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Chapter 34: Problem 21

The iron core of a solenoid has a length of \(40 \mathrm{~cm}\) and a cross section of \(5.0 \mathrm{~cm}^{2}\), and is wound with 10 turns of wire per \(\mathrm{cm}\) of length. Compute the inductance of the solenoid, assuming the relative permeability of the iron to be constant at 500 .

Short Answer

Expert verified
The inductance of the solenoid is approximately 0.251 H.

Step by step solution

01

Convert Measurements to Meters

First, we need to convert the measurements given in centimeters to meters. The length of the solenoid is \(40 \text{ cm}\) which is equivalent to \(0.4 \text{ m}\), and the cross-sectional area is \(5.0 \text{ cm}^2\) which is equivalent to \(5.0 \times 10^{-4} \text{ m}^2\).
02

Calculate the Number of Turns

The solenoid is wound with 10 turns per cm of length. The total number of turns \(N\) for a solenoid length of \(0.4 \text{ m}\) (or \(40 \text{ cm}\)) is calculated by multiplying the number of turns per cm by the length in cm. That is, \(N = 10 \times 40 = 400 \text{ turns}\).
03

Use the Formula for Inductance

The inductance \(L\) of a solenoid is given by the formula:\[L = \frac{\mu N^2 A}{l}\]where \(\mu\) is the magnetic permeability of the core, \(N\) is the number of turns, \(A\) is the cross-sectional area, and \(l\) is the length. The permeability \(\mu\) is given by \(\mu = \mu_0 \times \mu_r\), where \(\mu_0 = 4 \pi \times 10^{-7} \space \text{H/m}\) is the permeability of free space and \(\mu_r = 500\) is the relative permeability of the iron.
04

Calculate Magnetic Permeability \( \mu \)

Calculate \(\mu\) using the formula:\[ \mu = \mu_0 \cdot \mu_r = (4\pi \times 10^{-7}) \cdot 500 = 2\pi \times 10^{-4} \text{ H/m}\].
05

Compute the Inductance of the Solenoid

Substitute the calculated values into the inductance formula:\[L = \frac{(2\pi \times 10^{-4}) \times (400)^2 \times 5.0 \times 10^{-4}}{0.4} = \frac{(2\pi \times 10^{-4}) \times 160000 \times 5.0 \times 10^{-4}}{0.4}\]Computing the above expression gives: \[L \approx 0.251 \text{ Henries (H)}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solenoid
A solenoid is a long coil of wire, designed to generate a magnetic field when an electric current passes through it. This wire coil is usually wound into a helical shape, which helps to create a uniform magnetic field inside its core when the current flows. In practical applications, solenoids are used in devices that require controlled magnetic fields. They are quite common in electromagnets and inductors, and also play critical roles in various electronics and machinery.
The strength of the magnetic field depends on several factors including the number of turns of the wire, the current through the wire, and the core material used inside the solenoid. In simpler terms, the more loops in the coil and the stronger the current, the more intense the magnetic field in the solenoid. Moreover, the material inside plays a big part, as certain materials can amplify this magnetic field.
Magnetic Permeability
Magnetic permeability is a measure of how easily a material can become magnetized, or in other words, how well it can support the formation of a magnetic field within itself. This property is crucial because it determines the effectiveness of a material in concentrating magnetic field lines. Simply put, materials with high magnetic permeability are better at enhancing magnetic fields.
In the case of our solenoid, magnetic permeability is calculated by multiplying two permeability constants: the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \text{H/m} \)), which is a universal constant, and the relative permeability (\( \mu_r \)), which depends on the chosen core material. The result gives us a more accurate understanding of how the core material will enhance the magnetic field inside the solenoid.
Iron Core
An iron core is often used inside a solenoid to increase its inductance. Iron, being a ferromagnetic material, significantly enhances the magnetic field generated by a solenoid. This is due to its high relative permeability, which means that it can efficiently concentrate and channel the magnetic flux lines generated by the current running through the coil.
The iron core accounts for the solenoid's greater inductance compared to one with an air core. This higher inductance is beneficial for applications requiring significant electromagnetic forces or energy storage. Because of iron's properties, including its ability to be easily magnetized and demagnetized, it is chosen as the core material in many inductive devices.
Relative Permeability
Relative permeability (\( \mu_r \)) indicates how much better a material can conduct magnetic lines of force than a vacuum. This factor is a dimensionless value showing the ability of a material to support the formation of a magnetic field relative to a vacuum.
In this scenario, the given relative permeability of the iron is 500. This means that iron can conduct magnetic flux 500 times more effectively than a vacuum. When calculating inductance, knowing the relative permeability is essential because it directly affects how strongly a solenoid can concentrate magnetic fields, maximizing its inductance and making it much more efficient for use in electromagnetic applications.

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Most popular questions from this chapter

A \(2.0-\mu \mathrm{F}\) capacitor is charged through a \(30-\mathrm{M} \Omega\) resistor by a 45-V battery. Find \((a)\) the charge on the capacitor and \((b)\) the current through the resistor, both determined \(83 \mathrm{~s}\) after the charging process starts. The time constant of the circuit is \(R C=60 \mathrm{~s}\). Also, $$ q_{\infty}=V_{\infty} C=(45 \mathrm{~V})\left(2.0 \times 10^{-6} \mathrm{~F}\right)=9.0 \times 10^{-5} \mathrm{C} $$ (a) \(q=q_{\infty}\left(1-e^{-t / R C}\right)=\left(9.0 \times 10^{-5} \mathrm{C}\right)\left(1-e^{-83 / 60}\right)\) But \(e^{-83 / 60}=e^{-1.383}=0.25\) Then substitution gives $$ q=\left(9.0 \times 10^{-5} \mathrm{C}\right)(1-0.25)=67 \mu \mathrm{C} $$ (b) \(i=i_{0} e^{-t / R C}=\left(\frac{45 \mathrm{~V}}{30 \times 10^{6} \Omega}\right)\left(e^{-1.383}\right)=0.38 \mu \mathrm{A}\)

A \(5.0-\mu \mathrm{F}\) capacitor is charged to a potential difference of \(20 \mathrm{kV}\) across its plates. After being disconnected from the power source, it is connected across a \(7.0-\mathrm{M} \Omega\) resistor to discharge. What is the initial discharge current, and how long will it take for the capacitor voltage to decrease to 37 percent of the \(20 \mathrm{kV}\) ? The loop equation for the discharging capacitor is $$ v_{c}-i R=0 $$ where \(v_{c}\) is the p.d. across the capacitor. At the first instant, \(v_{c}=20 \mathrm{kV}\), so $$ i=\frac{v_{c}}{R}=\frac{20 \times 10^{3} \mathrm{~V}}{7.0 \times 10^{6} \Omega}=2.9 \mathrm{~mA} $$ The potential across the capacitor, as well as the charge on it, will decrease to \(0.37\) of its original value in one time constant. The required time is $$ R C=\left(7.0 \times 10^{6} \Omega\right)\left(5.0 \times 10^{-6} \mathrm{~F}\right)=35 \mathrm{~s} $$

A long air-core solenoid has cross-sectional area \(A\) and \(N\) loops of wire on its length \(d\). ( \(a\) ) Find its self-inductance. (b) What is its inductance if the core material has a permeability of \(\mu\) ? (a) We can write $$ |\mathscr{E}|=N\left|\frac{\Delta \Phi_{M}}{\Delta t}\right| \quad \text { and } \quad|\mathscr{}|=L\left|\frac{\Delta i}{\Delta t}\right| $$ Equating these two expressions for \(|\varepsilon|\) yields $$ L=N\left|\frac{\Delta \Phi_{M}}{\Delta i}\right| $$ If the current changes from zero to \(I\), then the flux changes from zero to \(\Phi_{M}\). Therefore, \(\Delta i=I\) and \(\Delta \Phi_{M}=\Phi_{M}\) in this case. The self-inductance, assumed constant for all cases, is then $$ L=N \frac{\Phi_{M}}{I}=N \frac{B A}{I} $$ But for an air-core solenoid, \(B=\mu_{0} n I=\mu_{0}(N / d) I\). Substitution gives \(L=\mu_{0} N^{2} A / d\). (b) If the material of the core has permeability \(\mu\) instead of \(\mu_{0}\), then \(B\), and therefore \(L\), will be increased by the factor \(\mu / \mu_{0}\). In that case, \(L=\mu N^{2} A / d\). An iron-core solenoid has a much higher self-inductance than an air-core solenoid has.

A steady current of \(2.5\) A creates a flux of \(1.4 \times 10^{-4} \mathrm{~Wb}\) in a coil of 500 turns. What is the inductance of the coil?

A 2000-loop solenoid is wound uniformly on a long rod with length \(d\) and cross-section \(A\). The relative permeability of the iron is \(k_{m}\). On top of this is wound a 50-loop coil which is used as a secondary. Find the mutual inductance of the system. The flux through the solenoid is $$ \Phi_{M}=B A=\left(k_{M} \mu_{0} n I_{p}\right) A=\left(k_{M} \mu_{0} n I_{p} A\right)\left(\frac{2000}{d}\right) $$ This same flux goes through the secondary. We have, then, $$ \left|\mathscr{E}_{s}\right|=N_{s}\left|\frac{\Delta \Phi_{M}}{\Delta t}\right| \quad \text { and } \quad\left|\varepsilon_{s}\right|=M\left|\frac{\Delta i_{p}}{\Delta t}\right| $$ from which $$ M=N_{s}\left|\frac{\Delta \Phi_{M}}{\Delta i_{p}}\right|=N_{s} \frac{\Phi_{M}-0}{I_{p}-0}=50 \frac{k_{M} \mu_{0} I_{p} A(2000 / d)}{I_{p}}=\frac{10 \times 10^{4} k_{M} \mu_{0} A}{d} $$
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