Solution: Resistivity from Field Energy Densities
Analysis: We relate the electric field energy density $u_E$ and magnetic field energy density $u_B$ via the current density $J$ and resistivity $\rho$.
Step 1: Electric Field Expression
The energy density of the electric field is given by:
$$ u_E = \frac{1}{2} \varepsilon_0 E^2 \implies E = \sqrt{\frac{2 u_E}{\varepsilon_0}} $$
Using Ohm’s law, the current density $J$ is:
$$ J = \frac{E}{\rho} = \frac{1}{\rho} \sqrt{\frac{2 u_E}{\varepsilon_0}} $$
Step 2: Magnetic Field Expression
Using Ampere’s Law for a cylinder of radius $r$, the magnetic field $B$ inside is:
$$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} \implies B(2\pi r) = \mu_0 (J \cdot \pi r^2) $$
$$ B = \frac{\mu_0 J r}{2} $$
Substituting $J$ from above:
$$ B = \frac{\mu_0 r}{2\rho} \sqrt{\frac{2 u_E}{\varepsilon_0}} $$
Step 3: Magnetic Energy Density
The magnetic energy density is $u_B = \frac{B^2}{2\mu_0}$. Substituting $B$:
$$ u_B = \frac{1}{2\mu_0} \left[ \frac{\mu_0 r}{2\rho} \sqrt{\frac{2 u_E}{\varepsilon_0}} \right]^2 $$
$$ u_B = \frac{1}{2\mu_0} \cdot \frac{\mu_0^2 r^2}{4\rho^2} \cdot \frac{2 u_E}{\varepsilon_0} = \frac{\mu_0 r^2 u_E}{4 \rho^2 \varepsilon_0} $$
Step 4: Calculation
Rearranging to solve for resistivity $\rho$:
$$ \rho^2 = \frac{\mu_0 r^2 u_E}{4 \varepsilon_0 u_B} \implies \rho = \frac{r}{2} \sqrt{\frac{\mu_0 u_E}{\varepsilon_0 u_B}} $$
Given:
$u_E = 2 \times 10^{-17} \text{ J/m}^3$,
$u_B = 0.4 \text{ J/m}^3$,
$r = 0.02 \text{ m}$,
$\mu_0 = 4\pi \times 10^{-7}$, $\varepsilon_0 = 8.85 \times 10^{-12}$.
$$ \frac{\mu_0}{\varepsilon_0} \approx \frac{4\pi \times 10^{-7}}{8.85 \times 10^{-12}} \approx 1.42 \times 10^5 \text{ (impedance squared)} $$
$$ \sqrt{\frac{2 \times 10^{-17}}{0.4}} = \sqrt{5 \times 10^{-17}} \approx 7.07 \times 10^{-9} $$
Plugging in the constants leads to the final result: