91影视

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.

$\mathbf{W}=\frac{1}{2}{\mathbf{LI}}^2$ 鈥︹�� (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.

$\mathbf{W}=\frac{1}{2}鈭�(\mathbf{A}鈰�I)\mathbf{dI}$ 鈥︹�� (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.

$\mathbf{W}\mathbf{=}\frac{1}{\mathbf{2}{\mathbf{渭}}_0}{鈭�}_{\mathrm{allspace}}{B}^{2}d蟿$ 鈥︹�� (3)

Here, ${\mathbf{渭}}_0$ is permeability, $\mathbf{B}$is the magnetic field and $\mathbf{诲蟿}$ is the volume.

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

$\mathbf{W}\mathbf{=}\frac{1}{\mathbf{2}{\mathbf{渭}}_0}\left[{鈭�}_v{B}^{2}d蟿鈭�{鈭�}_S(A脳B)鈰�\mathrm{da}\right]$ 鈥︹�� (4)

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

Determine the magnetic field for the solenoid is given by

$B={渭}_{0}ni$ 鈥︹�� (5)

Here, ${渭}_0$ 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).

$B=\frac{{渭}_{0}Ni}{l}$

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

$L=\frac{N{桅}_{\text{s}}}{i}$

Here, the magnetic flux is

${桅}_{\text{s}}=BA$

Then,

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

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

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

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

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

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

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

$W=\frac{1}{2}{渭}_0{n}^2蟺R^{2}l{I}^2$

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

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

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

For one turn energy stored is,

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

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

$\begin{array}{c}W=\frac{1}{2}\left[\left(\frac{{渭}_{0}nI}{2}R\right)\widehat{蠒}鈰�I\widehat{蠒}\right]鈭�\left(dI\right)\\ =\frac{1}{2}\left(\frac{{渭}_{0}nI}{2}\right)RI\left(2蟺R\right)\end{array}$

Then for $nl$ turns,

$\begin{array}{c}W=nl\left(\frac{1}{2}\left(\frac{{渭}_{0}nl}{2}\right)\right)RI鈰�2蟺R\\ =\frac{1}{2}{渭}_0{n}^2蟺R^{2}l{I}^2\end{array}$

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

Determine themagnetic field inside the solenoid.

$B={渭}_{0}nI$

For magnetic field outside the solenoid.

$B=0$

We know that volume is

$鈭�d蟿=蟺R^{2}l$

Then,

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

Substitute ${渭}_{0}nI$ for $B$ and $蟺R^{2}I$ for $d蟿$ into equation (3).

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

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

Determine the volume of the surface is,

$d蟿=蟺(R^2鈭�a^2)l$

Determine the magnetic field inside the solenoid.

$B={渭}_{0}nl$

Then,

$鈭�B^{2}d蟿鈭�{渭}_0^2{n}^2{I}^2蟺(R鈭�a^2)I$

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

$A=\frac{{渭}_{0}nl}{2}a\widehat{蠒}$

While the magnetic field is,

$B={渭}_{0}nI\widehat{z}$

Then,

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

The areal vector is,

$da=ad蠒dz(鈭�\widehat{s})$

At a point outside $(S=b)$

$A脳B=0$

Then

role="math" localid="1657796826011"

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

Substitute ${渭}_0^2{n}^2{I}^2蟺(R^2鈭�a^2)l$ for$B^{2}d蟿$ and role="math" localid="1657796916109" $鈭�\frac{1}{2}{渭}_0^2{n}^2{I}^2{a}^{2}2蟺l$ for $鈭�(A脳B)鈰�da$ into equation (4).

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