A long straight coaxial cable has a solid inner conductor of radius a = 2.0 mm and a thin outer…

Step 1. $J_0$ (a). $I = \displaystyle\int_0^a J_0\frac{r}{a}\,2\pi r\,dr = \frac{2\pi J_0a^2}{3}$, so $J_0 = \dfrac{3I}{2\pi a^2} = 17.9\ \text{A/mm}^2$.

Step 2. Inside (b, c). The current inside radius $r$ is $I\left(\dfrac{r}{a}\right)^3$, so $B = \dfrac{\mu_0 I r^2}{2\pi a^3}$. At $r = a/2$ only one eighth of the current is inside: $B = \dfrac{\mu_0 (I/8)}{2\pi(a/2)} = 3.75\ \text{mT}$. At $r = a$, $B = \dfrac{\mu_0 I}{2\pi a} = 15.0\ \text{mT}$.

Step 3. Between the conductors (d). $B = \dfrac{\mu_0 I}{2\pi r} = 5.00\ \text{mT}$ at $r = 6.0\ \text{mm}$. Outside the tube the enclosed current is zero, and so is the field.

Step 4. Energy (e). Inside, $\displaystyle\int_0^a \frac{B^2}{2\mu_0}2\pi r\,dr = \frac{\mu_0 I^2}{24\pi}$. Between, $\displaystyle\int_a^b \frac{B^2}{2\mu_0}2\pi r\,dr = \frac{\mu_0 I^2}{4\pi}\ln\frac{b}{a}$. Together:

Step 5. Push on the tube (f). Just inside the tube $B_b = \dfrac{\mu_0 I}{2\pi b} = 3.333\ \text{mT}$. The tube carries $\dfrac{I}{2\pi b}$ per metre of its width, against the inner current, so it is pushed outwards. Its current feels $B_b/2$, so the force per square metre is