Problem 18 - University Physics with Modern Physics

Question

An air-filled toroidal solenoid has 300 turns of wire, a mean radius of \(12.0 \mathrm{cm},\) and a cross-sectional area of 4.00 \(\mathrm{cm}^{2} .\) If the current is \(5.00 \mathrm{A},\) calculate: (a) the magnetic field in the solenoid; (b) the self-inductance of the solenoid; (c) the energy stored mean radius of \(12.0 \mathrm{cm},\) and a cross-sectional area of 4.00 \(\mathrm{cm}^{2} .\) If the current is \(5.00 \mathrm{A},\) calculate: (a) the magnetic field in the solenoid; (b) the self-inductance of the solenoid; (c) the energy stored

Step-by-Step Solution

Answer

(a) B ≈ 1.25 mT, (b) L ≈ 1.26 mH, (c) U ≈ 15.75 mJ.

Step 1: Calculate the Magnetic Field

To find the magnetic field inside the toroidal solenoid, we use the formula for the magnetic field inside a toroid: \( B = \frac{\mu_{0} N I}{2 \pi r} \), where \( B \) is the magnetic field, \( \mu_{0} \) is the permeability of free space \((4\pi \times 10^{-7} \, \text{T m/A})\), \( N \) is the number of turns, \( I \) is the current, and \( r \) is the mean radius of the toroid. Substitute the given values: \( N = 300 \), \( I = 5.00 \, \text{A} \), \( r = 12.0 \, \text{cm} = 0.12 \, m \). Substitute into the formula: \[ B = \frac{4\pi \times 10^{-7} \times 300 \times 5.00}{2 \pi \times 0.12} \] Calculate \( B \): \( B \approx 1.25 \times 10^{-3} \, \text{T} \).

Step 2: Calculate the Self-Inductance

The self-inductance \( L \) of a toroidal solenoid is given by the formula \( L = \frac{\mu_{0} N^{2} A}{2\pi r} \), where \( A \) is the cross-sectional area. Substitute the given values: \( N = 300 \), \( A = 4.00 \, \text{cm}^{2} = 4.00 \times 10^{-4} \, \text{m}^{2} \), and \( r = 0.12 \, m \). Substitute into the formula: \[ L = \frac{4\pi \times 10^{-7} \times 300^{2} \times 4.00 \times 10^{-4}}{2\pi \times 0.12} \] Calculate \( L \): \( L \approx 1.26 \times 10^{-3} \, \text{H} \).

Step 3: Calculate the Energy Stored

The energy \( U \) stored in the solenoid is given by \( U = \frac{1}{2} L I^{2} \). Use the value of \( L \) calculated in Step 2 and \( I = 5.00 \, \text{A} \). Substitute into the formula: \[ U = \frac{1}{2} \times 1.26 \times 10^{-3} \times (5.00)^{2} \] Calculate \( U \): \( U \approx 0.01575 \, \text{J} \).

Key Concepts

Magnetic Field

The magnetic field around and within a toroidal solenoid is crucial for understanding its behavior and functionality. A toroidal solenoid is essentially a coil wire wound into a donut-like shape, allowing for magnetic field generation in a closed loop. This type of solenoid confines its magnetic field within its core, which creates a strong magnetic field inside the coil and minimal field outside. This containment is why toroidal solenoids are often preferred in applications where a concentrated magnetic field is necessary.

To calculate the magnetic field within the toroid, we use the formula: \[ B = \frac{\mu_{0} N I}{2 \pi r} \] where \( B \) represents the magnetic field, \( \mu_{0} \) is the permeability of free space (known as \(4\pi \times 10^{-7} \, \text{T m/A} \)), \( N \) is the number of turns of wire, \( I \) is the current, and \( r \) is the mean radius. This formula helps us calculate how strong the magnetic field will be based on these factors. Ensuring accurate measurement and substitution into this formula allows physicists to predict the behavior of the solenoid effectively.

Self-Inductance

Self-inductance is a fundamental property of a toroidal solenoid, which describes its ability to induce a voltage in itself when the current flowing through it changes. This phenomenon is a result of electromagnetic induction, which is a cornerstone of electromagnetism.

The formula to calculate the self-inductance \( L \) of a toroidal solenoid is: \[ L = \frac{\mu_{0} N^{2} A}{2\pi r} \] where \( L \) is the self-inductance, \( N \) is the number of turns of wire, \( A \) is the cross-sectional area, and \( r \) is the mean radius. This equation highlights that a toroidal solenoid's self-inductance is directly proportional to the number of turns squared, the cross-sectional area, and inversely proportional to the mean radius.

Understanding self-inductance is essential when designing circuits with toroidal solenoids, as it influences the circuit's behavior concerning changing currents, particularly in AC applications.

Energy Stored

The energy stored within a toroidal solenoid is vital for understanding its potential for work in various applications. This stored energy arises from the magnetic field created by the current flowing through the solenoid. The amount of energy stored is quantified using the self-inductance of the solenoid and the current passing through it.

The energy \( U \) stored is calculated using the formula: \[ U = \frac{1}{2} L I^{2} \] where \( L \) is the self-inductance, and \( I \) is the current flowing through the solenoid. This relationship underscores that the energy is proportional to the self-inductance and the square of the current.

By storing energy, a toroidal solenoid can be used in various applications, from transformers to inductors, serving as energy reserves that can be released when needed.

Physics Calculations

Performing calculations in physics, like those for a toroidal solenoid, requires understanding both the theoretical concepts and the practical application of formulas. Each calculation, be it for magnetic field, self-inductance, or energy stored, correlates closely with the physical properties and dimensions of the solenoid, such as the number of turns, cross-sectional area, and mean radius.

To ensure accuracy, it is important to:

• Use correct units, converting cm to m where necessary.
• Accurately measure and input values into formulas.
• Understand the role of constants like \( \mu_0 \), the permeability of free space.

Having a structured approach to solve physics problems allows students to understand not only the calculation steps but also the underlying physics principles, enhancing their comprehension and application in real-world scenarios.

Magnetism Concepts

Magnetism in toroidal solenoids and other electromagnetic systems showcases the power and range of applications resulting from magnetic fields. Understanding these concepts is vital for those interested in physics and engineering.

Key magnetism concepts include:

• Magnetic fields, which result from moving electric charges and are fundamental to inductors and transformers.
• Electromagnetic induction, the principle by which current or voltage can be induced in a conductor by a changing magnetic field.
• Permeability, the measure of a material's ability to support the formation of a magnetic field within itself.

Each of these concepts becomes essential when designing and using devices that rely on toroidal solenoids, leading to innovations in technology and industry. Understanding these aspects ensures a solid foundation in magnetism and its applications.