Problem 53 - University Physics with Modern Physics

Question

A long solenoid with 60 turns of wire per centimeter carries a current of 0.15 A. The wire that makes up the solenoid is wrapped around a solid core of silicon steel (\(K_m =\) 5200). (The wire of the solenoid is jacketed with an insulator so that none of the current flows into the core.) (a) For a point inside the core, find the magnitudes of (i) the magnetic field \(\overrightarrow{B_0}\) due to the solenoid current; (ii) the magnetization \(\overrightarrow{M}\); (iii) the total magnetic field \(\overrightarrow{B}\). (b) In a sketch of the solenoid and core, show the directions of the vectors \(\overrightarrow{B}\), \(\overrightarrow{B_0}\), and \(\overrightarrow{M}\) inside the core.

Step-by-Step Solution

Answer

(a) (i) \( B_0 = 1.13 \times 10^{-3} \ \text{T} \); (ii) \( M = 4.67 \times 10^3 \ \text{A/m} \); (iii) \( B = 5.88 \ \text{T} \). (b) All vectors are aligned along the solenoid's axis.

Step 1: Calculate the Magnetic Field due to the Solenoid Current

To find the magnetic field inside a solenoid due to its current, we use the formula:\[ B_0 = \mu_0 n I \]where \( \mu_0 \) is the permeability of free space \((4\pi \times 10^{-7} \ \text{T m/A})\), \( n \) is the number of turns per unit length (60 turns/cm = 6000 turns/m), and \( I \) is the current (0.15 A). Substituting the values gives:\[ B_0 = (4\pi \times 10^{-7} \ \text{T m/A}) \times 6000 \ \text{turns/m} \times 0.15 \ \text{A} = 1.13 \times 10^{-3} \ \text{T} \].

Step 2: Determine the Magnetization

The magnetization \( M \) can be calculated using the formula:\[ M = \left( K_m - 1 \right) \frac{B_0}{\mu_0} \]where \( K_m \) is the relative permeability of silicon steel (5200). Substituting the values, we get:\[ M = (5200 - 1) \frac{1.13 \times 10^{-3} \ \text{T}}{4\pi \times 10^{-7} \ \text{T m/A}} \approx 4.67 \times 10^3 \ \text{A/m} \].

Step 3: Calculate the Total Magnetic Field

The total magnetic field \( B \) inside the core is given by:\[ B = \mu K_m H \]Here \( H \) is the magnetic field intensity, which is given by \( H = \frac{B_0}{\mu_0} \), so,\[ B = \mu_0 K_m \frac{B_0}{\mu_0} = K_m B_0 \]Substituting in our known values:\[ B = 5200 \times 1.13 \times 10^{-3} \ \text{T} \approx 5.88 \ \text{T} \].

Step 4: Sketch the Directions of Vectors

In the core of the solenoid, draw three vectors:1. \( \overrightarrow{B_0} \): This points along the axis of the solenoid and is due to the current.2. \( \overrightarrow{M} \): The magnetization vector points in the same direction as \( \overrightarrow{B_0} \).3. \( \overrightarrow{B} \): The total magnetic field \( \overrightarrow{B} \), which is the sum of \( \overrightarrow{B_0} \) and the field due to magnetization, also points in the same direction as \( \overrightarrow{B_0} \). All vectors are aligned along the solenoid's axis.

Key Concepts

Magnetization

Magnetization refers to the process where a material becomes magnetized when placed in a magnetic field. It is essentially the magnetic moment per unit volume of a material, representing how much the material contributes to the overall magnetic field. In this context, we're dealing with silicon steel, a material known for its high magnetic permeability, which enhances its ability to be magnetized.
When a solenoid, which is a coil of wire, carries an electric current, it creates a magnetic field along its axis. The core material in the solenoid, like silicon steel, significantly amplifies this magnetic field due to its magnetizing properties.
This can be measured by the formula: \[M = (K_m - 1) \frac{B_0}{\mu_0}\]Where:

• \(M\) is the magnetization.
• \(K_m\) is the relative permeability of the material.
• \(B_0\) is the initial magnetic field produced by the solenoid.
• \(\mu_0\) is the permeability of free space, which is a constant\( 4\pi \times 10^{-7} \ \text{T m/A}\).

This process results in a magnetized core with a magnetization \( M \approx 4.67 \times 10^3 \text{A/m}\), significantly boosting the magnetic characteristics of the solenoid.

Silicon Steel Core

A silicon steel core is used in electrical applications like transformers and solenoids due to its exceptional magnetic properties. Silicon steel is an alloy made mostly of iron with a small percentage of silicon added (usually around 3%). The inclusion of silicon in the iron drastically reduces energy loss by minimizing electrical currents induced inside the core itself, known as eddy currents.
The material has a very high relative permeability, denoted as \( K_m \), which indicates how easily it can become magnetized in response to an external magnetic field. In our problem, the silicon steel core has a \( K_m \) of 5200, meaning it is highly responsive compared to many other materials. This results in a strong magnetic field within the solenoid.
By enhancing the magnetic field intensity, a silicon steel core makes solenoids more efficient and effective, crucial for devices that rely on magnetic fields. Incidentally, the solid core allows for maintaining the direction and uniformity of magnetic fields without significant internal disturbances.

Magnetic Field Intensity

Magnetic field intensity, represented by \( H \), describes the strength of the magnetic field created by a magnetizing force, such as a current-carrying solenoid. It depends on the material's intrinsic properties and how it responds to an external magnetic influence.
This is calculated using the relation: \[H = \frac{B_0}{\mu_0}\]Where \( B_0 \) is the magnetic field generated by the solenoid, and \( \mu_0 \) remains the permeability of free space. In this context, it gauges the effectiveness of the current in generating a magnetic field within the solenoid itself.
The magnetic intensity of a solenoid showcasing high relative permeability, especially with a solid silicon steel core, leads to an intense and well-aligned field which means the solenoid efficiently channels magnetic lines along its path. Ultimately, the high magnetization within the core amplifies the initial intensity, shown as \[ B = K_m B_0 \], resulting in a total field magnitude of \( 5.88 \text{ T} \). This ensures the solenoid achieves its maximum potential in applications like electromagnets and inductors.