Respuesta :
Answer:
[tex] V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}[/tex]
[tex] V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}[/tex]
And as we can see we have that:
[tex] V_A = \frac{1}{4} V_B[/tex]
So then the best answer would be:
a. vA = vB/4
Explanation:
For this case we know the following conditions:
[tex] L_A = L_B =L [/tex] same length
[tex] I_A = I_B =I[/tex] both wires with the same current
Both wires are made of he same material, so then the number of electrons per cubic meter (n) are the same for both wires [tex] n_A = n_B =n[/tex]
We also know that [tex] r_A = 2 r_B[/tex] where r represent the radius.
Since we know that a wire have a cylindrical form we can find the area for each case:
[tex] A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B[/tex]
[tex] A_B = \pi r^2_B[/tex]
So then we have that [tex] A_A = 4 A_B[/tex]
Now we know that from the definition the drift velocity of electron in a wire is given by:
[tex] v_d = \frac{I}{neA}[/tex]
Where I is the current, n the number of electrons per cubic meter, e is the charge for the electron and A the area.
If we replace we have this:
[tex] V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}[/tex]
[tex] V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}[/tex]
And as we can see we have that:
[tex] V_A = \frac{1}{4} V_B[/tex]
So then the best answer would be:
a. vA = vB/4
