Problem 8 - An Introduction to Materials Engineering and Science for Chemical and Materials Engineers

Question

A coil of wire \(0.1 \mathrm{~m}\) long and having 15 turns carries a current of \(1.0 \mathrm{~A}\). (a) Compute the magnetic induction if the coil is within a vacuum. (b) A bar of molybdenum is now placed in the coil, and the current adjusted to maintain the same magnetic induction as in part (a). Calculate the magnetization.

Step-by-Step Solution

Answer

(a) Magnetic induction is \(1.884 \times 10^{-4} \ \mathrm{T}\). (b) Magnetization is \(1203.64 \ \mathrm{A/m}\).

Step 1: Understand the Formula for Magnetic Induction in a Vacuum

To compute the magnetic induction in a vacuum, we use the formula: \( B = \mu_0 \frac{N I}{L} \), where \( B \) is the magnetic induction, \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, \( I \) is the current, and \( L \) is the length of the coil.

Step 2: Calculate the Magnetic Induction in a Vacuum

We know \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T \cdot m/A} \), \( N = 15 \), \( I = 1.0 \ \mathrm{A} \), and \( L = 0.1 \ \mathrm{m} \). Substituting these into the formula gives: \[ B = \left( 4\pi \times 10^{-7} \right) \frac{15 \times 1.0}{0.1} = 1.884 \times 10^{-4} \ \mathrm{T} \].

Step 3: Calculate the Magnetization with Molybdenum

When a bar of molybdenum is placed in the coil, the magnetization \( M \) is calculated as\[ M = \frac{B - \mu_0 H}{\mu} \]. Since we want the same \( B \) as before, use the relation \( B = \mu_0(H + M) \). We already know that \( H = \frac{NI}{L} \) and hence \( H = \frac{15 \times 1.0}{0.1} = 150 \ \mathrm{A/m} \). Solve for \( M: \) \[ B = \mu_0 (H + M) = 1.884 \times 10^{-4} \ \mathrm{T}\]\[ M = \frac{B}{\mu_0} - H = \frac{1.884 \times 10^{-4}}{4\pi \times 10^{-7}} - 150 = 1203.64 \ \mathrm{A/m} \].

Key Concepts

Magnetization

Magnetization is a crucial concept in understanding how materials respond to magnetic fields. It represents the degree to which a material can become magnetized when exposed to a magnetic field. In simple terms, it is the measure of the magnetic moment per unit volume of a material.

The process of magnetization involves aligning the magnetic domains within a material in the direction of an applied magnetic field. These domains are tiny, independently magnetized regions within the substance. When a magnetic field is present, the domains align, resulting in a net magnetic moment.

• The formula for magnetization is given by: \( M = \frac{B - \mu_0 H}{\mu} \), where \( M \) is the magnetization, \( B \) is the magnetic induction, \( \mu_0 \) is the permeability of free space, and \( H \) is the magnetic field intensity.
• The concept of magnetization is essential for understanding how materials like molybdenum respond when subjected to a coil's magnetic field, as shown in the exercise.

Permeability of Free Space

The permeability of free space, also known as the magnetic constant, is a fundamental physical constant that characterizes how a magnetic field affects and propagates through a vacuum. It's a measure of the ability of a vacuum to support a magnetic field.

The symbol for the permeability of free space is \( \mu_0 \), and its approximate value is \( 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \). Understanding this concept is crucial for calculating magnetic fields in a vacuum, as it provides a baseline for the permeability of other materials.

• The formula for magnetic induction in a vacuum utilizes \( \mu_0 \) as: \( B = \mu_0 \frac{N I}{L} \).
• Knowing the permeability of free space is essential for predicting how a material will behave when placed in a magnetic field.

Magnetic Field in Vacuum

A magnetic field in a vacuum behaves distinctly because no material is present to affect the lines of the magnetic field. Calculating the magnetic field in such a situation provides a pure measure of the field produced by a given current in a coil.

For a wire coil, the magnetic field in a vacuum can be calculated using the formula \( B = \mu_0 \frac{N I}{L} \), where:

• \( N \) is the number of turns in the coil,
• \( I \) is the current flowing through the coil,
• \( L \) is the length of the coil.

In the context of the exercise, the magnetic field calculated in a vacuum serves as a benchmark to determine the effect of introducing materials like molybdenum, which have their specific permeability.

The value of \( B \), the magnetic induction in vacuum, is unhindered by material properties, making it a valuable reference in studying electromagnetic properties.

Molybdenum Bar in Magnetic Field

When introducing a molybdenum bar into a coil's magnetic field, it is essential to understand the impact this has on the overall magnetization process. Molybdenum is a paramagnetic material which means it becomes magnetized in an external magnetic field, but this magnetization is usually relatively weak and disappears once the external field is removed.

The placement of a molybdenum bar affects both the magnetization and the magnetic field strength in the coil. Here's how:

• The material's response to the field depends on its permeability, which determines how well it can be magnetized compared to the vacuum baseline.
• For molybdenum, you must adjust the current to maintain the same magnetic induction as when the coil was in a vacuum.
• The magnetization \( M \) in the presence of molybdenum is calculated to ensure the field strength remains consistent, using the relationship: \( B = \mu_0 (H + M) \).

Understanding how molybdenum interacts with a magnetic field enhances our insights into the behavior of different materials within varying magnetic environments.