Question

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm, and its resistance per unit length is 3.3 /km.

a) What is the energy density of the magnetic field at the surface of the wire in J/m^3 ?

b)Of the electric field?

Answer 1

a. ρ
`_\beta=1.996J/m^3`

b.
`U_E=9.445x10^(-15) J/m^3`

Step-by-step explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

`i=14A`,
`d=2.5mm`,
`R=3.3`Ω,
`l=1 km`,
`E_o=8.85x10^(-12)F/m`,
`u_o=4*x10^(-7)H/m`

ρ
`_\beta=(\beta^2)/(2*u_o)`

ρ
`_\beta=(1)/(2*u_o)*((u_o*i^2)/(2\pi *r))^2`

ρ
`_\beta=(u_o*i^2)/(8\pi*r)=(4\pi *10^(-7)H/m*(14A)^2)/(8\pi*(1.25x10^(-3)m)^2)`

ρ
`_\beta=1.996J/m^3`

b. The electric field can be find using the equation:

`U_E=(1)/(2)*E_o*E^2`

`E=((i*R)/(l))^2`

`U_E=(1)/(2)*8.85x10^(-12)*((14A*3.3)/(1000m))^2`

`U_E=9.445x10^(-15) J/m^3`