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The current in the windings of a toroidal solenoid is 2.400 A. There are 500 turns, and the mean radius is 25.00 \(\mathrm{cm}\) . The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 \(\mathrm{T}\) . Calculate (a) the relative permeability and \((b)\) the magnetic susceptibility of the material that fills the toroid.

Short Answer

Expert verified
(a) Relative permeability: approximately 1616.67. (b) Magnetic susceptibility: approximately 1615.67.

Step by step solution

01

Calculate the Magnetic Field without material

First, calculate the magnetic field inside the toroid if it were filled with a vacuum. Use the formula for the magnetic field inside a toroid: \[ B = \frac{\mu_0 \cdot n \cdot I}{2\pi r} \]where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \) is the permeability of free space, \(n\) is the number of turns, \(I\) is the current, and \(r\) is the mean radius in meters. Substitute the values: \(n = 500\), \(I = 2.400 \, \text{A}\), and \(r = 0.25 \, \text{m}\). Calculate:\[ B = \frac{(4\pi \times 10^{-7}) \cdot 500 \cdot 2.400}{2\pi \times 0.25} \approx 0.0012 \, \text{T} \]
02

Calculate Relative Permeability

Given the magnetic field inside the toroid with the material is \(B_{\text{material}} = 1.940 \, \text{T}\). The relative permeability \(\mu_r\) is the ratio of the magnetic field with the material to the magnetic field in a vacuum, given by:\[ \mu_r = \frac{B_{\text{material}}}{B} \].Substitute the known values:\[ \mu_r = \frac{1.940}{0.0012} \approx 1616.67 \]
03

Calculate Magnetic Susceptibility

Magnetic susceptibility \(\chi_m\) is related to the relative permeability by the formula:\[ \chi_m = \mu_r - 1 \]. Using the value from the previous step:\[ \chi_m = 1616.67 - 1 = 1615.67 \]
04

Conclusion

The relative permeability of the material filling the toroid is approximately \(1616.67\), and the magnetic susceptibility of the material is approximately \(1615.67\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field
A toroidal solenoid is essentially a coil of wire shaped like a donut. It creates a magnetic field when an electric current passes through it. The magnetic field inside a toroidal solenoid is determined by the current flowing through the wire and the characteristics of the coil itself, such as the number of turns and the radius. For a toroid filled with vacuum, the magnetic field can be calculated using the formula:
\[ B = \frac{\mu_0 \cdot n \cdot I}{2\pi r} \]where:
  • \( \mu_0 \) is the permeability of free space, \(4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \).
  • \( n \) is the number of turns in the solenoid.
  • \( I \) is the current passing through the windings.
  • \( r \) is the mean radius of the toroid in meters.
The formula shows how these parameters affect the strength of the magnetic field. A larger current or more turns increases the magnetic field, while a larger radius decreases it.
In cases where the toroid is filled with a material other than vacuum, you need to consider the material's permeability, which is where the next two concepts come into play.
Relative Permeability
Relative permeability \( \mu_r \) is a measure of how much better a material can conduct magnetic lines of flux compared to a vacuum. It reflects the material's ability to enhance the magnetic field inside the solenoid. In simple terms, relative permeability indicates how a material increases the magnetic field compared to when the space inside the toroid is filled with vacuum.
To find the relative permeability, we use the magnetic field inside the toroid when it is filled with the material \( B_{\text{material}} \) and divide it by the magnetic field with a vacuum \( B \):
\[ \mu_r = \frac{B_{\text{material}}}{B} \]With this formula, you can calculate how much stronger the magnetic field is in a given material environment compared to a vacuum condition.
In our example, the magnetic field with the material is given as \(1.940 \, \text{T}\), while the field with a vacuum was calculated to be approximately \(0.0012 \, \text{T}\). This results in a relative permeability of about \(1616.67\), meaning the magnetic material enhances the magnetic field by this factor.
Magnetic Susceptibility
Magnetic susceptibility, \( \chi_m \), is closely related to relative permeability. It quantifies a material's tendency to become magnetized in response to an external magnetic field. Essentially, it tells us how readily a material will polarize in response to the magnetic influence.
The relationship between magnetic susceptibility and relative permeability is expressed by the equation:
\[ \chi_m = \mu_r - 1 \]This formula signifies that susceptibility is one less than the relative permeability. This subtraction occurs because susceptibility zeroes in on the increase in the field inside the material compared to the field it would have without the material.
In the exercise example, with a calculated relative permeability of \(1616.67\), the magnetic susceptibility is \(1615.67\). This sizable number indicates that the material significantly enhances the magnetic field, showing a strong response to the external magnetic environment within the toroid.

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Most popular questions from this chapter

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 1.40 \(\mathrm{m}\) long and 20.0 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

Two long. parallel transmission lines, 40.0 \(\mathrm{cm}\) apart, carry \(25.0-\mathrm{A}\) and 75.0 - A currents. Find all locations where the net magnetic field of the two wires is zero if these currents are in (a) the same direction and (b) the opposite direction.

A toroidal solenoid (see Example 28.10 ) has inner radius \(r_{1}=15.0 \mathrm{cm}\) and ourer radius \(r_{2}=18.0 \mathrm{cm} .\) The solenoid has 250 turns and carries a current of 8.50 A. What is the magnitude of the magnetic field at the following distances from the center of the torus: (a) \(12.0 \mathrm{cm} ;(\mathrm{b}) 16.0 \mathrm{cm} ;(\mathrm{c}) 20.0 \mathrm{cm} ?\)

A wooden ring whose mean diameter is 14.0 \(\mathrm{cm}\) is wound with a closely spaced toroidal winding of 600 turns. Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.650 \(\mathrm{A}\) .

A magnetic field of 37.2 T has been achieved at the MTT Francis Bitter National Magnetic Laboratory. Find the current needed to achieve such a field (a) 2.00 \(\mathrm{cm}\) from a long, straight wire; \((b)\) at the center of a circular coil of radius 42.0 \(\mathrm{cm}\) that has 100 turns; \((\mathrm{c})\) near the center of a solenoid with radius \(2.40 \mathrm{cm},\) length \(32.0 \mathrm{cm},\) and \(40,000\) turns.

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