91Ó°ÊÓ

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.

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.

$\mathbf{W}=\frac{1}{2}âˆ\circledR (\mathbf{A}â‹\dots I)\mathbf{dI}$ …… (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.

$\mathbf{W}\mathbf{=}\frac{1}{\mathbf{2}{\mathbf{μ}}_0}{∫}_{\mathrm{allspace}}{B}^{2}d\mathrm{τ}$ …… (3)

Here, ${\mathbf{μ}}_0$ is permeability, $\mathbf{B}$is the magnetic field and $\mathbf{»åÏ„}$ is the volume.

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

$\mathbf{W}\mathbf{=}\frac{1}{\mathbf{2}{\mathbf{μ}}_0}\left[{∫}_v{B}^{2}d\mathrm{Ï„}−{âˆ\circledR }_S(A×B)â‹\dots \mathrm{da}\right]$ …… (4)

Here, ${\mathbf{μ}}_0$ is permeability, $\mathbf{B}$ is the magnetic field, $\mathbf{A}$ is the vector potential at the surface of the solenoid and $\mathbf{»åÏ„}$ is the volume.

Determine the magnetic field for the solenoid is given by

$B={\mathrm{μ}}_{0}ni$ …… (5)

Here, ${\mathrm{μ}}_0$ 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).

$B=\frac{{\mathrm{μ}}_{0}Ni}{l}$

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

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

Here, the magnetic flux is

${\mathrm{Φ}}_{\text{s}}=BA$

Then,

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

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

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

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

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

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

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

$W=\frac{1}{2}{\mathrm{μ}}_0{n}^2\mathrm{π}{R}^{2}l{I}^2$

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

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

$A=\left(\frac{{\mathrm{μ}}_{0}nl}{2}\right)R\widehat{\mathrm{ϕ}}$

For one turn energy stored is,

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

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

$\begin{array}{c}W=\frac{1}{2}\left[\left(\frac{{\mathrm{μ}}_{0}nI}{2}R\right)\widehat{\mathrm{Ï•}}â‹\dots I\widehat{\mathrm{Ï•}}\right]âˆ\circledR \left(dI\right)\\ =\frac{1}{2}\left(\frac{{\mathrm{μ}}_{0}nI}{2}\right)RI\left(2\mathrm{Ï€}R\right)\end{array}$

Then for $nl$ turns,

$\begin{array}{c}W=nl\left(\frac{1}{2}\left(\frac{{\mathrm{μ}}_{0}nl}{2}\right)\right)RIâ‹\dots 2\mathrm{Ï€}R\\ =\frac{1}{2}{\mathrm{μ}}_0{n}^2\mathrm{Ï€}{R}^{2}l{I}^2\end{array}$

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

Determine themagnetic field inside the solenoid.

$B={\mathrm{μ}}_{0}nI$

For magnetic field outside the solenoid.

$B=0$

We know that volume is

$∫d\mathrm{τ}=\mathrm{π}{R}^{2}l$

Then,

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

Substitute ${\mathrm{μ}}_{0}nI$ for $B$ and $\mathrm{π}{R}^{2}I$ for $d\mathrm{τ}$ into equation (3).

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

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

Determine the volume of the surface is,

$d\mathrm{τ}=\mathrm{π}(R^2−a^2)l$

Determine the magnetic field inside the solenoid.

$B={\mathrm{μ}}_{0}nl$

Then,

$∫B^{2}d\mathrm{τ}−{\mathrm{μ}}_0^2{n}^2{I}^2\mathrm{π}(R−a^2)I$

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

$A=\frac{{\mathrm{μ}}_{0}nl}{2}a\widehat{\mathrm{ϕ}}$

While the magnetic field is,

$B={\mathrm{μ}}_{0}nI\widehat{z}$

Then,

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

The areal vector is,

$da=ad\mathrm{ϕ}dz(−\widehat{s})$

At a point outside $(S=b)$

$A×B=0$

Then

role="math" localid="1657796826011"

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

Substitute ${\mathrm{μ}}_0^2{n}^2{I}^2\mathrm{Ï€}(R^2−a^2)l$ for$B^{2}d\mathrm{Ï„}$ and role="math" localid="1657796916109" $−\frac{1}{2}{\mathrm{μ}}_0^2{n}^2{I}^2{a}^{2}2\mathrm{Ï€}l$ for $âˆ\circledR (A×B)â‹\dots da$ into equation (4).

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