Chapter 30 Problem 18

Problem

Calculate the magnetic and electric energy densities at the surface of a 3.0-mm-diameter copper wire carrying a 15-A current.

Solution

$u_{B}={\frac {B^{2}}{2\mu _{0}}}={\frac {1}{2\mu _{0}}}\left({\frac {\mu _{0}I}{2\pi R}}\right)^{2}$ $={\frac {\mu _{0}I^{2}}{8\pi ^{2}R^{2}}}$ $={\frac {\left(4\pi \times 10^{-7}{\text{ T}}\cdot {\text{m}}{\big /}{\text{A}}\right)\left(15{\text{ A}}\right)^{2}}{8\pi ^{2}\left(1.5\times 10^{-3}{\text{ m}}\right)^{2}}}=1.6{\text{ J}}{\big /}{\text{m}}^{3}$

To calculate the electric energy density, we need the electric field at the surface of the wire. The electric field will equal the voltage difference along the wire divided by the length of the wire. We can calculate the voltage drop using Ohm’s law and the resistance from the resistivity and diameter of the wire.

$E={\frac {V}{l}}={\frac {IR}{l}}={\frac {I\rho l}{l\pi r^{2}}}={\frac {I\rho }{\pi r^{2}}}$

$u_{E}={\frac {1}{2}}\epsilon _{0}E^{2}={\frac {1}{2}}\epsilon _{0}\left({\frac {I\rho }{\pi r^{2}}}\right)^{2}$ ${\frac {1}{2}}\left(8.85\times 10^{-12}{\text{ C}}^{2}{\big /}{\text{N}}\cdot {\text{m}}^{2}\right)\left[{\frac {\left(15{\text{ A}}\right)\left(1.68\times 10^{-8}\Omega \cdot {\text{m}}\right)^{2}}{\pi \left(1.5\times 10^{-3}{\text{ m}}\right)^{2}}}\right]^{2}$ $=5.6\times 10^{-15}{\text{ J}}{\big /}{\text{m}}^{3}$