Chapter 34, Problem 32SP

(a)

To determine

The relative permeability of the iron when the long iron-core solenoid is connected across $6\text{ V}$ battery, the current rises to $0.63$ of its maximum value after a time of $0.75\text{s}$ andwhen solenoid is air cored, the current rises to $0.63$ of its maximum value after a time of $0.0025\text{s}$.

Answer to Problem 32SP

Solution:

$0.3\times {10}^3$

Explanation of Solution

Given data:

The voltage across battery is $6\text{ V}$.

The current rises to $0.63$ of its maximum value after a time of $0.75\text{s}$ in iron-core solenoid.

The current rises to $0.63$ of its maximum value after a time of $0.0025\text{s}$ in air-core solenoid.

Formula used:

The expression of current through the coil in case of charging is written as,

$i=i_0\left(1-e^{\left(-\frac{t}{\tau }\right)}\right)$

Here, $i_0$ is the maximum current in the circuit, $t$ is the time, and $\tau$ is the time constant.

The expression of time constant for the coil is written as,

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

Here, $L$ is the inductance and $R$ is the resistance of the coil.

The expression of inductance of the inductor is written as,

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

Here, $\mu$ is the permeability of the material, $N$ is the number of turns in the coil, $A$ is the area of the coil, and $l$ is the length of the coil.

The relative permeability of the iron is expressed as,

${\mu }_r=\frac{{\mu }_i}{{\mu }_0}$

Here, ${\mu }_i$ is the permeability of iron and ${\mu }_0$ is the permeability of air.

Explanation:

Consider the expression of current through the coil when iron core is present.

$i_1=i_0\left(1-e^{\left(-\frac{t_1}{{\tau }_1}\right)}\right)$

Here, $t_1$ is the time and ${\tau }_1$ is the time constant for the iron core solenoid.

Substitute $0.63i_0$ for $i_1$ and $0.75\text{ s}$ for $t_1$

$0.63i_0=i_0\left(1-e^{\left(-\frac{0.75\text{s}}{{\tau }_1}\right)}\right)$

Simplify the expression.

$\begin{array}{c}{i}_0{e}^{\left(-\frac{0.75\text{ s}}{{\tau }_1}\right)}=\left(1-0.63\right)i_0\\ =0.37\end{array}$

Take $\log$ on both sides.

$\begin{array}{c}\ln \left(e^{\left(-\frac{0.75\text{ s}}{{\tau }_1}\right)}\right)=\ln \left(0.37\right)\\ -\frac{0.75\text{ s}}{{\tau }_1}=-0.994\end{array}$

Solve for ${\tau }_1$.

${\tau }_1=0.754\text{ s}$

Write the expression of time constant for the iron core solenoid.

${\tau }_1=\frac{L_1}{R}$

Here, $L_1$ is the inductance for the iron core solenoid.

Substitute $0.754\text{ s}$ for ${\tau }_1$

$\frac{L_1}{R}=0.754\text{s}$ …… (1)

Consider the expression of current through the coil when iron core is removed.

$i_2=i_0\left(1-e^{\left(-\frac{t_2}{{\tau }_2}\right)}\right)$

Here, $t_2$ is the time and ${\tau }_2$ is the time constant when iron core is removed.

Substitute $0.63i_0$ for $i_2$ and $0.0025\text{s}$ for $t_2$

$0.63i_0=i_0\left(1-e^{\left(-\frac{0.0025\text{s}}{{\tau }_2}\right)}\right)$

Simplify the expression.

$\begin{array}{c}{i}_0{e}^{\left(-\frac{0.0025\text{ s}}{{\tau }_2}\right)}=\left(1-0.63\right)i_0\\ =0.37\end{array}$

Take $\log$ on both sides.

$\begin{array}{c}\ln \left(e^{\left(-\frac{0.0025\text{s}}{{\tau }_2}\right)}\right)=\ln \left(0.37\right)\\ -\frac{0.0025\text{s}}{{\tau }_2}=-0.994\end{array}$

Solve for ${\tau }_2$.

${\tau }_2=2.5\times {10}^{-3}\text{ s}$

Write the expression of time constant when the iron core is removed.

${\tau }_2=\frac{L_2}{R}$

Here, $L_2$ is the inductance for the iron core solenoid.

Substitute $2.5\times {10}^{-3}\text{ s}$ for ${\tau }_2$

$\frac{L_2}{R}=2.5\times {10}^{-3}\text{s}$ …… (2)

Divide equation (1) and (2).

$\frac{\left(\frac{L_1}{R}\right)}{\left(\frac{L_2}{R}\right)}=\frac{0.754\text{s}}{2.5\times {10}^{-3}\text{s}}$

Simplify the expression.

$\frac{L_1}{L_2}=0.3\times {10}^3$

Consider the expression of inductance of the inductor when iron core is present.

$L_1=\frac{{\mu }_i{N}^{2}A}{l}$ …… (3)

Here, ${\mu }_i$ is the permeability of the iron.

Consider the expression of inductance of the inductor when iron core is not present.

$L_2=\frac{{\mu }_0{N}^{2}A}{l}$ …… (4)

Here, ${\mu }_0$ is the permeability of the air.

From equation (3) and (4)

$\begin{array}{c}\frac{L_1}{L_2}=\frac{\left(\frac{{\mu }_i{N}^{2}A}{l}\right)}{\left(\frac{{\mu }_0{N}^{2}A}{l}\right)}\\ =\frac{{\mu }_i}{{\mu }_0}\end{array}$

Recall the expression of relative permeability of the iron.

${\mu }_r=\frac{{\mu }_i}{{\mu }_0}$

Substitute ${\mu }_r$ for $\frac{{\mu }_i}{{\mu }_0}$

$\frac{L_1}{L_2}={\mu }_r$

Substitute $0.3\times {10}^3$ for $\frac{L_1}{L_2}$

${\mu }_r=0.3\times {10}^3$

Conclusion:

Therefore, the relative permeability of the iron is $0.3\times {10}^3$.

(b)

To determine

The value of $L$ for the air-core solenoid, which when connected across $6\text{ V}$ battery gives a maximum current of $0.5\text{A}$.

Answer to Problem 32SP

Solution:

$0.03\text{ H}$

Explanation of Solution

Given data:

The voltage across battery is $6\text{ V}$.

From previous part, the value of time constant for air-cored solenoid is $2.5\times {10}^{-3}\text{ s}$.

The value of maximum current is $0.5\text{ A}$.

Formula used:

The Ohm’s Law states that the current through the resistor is directly proportional to the applied voltage across its terminal.

Mathematically,

$\begin{array}{l}I\propto V\\ I=\frac{1}{R}V\end{array}$

Rearrange for $V$.

$V=IR$

Here, $R$ is the resistance, $V$ is the voltage, and $I$ is the current through the resistors.

The expression of time constant for the coil is written as,

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

Here, $L$ is the inductance and $R$ is the resistance of the coil.

Explanation:

Recall the expression of Ohm’s law.

$V=IR$

Solve for $R$.

$R=\frac{V}{I}$

Substitute $6\text{V}$ for $V$ and $0.5\text{A}$ for $I$

$\begin{array}{c}R=\frac{6\text{V}}{0.5\text{A}}\\ =12\Omega \end{array}$

Consider the expression of time constant for the coil.

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

Solve for $L$.

$L=\tau R$

Substitute $2.5\times {10}^{-3}\text{ s}$ for $\tau$ and $12\Omega$ for $R$

$\begin{array}{c}L=\left(2.5\times {10}^{-3}\text{ s}\right)\left(12\Omega \right)\\ =0.03\text{ H}\end{array}$

Conclusion:

Therefore, the inductance of the air-cored coil is $0.03\text{ H}$.