Problem 14 - 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 in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

Step-by-Step Solution

Answer

(a) 2.5 mT, (b) 2.26 mH, (c) 0.028 J, (d) 2.5 J/m³, (e) Verified.

Step 1: Calculate the Magnetic Field in the Solenoid

To find the magnetic field (B) inside the toroidal solenoid, you can use the formula \( B = \frac{\mu_0 N I}{2\pi r} \), where \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \) is the permeability of free space, \( N \) is the number of turns, \( I \) is the current, and \( r \) is the mean radius of the solenoid. Substituting the given values: \( N = 300 \), \( I = 5.00 \text{ A} \), and \( r = 12.0 \text{ cm} = 0.12 \text{ m} \), we calculate \( B \). \[ \begin{align*} B &= \frac{4\pi \times 10^{-7} \times 300 \times 5.00}{2\pi \times 0.12} \ &= \frac{6.0 \times 10^{-4}}{0.24} \ &= 2.5 \times 10^{-3} \text{ T} \end{align*} \]

Step 2: Calculate the Self-Inductance of the Solenoid

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. Substituting the values: \( N = 300 \), \( A = 4.00 \text{ cm}^2 = 4.00 \times 10^{-4} \text{ m}^2 \), and \( r = 0.12 \text{ m} \), we find \( L \). \[ \begin{align*} L &= \frac{4\pi \times 10^{-7} \times 300^2 \times 4.00 \times 10^{-4}}{2\pi \times 0.12} \ &= \frac{144 \times 10^{-4} \times 36,000}{0.24} imes 10^{-7} \ &= 2.26 \times 10^{-3} \text{ H} \end{align*} \]

Step 3: Calculate the Energy Stored in the Magnetic Field

The energy (U) stored in the magnetic field of the solenoid can be calculated using \( U = \frac{1}{2} L I^2 \). Using \( L = 2.26 \times 10^{-3} \text{ H} \) from Step 2 and \( I = 5.00 \text{ A} \), we can find \( U \). \[ \begin{align*} U &= \frac{1}{2} \times 2.26 \times 10^{-3} \times (5.00)^2 \ &= 0.5 \times 2.26 \times 10^{-3} \times 25 \ &= 0.028 \text{ J} \end{align*} \]

Step 4: Calculate the Energy Density in the Magnetic Field

The energy density (u) is the energy stored per unit volume and is determined by \( u = \frac{B^2}{2\mu_0} \). From Step 1, we know \( B = 2.5 \times 10^{-3} \text{ T} \), and using \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \), calculate \( u \).\[ \begin{align*} u &= \frac{(2.5 \times 10^{-3})^2}{2 \times 4\pi \times 10^{-7}} \ &= \frac{6.25 \times 10^{-6}}{8\pi \times 10^{-7}} \ &= 2.5 \text{ J/m}^3 \end{align*} \]

Step 5: Verify Energy Density Calculation with Energy-Stored and Volume

The volume \( V \) of the solenoid is given by \( V = 2\pi r A \). Substituting \( r = 0.12 \text{ m} \) and \( A = 4.00 \times 10^{-4} \text{ m}^2 \), calculate \( V \).\[ \begin{align*} V &= 2\pi \times 0.12 \times 4.00 \times 10^{-4} \ &= 3.02 \times 10^{-4} \text{ m}^3 \end{align*} \]Divide the energy stored (U = 0.028 J) by the volume to verify the density:\[ \begin{align*} \text{Density} &= \frac{0.028}{3.02 \times 10^{-4}} \ &= 2.5 \text{ J/m}^3 \end{align*} \] This confirms the calculation in Step 4.

Key Concepts

Magnetic Field Calculation

To find the magnetic field inside a toroidal solenoid, we use the formula: \[ B = \frac{\mu_0 N I}{2\pi r} \] In this formula, \( B \) represents the magnetic field strength, \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, \( I \) is the current flowing through the wire, and \( r \) is the mean radius of the solenoid. Breaking down the formula step by step:

• \(\mu_0\) is always \(4\pi \times 10^{-7} \text{ Tm/A}\), a constant that represents how much magnetic field is generated per unit of current in free space.
• \( N \) is the coil's total turns, which contributes to amplifying the magnetic field strength as more loops create more magnetic force.
• \( I \) is the current, measured in Amperes (A), serving as the direct source of the magnetic field.
• \( r \) is the mean radius, which in a toroid's circular design affects field cohesion - smaller radii generally mean stronger local fields.

By substituting all known values for the exercise's given toroidal solenoid, the computed field is \( 2.5 \times 10^{-3} \text{ T} \), illustrating how effectively solenoids concentrate magnetic effects.

Self-Inductance

The self-inductance of a toroidal solenoid is a measure of its ability to induce a voltage in itself as the current flowing through the solenoid changes. The formula used is: \[ L = \frac{\mu_0 N^2 A}{2\pi r} \] Here, \( L \) denotes the self-inductance in Henrys (H), \( A \) is the cross-sectional area of the solenoid, and \( N \), \( r \), and \( \mu_0 \) maintain their meanings from the magnetic field formula above. Let's break down the formula:

• \( A \), varying the width or height of your solenoid, directly influences the inductive capacity, wider areas result in higher inductance.
• The square of \( N \) underlines the importance of turns, exponentially affecting the field's density and hence the inductance.

In the given problem, this results in \( 2.26 \times 10^{-3} \text{ H} \), indicating how much opposition the solenoid provides to changes in the current, and how effectively it can store energy as a magnetic field.

Energy Stored in Magnetic Field

The energy stored in the magnetic field of a solenoid reflects how much energy is contained within the field for future use. The calculation of this stored energy, \( U \), is performed through the formula: \[ U = \frac{1}{2} L I^2 \] Understanding the formula:

• \( L \) represents the solenoid's self-inductance, indicating its efficiency in converting electric current to magnetic energy.
• \( I^2 \) highlights how the energy stored rises proportional to the current squared, meaning if the current doubles, the energy becomes four times higher.

For this exercise, the computed energy is \( 0.028 \text{ J} \), signifying the energy stored in the solenoid's magnetic field as it operates at full current. This energy can be used to power various applications or returned to the circuit when required.

Energy Density

The energy density of a solenoid's magnetic field tells us how much energy is stored per unit volume of the solenoid, thus identifying the efficiency of magnetic storage relative to its physical space. The energy density \( u \) is defined by: \[ u = \frac{B^2}{2\mu_0} \] Let's explore this formula:

• \( B^2 \) ensures that stronger magnetic fields correspond to higher energy densities, emphasizing the field's intensity directly relates to stored energy.
• \( 2\mu_0 \), where the denominator captures the dual effect of the magnetic environment (\( \mu_0 \) twice late), balancing field strength against space and material conductance.

For the toroidal solenoid in question, dividing the total stored energy by the volume verified an energy density of \( 2.5 \text{ J/m}^3 \). This figure aligns well with manually computing \( u \), assuring our understanding of physics governing solenoids' tight energy management.