Chapter 20, Problem 52P

(a)

To determine

The resistance of the wire.

Answer to Problem 52P

Solution: The resistance of the wire is $1.3\Omega$.

Explanation of Solution

Given Info: The length of insulated copper wire is $60.0\text{m}$ and radius of the copper wire is $0.50\text{mm}$.

Use the table 17.1 to obtain the resistivity of the copper wire is $1.7\times {10}^{-8}\Omega \cdot \text{m}$.

Formula to calculate the resistance of the wire is,

$R_{\text{wire}}=\frac{{\rho }_{\text{Cu}}L}{A_{\text{wire}}}$

• $A_{\text{wire}}$ is the cross sectional area of the wire,
• $R_{\text{wire}}$ is the resistance of the wire,
• ${\rho }_{\text{Cu}}$ is the resistivity of the copper wire,
• L is the length of the wire,

Use $\pi r^2$ for $A_{\text{wire}}$ in the above relation.

$R_{\text{wire}}=\frac{{\rho }_{\text{Cu}}L}{\pi r^2}$

• r is the radius of the copper wire,

Substitute $1.7\times {10}^{-8}\Omega \cdot \text{m}$ for ${\rho }_{\text{Cu}}$, $60.0\text{m}$ for L, and $0.50\text{mm}$ for $r$.

$\begin{array}{c}{R}_{\text{wire}}=\frac{\left(1.7\times {10}^{-8}\Omega \cdot \text{m}\right)\left(60.0\text{m}\right)}{\left(3.14\right){\left[\left(0.50\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right]}^2}\\ =1.3\Omega \end{array}$

Conclusion:

Therefore, the resistance of the wire is $1.3\Omega$.

(b)

To determine

The number of turns will be made in the wire.

Answer to Problem 52P

Solution: The number of turn can be made with the wire is $4.8\times {10}^2\text{turns}$.

Explanation of Solution

Given Info: The length of insulated copper wire is $60.0\text{m}$ and radius of the solenoid is $2.0\text{cm}$.

Formula to calculate the number of turns in the wire is,

$N=\frac{L}{C}$

• $N$ is the number of turns in the wire,
• C is the circumference of the loop,
• $L$ is the length of insulated copper wire,

Use $2\pi r_s$ for C in the above relation,

$N=\frac{L}{2\pi r_s}$

• $r_s$ is the radius of the solenoid,

Substitute $60.0\text{m}$ for L and $2.0\text{cm}$ for $r_s$.

$\begin{array}{c}N=\frac{\left(60.0\text{m}\right)}{2\left(3.14\right)\left(2.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =4.8\times {10}^2\text{turns}\end{array}$

Conclusion:

Therefore, the number of turn can be made with the wire is $4.8\times {10}^2\text{turns}$.

(c)

To determine

The length of the resulting solenoid.

Answer to Problem 52P

Solution: The length of the resulting solenoid is $0.48\text{m}$.

Explanation of Solution

Given Info: The number of turn in the wire is $4.8\times {10}^2\text{turns}$ and radius of the copper wire is $0.50\text{mm}$.

Formula to calculate the length of the resulting solenoid is,

$l=ND$

• $N$ is the number of turns in the wire,
• D is the diameter of the wire,
• $l$ is the length of the solenoid,

Use $2r$ for D in the above relation,

$l=N\left(2r\right)$

• $r$ is the radius of the copper wire,

Substitute $4.8\times {10}^2\text{turns}$ for N and $0.50\text{mm}$ for $r$.

$\begin{array}{c}l=\left(4.8\times {10}^2\text{turns}\right)2\left(0.50\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\\ =0.48\text{m}\end{array}$

Conclusion:

Therefore, the length of the resulting solenoid is $0.48\text{m}$.

(d)

To determine

The self inductance of the solenoid.

Answer to Problem 52P

Solution: The self inductance of the solenoid is $0.76\text{mH}$.

Explanation of Solution

Given Info: The permeability of free space is $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$, number of turn in the wire is $4.8\times {10}^2\text{turns}$, radius of the solenoid is $2.0\text{cm}$.and length of the resulting solenoid is 0.48 m.

Formula to calculate the self inductance of the solenoid is,

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

• $N$ is the number of turns in the wire,
• ${\mu }_0$ is the permeability of free space,
• $L$ is the inductance of the solenoid,

Use $\pi r_s^2$ for $A_s$ in the above relation,

$L=\frac{{\mu }_0{N}^2\left(\pi r_s^2\right)}{l}$

• $r_s$ is the radius of the solenoid,

Substitute $4.8\times {10}^2\text{turns}$ for N, $2.0\text{cm}$ for $r_s$, 0.48 m for l and $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$.

$\begin{array}{c}L=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right){\left(4.8\times {10}^2\text{turns}\right)}^2\pi {\left[\left(2.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}{\left(0.48\text{m}\right)}\\ =7.6\times {10}^{-4}\text{H}\\ =0.76\text{mH}\end{array}$

Conclusion:

Therefore, the self inductance of the solenoid is $0.76\text{mH}$.

(e)

To determine

The time constant of the circuit.

Answer to Problem 52P

Solution: The time constant of the circuit is $0.46\text{ms}$.

Explanation of Solution

Given Info: The inductance of the solenoid is $0.76\times {10}^{-3}\text{H}$, resistance of the wire is $1.3\Omega$ and internal resistance is $350\text{m}\Omega$.

Formula to calculate the time constant of the RL circuit is,

$\tau =\frac{L}{R_{\text{total}}}$

• $\tau$ is the time constant,
• $R_{\text{total}}$ is the total resistance,
• $L$ is the inductance of the solenoid,

Use $R_{\text{w}}+r_{\text{in}}$ for $R_{\text{total}}$ in the above relation,

$\tau =\frac{L}{R_{\text{w}}+r_{\text{in}}}$

• $R_{\text{w}}$ is the resistance of the wire,
• $r_{\text{in}}$ is the internal resistance

Substitute $0.76\times {10}^{-3}\text{H}$ for L, $1.3\Omega$ for $R_{\text{w}}$ and $350\text{m}\Omega$ for $r_{\text{in}}$.

$\begin{array}{c}\tau =\frac{\left(0.76\times {10}^{-3}\text{H}\right)}{\left(1.3\Omega +350\text{m}\Omega \right)}\\ =\frac{\left(0.76\times {10}^{-3}\text{H}\right)}{\left(1.3\Omega +0.350\Omega \right)}\\ =0.46\text{ms}\end{array}$

Conclusion:

Therefore, the time constant of the circuit is $0.46\text{ms}$.

(f)

To determine

The maximum current attained in the circuit.

Answer to Problem 52P

Solution: The maximum current attained in the circuit is $3.6\text{A}$.

Explanation of Solution

Given Info: The emf of the battery is $6.0\text{V}$ and total resistance is $1.65\Omega$.

Formula to calculate the maximum current is,

$I_{\max }=\frac{\epsilon }{R_{\text{total}}}$

• $I_{\max }$ is the maximum current,
• $R_{\text{total}}$ is the total resistance,
• $\epsilon$ is the emf,

Substitute $6.0\text{V}$ for $\epsilon$ and $1.65\Omega$ for $R_{\text{total}}$.

$\begin{array}{c}{I}_{\max }=\frac{\left(6.0\text{V}\right)}{\left(1.65\Omega \right)}\\ =3.6\text{A}\end{array}$

Conclusion:

Therefore, the maximum current attained in the circuit is $3.6\text{A}$.

(g)

To determine

The time taken to reach 99.9% of maximum current.

Answer to Problem 52P

Solution: The time taken to reach 99.9% of maximum current is $3.2\text{ms}$.

Explanation of Solution

Given Info: The time constant of the circuit is $0.46\text{ms}$.

Formula to calculate the current is,

$I=I_{\max }\left(1-e^{-t/\tau }\right)$ (1)

• $I_{\max }$ is the maximum current,
• $I$ is the current
• $\tau$ is the time constant,
• t is the time,

Since the time taken to reach 99.9% of maximum current is,

$I=0.999I_{\max }$ (2)

Compare equation (1) and (2).

$\begin{array}{c}1-e^{-t/\tau }=0.999\\ e^{-t/\tau }=1-0.999\\ e^{-t/\tau }=0.001\end{array}$

Rewrite the above relation in terms of t.

$\begin{array}{c}{e}^{-t/\tau }=0.001\\ -\frac{t}{\tau }=\ln \left(0.001\right)\\ t=-\tau \ln \left(0.001\right)\end{array}$

Substitute $0.46\text{ms}$ for $\tau$ to find t.

$\begin{array}{c}t=-\left(0.46\text{ms}\right)\ln \left(0.001\right)\\ =3.2\text{ms}\end{array}$

Conclusion:

Therefore, the time taken to reach 99.9% of maximum current is $3.2\text{ms}$.

(h)

To determine

The maximum energy stored in the inductor.

Answer to Problem 52P

Solution: The maximum energy stored in the inductor is $4.9\text{mJ}$.

Explanation of Solution

Given Info: The inductance of the solenoid is $0.76\times {10}^{-3}\text{H}$ and the maximum current attained in the circuit is $3.6\text{A}$.

Formula to calculate the maximum energy stored in the inductor is,

$E_L=\frac{1}{2}L{I}_{\max }^2$

• $I_{\max }$ is the maximum current,
• $E_L$ is the energy stored in the inductor,
• $L$ is the inductance,

Substitute $0.76\times {10}^{-3}\text{H}$ for $L$ and $3.6\text{A}$ for $I_{\max }$.

$\begin{array}{c}{E}_L=\frac{1}{2}\left(0.76\times {10}^{-3}\text{H}\right){\left(3.6\text{A}\right)}^2\\ =4.9\text{mJ}\end{array}$

Conclusion:

Therefore, the maximum energy stored in the inductor is $4.9\text{mJ}$.