Physics, 12.01.2021 17:50 Nayblue2002

In a copper wire that has a diameter of 1.63 mm, the drift velocity is 7.08x10^-4 m/s. Assume that there is one free electron for conduction per copper atom. Copper has an atomic mass of 1.055x10^-25 kg and a density of about 9 g/cm^3. Required:
a. What is the current in the wire?
b. What is the current density?

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Answer from: Jolenesopalski

(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 $v_{d}=7.08\times10^{-4}\ m/s$

Atomic mass of copper $m=1.055\times10^{-25}\ kg$

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

$n=\dfrac{N\times\rho}{w}$

Where, w= atomic weight

Put the value into the formula

$n=\dfrac{6.023\times10^{23}\times9\times10^{6}}{63.5}$

$n=0.853\times10^{29}\ electron/m^3$

(a).We need to calculate the current in the wire

Using formula of drift velocity

$v_{d}=\dfrac{I}{nqA}$

$I=v_{d}nqA$

Where, A = cross section area

q = charge of electron

n = number density of electron

$v_{d}$ = drift velocity

Put the value into the formula

$I=7.08\times10^{-4}\times0.853\times10^{29}\times1.6\times10^{-19}\times\pi\times(\dfrac{1.63\times10^{-3}}{2})^2$

$I=20.16\ A$

(b). We need to calculate the current density

Using formula of current density

$J=nqv_{d}$

Put the value into the formula

$J=0.853\times10^{29}\times1.6\times10^{-19}\times7.08\times10^{-4}$

$J=9662784\ A/m^2$

$J=9.66\ A/mm^2$

Hence, (a). The current in the wire is 20.16 A.

(b). The current density is 9.66 A/mm².

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In a copper wire that has a diameter of 1.63 mm, the drift velocity is 7.08x10^-4 m/s. Assume that t...

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