Problem 15 - University Physics with Modern Physics

Question

A solenoid 25.0 \(\mathrm{cm}\) long and with a cross-sectional area of 0.500 \(\mathrm{cm}^{2}\) contains 400 turns of wire and carries a current of 80.0 A. Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic fleld if the solenoid is filled with air; (c) the total energy contained in the coil's magnetic field (assume the field is uniform); (d) the inductance of the solenoid.

Step-by-Step Solution

Answer

(a) 0.402 T; (b) 64.4 J/m³; (c) 8.05×10⁻⁴ J; (d) 2.53×10⁻⁵ H.

Step 1: Calculate the Magnetic Field

The magnetic field inside a solenoid can be calculated using the formula for a solenoid: \( B = \mu_0 n I \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space \((4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A})\), \( n \) is the number of turns per unit length \( (\text{turns/m}) \), and \( I \) is the current. First, we calculate \( n = \frac{N}{L} = \frac{400}{0.25} = 1600 \, \text{turns/m}. \) Then, substitute into the formula: \( B = 4\pi \times 10^{-7} \times 1600 \times 80 = 0.402 \, \text{T} \).

Step 2: Calculate the Energy Density in the Magnetic Field

The energy density \( u \) in a magnetic field is given by \( u = \frac{1}{2} \frac{B^2}{\mu_0} \). Use \( B = 0.402 \text{ T} \) from Step 1: \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \). Substituting values, \( u = \frac{1}{2} \frac{(0.402)^2}{4\pi \times 10^{-7}} = 64.4 \, \text{J/m}^3 \).

Step 3: Calculate the Total Energy in the Coil's Magnetic Field

The total energy \( U \) in the magnetic field is given by \( U = u \times \text{Volume} \). The volume \( V \) is the cross-sectional area \( A \) times the length \( L \): \( V = 0.500 \times 10^{-4} \times 0.25 = 1.25 \times 10^{-5} \text{ m}^3 \). Now calculate \( U = 64.4 \times 1.25 \times 10^{-5} = 8.05 \times 10^{-4} \text{ J} \).

Step 4: Calculate the Inductance of the Solenoid

The inductance \( L \) of a solenoid is given by \( L = \mu_0 \frac{N^2 A}{l} \). Substitute \( \mu_0 = 4\pi \times 10^{-7} \), \( N = 400 \), \( A = 0.500 \times 10^{-4} \), and \( l = 0.25 \): \( L = 4\pi \times 10^{-7} \times \frac{(400)^2 \times 0.500 \times 10^{-4}}{0.25} = 2.53 \times 10^{-5} \text{ H} \).

Key Concepts

Energy Density in a Solenoid

When dealing with solenoids, energy density is a key concept in understanding how the magnetic field stores energy within the solenoid. The energy density, denoted as \( u \), represents the amount of energy stored per unit volume in the magnetic field. This is important because solenoids are often used to create uniform magnetic fields in scientific and industrial applications. To calculate the energy density of a solenoid's magnetic field, we use the formula for energy density:

• \( u = \frac{1}{2} \frac{B^2}{\mu_0} \).
• Here, \( B \) is the magnetic field strength and \( \mu_0 \) is the permeability of free space, a constant value \( (4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}) \).

In our specific example, substituting the values yields an energy density of \( 64.4 \, \text{J/m}^3 \). This simply means that each cubic meter of the magnetic field within the solenoid has 64.4 joules of energy. Remember, having high energy density means efficient energy storage, which can be advantageous in applications where space is limited.

Inductance of a Solenoid

Inductance is a fundamental property of solenoids that describes their ability to store energy in the magnetic field when an electric current passes through them. It essentially measures the efficiency with which the solenoid can convert electrical energy into magnetic energy. Inductance, denoted by the letter \( L \), depends on several factors: the number of turns in the solenoid, the cross-sectional area, its length, and the material inside. The formula to calculate the inductance of a solenoid is given by:

• \( L = \mu_0 \frac{N^2 A}{l} \)
• where \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( l \) is the length of the solenoid.

In the exercise example, substituting the values results in an inductance of \( 2.53 \times 10^{-5} \, \text{H} \) (henrys). The henry (H) is the SI unit for inductance, and it reflects the robustness of the solenoid to store magnetic energy. Higher inductance indicates greater capacity to store energy, making it vital for efficient circuits, especially those involving alternating currents, like transformers and inductors.

Total Magnetic Energy in a Solenoid

The total magnetic energy contained in a solenoid is an extension of the energy density concept. While energy density gives energy information per unit volume, total magnetic energy considers the entire space that the solenoid's magnetic field fills. This overall energy is crucial for applications that depend on the solenoid's ability to function over its entire volume, such as magnetic resonance imaging (MRI) machines. To find the total magnetic energy, use the formula for the total energy \( U \):

• \( U = u \times \text{Volume} \)
• Here, \( u \) is the energy density we've calculated, and \( \text{Volume} \) is the product of the solenoid's length and the cross-sectional area.

For our problem, with a calculated energy density of \( 64.4 \, \text{J/m}^3 \) and a volume of \( 1.25 \times 10^{-5} \, \text{m}^3 \), the total energy is \( 8.05 \times 10^{-4} \, \text{J} \). This tells us how much energy is entirely stored in the solenoid's magnetic field, illustrating the capability of the solenoid to contain and maintain energy in a steady state.