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Chapter 28: Problem 3

The number of turns in a solenoid is doubled, and its length is halved. How does its magnetic field change? a) it doubles b) it is halved c) it quadruples d) it remains unchanged

Short Answer

Expert verified
Answer: The magnetic field quadruples.

Step by step solution

01

Identify the formula for the magnetic field of a solenoid

To determine how the magnetic field changes, we first need to know the formula for the magnetic field of a solenoid. The formula is given by: B = 渭鈧� * n * I where B is the magnetic field, 渭鈧� is the magnetic constant (also known as the permeability of free space), n is the number of turns per unit length, and I is the current flowing through the solenoid. We'll assume that the current I stays the same because the exercise doesn't mention it changing.
02

Analyze the changes in the number of turns and length of the solenoid

The exercise states that the number of turns in the solenoid is doubled, so we will multiply the number of turns (n) by 2: n_new = 2 * n It also says that the length of the solenoid is halved, so we will divide the length (L) by 2: L_new = L / 2
03

Calculate the new number of turns per unit length

Since the number of turns per unit length is given by n = N / L, we can express the new number of turns per unit length (n_new) with the new values of N and L: n_new = N_new / L_new n_new = (2 * N) / (L / 2) Simplifying the above expression, we get: n_new = 4 * (N / L) Comparing this to the original number of turns per unit length: n_new = 4 * n
04

Calculate the new magnetic field and compare to the original

Now that we have the new number of turns per unit length, we can calculate the new magnetic field (B_new) using the original formula: B_new = 渭鈧� * n_new * I Replacing n_new with the expression we found in Step 3: B_new = 渭鈧� * (4 * n) * I Comparing this to the original magnetic field: B_new = 4 * (渭鈧� * n * I) We find that the new magnetic field is four times the original magnetic field: B_new = 4 * B Therefore, the correct answer is (c) the magnetic field quadruples.

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Key Concepts

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

Number of Turns in Solenoid
A solenoid is essentially a coil of wire, and the number of turns refers to how many loops or coils the wire makes. This is critical because each loop generates its own magnetic field. When the wire is coiled, these individual magnetic fields combine to create a stronger total magnetic field inside the solenoid.
  • More turns mean more loops, hence a stronger magnetic field.
  • Doubling the number of turns in a solenoid doubles the contribution each loop adds to the total magnetic field.
Thus, increasing the number of turns amplifies the overall magnetic field within a solenoid, especially when other factors, such as current, are kept constant.
Length of Solenoid
The length of a solenoid is another key factor that affects its magnetic field. The length determines the distribution of the turns within the solenoid, which in turn affects how concentrated the magnetic field is.
  • By halving the length, you effectively concentrate the same number of turns in a smaller space.
  • This concentration increases the number of turns per unit length.
The increase in turns per unit length when the solenoid's length is reduced results in a stronger magnetic field. So, even with the same overall number of turns, a shorter solenoid creates a more intense magnetic environment inside.
Magnetic Constant
The magnetic constant, often denoted as \( \mu_0 \), is an integral part of understanding solenoids. It represents the permeability of free space, which is a measure of how easily a magnetic field can form in a vacuum.
  • The magnetic constant provides a baseline to calculate how a material will interact with magnetic fields.
  • It remains unchanged regardless of changes to the solenoid's physical characteristics, like length or the number of turns.
When calculating the magnetic field inside a solenoid, the magnetic constant scales the effect of the current and turn density. It ensures that the calculated magnetic field is accurate and reflects the true strength of the solenoid's magnetic field in practical applications.

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Most popular questions from this chapter

What is a good rule of thumb for designing a simple magnetic coil? Specifically, given a circular coil of radius \(\sim 1 \mathrm{~cm},\) what is the approximate magnitude of the magnetic field, in gauss per amp per turn? (Note: \(1 \mathrm{G}=0.0001 \mathrm{~T}\).) a) \(0.0001 \mathrm{G} /(\mathrm{A}\) -turn \()\) b) \(0.01 \mathrm{G} /(\) A-turn \()\) c) \(1 \mathrm{G} /(\mathrm{A}\) -turn \()\) d) \(100 \mathrm{G} /(\mathrm{A}\) -turn \()\)

If you want to construct an electromagnet by running a current of 3.0 A through a solenoid with 500 windings and length \(3.5 \mathrm{~cm}\) and you want the magnetic field inside the solenoid to have magnitude \(B=2.96 \mathrm{~T}\), you can insert a ferrite core into the solenoid. What value of the relative magnetic permeability should this ferrite core have in order to make this work?

The magnetic force cannot do work on a charged particle since the force is always perpendicular to the velocity. How then can magnets pick up nails? Consider two parallel current-carrying wires. The magnetic fields cause attractive forces between the wires, so it appears that the magnetic field due to one wire is doing work on the other wire. How is this explained? a) The magnetic force can do no work on isolated charges; this says nothing about the work it can do on charges confined in a conductor. b) Since only an electric field can do work on charges, it is actually the electric fields doing the work here. c) This apparent work is due to another type of force.

A current element produces a magnetic field in the region surrounding it. At any point in space, the magnetic field produced by this current element points in a direction that is a) radial from the current element to the point in space. b) parallel to the current element. c) perpendicular to the current element and to the radial direction.

A wire of radius \(R\) carries current \(i\). The current density is given by \(J=J_{0}(1-r / R),\) where \(r\) is measured from the center of the wire and \(J_{0}\) is a constant. Use Ampere's Law to find the magnetic field inside the wire at a distance \(r
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