Question

An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temperature of the wire is 55.0°C. Assume one free electron per atom. Given that at 20 degrees, rhoo = 2.82x10-8 Ωm and α = 3.9x10-3 /C. Determine:

a) the resistivity of the wire.
b) the current density in the wire.
c) the total current in the wire.
d) the potential different that must exist between the ends of a 2m length of wire if the given electric field is to be produced.

Answer 1

Answer with Explanation:

We are given that

Diameter of coil=d=0.115mm

Radius, r=
`(d)/(2)=(0.115)/(2)=0.0575mm=0.0575* 10^(-3) m`

Using
`1mm=10^(-3) m`

Electric field=E=0.235V/m

T=55 degree C

`T_0=20^(\circ) C`

`\rho_0=2.82* 10^(-8)\Omega m`

`\alpha=3.9* 10^(-3)/C`

(a).We know that

`\rho=\rho_0(1+\alpha(T-T_0))`

Substitute the values

`\rho=2.82* 10^(-8)(1+3.9* 10^(-3)(55-20))`

`\rho=3.2* 10^(-8)\Omega m`

(b).Current density,
`J=(E)/(\rho)`

Using the formula

`J=(0.235)/(3.2* 10^(-8))=7.3* 10^6A/m^2`

c.Total current,I=JA

Where
`A=\pi r^2`

`\pi=3.14`

Using the formula

`I=7.3* 10^6* 3.14* (0.0575* 10^(-3))^2`

I=0.076A

d.Length of wire=l=2m

`V=El`

Substitute the values

`V=0.235* 2=0.47 V`