^{-19} C) (0.0013m)^2}= 1.78 x10^{-6} m/s[/tex]
Explanation:
For this case we have the following data given:
[tex[ d = 2.6 mm= 0.0026 m[/tex] represent the diameter
[tex] r = \frac{d}{2}= 1.3 mm = 0.0013 m[/tex] represent the radius
[tex] Q = 420 C[/tex] represent the charge
[tex] t = 80 min *\frac{60 s}{1 min}= 4800 s[/tex] represent the time
Part a
For this case we want to find the current flow I, and by definition we know that :
[tex] I = \fra{Q}{t} [/tex]
And replacing we have:
[tex] I =\frac{Q}{t}= \frac{420 C}{4800 s}= 0.0875 A[/tex]
Part b
For this case we want to find the velocity [tex] v_d[/tex] associated to the drift velocity, and for this case we can use the following formula:
[tex] I = n |q| v_d A[/tex]
And solving for vd we got:
[tex] v_d = \frac{I}{n |q| A}[/tex]
We know that [tex] A = \pi r^2[/tex] and replacing we have:
[tex] v_d = \frac{I}{n |q| \pi r^2}[/tex]
And replacing we got:
[tex] v_d = \frac{0.0875 A}{\pi (5.8x10^{28}) (1.6x10^{-19} C) (0.0013m)^2}= 1.78 x10^{-6} m/s[/tex]
