Chapter 20, Problem 39E

(a)

To determine

The magnetic field inside the solenoid.

Answer to Problem 39E

The magnetic field inside the solenoid is $4\pi k_m{k}^{\prime }nI$.

Explanation of Solution

Write the expression for Ampere’s law.

    ${\displaystyle \oint B}\cdot dl=4\pi k^{\prime }NI$        (1)

Here, $B$ is the magnetic field, $dl$ is small elemental length, $k^{\prime }$ is constant, $n$ is the number of and $I$ is the current.

Conclusion:

The magnetic field and the length of the solenoid is parallel to each other.

Rewrite equation (1).

    $B{\displaystyle \oint dl}=4\pi k^{\prime }{k}_{m}NI$

Here, $k_m$ is the magnetic constant for coil.

Substitute $l$ for the closed integral and $\left(nl\right)$ for $N$ in the above equation.

    $\begin{array}{c}B\left(l\right)=4\pi k_m{k}^{\prime }\left(nl\right)I\\ B=4\pi k_m{k}^{\prime }nI\end{array}$

Here, $l$ is the length of the rectangle and $n$ is the number of turns per unit length.

Thus, the magnetic field inside the solenoid is $4\pi k_m{k}^{\prime }nI$.

(b)

To determine

The amount of energy stored per unit volume in the solenoid.

Answer to Problem 39E

The energy stored per unit volume of the solenoid is $\frac{2\pi N^2{k}_m{k}^{\prime }{I}^2}{l^2}$.

Explanation of Solution

Write the expression for magnetic field inside the solenoid.

    $B=4\pi k_m{k}^{\prime }nI$

Write the expression for the magnitude of the inductance.

    $L=\frac{N\varphi }{I}$

Here, $L$ is the inductance and $\varphi$ is the induced magnetic flux.

Substitute $\left(BA\right)$ for $\varphi$ in the above expression.

    $L=\frac{N\left(BA\right)}{I}$        (2)

Here, $A$ is the area of the solenoid.

Write the expression of magnetic field per unit volume of the solenoid.

    $U=\frac{1}{2Al}L{I}^2$        (3)

Here, $U$ is the energy per unit volume.

Conclusion:

Substitute $4\pi k_m{k}^{\prime }nI$ for $B$ in equation (2).

    $\begin{array}{c}L=\frac{N\left(4\pi k_m{k}^{\prime }nI\right)A}{I}\\ =4\pi N{k}_m{k}^{\prime }nA\end{array}$

Substitute $4\pi N{k}_m{k}^{\prime }nA$ for $L$ in equation (3).

    $\begin{array}{c}U=\frac{1}{2Al}\left(4\pi N{k}_m{k}^{\prime }nA\right)I^2\\ =\frac{2\pi N{k}_m{k}^{\prime }n{I}^2}{l}\end{array}$

Substitute $\left(\frac{N}{l}\right)$ for $n$ in the above equation.

    $U=\frac{2\pi N^2{k}_m{k}^{\prime }{I}^2}{l^2}$

Thus, the energy stored per unit volume of the solenoid is $\frac{2\pi N^2{k}_m{k}^{\prime }{I}^2}{l^2}$.

(c)

To determine

The value of a constant..

Answer to Problem 39E

The value of the constant is $\frac{1}{8\pi k^{\prime }}$.

Explanation of Solution

Write the expression for the energy stored in a magnetic field.

    $U=c\frac{B^2}{k_m}$        (4)

Here, $c$ is the constant.

Write the expression for magnetic field inside the solenoid.

    $B=\frac{4\pi k_m{k}^{\prime }NI}{l}$

Substitute $\left(\frac{4\pi k_m{k}^{\prime }NI}{l}\right)$ for $B$ in equation (4).

    $U=c\frac{{\left(\frac{4\pi k_m{k}^{\prime }NI}{l}\right)}^2}{k_m}$        (5)

Write the expression of the energy stored in a solenoid.

    $U=\frac{2\pi N^2{k}_m{k}^{\prime }{I}^2}{l^2}$        (6)

Conclusion:

Compare equation (5) and (6).

    $c\frac{{\left(\frac{4\pi k_m{k}^{\prime }NI}{l}\right)}^2}{k_m}=\frac{2\pi N^2{k}_m{k}^{\prime }{I}^2}{l^2}$

Solve the above equation.

    $\begin{array}{c}c\frac{{\left(4\pi k_m{k}^{\prime }NI\right)}^2}{l^2{k}_m}=\frac{2\pi N^2{k}_m{k}^{\prime }{I}^2}{l^2}\\ c=\frac{1}{8\pi k^{\prime }}\end{array}$

Thus, the value of the constant is $\frac{1}{8\pi k^{\prime }}$.