/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 You have a fixed length of wire ... [FREE SOLUTION] | 91Ó°ÊÓ

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You have a fixed length of wire to wind into an inductor. Will you get more inductance if you wind a short coil with large diameter, or a long coil with small diameter?

Short Answer

Expert verified
The inductance will be higher with a short coil with a large diameter.

Step by step solution

01

Inductance Formula Analysis

The inductance is given by \(L = \mu_0 n^2 A l^{-1}\), with \(n\) being the number of turns per unit length, \(A\) is the cross-sectional area of each turn, and \(l\) is the length of the coil. When the length of the wire is fixed, the number of turns of the coil, \(n\), remains constant. Since \(A\) is proportional to the square of the coil's radius (or diameter), and \(l\) is inversely proportional to the coil's diameter, a larger diameter coil will decrease \(l\) and increase \(A\), which should increase \(L\).
02

Short Coil with Large Diameter

In this case, the cross-sectional area \(A\) is large (since \(A\) is proportional to the square of the coil's diameter), and the length \(l\) of the coil is short. Thus, the denominator of the formula \(l^{-1}\) is small, and \(A\) in the numerator is large. The overall inductance \(L\) will therefore be higher.
03

Long Coil with Small Diameter

In this case, the cross-sectional area \(A\) is small, and the length \(l\) of the coil is long. The denominator of the formula \(l^{-1}\) is large, and \(A\) in the numerator is small. The overall inductance \(L\) will therefore be lower.
04

Comparison

By comparing the two cases, we can conclude that a short coil with a large diameter will have a higher inductance than a long coil with a small diameter, assuming the number of turns \(n\) remains constant.

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Key Concepts

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

Inductance Formula Analysis
Inductance in electrical circuits is a fundamental property that measures the ability of a coil to store energy in the form of a magnetic field when an electric current passes through it. The main formula to analyze inductance is given by \( L = \mu_0 n^2 A l^{-1} \), where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, \( A \) is the cross-sectional area of each turn, and \( l \) is the length of the coil. Understanding this formula is crucial as each variable greatly impacts the coil's inductance. For example, increasing the number of turns, \( n \) or the cross-sectional area, \( A \) can result in a higher inductance, whereas increasing the coil's length, \( l \) will decrease it. Keeping the wire length fixed means \( n \) stays constant, but manipulating diameter and length of the coil can change the inductance value.

When faced with problems involving inductance, it's important to have a clear grasp of each component in the formula. The inductance formula analysis can guide users in predicting how changes to the coil will affect the overall magnetic properties.
Effects of Coil Geometry on Inductance
The geometry of a coil, including its diameter and the length, significantly affects its inductance. From the formula \( L = \mu_0 n^2 A l^{-1} \), it is noticeable that the cross-sectional area \( A \) plays a vital role. Since \( A \) is proportional to the square of the coil's diameter, a wider coil will have a greater \( A \) and consequently a higher inductance. On the other hand, the length \( l \) of the coil influences inductance inversely. A shorter coil lessens the path the magnetic lines of force have to traverse, which concentrates the magnetic field, boosting the inductance as well.

Understanding these relationships is central to designing coils for specific inductive properties. For instance, if a high inductance is required, manufacturers might opt for a coil with a large diameter and a shorter length, provided other conditions such as the total number of turns remain constant.
Electromagnetic Principles in Physics
Inductance is deeply rooted in the electromagnetic principles governed by Maxwell's equations. These principles explain how electric currents and magnetic fields are interrelated. A changing electric current in a coil will produce a changing magnetic field, and conversely, a changing magnetic field induces an electromotive force (EMF) in a conductor. This phenomenon is known as electromagnetic induction, and it's the cornerstone of how inductors work in circuits.

The principle of electromagnetic induction is key to understanding and utilizing inductance in technology, from simple transformers to complex electrical components in high-tech devices. Recognizing this interplay between electric currents and magnetic fields equips students and professionals with the foundational knowledge to innovate and solve intricate physics problems.
Physics Problem Solving
Problem solving in physics, especially when it comes to electromagnetic phenomena like inductance, requires a systematic approach. It begins by identifying known and unknown variables and then applying relevant physical laws or formulas. In our case, we analyze the inductance of a coil, which involves applying the inductance formula and understanding how variables such as coil geometry affect the outcome.

Effective problem solving often involves comparing scenarios. As done in our exercise, comparing a short coil with a large diameter to a long coil with a small diameter under the constants of wire length and number of turns illustrates a practical application of theoretical knowledge. This not only deepens comprehension but also equips learners with the analytical skills necessary to tackle real-world physics problems.

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Most popular questions from this chapter

During lab, you're given a circular wire loop of resistance \(R\) and radius \(a\) with its plane perpendicular to a uniform magnetic field. You're supposed to increase the field strength from \(B_{1}\) to \(B_{2}\) and measure the total charge that moves around the loop. Your lab partner claims that the details of how you vary the field will make a difference in the total charge; your hunch is that it won't. By integrating the loop current over time, determine who's right.

The current in an inductor is changing at \(100 \mathrm{A} / \mathrm{s}\) and the inductor emf is 40 \(\mathrm{V}\). What's the self-inductance?

A magnetic field is given by \(\vec{B}=B_{0}\left(x / x_{0}\right)^{2} \hat{k},\) where \(B_{0}\) and \(x_{0}\) are constants. Find an expression for the magnetic flux through a square of side \(2 x_{0}\) that lies in the \(x\) -y plane with one corner at the origin and sides coinciding with the positive \(x\) - and \(y\) -axes.

A flip coil is used to measure magnetic fields. It's a small coil placed with its plane perpendicular to a magnetic field, and then flipped through \(180^{\circ} .\) The coil is connected to an instrument that measures the total charge \(Q\) that flows during this process. If the coil has \(N\) turns, area \(A,\) and resistance \(R,\) show that the field strength is \(B=Q R / 2 N A.\)

A circular wire loop \(40 \mathrm{cm}\) in diameter has resistance \(100 \Omega\) and lies in a horizontal plane. A uniform magnetic field points vertically downward, and in 25 ms it increases linearly from \(5.0 \mathrm{mT}\) to \(55 \mathrm{mT}\). Find the magnetic flux through the loop at (a) the beginning and (b) the end of the 25 -ms period. (c) What's the loop current during this time? (d) Which way does this current flow?

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