Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77\times {10}^{-3}\text{T}$

Explanation of Solution

Given Info: current in the coil is 2A, number of turns is 300, length of the coil is $20\text{cm}$.

Formula for initial magnetic field in the solenoid,

$B_i=N\frac{{\mu }_{o}I}{l}$

• l is the length of the coil
• N is the number of turns in the coil
• I is the current in the coil

Substitute 2A for I, 300 for N, 20cm for l and $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for ${\mu }_o$ in the above expression to get $B_i$

$\begin{array}{c}{B}_i=\frac{4\pi \times {10}^{-7}\text{T}\text{.m/A}\times 300\times 2\text{A}}{\left(20\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =3.77\times {10}^{-3}\text{T}\end{array}$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77\times {10}^{-3}\text{T}$

(b)

To determine

Final magnetic field in the solenoid

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42\times {10}^{-3}\text{T}$

Explanation of Solution

Given Info: current in the coil is 5A, number of turns is 300, length of the coil is $20\text{cm}$

Formula for final magnetic field in the solenoid,

$B_f=N\frac{{\mu }_{o}I}{l}$

Substitute 5A for I, , 300 for N, 20cm for l and $4\pi \times {10}^{-7}\text{T}\text{.m/A}$ for $\mu$ in the above expression to get $B_i$

$\begin{array}{c}{B}_i=\frac{4\pi \times {10}^{-7}\text{T}\text{.m/A}\times 300\times 5\text{A}}{\left(20\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =9.42\times {10}^{-3}\text{T}\end{array}$

Conclusion:

Final magnetic field in the solenoid is $9.42\times {10}^{-3}\text{T}$

(c)

To determine

Area of 4-turn coil

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07\times {10}^{-4}{\text{m}}^{\text{2}}$

Explanation of Solution

Given Info: radius of coil is $1.50\times {10}^{-2}\text{m}$

Formula for area of 4-turn coil,

$A=\pi r^2$

Substitute $1.50\times {10}^{-2}\text{m}$ for r in the above expression to get A.

$\begin{array}{c}A=\pi {\left(1.50\times {10}^{-2}\text{m}\right)}^2\\ =7.07\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

$$

Conclusion:

Area of 4-turn coil is $7.07\times {10}^{-4}{\text{m}}^{\text{2}}$

(d)

To determine

Change in magnetic flux through 4- turn coil.

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99\times {10}^{-6}\text{Wb}$

Explanation of Solution

Given Info: initial magnetic field is $3.77\times {10}^{-3}\text{T}$, final magnetic field is $9.42\times {10}^{-3}\text{T}$ and area of coil is  $7.07\times {10}^{-4}{\text{m}}^{\text{2}}$

Formula for change in magnetic flux,

$\Delta \varphi =\Delta BA$

Simplify above expression

$\Delta \varphi =\left(B_f-B_i\right)A$

Substitute $9.42\times {10}^{-3}\text{T}$ for $B_f$ , $3.77\times {10}^{-3}\text{T}$ for $B_i$ and $7.07\times {10}^{-4}{\text{m}}^{\text{2}}$ for A in the above expression to get $\Delta \varphi$

$\begin{array}{c}\Delta \varphi =\left(9.42\times {10}^{-3}\text{T}-3.77\times {10}^{-3}\text{T}\right)7.07\times {10}^{-4}{\text{m}}^{\text{2}}\\ =3.99\times {10}^{-6}\text{Wb}\end{array}$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99\times {10}^{-6}\text{Wb}$

(e)

To determine

Average Induced emf in 4-turn coil.

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77\times {10}^{-5}\text{V}$

Explanation of Solution

Given Info: magnetic flux is  $3.99\times {10}^{-6}\text{Wb}$

Formula for Average Induced emf in 4-turn coil

$\epsilon =N\left(\frac{\Delta {\varphi }_B}{\Delta t}\right)$

• $\Delta {\varphi }_B$ is the change in magnetic flux
• $\Delta t$ is the change in time

Substitute $3.99\times {10}^{-6}\text{Wb}$ for $\Delta \varphi$ , 4 for N , and $0.900\text{s}$ for $\Delta t$ in the above expression to get $\epsilon$

$\begin{array}{c}\epsilon =4\left(\frac{3.99\times {10}^{-6}\text{Wb}}{0.900\text{s}}\right)\\ =1.77\times {10}^{-5}\text{V}\end{array}$

Conclusion:

Average Induced emf in 4-turn coil is $1.77\times {10}^{-5}\text{V}$

(f)

To determine

Why the contribution of the magnetic field by the current in the 4-turn coil be neglected.

Answer to Problem 19P

Contributions to the magnetic field by the current in the 4-turn coil is negligible

Explanation of Solution

The magnitude of the magnetic field depends on the amount of induced current in the coil. From above calculations, the emf induced in the coil is small according to part (e). When the emf is small, the current in the 4-turn will also be small. Hence, the magnetic field generated by the current is negligible.

Conclusion:

Contributions to the magnetic field by the current in the 4-turn coil is negligible