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Chapter 34: Problem 23

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?

Short Answer

Expert verified
The inductance of the coil is 0.028 H.

Step by step solution

01

Understanding the Formula

The inductance of a coil can be calculated using the formula: \[ L = \frac{N \cdot \Phi}{I} \]where \( L \) is the inductance, \( N \) is the number of turns, \( \Phi \) is the magnetic flux, and \( I \) is the current.
02

Substituting the Given Values

Substitute the given values into the formula: number of turns \( N = 500 \), magnetic flux \( \Phi = 1.4 \times 10^{-4} \text{ Wb} \), and current \( I = 2.5 \text{ A} \).
03

Calculating Inductance

Plug the values into the formula and calculate:\[ L = \frac{500 \cdot 1.4 \times 10^{-4}}{2.5} \]First, calculate the numerator:\[ 500 \times 1.4 \times 10^{-4} = 0.07 \text{ Wb} \]Then, divide by the current:\[ L = \frac{0.07}{2.5} = 0.028 \text{ H} \]
04

Final Answer

The inductance of the coil is \( 0.028 \text{ H} \).

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Key Concepts

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

Magnetic Flux
Magnetic flux, often symbolized by \( \Phi \), is an essential concept in electromagnetism. It represents the amount of magnetic field passing through a certain area. Imagine it like shower water passing through a window screen - the water represents the magnetic field, and the screen is the area. The key factors that influence magnetic flux are:
  • Strength of the magnetic field: A stronger field means more flux.
  • Area the field passes through: A larger area collects more field lines.
  • Orientation of the field: If the field is perpendicular to the area, more field lines pass through, increasing flux.
The unit of magnetic flux is the Weber (Wb). In the given exercise, the coil has a flux of \( 1.4 \times 10^{-4} \text{ Wb} \), indicating the strength and extent of the magnetic field within it. Understanding magnetic flux helps in analyzing how much of the magnetic field interacts with a coil, directly impacting the inductance.
Current
Current, denoted as \( I \), is the flow of electric charge and is measured in amperes (A). It plays a critical role in electromagnetism and circuits by generating a magnetic field around a conductor. Think of water flowing through a pipe; the current is like the water flow, powering devices and creating fields around wires. In the exercise, the current is given as \( 2.5 \text{ A} \), which means a steady flow of 2.5 amperes circulates through the coil, enabling the creation of a magnetic field. This current contributes to establishing magnetic flux when it interacts with the coil's turns. Moreover, it is crucial for calculating inductance, as inductance involves how effectively this current can induce magnetic flux across the coil. An important point to remember is that any variation in current will impact the coil's magnetic properties, affecting both the magnetic flux and the overall inductance.
Number of Turns
The number of turns in a coil, represented by \( N \), refers to how many loops or windings the coil comprises. It significantly influences the coil's magnetic properties. Consider a single loop as having limited ability to interact with the magnetic field. Adding more loops increases this interaction, dramatically affecting phenomena like magnetic flux and inductance.Key points to consider about the number of turns include:
  • More turns mean higher potential for building magnetic flux.
  • Increased turns lead to greater inductance, as it augments the coil's ability to store energy in the form of a magnetic field.
  • The coil's inductance is directly proportional to the number of turns, making \( N \) a pivotal factor in inductance calculations.
In our exercise, the coil has \( 500 \) turns. This large number of turns amplifies the magnetic flux produced for a given current, increasing the inductance. By understanding how the number of turns affects a coil, learners can better grasp its role in building efficient electromagnetic components.

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Most popular questions from this chapter

[II] When a long iron-core solenoid is connected across a 6-V battery, the current rises to \(0.63\) of its maximum value after a time of \(0.75 \mathrm{~s}\). The experiment is then repeated with the iron core removed. Now the time required to reach \(0.63\) of the maximum is \(0.0025 \mathrm{~s}\). Calculate ( \(a\) ) the relative permeability of the iron and \((b) L\) for the aircore solenoid if the maximum current is \(0.5 \mathrm{~A}\).

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 .

A steady current of \(2 \mathrm{~A}\) in a coil of 400 turns causes a flux of \(10^{-4}\) \(\mathrm{Wb}\) to link (pass through) the loops of the coil. Compute ( \(a\) ) the average back emf induced in the coil if the current is stopped in \(0.08 \mathrm{~s},(b)\) the inductance of the coil, and \((c)\) the energy stored in the coil. (a) \(|\varepsilon|=N\left|\frac{\Delta \Phi_{M}}{\Delta t}\right|=400 \frac{\left(10^{-4}-0\right) \mathrm{Wb}}{0.08 \mathrm{~s}}=0.5 \mathrm{~V}\) (b) \(|\boldsymbol{E}|=N\left|\frac{\Delta i}{\Delta t}\right| \quad\) or \(\quad L=\left|\frac{\delta \Delta t}{\Delta i}\right|=\frac{(0.5 \mathrm{~V})(0.08 \mathrm{~s})}{(2-0) \mathrm{A}}=0.02 \mathrm{H}\) (c) Energy \(=\frac{1}{2} L I^{2}=\frac{1}{2}(0.02 \mathrm{H})(2 \mathrm{~A})^{2}=0.04 \mathrm{~J}\)

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 $$ |\varepsilon|=N\left|\frac{\Delta \Phi_{M}}{\Delta t}\right| \text { and } \quad|\varepsilon|=L\left|\frac{\Delta i}{\Delta t}\right| $$ 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 selfinductance, assumed constant for all cases, is then $$ L=N \frac{\Phi_{M}}{I}=N \frac{B A}{I} $$ But for an air-core solenoid, \(B=\mu_{0} \mathrm{nI}=\mu(N / d) I\). Substitution gives (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.

The current in a coil starts out at \(6.00 \mathrm{~A}\) and drops uniformly to zero in a time of \(6.00 \mathrm{~ms}\). Determine the self-inductance, given that there is a measured emf of \(200 \mathrm{~V}\) across the coil while the current is dropping.
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