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Chapter 23: Problem 58

CE The number of turns per meter in a solenoid of fixed length is doubled. At the same time, the current in the solenoid is halved. Does the energy stored in the inductor increase, decrease, or stay the same? Explain.

Short Answer

Expert verified
The energy stored in the solenoid stays the same.

Step by step solution

01

Understand the Energy Stored in Inductor Formula

The energy stored in a solenoid is given by the formula: \( U = \frac{1}{2} L I^2 \), where \( U \) is the energy stored, \( L \) is the inductance, and \( I \) is the current. This formula suggests that any change in \( L \) or \( I \) will affect the energy stored in the solenoid.
02

Understand the Inductance of a Solenoid

The inductance \( L \) of a solenoid is given by the equation \( L = \mu_0 n^2 A l \), where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, \( A \) is the cross-sectional area, and \( l \) is the length of the solenoid. Increasing the number of turns per meter \( n \) by a factor of two will increase \( n^2 \) by a factor of four.
03

Calculate the New Inductance

Doubling \( n \) increases \( n^2 \) by a factor of four. Therefore, the inductance \( L \) becomes \( 4L \) since other factors such as \( A \) and \( l \) are unchanged by the problem statement.
04

Calculate the New Energy Stored

With the inductance \( L \) quadrupling to \( 4L \) and the current \( I \) halved to \( \frac{I}{2} \), we find the new energy stored using the formula: \( U' = \frac{1}{2} (4L) \left(\frac{I}{2}\right)^2 \). This simplifies to \( U' = \frac{1}{2} (4L) \frac{I^2}{4} = \frac{1}{2}L I^2 = U \).
05

Determine the Final Outcome

Since \( U' = U \), the energy stored in the solenoid remains the same.

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Key Concepts

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

Inductance Formula
Inductance is a key concept in understanding how energy is stored in a solenoid. The formula for the energy stored in a solenoid is represented as \( U = \frac{1}{2} L I^2 \). Here, \( U \) stands for the energy stored, \( L \) is the inductance, and \( I \) is the current flowing through the solenoid.

This formula indicates a direct relationship between the inductance \( L \) and the square of the current \( I^2 \). Any change in either \( L \) or \( I \) will result in a change in the energy stored \( U \).

- It's important to note that the inductance \( L \) itself depends on certain physical properties of the solenoid.- The formula lays down the foundation: if either the inductance or the current increases, the energy stored also tends to increase, and vice versa.
Current in Solenoid
The current \( I \) in a solenoid plays a crucial role in determining the energy stored within. The energy stored is proportional to the square of the current, as seen in the formula: \( U = \frac{1}{2} L I^2 \). This means that if the current is halved, the effect on the energy stored is significant because \( I^2 \) decreases by a factor of 4.

- In the provided exercise, the current in the solenoid is halved. Consequently, the energy contribution from \( I^2 \) changes drastically, reducing the stored energy unless compensated by an increase in inductance.- Understanding this relationship is essential, as the energy stored is not only affected by how much current flows but also how consistently it flows through the solenoid's windings.
Solenoid Inductance
The inductance \( L \) of a solenoid is determined by the formula \( L = \mu_0 n^2 A l \). Here:
  • \( \mu_0 \) is the permeability of free space.
  • \( n \) is the number of turns per unit length, which is crucial since it impacts the density of the magnetic field lines.
  • \( A \) is the cross-sectional area, and \( l \) is the length of the solenoid.
The given exercise illustrates that doubling the turns per meter \( n \) increases \( n^2 \) by a factor of four, leading to the inductance \( L \) becoming four times larger, provided other factors like \( A \) and \( l \) remain constant.

- This increase in inductance (resulting from the change in \( n^2 \)) compensates for the decrease in energy caused by reducing the current, resulting in the energy stored remaining unchanged.- It highlights how changes in the solenoid's physical structure directly impact its magnetic properties, thereby affecting energy storage.

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Most popular questions from this chapter

A bar magnet with its north pole pointing downward is falling toward the center of a horizontal conducting ring. As viewed from above, is the direction of the induced current in the ring clockwise or counterclockwise? Explain.

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MRI Solenoid The magnetic field produced by an MRI solenoid \(2.5 \mathrm{m}\) long and \(1.2 \mathrm{m}\) in diameter is \(1.7 \mathrm{T}\). Find the magnitude of the magnetic flux through the core of this solenoid.

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IP A single-turn rectangular loop of width \(W\) and length \(L\) moves parallel to its length with a speed v. The loop moves from a region with a magnetic field \(\overrightarrow{\mathrm{B}}\) perpendicular to the plane of the loop to a region where the magnetic field is zero, as shown in Figure \(23-47 .\) Find the rate of change in the magnetic flux through the loop (a) before it enters the region of zero field, (b) just after it enters the region of zero field, and (c) once it is fully within the region of zero field. (d) For each of the cases considered in parts (a), (b), and (c), state whether the induced current in the loop is clockwise, counterclockwise, or zero. Explain in each case.
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