Problem 52 - University Physics with Modern Physics

Question

A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80. (a) What is the magnetic field in the core? (b) What part of the magnetic field is due to atomic currents?

Step-by-Step Solution

Answer

(a) The magnetic field in the core is about 3.35 mT. (b) The part due to atomic currents is approximately 3.30 mT.

Step 1: Understand the Formula for the Magnetic Field

The magnetic field inside a toroidal solenoid is given by the formula: \( B = \mu N \frac{I}{l} \), where \( B \) is the magnetic field, \( \mu \) is the permeability of the solenoid (which depends on the vacuum permeability and the relative permeability), \( N \) is the total number of turns, \( I \) is the current, and \( l \) is the magnetic path length.

Step 2: Calculate the Magnetic Path Length

The magnetic path length \( l \) in the toroid is the circumference of the mean radius of the solenoid: \( l = 2 \pi R \), where \( R = 6.0 \) cm = 0.06 m. So, \( l = 2 \pi \times 0.06 \) m.

Step 3: Determine the Relative Permeability

The relative permeability of the core is given as 80. The permeability of the core is \( \mu = \mu_0 \times \text{relative permeability} \), where \( \mu_0 \) (permeability of free space) is \( 4\pi \times 10^{-7} \) Tm/A. So, \( \mu = 4\pi \times 10^{-7} \times 80 \).

Step 4: Calculate the Magnetic Field

Substitute the values into the formula \( B = \mu N \frac{I}{l} \):\( N = 400 \), \( I = 0.25 \) A, \( \mu = 4\pi \times 10^{-7} \times 80 \), and \( l = 2 \pi \times 0.06 \). Calculate the magnetic field \( B \).

Step 5: Calculate the Magnetic Field from Atomic Currents

The part of the magnetic field due to atomic currents, also known as magnetization, is the additional magnetic field provided by the material of the core and is given by the equation: \( B_m = \mu_0 M \), where \( M \) is the magnetization.Using the relation \( M = \chi_m H \) (where \( \chi_m = \text{relative permeability} - 1 \) and \( H \) is the magnetic field strength, given by \( H = \frac{NI}{l} \)), calculate \( B_m \).

Step 6: Final Step: Calculate the Part of the Field Due to Atomic Currents

Calculate the magnetic field strength \( H = \frac{NI}{l} \), then use \( M = (80 - 1)H \). Finally, calculate \( B_m = \mu_0 M \) to find the magnetic field due to the atomic currents.

Key Concepts

Toroidal Solenoid

A toroidal solenoid is a coil of wire shaped like a doughnut. It is often used in applications where a strong, uniform magnetic field is needed, like in transformers or inductors. The special doughnut shape ensures that the magnetic field lines are contained within the core, making it very efficient.
This efficiency is heightened by the solenoid's characteristic of minimizing stray magnetic fields outside the core. The magnetic field inside a toroidal solenoid can be calculated using the formula: \[B = \mu N \frac{I}{l},\]where:

• \(B\) is the magnetic field
• \(\mu\) is the permeability of the material
• \(N\) is the number of turns of the wire
• \(I\) is the current flowing through the solenoid
• \(l\) is the magnetic path length

This formula shows how the magnetic field is directly influenced by the material's properties and the design of the solenoid itself.

Relative Permeability

Relative permeability is a measure of how much better a material can support the formation of a magnetic field compared to a vacuum. It is a dimensionless number that compares the permeability of the material to the permeability of free space, represented by \(\mu_0\).
For the core of the toroidal solenoid, relative permeability is given as 80. This means that the core material can support a magnetic field 80 times more effectively than a vacuum.
The formula to find the permeability \(\mu\) of the core is:\[\mu = \mu_0 \times \text{relative permeability}\]This highlights how materials like ferromagnetic substances improve magnetic performance due to their high relative permeability, vital for efficient electromagnetic applications.

Atomic Currents

Atomic currents are the microscopic motions of electrons at the atomic level that contribute to a material's magnetization. When an external magnetic field is applied, these tiny loops of current align with the field, strengthening the overall magnetic field inside the material.
The additional magnetic field produced by atomic currents in a material is termed magnetization and is represented by \(M\). This aligns with the external field to create a larger effective magnetic field.
In our example, to find the portion of the magnetic field arising from atomic currents, we use:\[B_m = \mu_0 M\]where \(M\) can be calculated using the magnetic susceptibility \(\chi_m\) and the magnetic field strength \(H\). Recognizing atomic currents’ impact is integral for designing materials with desired magnetic properties.

Magnetization

Magnetization represents the extent to which a material can be magnetized. It describes the alignment of atomic magnetic moments within a material, leading to the material exhibiting a magnetic field. This is intensified when materials with high relative permeability are used, as they can support these internal alignments effectively.
The magnetization \(M\) is calculated by:\[M = \chi_m H,\]where \(\chi_m\) is the magnetic susceptibility \((\text{relative permeability} - 1)\) and \(H\) is the magnetic field strength, \(H = \frac{NI}{l}\).
Magnetization contributes a significant portion of the internal magnetic field, especially in materials like iron, making them powerful components in electromagnetic devices. By understanding the relationship between \(M\), the external magnetic field, and the material's intrinsic properties, one can design effective electromagnetic systems.