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Chapter 15: Problem 27

What happens to the reluctance of a magnetic path if its length is doubled? If the cross-sectional area is doubled? If the relative permeability is doubled?

Short Answer

Expert verified
Doubling the length doubles reluctance. Doubling the area or relative permeability halves it.

Step by step solution

01

Understanding Reluctance

Reluctance (denoted as \( ext{R} \)) is the property of a magnetic circuit which opposes the passage of magnetic flux. It is analogous to resistance in an electrical circuit. The formula for reluctance is given by: \[ ext{R} = rac{l}{ ext{A} imes ext{碌} imes ext{碌}_0} \] where \( l \) is the length of the magnetic path, \( ext{A} \) is the cross-sectional area, \( ext{碌} \) is the relative permeability of the material, and \( ext{碌}_0 \) is the permeability of free space.
02

Analyzing the Effect of Length

If the length of the magnetic path is doubled, \( l' = 2l \). The reluctance becomes: \[ ext{R}' = rac{2l}{ ext{A} imes ext{碌} imes ext{碌}_0} = 2 imes rac{l}{ ext{A} imes ext{碌} imes ext{碌}_0} = 2 ext{R} \] Thus, doubling the length of the path doubles the reluctance.
03

Analyzing the Effect of Cross-Sectional Area

If the cross-sectional area of the magnetic path is doubled, \( ext{A}' = 2 ext{A} \). The reluctance becomes: \[ ext{R}' = rac{l}{(2 ext{A}) imes ext{碌} imes ext{碌}_0} = rac{1}{2} imes rac{l}{ ext{A} imes ext{碌} imes ext{碌}_0} = rac{ ext{R}}{2} \] Thus, doubling the area halves the reluctance.
04

Analyzing the Effect of Relative Permeability

If the relative permeability is doubled, \( ext{碌}' = 2 ext{碌} \). The reluctance becomes: \[ ext{R}' = rac{l}{ ext{A} imes (2 ext{碌}) imes ext{碌}_0} = rac{1}{2} imes rac{l}{ ext{A} imes ext{碌} imes ext{碌}_0} = rac{ ext{R}}{2} \] Thus, doubling the relative permeability halves the reluctance.

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Key Concepts

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

Magnetic Path Length
The magnetic path length, denoted in formulas as \( l \), is a critical factor in determining the magnetic circuit's reluctance. Think of it as the distance the magnetic field lines travel within the magnetic material. This concept is quite similar to the length of wire in an electrical circuit, which determines the resistance. When the length of the magnetic path is increased, the magnetic flux encounters more resistance, much like how extending a wire increases its electrical resistance.
  • When the length of the magnetic path is doubled, the reluctance, which opposes magnetic flux, also doubles.
  • This is because reluctance is directly proportional to the path length.
Thus, understanding and controlling the length is crucial in designing magnetic circuits efficiently.
Cross-Sectional Area in Magnetism
The cross-sectional area of a magnetic path, represented as \( A \) in formulas, plays a pivotal role in the behavior of a magnetic circuit. It is analogous to the width of a pipe in water flow systems, influencing how much magnetic field can pass through.An increase in the cross-sectional area implies that more space is available for magnetic field lines to spread, which effectively reduces the reluctance.
  • Doubling the cross-sectional area results in halving the reluctance.
  • This is because reluctance is inversely proportional to the cross-sectional area.
As a result, to minimize reluctance, a greater cross-sectional area is beneficial.
Relative Permeability
Relative permeability, noted as \( \mu \), indicates how effectively a material can carry magnetic flux compared to a vacuum. In essence, it measures the material's magnetic responsiveness and is a key factor in calculating reluctance.
  • Materials with higher relative permeability offer less opposition to magnetic flux, acting almost like lubricated paths for magnetic lines.
  • If the relative permeability of a material is doubled, the reluctance is halved.
Thus, selecting materials with high relative permeability is vital for efficient magnetic circuit design, as it reduces the opposition to magnetic flux passage.

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Most popular questions from this chapter

Two coils wound on a common core have \(L_{1}=1 \mathrm{H}, L_{2}=2 \mathrm{H}\), and \(M=0.5 \mathrm{H}\). The currents are \(i_{1}=\cos (377 t) \mathrm{A}\) and \(i_{2}=\) \(0.5 \cos (377 t)\) A. Both of the currents enter dotted terminals. Find expressions for the voltages across the coils.

A \(60-\mathrm{Hz} 40-\mathrm{kVA} 8000 / 240-\mathrm{V}-\mathrm{rms}\) transformer has the following equivalent-circuit parameters: $$ \begin{array}{lcl} \hline \text { Primary resistance } & R_{1} & 20 \Omega \\ \text { Secondary resistance } & R_{2} & 0.04 \Omega \\ \text { Primary leakage reactance } & X_{1}=\omega L_{1} & 100 \Omega \\ \text { Secondary leakage reactance } & X_{2}=\omega L_{2} & 0.2 \Omega \\ \text { Magnetizing reactance } & X_{m}=\omega L_{m} & 40 \mathrm{k} \Omega \\\ \text { Core-loss resistance } & R_{c} & 200 \mathrm{k} \Omega \\ \hline \end{array} $$ Find the percentage regulation and power efficiency for the transformer for a rated load having a lagging power factor of \(0.9\).

What are the physical units of reluctance in terms of kilograms, coulombs, meters, and seconds?

At a frequency of \(60 \mathrm{~Hz}\), the core loss of a certain coil with an iron core is \(3.6 \mathrm{~W}\), and at a frequency of \(120 \mathrm{~Hz}\), it is \(11.2 \mathrm{~W}\). The peak flux density is the same for both cases. Determine the power loss due to hysteresis and that due to eddy currents for \(60-\mathrm{Hz}\) operation.

A symmetrical toroidal coil is wound on a plastic core \(\left(\mu_{r} \cong 1\right)\) and is found to have an inductance of \(1 \mathrm{mH}\). What inductance will result if the core material is changed to a ferrite having \(\mu_{r}=200\) ? Assume that the entire magnetic path is composed of ferrite.
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