Answer:
a. ρ[tex]_\beta=1.996J/m^3[/tex]
b. [tex]U_E=9.445x10^{-15} J/m^3[/tex]
Explanation:
a. To find the density of magnetic field given use the gauss law and the equation:
[tex]i=14A[/tex], [tex]d=2.5mm[/tex], [tex]R=3.3[/tex]Ω, [tex]l=1 km[/tex], [tex]E_o=8.85x10^{-12}F/m[/tex], [tex]u_o=4*x10^{-7}H/m[/tex]
ρ[tex]_\beta=\frac{\beta^2}{2*u_o}[/tex]
ρ[tex]_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2[/tex]
ρ[tex]_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}[/tex]
ρ[tex]_\beta=1.996J/m^3[/tex]
b. The electric field can be find using the equation:
[tex]U_E=\frac{1}{2}*E_o*E^2[/tex]
[tex]E=(\frac{i*R}{l})^2[/tex]
[tex]U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2[/tex]
[tex]U_E=9.445x10^{-15} J/m^3[/tex]
