Problem 11.3.10^∗

Solution

Magnetic field inside the wire

Enclosed current at radius r:

$I_{\text{enc}} = I \frac{r^2}{r_2^2}$

Ampère's law:

$H_1 \cdot 2\pi r = I \frac{r^2}{r_2^2} \;\Longrightarrow\; H_1 = \frac{I r}{2\pi r_2^2}$

Magnetic flux density:

$B_1 = \mu_1 \mu_0 H_1 = \frac{\mu_1 \mu_0 I r}{2\pi r_2^2}$

Magnetic field between the wire and the casing

The enclosed current is the total current of the wire (I)

$H_2 \cdot 2\pi r = I \;\Longrightarrow\; H_2 = \frac{I}{2\pi r}$

$B_2 = \mu_2 \mu_0 H_2 = \frac{\mu_2 \mu_0 I}{2\pi r}$

For$r > r_1$, the net current is zero and the field is zero

Magnetic flux per unit length

Internal flux$(\Phi'_{\text{int}})$
According to the convention adopted in the problem, $B_1$ is integrated directly over the entire cross-section of the wire:

$\Phi'_{\text{int}} = \int_0^{r_2} B_1 \, dr = \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \int_0^{r_2} r \, dr = \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \cdot \frac{r_2^2}{2} = \frac{\mu_1 \mu_0 I}{4\pi}$

External flux $(\Phi'_{\text{ext}})$
Between the two conductors:

$\Phi'_{\text{ext}} = \int_{r_2}^{r_1} B_2 \, dr = \frac{\mu_2 \mu_0 I}{2\pi} \int_{r_2}^{r_1} \frac{dr}{r} = \frac{\mu_2 \mu_0 I}{2\pi} \ln \frac{r_1}{r_2}$

Total flux per unit length:

$\Phi' = \Phi'_{\text{int}} + \Phi'_{\text{ext}} = \frac{\mu_0 I}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)$

Inductance per unit length

By definition have $L' = \Phi'/I$

$\boxed{L' = \frac{\mu_0}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)}$