Question

2- A copper wire of 3mm diameter with conductivity of 6.7 10' (0.M), and electron mobility of 0.0064 m2 /V sec. Is subjected to an electric field of 30 mV/m. Find (a) the charge density of free electrons, (b) the current density, (c) current flowing in the wire, (d) the electron draft velocity.

Answer 1

a)
`n=5.98*10^(26)/m^3`

b)
`i=2010000A/m^2`

c)
`I_w=14.207A`

d)
`V_e=1.92*10^(-4)m/s`

Step-by-step explanation:

From the question we are told that:

Diameter
`d=3mm=>3*10^(-3)`

Conductivity
`\sigma= 6.7 10^7 (0.M),`

Electron mobility
`\phi= 0.0064 m2 /V sec`

Electric field
`E= 30 mV/m`

a)

Generally the equation for Charge Density is mathematically given by

`\phi=(\sigma)/(n e)`

Therefore

`n=(\sigma)/(\phi e)`

`n=(6.7 10^7)/(1.6*10^(-19) *0.0064)`

`n=5.98*10^(26)/m^3`

b)

Generally the equation for current density is mathematically given by

`i=\sigma*E`

`i= 30*10^{-3] *6.7 10^7`

`i=2010000A/m^2`

c)

Generally the equation for current in wire is mathematically given by

`I_w=iA`

`I_w=i*\pi r^2`

`I_w=(2010000)*\pi( 1.5*10^(-3))^2`

`I_w=14.207A`

d)

Generally the equation for electron draft velocity. is mathematically given by

`V_e=\phi E`

`V_e=(0.0064)*(30*10^(-3))`

`V_e=1.92*10^(-4)m/s`