Chapter 31, Problem 37PQ

(a)

To determine

Find the magnetic field inside the solenoid when it is connected to emf of $11.6\text{ V}$

Answer to Problem 37PQ

The magnetic field inside the solenoid is $\underset{\_}{9.1\times {10}^{-3}\text{ T}}$.

Explanation of Solution

Number of turns in solenoid is the ratio of length of solenoid to the diameter of the copper wire. Diameter of the copper is $2r$.

Write the expression for number of turns in solenoid as.

    $N=\frac{L}{2r}$                                                                                                           (I)

Here, $N$ is number of turns, $L$ is length and $r$ is radius of copper wire.

Write the expression for total length of the copper wire as.

  $l=N\times 2D$                                                                                                   (II)

Here, $l$ is length of copper wire, $N$ is number of turns of solenoid and $D$ is diameter of solenoid.

Write the expression for resistance of copper wire as.

    $R=\frac{\rho l}{A}$                                                                                                        (III)

Here, $R$ is resistance, $\rho$ is resistivity, $l$ is length of copper wire and $A$ is area of copper wire.

Area of the copper wire is $\pi r^2$.

Substitute $\pi r^2$ for $A$ in above equation (III).

    $R=\frac{\rho l}{\pi r^2}$                                                                                                      (IV)

Here, $r$ is radius of copper wire.

Write the expression for current in copper wire as.

    $I=\frac{\epsilon }{R}$                                                                                                           (V)

Here, $I$ is current in wire, $\epsilon$ is emf and $R$ is resistance of copper wire.

Write the expression for magnetic field as.

    $B=\frac{{\mu }_{0}NI}{L}$                                                                                                   (VI)

Here, $B$ is magnetic field, $I$ is current in wire, $L$ is length of solenoid and ${\mu }_0$ is permeability of free space.

Conclusion:

Substitute $33.0\text{ cm}$ for $L$ and $0.51\text{ mm}$ for $r$ in equation (I).

    $\begin{array}{c}N=\frac{33\text{ cm}}{2\times 0.51\text{ mm}}\\ =\frac{33\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{2\times 0.51\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)}\\ =323.5\\ \approx 324\text{ turns}\end{array}$

Substitute $324$ for $N$ and $7.5\text{ cm}$ for $D$ in equation (II).

    $\begin{array}{c}l=324\times \pi \times 7.5\text{ cm}\\ \text{=}324\times \pi \times 7.5\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\\ \text{=76}\text{.34 m}\end{array}$

Substitute $76.34\text{ m}$ for $l$, $1.68\times {10}^{-8}\Omega \text{-m}$ for $\rho$ and $0.51\text{ mm}$ for $r$ in equation (IV).

    $\begin{array}{c}R=\frac{1.68\times {10}^{-8}\Omega \text{-m}\times 76.34\text{ m}}{\pi \times {\left(0.51\text{ mm}\right)}^2}\\ =\frac{1.68\times {10}^{-8}\Omega \text{-m}\times 76.302\text{ m}}{\pi \times {\left(0.51\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)}^2}\\ =1.5695\Omega \end{array}$

Substitute $1.5695\Omega$ for $R$ and $11.6\text{ V}$ for $\epsilon$ in equation (V).

    $\begin{array}{c}I=\frac{11.6\text{ V}}{1.5695\Omega }\\ =7.39\text{ A}\end{array}$

Substitute $7.39\text{ A}$ for $I$, $324$ for $N$, ${\text{4π×10}}^{\text{-7}}\frac{\text{T-m}}{\text{A}}$ for ${\mu }_0$, and $33\text{ cm}$ for $L$ in equation (VI)

    $\begin{array}{c}B=\frac{\left(\text{4}\pi {\text{×10}}^{-7}\frac{\text{T-m}}{\text{A}}\right)324\times 7.39\text{ A}}{33\text{ cm}}\\ =\frac{\left(\text{4}\pi {\text{×10}}^{-7}\frac{\text{T-m}}{\text{A}}\right)\left(2394.36\text{ A}\right)}{33\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =9.1\times {10}^{-3}\text{ T}\end{array}$

Thus, the magnetic field inside the solenoid is $\underset{\_}{9.1\times {10}^{-3}\text{ T}}$.

(b)

To determine

The magnetic field inside the solenoid when turns have two layers.

Answer to Problem 37PQ

The magnetic field inside the solenoid is same.

Explanation of Solution

When turns have two layers, it means number of turns will be double that is $N=2N$.

Substitute $2N$ for $N$ in equation (VI).

    $B=\frac{{\mu }_{0}2NI}{L}$

Substitute $\frac{L}{2r}$ for $N$ in above equation as.

    $B=\frac{{\mu }_{0}2\left(\frac{L}{2r}\right)I}{L}$

Rearrange the above expression as.

  $B=\frac{{\mu }_{0}I}{R}$                                                                                                    (VII)

Conclusion:

The new magnetic field is independent of number of turns.

Thus, there will be no change in magnetic field inside the solenoid.