Energy stored in a magnetic field!

The magnetic energy stored in the cylinder is given by:

$E=\displaystyle\iiint_{\text{cylinder}}\frac{\mathbf{B}\cdot\mathbf{B}}{2\mu_{0}}\text{d}V$

Now the trick here to avoid difficult calculations is to split the field at any point inside the cylinder into two components:

The first of these components arises only due to the current flowing in the cylindrical region. Let us call it $\mathbf{B}_{\text{cylinder}}$ .

The second component is the field due to the remaining part of the wire assuming the cylinder is removed. Let us call it $\mathbf{B}_{\text{remaining}}$ .

Thus $\mathbf{B}=\mathbf{B}_{\text{cylinder}}+\mathbf{B}_{\text{remaining}}$

Therefore $E=\displaystyle\iiint_{\text{cylinder}}\frac{\mathbf{B}_{\text{cylinder}}\cdot\mathbf{B}_{\text{cylinder}}}{2\mu_{0}}\text{d}V+\displaystyle\iiint_{\text{cylinder}}\frac{\mathbf{B}_{\text{remaining}}\cdot\mathbf{B}_{\text{remaining}}}{2\mu_{0}}\text{d}V+\displaystyle\iiint_{\text{cylinder}}\frac{2\mathbf{B}_{\text{cylinder}}\cdot\mathbf{B}_{\text{remaining}}}{2\mu_{0}}\text{d}V$

$=E_{1}+E_{2}+E_{3}$

$\text{Computing }E_{1}\text{:}$

It can be easily shown that $\mathbf{B}_{\text{cylinder}}=\dfrac{\mu_{0}Jx}{2}$ where $x$ is the distance from the axis of the cylinder.

Consider a small cylindrical shell of thickness $\text{d}x$ at a radial distance of $x$ :

$\text{d}E_{1}=\dfrac{1}{2\mu_{0}}\dfrac{\mu_{0}^{2}J^{2}x^{2}}{4}(2\pi xl\text{d}x)$

$\therefore E_{1}=\dfrac{\mu_{0}J^{2}\pi l}{4}\displaystyle\int_{0}^{R} x^{3}\text{d}x=\dfrac{\mu_{0}J^{2}\pi lR^{4}}{16}$

$\text{Computing }E_{2}\text{:}$

First we shall find $\mathbf{B}_{\text{remaining}}$ .

Refer the diagram above. We want to find the field at a point whose position vector w.r.t. the center of the one of the cross sections of the cylinder is $\mathbf{b}$ . The position vector of the center of this cross section is $\mathbf{a}$ . Assume that the plane of the cylinder is the $xy$ plane.

Thus $\mathbf{B}_{\text{remaining}}=\mathbf{B}-\mathbf{B}_{\text{cylinder}}$

$=\dfrac{\mu_{0}J}{2}(\hat{k}\times(\mathbf{a}+\mathbf{b}))-\dfrac{\mu_{0}J}{2}(\hat{k}\times\mathbf{b})$

$=\dfrac{\mu_{0}J}{2}(\hat{k}\times\mathbf{a})$

Thus $\mathbf{B}_{\text{remaining}}=\dfrac{\mu_{0}J}{2}(\hat{k}\times\mathbf{a})$ which is a constant!

Therefore $E_{2}=\displaystyle\iiint_{\text{cylinder}}\frac{\mathbf{B}_{\text{remaining}}\cdot\mathbf{B}_{\text{remaining}}}{2\mu_{0}}\text{d}V$

$=\dfrac{1}{2\mu_{0}}\dfrac{\mu_{0}^{2}J^{2}d^{2}}{4}\times\text{Volume of cylinder}$

$\therefore E_{2}=\dfrac{\mu_{0}J^{2}\pi lR^{2}d^{2}}{8}$

$\text{Computing }E_{3}\text{:}$

Refer the diagram above. Say we want to compute $\mathbf{B}_{\text{cylinder}}\cdot\mathbf{B}_{\text{rremaining}}$ at a point $A$ . Now consider the diametrically opposite point $B$ in the diagram. If the angle between $\mathbf{B}_{\text{cylinder}}$ and $\mathbf{B}_{\text{remaining}}$ at $A$ is $\theta$ then the angle between them at point $B$ is $180-\theta$ . Thus all the contributions due to $\mathbf{B}_{\text{cylinder}}\cdot\mathbf{B}_{\text{rremaining}}$ cancel out.

So $E_{3}=\displaystyle\iiint_{\text{cylinder}}\frac{2\mathbf{B}_{\text{cylinder}}\cdot\mathbf{B}_{\text{remaining}}}{2\mu_{0}}\text{d}V=0$

Thus $E=E_{1}+E_{2}=\boxed{\dfrac{\mu_{0}J^{2}\pi R^{2}l} {8}\left( \dfrac{R^{2}}{2}+d^{2}\right)}$

Using Ampere's Law, we get :

$B(r) = \dfrac{\mu_{0} J r}{2}$

Consider a cross-section. On that , consider a circle of radius $r$ centered at axis of infinite cylinder. Consider the region where it intersects the given cylinder, as shown:

Clearly, using cosine rule,

$\cos \phi = \dfrac{ r^2+d^2-R^2}{2dr}$

Hence, $dV = 2r \phi dr \times L$

Hence,

$dE_{B} = \dfrac{B^2}{2 \mu_{0}} dV = \dfrac{1}{2 \mu_{0}}\bigg(\dfrac{\mu_{0}J r}{2}\bigg)^2 2Lr \arccos\bigg(\dfrac{r^2+d^2-R^2}{2dr}\bigg) dr$

= $\dfrac{\mu_{0} J^2}{4} r^3 \text{ } \arccos\bigg(\dfrac{r^2+d^2-R^2}{2dr}\bigg) dr$

Hence, $E = \dfrac{\mu_{0}J^2L}{4} \displaystyle \int_{d-R}^{d+R} r^3 \text{ } \arccos \bigg(\dfrac{r^2+d^2-R^2}{2dr}\bigg) {\mathrm dr}$

= $\dfrac{1}{5 \pi} \displaystyle \int_{2\sqrt{2}-2}^{2\sqrt{2}+2} r^3 \text{ } \arccos \bigg(\dfrac{r^2+4}{4 \sqrt{2} r} \bigg) {\mathrm dr}$

Use the substitution $r = 2x$ , and then $x+1/x = 2 \sqrt{2} \cos \theta$ to get the answer as 4