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Chapter 28: Problem 45

A wooden ring whose mean diameter is 14.0 \(\mathrm{cm}\) is wound with a closely spaced toroidal winding of 600 turns. Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.650 \(\mathrm{A}\) .

Short Answer

Expert verified
The magnetic field at the center of the cross section of the windings is approximately 1.114 mT.

Step by step solution

01

Identify the Known Values

We are given the mean diameter of the toroid, the number of turns in the winding, and the current flowing through the winding. These values are:- Mean diameter, \( D = 14.0 \; \mathrm{cm} = 0.14 \; \mathrm{m} \) (since 1 cm = 0.01 m)- Number of turns, \( N = 600 \)- Current, \( I = 0.650 \; \mathrm{A} \)
02

Calculate the Radius

The mean radius \( r \) of the toroid is half of the mean diameter. Calculate \( r \):\[ r = \frac{D}{2} = \frac{0.14 \; \mathrm{m}}{2} = 0.07 \; \mathrm{m} \]
03

Use the Toroid Magnetic Field Formula

The magnetic field \( B \) inside a toroid can be calculated using the formula:\[ B = \frac{\mu_0 N I}{2 \pi r} \]where \( \mu_0 = 4\pi \times 10^{-7} \; \mathrm{T}\cdot\mathrm{m/A} \) (permeability of free space).
04

Substitute the Known Values into the Formula

Substitute the known values into the magnetic field equation:\[ B = \frac{(4\pi \times 10^{-7} \; \mathrm{T}\cdot\mathrm{m/A}) \cdot 600 \cdot 0.650 \; \mathrm{A}}{2 \pi \cdot 0.07 \; \mathrm{m}} \]
05

Simplify and Calculate

Cancel out \( \pi \) in the numerator and denominator:\[ B = \frac{(4 \times 10^{-7} \; \mathrm{T}\cdot\mathrm{m/A}) \cdot 600 \cdot 0.650}{2 \cdot 0.07} \]Calculate the result:- Numerator: \( (4 \times 10^{-7}) \cdot 600 \cdot 0.650 = 1.56 \times 10^{-4} \)- Denominator: \( 2 \cdot 0.07 = 0.14 \ \)- \[ B = \frac{1.56 \times 10^{-4}}{0.14} = 1.114 \times 10^{-3} \]Therefore, \( B \approx 1.114 \; \mathrm{mT} \).

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Key Concepts

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

Toroid Magnetic Field Formula
A toroid is a doughnut-shaped object, allowing a magnetic field to be formed along the inside of the ring. To determine the magnetic field strength within a toroid, we use a specific formula. This formula is crucial in physics to understand the behavior of magnetic fields in toroidal structures.

The formula for calculating the magnetic field inside a toroid is given by:
  • \( B = \frac{\mu_0 N I}{2 \pi r} \)
In this formula:
  • \( B \) represents the magnetic field inside the toroid.
  • \( \mu_0 \) is the permeability of free space, a constant with a value of \( 4\pi \times 10^{-7} \, \mathrm{T}\cdot\mathrm{m/A} \).
  • \( N \) is the number of turns in the winding.
  • \( I \) represents the current flowing through the windings.
  • \( r \) is the mean radius of the toroid.
Understanding each component helps in applying this formula effectively. This straightforward expression highlights the direct relationship between the magnetic field and its influencing factors like current and number of turns.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a fundamental constant used in the calculation of magnetic fields. It is a measure of how much resistance is encountered when forming a magnetic field in a vacuum.

In the context of a toroid:
  • \( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T}\cdot\mathrm{m/A} \)
This constant is essential because it allows us to quantify the magnetic response of an idealized vacuum and provides a baseline for calculating magnetic fields in various mediums.
The concept of permeability is not just limited to free space. Different materials have different permeabilities which affect how magnetic fields behave within them. However, for many practical calculations, especially in education, the permeability of free space serves as an ideal standard.
Current in Windings
The current that passes through the windings of a toroid significantly impacts the magnetic field produced within it. Essentially, the magnetic field strength is directly proportional to the amount of current flowing through the coil.

In our example:
  • The current, \( I \), was given as 0.650 A.
This means as more electrical current is passed through the coil, the stronger the magnetic field inside the toroid will be.
The relationship can be visualized by the toroid magnetic field formula, where the magnetic field \( B \) increases along with increased current \( I \). It's a simple yet powerful principle that demonstrates the dynamic nature of electromagnetism and is a core concept in electrical engineering and physics.
Mean Diameter and Radius Calculation
When calculating the magnetic field in a toroid, understanding how to determine the mean diameter and radius is crucial. The mean diameter is the average measurement across the open space in the middle of the toroid, along its entirety.

Here's how you can determine these critical dimensions:
  • Mean Diameter: Given as 14 cm or 0.14 meters (conversion from centimeters to meters).
  • Mean Radius: The radius, \( r \), is half of the mean diameter.
Mathematically, the radius can be calculated using:
  • \[ r = \frac{D}{2} = \frac{0.14 \, \mathrm{m}}{2} = 0.07 \, \mathrm{m} \]
Having the right radius is vital because it's directly used in the toroide magnetic field formula. A precise understanding of these terms ensures accurate calculations and makes interpreting the results much simpler for students.

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Most popular questions from this chapter

Two identical circular, wire loops 40.0 \(\mathrm{cm}\) in diameter each carry a current of 1.50 \(\mathrm{A}\) in the same direction. These loops are parallel to each other and are 25.0 \(\mathrm{cm}\) apart. Line \(a b\) is normal to the plane of the loops and passes through their centers. A proton is fired at 2400 \(\mathrm{km} / \mathrm{sperpendicular}\) to line \(a b\) from a point midway between the centers of the loops. Find the magnitude and direction of the magnetic force these loops exert on the proton just after it is fired.

Two long, parallel wires are separated by a distance of \(2.50 \mathrm{cm} .\) The force per unit length that each wire exerts on the other is 4.00 \(\times 10^{-5} \mathrm{N} / \mathrm{m}\) , and the wires repel each other. The current in one wire is 0.600 \(\mathrm{A}\) . (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions?

A long, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}}\) . The current density, although symmetrical about the cylinder axis, is not constant and varies according to the relationship $$ \begin{array}{rlrl}{\overrightarrow{\boldsymbol{J}}} & {=\left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{\boldsymbol{k}}} & {} & {\text { for } r \leq a} \\ {} & {=\mathbf{0}} & {} & {\text { for } \boldsymbol{r} \geq a}\end{array} $$ where the radius of the cylinder is \(a=5.00 \mathrm{cm}, r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to \(600 \mathrm{A} / \mathrm{m},\) and \(\delta\) is a constant equal to \(2.50 \mathrm{cm} .\) (a) Let \(I_{0}\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_{0}\) in terms of \(b, \delta,\) and \(a .\) Evaluate your expression to obtain a numerical value for \(I_{0}\) . \((b)\) Using Ampere's law, derive an expression for the magnetic field \(\vec{B}\) in the region \(r \geq a\) . Express your answer in terms of \(I_{0}\) rather than \(b\) . (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. Express your answer in terms of \(I_{0}\) rather than \(b\) . (d) Using Ampere's law, derive an expression for the magnetic field \(\vec{B}\) in the region \(r \leq a\) (e) Evaluate the magnitude of the magnetic field at \(r=\delta, r=a,\) and \(r=2 a\) .

Along, straight wire lies along the \(z\) -axis and carries a \(4.00-\mathrm{A}\) current in the \(+z\) -direction. Find the magnetic field (magnitude and direction) produced at the following points by a \(0.500-\mathrm{mm}\) segment of the wire centered at the origin: (a) \(x=2.00 \mathrm{m}, y=0\) , \(z=0 ;(b) x=0, y=2.00 \mathrm{m}, z=0 ;(\mathrm{c}) x=2.00 \mathrm{m}, y=2.00 \mathrm{m}\) \(z=0 ;(\mathrm{d}) x=0, y=0, z=2.00 \mathrm{m}\)

Two long. parallel transmission lines, 40.0 \(\mathrm{cm}\) apart, carry \(25.0-\mathrm{A}\) and 75.0 - A currents. Find all locations where the net magnetic field of the two wires is zero if these currents are in (a) the same direction and (b) the opposite direction.
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