Chapter 20, Problem 57AP

An 820-turn wire coil of resistance 24.0 Ω is placed on lop of a 12 500-turn, 7.00-cm-long solenoid, as in Figure P20.57. Both coil and solenoid have cross-sectional area of 1.00 × 10⁻⁴ m². (a) How long does it take the solenoid current to reach 0.632 times its maximum value? (b) Determine the average back emf caused by the self-inductance of the solenoid during this interval. The magnetic field produced by the solenoid at the location of the coil is one-half as strong as the field at the center of the solenoid. (c) Determine the average rate of change in magnetic flux through each turn of the coil during the stated interval. (d) Find the magnitude of the average induced current in the coil.

Figure P20.57

To determine

The time taken by the solenoid current to reach 0.632 times its maximum value.

Answer to Problem 57AP

Solution: The time taken by the solenoid current to reach 0.632 times its maximum value is $20.0\text{ms}$.

Explanation of Solution

Given Info: The number of turns is 12500, cross sectional area of the solenoid is $1.00\times {10}^{-4}{\text{m}}^2$, and length of the solenoid is $7.00\text{cm}$.

Formula to calculate the self-inductance of the solenoid coil is,

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

• L is the self-inductance,
• ${\mu }_0$ 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 12500-turn for N, $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, $1.00\times {10}^{-4}{\text{m}}^2$ for A and $7.00\text{cm}$ for l.

$\begin{array}{c}L=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right){\left(12500\right)}^2\left(1.00\times {10}^{-4}{\text{m}}^2\right)}{\left(7.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =0.280\text{H}\end{array}$

Thus, the self-inductance of the solenoid coil is 0.280 H.

Conclusion:

Since the current in the solenoid reaches a maximum current of $I_s=0.632I_{\max }$ in a time of $t=\tau$, so that,

The formula to calculate the time period is,

$t=\frac{L}{R}$

• t is the time taken by the solenoid current to reach 0.632 times its maximum value,
• R is the resistance of the circuit,

Substitute 0.280 H for L and $14.0\Omega$ for R.

$\begin{array}{c}t=\frac{\left(0.280\text{H}\right)}{\left(14.0\Omega \right)}\\ =0.02\text{s}\\ =2.00\times {10}^{-2}\text{s}\\ \approx 20.0\text{ms}\end{array}$

Therefore, the time taken by the solenoid current to reach 0.632 times its maximum value is $20.0\text{ms}$.

To determine

The average back emf caused by the self-inductance of the solenoid.

Answer to Problem 57AP

Solution: The average back emf caused by the self-inductance of the solenoid is

37.9 V.

Explanation of Solution

Given Info: The self-inductance of the solenoid coil is 0.280 H, potential difference across the battery is $60.0\text{V}$, and resistance of the circuit is $14.0\Omega$.

Formula to calculate the change in the solenoid current at the 0.632 times is,

$\Delta I_s=0.632I_{\max }-0$

• $\Delta I_s$ is the change in the solenoid current,
• $I_{\max }$ is the maximum value of the current,

Use $\Delta V/R$ for $I_{\max }$ in the above relation.

$\Delta I_s=0.632\left(\frac{\Delta V}{R}\right)$

• $\Delta V$ is the potential difference across the battery,

Substitute $60.0\text{V}$ for $\Delta V$ and $14.0\Omega$ for R to find $\Delta I_s$.

$\begin{array}{c}\Delta I_s=0.632\left(\frac{60.0\text{V}}{14.0\Omega }\right)\\ =2.71\text{A}\end{array}$

Thus, the change in the solenoid current is 2.71 A.

Conclusion:

Formula to calculate the average back emf is,

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

• ${\epsilon }_{\text{back}}$ is the back emf,
• L is the self-inductance,
• $\Delta t$ is the time taken by the solenoid current,

Substitute $0.280\text{H}$ for L, $2.71\text{A}$ for $\Delta I_s$, and $20.0\text{ms}$ for $\Delta t$.

$\begin{array}{c}{\epsilon }_{\text{back}}=\left(0.280\text{H}\right)\left[\frac{\left(2.71\text{A}\right)}{\left(20.0\text{ms}\right)\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)}\right]\\ =0.0379\times {10}^3\text{V}\\ =37.9\text{V}\end{array}$

Therefore, the average back emf caused by the self-inductance of the solenoid is

37.9 V.

To determine

The average rate of change in magnetic flux through each turn of the coil.

Answer to Problem 57AP

Solution: The rate of change in magnetic flux through each turn of the coil is $1.52\times {10}^{-3}\text{V}$.

Explanation of Solution

Given Info: The number of turns is 12500, cross sectional area of the solenoid is $1.00\times {10}^{-4}{\text{m}}^2$, the time taken by the solenoid current to reach 0.632 times its maximum value is $2.00\times {10}^{-2}\text{s}$ and length of the solenoid is $7.00\text{cm}$.

Since the change in the magnetic field at the location of the coil is one half of the change in the field at the center of the solenoid, so that $\Delta B_{\text{coil}}=\frac{1}{2}\left[{\mu }_0{n}_s\left(\Delta I_s\right)\right]$.

Formula to calculate the average rate of change of flux through each turn of the coil is, $\frac{\Delta {\Phi }_B}{\Delta t}=\frac{\left(\Delta B_{\text{coil}}\right)A_{\text{coil}}}{\Delta t}$

• $\Delta {\Phi }_B/\Delta t$ is the rate of change of flux,
• $\Delta B_{\text{coil}}/\Delta t$ is the rate of change of magnetic field of the coil,
• $A_{\text{coil}}$ is the cross section area of the coil,

Use $1/2\left[{\mu }_0{n}_s\left(\Delta I_s\right)\right]$ for $\Delta B_{\text{coil}}$ and $N_s/l_s$ for $n_s$ in the above relation,

$\begin{array}{c}\frac{\Delta {\Phi }_B}{\Delta t}=\frac{\frac{1}{2}\left[{\mu }_0{N}_s\left(\Delta I_s\right)\right]A_{\text{coil}}}{l_s\Delta t}\\ =\frac{{\mu }_0{N}_s\left(\Delta I_s\right)A_{\text{coil}}}{2l_s\left(\Delta t\right)}\end{array}$

• ${\mu }_0$ is the permeability of free space,
• $N_s$ is the number of turns in the solenoid,
• $\Delta I_s$ is the is the change in the solenoid current,
• $l_s$ is the length of the solenoid,

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, 12500 for $N_s$, $2.71\text{A}$ for $\Delta I_s$, $1.00\times {10}^{-4}{\text{m}}^2$ for $A_{\text{coil}}$, $7.00\text{cm}$ for $l_s$, and $2.00\times {10}^{-2}\text{s}$ for $\Delta t$.

$\begin{array}{c}\frac{\Delta {\Phi }_B}{\Delta t}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(12500\right)\left(2.71\text{A}\right)\left(1.00\times {10}^{-4}{\text{m}}^2\right)}{2\left(7.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(2.00\times {10}^{-2}\text{s}\right)}\\ =1.52\times {10}^{-3}\text{V}\end{array}$

Conclusion:

Therefore, the rate of change in magnetic flux through each turn of the coil is $1.52\times {10}^{-3}\text{V}$.

The number of turns is 12500, cross sectional area of the solenoid is $1.00\times {10}^{-4}{\text{m}}^2$, the time taken by the solenoid current to reach 0.632 times its maximum value is $2.00\times {10}^{-2}\text{s}$ and length of the solenoid is $7.00\text{cm}$.

Since the change in the magnetic field at the location of the coil is one half of the change in the field at the center of the solenoid, so that $\Delta B_{\text{coil}}=\frac{1}{2}\left[{\mu }_0{n}_s\left(\Delta I_s\right)\right]$.

Formula to calculate the average rate of change of flux through each turn of the coil is, $\frac{\Delta {\Phi }_B}{\Delta t}=\frac{\left(\Delta B_{\text{coil}}\right)A_{\text{coil}}}{\Delta t}$

• $\Delta {\Phi }_B/\Delta t$ is the rate of change of flux,
• $\Delta B_{\text{coil}}/\Delta t$ is the rate of change of magnetic field of the coil,
• $A_{\text{coil}}$ is the cross section area of the coil,

Use $1/2\left[{\mu }_0{n}_s\left(\Delta I_s\right)\right]$ for $\Delta B_{\text{coil}}$ and $N_s/l_s$ for $n_s$ in the above relation,

$\begin{array}{c}\frac{\Delta {\Phi }_B}{\Delta t}=\frac{\frac{1}{2}\left[{\mu }_0{N}_s\left(\Delta I_s\right)\right]A_{\text{coil}}}{l_s\Delta t}\\ =\frac{{\mu }_0{N}_s\left(\Delta I_s\right)A_{\text{coil}}}{2l_s\left(\Delta t\right)}\end{array}$

• ${\mu }_0$ is the permeability of free space,
• $N_s$ is the number of turns in the solenoid,
• $\Delta I_s$ is the is the change in the solenoid current,
• $l_s$ is the length of the solenoid,

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$, 12500 for $N_s$, $2.71\text{A}$ for $\Delta I_s$, $1.00\times {10}^{-4}{\text{m}}^2$ for $A_{\text{coil}}$, $7.00\text{cm}$ for $l_s$, and $2.00\times {10}^{-2}\text{s}$ for $\Delta t$.

$\begin{array}{c}\frac{\Delta {\Phi }_B}{\Delta t}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(12500\right)\left(2.71\text{A}\right)\left(1.00\times {10}^{-4}{\text{m}}^2\right)}{2\left(7.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(2.00\times {10}^{-2}\text{s}\right)}\\ =1.52\times {10}^{-3}\text{V}\end{array}$

Conclusion:

Therefore, the rate of change in magnetic flux through each turn of the coil is $1.52\times {10}^{-3}\text{V}$.

To determine

The magnitude of the average induced current in the coil.

Answer to Problem 57AP

Solution: The magnitude of the average induced current in the coil is $51.9\text{mA}$.

Explanation of Solution

Given Info: The rate of change in magnetic flux through each turn of the coil is $1.52\times {10}^{-3}\text{V}$ number of turns in the coil is 820, and resistance of the coil is $24.0\Omega$.

Formula to calculate the average induced current in the coil is,

$I_{\text{coil}}=\frac{{\epsilon }_{\text{coil}}}{R_{\text{coil}}}$

• $I_{\text{coil}}$ is the induced current in the coil,
• ${\epsilon }_{\text{coil}}$ is the emf across the coil,
• $R_{\text{coil}}$ is the resistance of the coil,

Use $N_{\text{coil}}{\left(\Delta {\Phi }_B/\Delta t\right)}_{\text{coil}}$ for ${\epsilon }_{\text{coil}}$ in the above relation.

$I_{\text{coil}}=\frac{N_{\text{coil}}{\left(\Delta {\Phi }_B/\Delta t\right)}_{\text{coil}}}{R_{\text{coil}}}$

• $N_{\text{coil}}$ is the number of turns in coil,

Substitute 820 turns for $N_{\text{coil}}$, $1.52\times {10}^{-3}\text{V}$ for ${\left(\Delta {\Phi }_B/\Delta t\right)}_{\text{coil}}$, and $24.0\Omega$ for $R_{\text{coil}}$.

$\begin{array}{c}{I}_{\text{coil}}=\frac{\left(820\right)\left(1.52\times {10}^{-3}\text{V}\right)}{\left(24.0\Omega \right)}\\ =51.9\times {10}^{-3}\text{A}\\ =51.9\text{mA}\end{array}$

Conclusion:

Therefore, the magnitude of the average induced current in the coil is $51.9\text{mA}$.