Q7.28P - Introduction to Electrodynamics

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 $\mathbf{b}\mathbf{>}\mathbf{R}$ ). 

Step-by-Step Solution

Answer

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

(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 1: 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.

 

Step 2: Determine the formula of the energy stored in the inductor.

Write the formula of the 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 of the 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.

Step 3: (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 the energy 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  $W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}L{I}^2$. 

Step 4: (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 $W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2$ .

Step 5: (c) Determine the value of the energy stored in the inductor for magnetic field inside the solenoid.

Determine the magnetic 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 the energy 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 $W=\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2$  .

Step 6: (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,

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

The areal vector is,

  $da=ad\varphi dz(-\widehat{s})$

At a point outside   $(S=b)$

  $A\times B=0$

Then

$\begin{array}{c}\oint (A\times B)\cdot da=\int \frac{1}{2}{\mu }_0^2{n}^2{I}^{2}a\left(ad\varphi dz\right)\\ =-\frac{1}{2}{\mu }_0^2{n}^2{I}^2{a}^{2}2\pi I\end{array}$

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 $-\frac{1}{2}{\mu }_0^2{n}^2{I}^2{a}^{2}2\pi l$  for $\oint (A\times B)\cdot da$  into equation (4).

  $\begin{array}{c}W=\frac{1}{2}{\mu }_0\left[{\mu }_0^2{n}^2{I}^2\pi (R^2-a^2)l-\frac{1}{2}{\mu }_0^2{n}^2{l}^2{a}^{2}2\pi l\right]\\ =\frac{1}{2}{\mu }_0{n}^2\pi R^{2}l{I}^2\end{array}$

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