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Suppose you wish to make a solenoid whose self-inductance is 1.4 \(\mathrm{mH}\). The inductor is to have a cross-sectional area of \(1.2 \times 10^{-3} \mathrm{m}^{2}\) and a length of \(0.052 \mathrm{m}\). How many turns of wire are needed?

Short Answer

Expert verified
220 turns are needed.

Step by step solution

01

Understanding the Inductance Formula

The self-inductance (L) of a solenoid is given by the formula: \( L = \mu_0 \frac{N^2 A}{l} \), where \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T \cdot m/A} \)), \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( l \) is the length of the solenoid.
02

Rearranging the Formula for N

We need to solve for the number of turns, \( N \). Rearranging the formula for \( L \) gives: \( N^2 = \frac{L \cdot l}{\mu_0 \cdot A} \). Thus, \( N = \sqrt{\frac{L \cdot l}{\mu_0 \cdot A}} \).
03

Substitute Known Values

Substitute the given values into the formula: \( N = \sqrt{\frac{1.4 \times 10^{-3} \cdot 0.052}{4\pi \times 10^{-7} \cdot 1.2 \times 10^{-3}}} \).
04

Calculating Intermediate Steps

Calculate the components inside the square root. First, calculate the numerator: \( L \cdot l = 1.4 \times 10^{-3} \cdot 0.052 = 7.28 \times 10^{-5} \). Then the denominator: \( \mu_0 \cdot A = 4\pi \times 10^{-7} \cdot 1.2 \times 10^{-3} = 1.51 \times 10^{-9}s \).
05

Final Calculation for N

Now calculate \( N \): \( N = \sqrt{\frac{7.28 \times 10^{-5}}{1.51 \times 10^{-9}}} = \sqrt{48278.1} \approx 219.6 \). Thus, the number of turns required is approximately 220 (since \( N \) must be an integer).

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Key Concepts

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

Self-Inductance
Self-inductance is a fundamental concept in electromagnetism, especially when dealing with coils and solenoids. It is the property of a solenoid that allows it to induce voltage in itself when the current passing through it changes. The self-inductance of a solenoid "L" is a measure of how effectively it can convert a change in current into electromagnetic force. This is a result of the electromagnetic field created around the coil.
\(L\) is measured in henrys (H), and it gives us the ability to predict how a solenoid will react to changes in current. In our exercise, we wanted the solenoid to have a self-inductance of 1.4 mH (millihenrys). Remember, understanding self-inductance helps you comprehend how solenoids function in circuits and how they are essential in creating inductive cells in various devices.
Number of Turns
The number of turns in a solenoid greatly influences its inductive properties. Essentially, this is the count of how many coils of wire are wrapped around the solenoid's magnetic core. More turns mean a greater magnetic field and higher self-inductance.
In our problem, we calculated the exact number of wire turns needed to reach the desired 1.4 mH inductance given certain structural properties. When manipulating formulae involving solenoids, rearranging to solve for the number of turns is often necessary. Through calculations, we found that approximately 220 turns are required. This solution demonstrates the practical aspect of designing circuits where specific inductance is needed.
Solenoid Formula
The formula for the self-inductance of a solenoid is pivotal for solving many practical physics and engineering problems. The relationship is given by\[ L = \mu_0 \frac{N^2 A}{l} \]where:
  • \(L\) is the inductance in henrys,
  • \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \ \mathrm{T \cdot \mathrm{m/A}}\)),
  • \(N\) is the number of turns,
  • \(A\) is the cross-sectional area, and
  • \(l\) is the length of the solenoid.
In our context, the formula was rearranged in terms of \(N\) to calculate the exact number of turns needed. This formula is a cornerstone of electromagnetic theory and highlights the interrelationship between the physical configuration of a solenoid and its electrical properties.
Electromagnetism Concepts
Understanding the broader picture, electromagnetism is a field in physics that studies the interaction of electric currents and magnetic fields. Solenoids are perfect examples of devices influenced by electromagnetic principles.
These devices create a uniform magnetic field when an electric current flows through the wire coil. The strength and efficiency of this magnetic field depend on several factors, such as the number of turns and the current flowing through the wire.
By altering the number of turns in a solenoid, as seen in our problem, you control its self-inductance and the generated magnetic field. Hence, mastering electromagnetism concepts like self-inductance, permeability, and electrical properties of materials is crucial for designing effective electronic components.

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Most popular questions from this chapter

The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is \(50: 1 .\) The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3}\) A. Find the power consumed by the air filter.

A conducting coil of 1850 turns is connected to a galvanometer, and the total resistance of the circuit is \(45.0 \Omega .\) The area of each turn is \(4.70 \times 10^{-4} \mathrm{m}^{2} .\) This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced to flow around the circuit is measured to be \(8.87 \times 10^{-3} \mathrm{C} .\) Find the magnitude of the magnetic field.

A long, current-carrying solenoid with an air core has 1750 turns per meter of length and a radius of \(0.0180 \mathrm{m}\). A coil of 125 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system?

During a 72 -ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a \(6.0-\mathrm{mA}\) current in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is \(12 \Omega .\) The mutual inductance between the two coils is 3.2 mH. What is the change in the primary current?

A constant current of \(I=15 \mathrm{A}\) exists in a solenoid whose in- ductance is \(L=3.1 \mathrm{H}\). The current is then reduced to zero in a certain amount of time. (a) If the current goes from 15 to 0 A in a time of 75 ms, what is the emf induced in the solenoid? \(?\) (b) How much electrical energy is stored in the solenoid? (c) At what rate must the electrical energy be removed from the solenoid when the current is reduced to \(0 \mathrm{A}\) in a time of \(75 \mathrm{ms} ?\) Note that the rate at which energy is removed is the power.

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