Chapter 20, Problem 43P

(a)

To determine

The expression for the magnetic field inside the solenoid.

Answer to Problem 43P

The expression for the magnetic field inside the solenoid is $\underset{\_}{B={\mu }_{0}nI}$.

Explanation of Solution

Write the equation of magnetic field from Ampere’s law.

  ${\displaystyle \oint \overrightarrow{B}.d\overrightarrow{l}}={\mu }_{0}NI$                                                                                                     (I)

Here, $B$ is the magnetic field, $dl$ is the elementary length, ${\mu }_0$ is the permeability of free space, $N$ is the number of loop and $I$ is the current.

Write the equation of the elementary length.

  ${\displaystyle \oint dl}=l$                                                                                                     .................. (II)

Here, $l$ is the length.

Write the expression for the number of turns per unit length.

  $n=\frac{N}{l}$                                                                                                                 (III)

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

Conclusion:

Substitute $l$ for ${\displaystyle \oint dl}$ in equation (I) to find $B$.

  $\begin{array}{c}B\left(l\right)={\mu }_{0}NI\\ B={\mu }_0\left(\frac{N}{l}\right)I\\ ={\mu }_{0}nI\end{array}$

Thus, the expression for the magnetic field inside the solenoid is $\underset{\_}{B={\mu }_{0}nI}$.

(b)

To determine

The expression of magnetic flux through one turn.

Answer to Problem 43P

The expression of magnetic flux through one turn is $\underset{\_}{\varphi =B\pi r^2}$.

Explanation of Solution

Write the equation of the magnetic flux.

  $\varphi =BA$                                                                                                              (IV)

Here, $\varphi$ is the magnetic flux and $A$ area of the loop.

Write the expression for the area of the loop.

  $A=\pi r^2$                                                                                                                (V)

Here, $r$ is the radius.

Conclusion:

Substitute, $\pi r^2$ for $A$ in equation (IV) to find $\varphi$.

  $\begin{array}{c}\varphi =B\left(\pi r^2\right)\\ =B\pi r^2\end{array}$

Thus, the expression of magnetic flux through one turn is $\underset{\_}{\varphi =B\pi r^2}$.

(c)

To determine

The total flux linkage through all turns of the solenoid.

Answer to Problem 43P

The total flux linkage through all turns of the solenoid is $\underset{\_}{N\varphi =NB\pi r^2}$.

Explanation of Solution

Considering the total number of turns of the coil is $N$.

Write the equation for the total flux.

  ${\varphi }_{\text{tot}}=N\varphi$                                                                                                           (VI)

Here, ${\varphi }_{\text{tot}}$ is the total number of flux.

Conclusion:

Substitute, $B\pi r^2$ for $\varphi$ in equation (VI).

  $\begin{array}{c}{\varphi }_{\text{tot}}=N\left(B\pi r^2\right)\\ =NB\pi r^2\end{array}$

Therefore, the total flux linkage through all turns of the solenoid is $\underset{\_}{N\varphi =NB\pi r^2}$.

(d)

To determine

The expression for the self inductance of the solenoid.

Answer to Problem 43P

The expression for the self inductance of the solenoid is $\underset{\_}{L={\mu }_0{n}^2\pi r^{2}l}$.

Explanation of Solution

Write the equation from definition for the self inductance.

  $N\varphi =LI$                                                                                                          (VII)

Here, $L$ is the self-inductance.

Conclusion:

Substitute, $B\pi r^2$ for $\varphi$, ${\mu }_{0}nI$ for $B$ and $nl$ for $N$ in equation (VII).

  $\begin{array}{c}\left(nl\right)\left(B\pi r^2\right)=LI\\ \left(nl\right)\left({\mu }_{0}nI\right)\pi r^2=LI\end{array}$

Rewrite the equation to find $L$.

    $L={\mu }_0{n}^2\pi r^{2}l$

Therefore, the expression for the self inductance of the solenoid is $\underset{\_}{L={\mu }_0{n}^2\pi r^{2}l}$.