Question

Find the energy stored in a section of length $\mathit{l}$of a long solenoid (radius$\mathit{R}$, current$\mathit{I}$, $\mathit{n}$ turns per unit length),

(a) using Eq. 7.30 (you found $\mathit{L}$ in Prob. 7.24);

(b) using Eq. 7.31 (we worked out $\mathit{A}$ in Ex. 5.12);

(c) using Eq. 7.35;

(d) using Eq. 7.34 (take as your volume the cylindrical tube from radius $\mathbf{a}\mathbf{<}\mathbf{R}$ out to radius$\mathbf{b}\mathbf{>}\mathbf{R}$).

Short answer

(a) The value ofthe energy stored in the inductor is $\mathrm{W}=\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{n}}^2{\mathrm{\pi R}}^2{\mathrm{LI}}^2$.

(b) The value of the energy stored in the inductor is $\mathrm{W}=\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{n}}^2{\mathrm{\pi R}}^2{\mathrm{LI}}^2$.

(c) The value of the energy stored in the inductor is$\mathrm{W}=\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{n}}^2{\mathrm{\pi R}}^2{\mathrm{LI}}^2$ .

(d) The value of the energy stored in the inductor is $\mathrm{W}=\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{n}}^2{\mathrm{\pi R}}^2{\mathrm{LI}}^2$.

Step-by-step solution

Write the given data from the question.

Consider the energy stored in a section of length $l$ of a long solenoid (radius$R$, current$I$ , $n$ turns per unit length)

Consider a cylindrical tube of inner radius $\mathrm{a}$ and outer radius$b$, imagine $a$ Gaussian surface of length$l$ inside the solenoid.

Determine the formula of the energy stored in the inductor.

Write the formula ofthe energy stored in the inductor of self-inductance.

$\mathbf{W}=\frac{1}{2}{\mathbf{LI}}^2$ …… (1)

Here, $\mathbf{L}$ is the self-inductance and $\mathbf{l}$is the length of the solenoid.

Write the formula ofthe energy stored in the inductor for vector potential at the surface of the solenoid.

$\mathbf{W}=\frac{1}{2}\oint (\mathbf{A}\cdot I)\mathbf{dI}$ …… (2)

Here, $\mathbf{A}$is the vector potential at the surface of the solenoid and $\mathbf{l}$is the length of the solenoid.

Write the formula of the energy stored in the inductor for magnetic field inside the solenoid.

$\mathbf{W}\mathbf{=}\frac{1}{\mathbf{2}{\mathbf{\mu }}_0}{\int }_{\mathrm{allspace}}{B}^{2}d\tau$ …… (3)

Here, ${\mathbf{\mu }}_0$ is permeability, $\mathbf{B}$is the magnetic field and $\mathbf{d\tau }$ is the volume.

Write the formula of the energy stored in the inductor for volume of the surface.

$\mathbf{W}\mathbf{=}\frac{1}{\mathbf{2}{\mathbf{\mu }}_0}\left[{\int }_v{B}^{2}d\tau -{\oint }_S(A\times B)\cdot \mathrm{da}\right]$ …… (4)

Here, ${\mathbf{\mu }}_0$ is permeability, $\mathbf{B}$ is the magnetic field, $\mathbf{A}$ is the vector potential at the surface of the solenoid and $\mathbf{d\tau }$ is the volume.

(a) Determine the value of the energy stored in the inductor of self-inductance L.

Determine the magnetic field for the solenoid is given by

$B={\mu }_{0}ni$ …… (5)

Here, ${\mu }_0$ is permeability of the free space, $n$is the number of turns of the solenoid per unit length, and $i$ is the current passing through the solenoid.

The number of turns in length $l$ of the solenoid is

$N=nl$

Therefore, the number of turns of the solenoid per unit length is

$n=\frac{N}{l}$

Substitute $n=\frac{N}{l}$into equation (5).

$B=\frac{{\mu }_{0}Ni}{l}$

Determine the self-inductance $L$of the coil of turn $N$ is

$L=\frac{N{\Phi }_{\text{s}}}{i}$

Here, the magnetic flux is

${\Phi }_{\text{s}}=BA$

Then,

$L=\frac{NBA}{i}$

Substitute $\frac{{\mu }_{0}Ni}{l}$for $B$ into above equation.

$\begin{array}{c}L=\frac{NA}{i}\left(\frac{{\mu }_{0}Ni}{l}\right)\\ =\frac{{\mu }_0{N}^{2}A}{l}\end{array}$

Substitute $\pi R^2$ is the area of the solenoid $\left(A\right)$and $N=nl$in the above equation and simplify.

$\begin{array}{c}L=\frac{{\mu }_0{\left(nl\right)}^2\left(\pi R^2\right)}{l}\\ ={\mu }_0{n}^{2}l\left(\pi R^2\right)\end{array}$

Determine theenergy stored in the inductor of self-inductance$L$ .

Substitute${\mu }_0{n}^{2}l\left(\pi R^2\right)$ for $L$ into equation (1).

$W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2$

Therefore, the value of the energy stored in the inductor is $W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}L{I}^2$.

(b) Determine the value of the energy stored in the inductor for vector potential at the surface of the solenoid.

Determine the vector potential at the surface of the solenoid is,

$A=\left(\frac{{\mu }_{0}nl}{2}\right)R\widehat{\varphi }$

For one turn energy stored is,

Now determine the energy stored in the inductor for vector potential at the surface of the solenoid.

Substitute $\left(\frac{{\mu }_{0}nl}{2}\right)R\widehat{\varphi }$ for $A$ into equation (2).

$\begin{array}{c}W=\frac{1}{2}\left[\left(\frac{{\mu }_{0}nI}{2}R\right)\widehat{\varphi }\cdot I\widehat{\varphi }\right]\oint \left(dI\right)\\ =\frac{1}{2}\left(\frac{{\mu }_{0}nI}{2}\right)RI\left(2\pi R\right)\end{array}$

Then for $nl$ turns,

$\begin{array}{c}W=nl\left(\frac{1}{2}\left(\frac{{\mu }_{0}nl}{2}\right)\right)RI\cdot 2\pi R\\ =\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2\end{array}$

Therefore, the value of the energy stored in the inductor is$W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2$ .

(c) Determine the value of the energy stored in the inductor for magnetic field inside the solenoid.

Determine themagnetic field inside the solenoid.

$B={\mu }_{0}nI$

For magnetic field outside the solenoid.

$B=0$

We know that volume is

$\int d\tau =\pi R^{2}l$

Then,

Determine theenergy stored in the inductor for magnetic field inside the solenoid.

Substitute ${\mu }_{0}nI$ for $B$ and $\pi R^{2}I$ for $d\tau$ into equation (3).

$\begin{array}{c}W=\frac{1}{2{\mu }_0}{\left({\mu }_{0}nI\right)}^2\pi R^{2}I\\ =\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2\end{array}$

Therefore, the value of the energy stored in the inductor is$W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2$ .

(d) Determine the energy stored in the inductor for volume of the surface.

Determine the volume of the surface is,

$d\tau =\pi (R^2-a^2)l$

Determine the magnetic field inside the solenoid.

$B={\mu }_{0}nl$

Then,

$\int B^{2}d\tau -{\mu }_0^2{n}^2{I}^2\pi (R-a^2)I$

At a point inside$(S=a)$, the vector potential is

$A=\frac{{\mu }_{0}nl}{2}a\widehat{\varphi }$

While the magnetic field is,

$B={\mu }_{0}nI\widehat{z}$

Then,

role="math" localid="1657796736019" $\begin{array}{c}A\times B=\frac{1}{2}{\mu }_0^2{n}^2{I}^{2}a(\widehat{\varphi }\times \widehat{z})\\ =\frac{1}{2}{\mu }_0^2{n}^2{I}^{2}a\left(\widehat{s}\right)\end{array}$

The areal vector is,

$da=ad\varphi dz(-\widehat{s})$

At a point outside $(S=b)$

$A\times B=0$

Then

role="math" localid="1657796826011"

Determine the energy stored in the inductor for volume of the surface.

Substitute ${\mu }_0^2{n}^2{I}^2\pi (R^2-a^2)l$ for$B^{2}d\tau$ and role="math" localid="1657796916109" $-\frac{1}{2}{\mu }_0^2{n}^2{I}^2{a}^{2}2\pi l$ for $\oint (A\times B)\cdot da$ into equation (4).

Therefore, the value of the energy stored in the inductor is $W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}L{I}^2$.