Respuesta :
Answer:
(a). The current in the wire is 20.16 A.
(b). The current density is 9.66 A/mm².
Explanation:
Given that,
Diameter = 1.63 mm
Drift velocity [tex]v_{d}=7.08\times10^{-4}\ m/s[/tex]
Atomic mass of copper [tex]m=1.055\times10^{-25}\ kg[/tex]
Atomic weight of copper =63.5 g
Density = 9 g/cm³
We need to calculate the number density of electron
Using formula of number density
[tex]n=\dfrac{N\times\rho}{w}[/tex]
Where, w= atomic weight
Put the value into the formula
[tex]n=\dfrac{6.023\times10^{23}\times9\times10^{6}}{63.5}[/tex]
[tex]n=0.853\times10^{29}\ electron/m^3[/tex]
(a).We need to calculate the current in the wire
Using formula of drift velocity
[tex]v_{d}=\dfrac{I}{nqA}[/tex]
[tex]I=v_{d}nqA[/tex]
Where, A = cross section area
q = charge of electron
n = number density of electron
[tex]v_{d}[/tex] = drift velocity
Put the value into the formula
[tex]I=7.08\times10^{-4}\times0.853\times10^{29}\times1.6\times10^{-19}\times\pi\times(\dfrac{1.63\times10^{-3}}{2})^2[/tex]
[tex]I=20.16\ A[/tex]
(b). We need to calculate the current density
Using formula of current density
[tex]J=nqv_{d}[/tex]
Put the value into the formula
[tex]J=0.853\times10^{29}\times1.6\times10^{-19}\times7.08\times10^{-4}[/tex]
[tex]J=9662784\ A/m^2[/tex]
[tex]J=9.66\ A/mm^2[/tex]
Hence, (a). The current in the wire is 20.16 A.
(b). The current density is 9.66 A/mm².
