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

Answer:

a. ρ$_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

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$_\beta=\frac{\beta^2}{2*u_o}$

ρ$_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

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

ρ$_\beta=1.996J/m^3$

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

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$