Answer:
t= 24080 s
Explanation:
Given that
Current in the wire ,I = 4 A
The charge ,q = 6.02 x 10²³ e C
We know that
[tex]I=\dfrac{q}{t}[/tex]
I=Current
q=Charge
t=time
[tex]t=\dfrac{q}{I}[/tex]
Now by putting the values in the above equation we get'
[tex]t=\dfrac{6.02\times 10^{23}\times 1.6\times 10^{-19}}{4}\ s[/tex]
t= 24080 s
