Chapter 20, Problem 38P

(a)

To determine

The Solenoid’s inductance $\left(L\right)$

Answer to Problem 38P

Solution: The Solenoid’s inductance $\left(L\right)$ is $1.32\times {10}^{-8}\text{H}$

Explanation of Solution

Given Info: Number of turns in the coil N is $475\text{turns}$, Cross sectional area A is $2.80\times {10}^{-9}{\text{m}}^2$

Permeability of Free Space ${\mu }_0$ is $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ length l is $6.00\text{cm}$

Formula to calculate the Solenoid’s inductance $\left(L\right)$ is

$L=\frac{{\mu }_0{N}^{2}A}{l}$

• • L is the inductance of the Solenoid,
  • ${\mu }_0$ is the Permeability of Free Space,
  • N is the Number of turns,
  • A is the Cross sectional area,
  • L is the length of the solenoid,

Substitute $475\text{turns}$ for N, $2.80\times {10}^{-9}{\text{m}}^2$ for A, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $6.00\text{cm}$ for l to find L

$\begin{array}{c}L=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right){\left(475\right)}^2\left(2.80\times {10}^{-9}{\text{m}}^2\right)}{\left(6.00\text{cm}\right)\left(\frac{1\text{cm}}{{10}^2\text{m}}\right)}\\ =\frac{7.934\times {10}^{-10}}{6\times {10}^{-2}}\text{H}\\ =1.32\times {10}^{-8}\text{H}\end{array}$

Conclusion:

The Solenoid’s inductance $\left(L\right)$ is $1.32\times {10}^{-8}\text{H}$

(b)

To determine

The average emf across the around the solenoid ${\epsilon }_{\text{L}}$

Answer to Problem 38P

Solution: The average emf across the around the solenoid ${\epsilon }_{\text{L}}$ is $6.34\times {10}^{-6}\text{V}$

Explanation of Solution

Given Info: Solenoid’s inductance L is $1.32\times {10}^{-8}\text{H}$, the rate of change of current $\Delta I$ is $\left(-2.0\text{A}-\left(\text{+2}\text{.0}\text{A}\right)\right)$, the change in time $\Delta T$ is $8.33\times {10}^{-3}\text{s}$.

Formula to calculate the average emf across the around the solenoid,

${\epsilon }_{\text{L}}=-L\left(\frac{\Delta I}{\Delta t}\right)$

• ${\epsilon }_{\text{L}}$ is the average induced emf in the solenoid
• L is the self inductance of the solenoid
• $\Delta I$ is the rate of change of current
• $\Delta t$ is the change in time

Formula to rate of change in current is,

$\Delta I=I_f-I_i$

• $I_f$ is the final amount of current,
• $I_i$ is the initial amount of current,

Use $I_f-I_i$ for $\Delta I$ in ${\epsilon }_{\text{L}}=-L\left(\frac{\Delta I}{\Delta t}\right)$ to rewrite ${\epsilon }_{\text{L}}$

${\epsilon }_{\text{L}}=-L\left(\frac{I_f-I_i}{\Delta t}\right)$

Substitute $1.32\times {10}^{-8}\text{H}$ for L, $8.33\times {10}^{-3}\text{s}$ for $\Delta t$, $\text{2}\text{.0}\text{A}$ for $I_i$ and $-2.00\text{A}$ for $I_f$ in the above equation to find ${\epsilon }_{\text{L}}$.

$\begin{array}{c}{\epsilon }_{\text{L}}=-\left(1.32\times {10}^{-8}\text{H}\right)\left(\frac{\left(-2.0\text{A}-\left(\text{+2}\text{.0}\text{A}\right)\right)}{8.33\times {10}^{-3}\text{s}}\right)\\ =-\left(1.32\times {10}^{-8}\text{H}\right)\left(\frac{-4\text{A}}{8.33\times {10}^{-3}\text{s}}\right)\\ =6.34\times {10}^{-6}\text{V}\end{array}$

Conclusion:

The average emf across the around the solenoid ${\epsilon }_{\text{L}}$ is $6.34\times {10}^{-6}\text{V}$