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Chapter 28: Problem 53

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.

Short Answer

Expert verified
(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 by step solution

01

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:\[ B_0 = (4\pi \times 10^{-7} \ \text{T m/A}) \times 6000 \ \text{turns/m} \times 0.15 \ \text{A} = 1.13 \times 10^{-3} \ \text{T} \].
02

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:\[ M = (5200 - 1) \frac{1.13 \times 10^{-3} \ \text{T}}{4\pi \times 10^{-7} \ \text{T m/A}} \approx 4.67 \times 10^3 \ \text{A/m} \].
03

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:\[ B = 5200 \times 1.13 \times 10^{-3} \ \text{T} \approx 5.88 \ \text{T} \].
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

The magnetic field around the head has been measured to be approximately 3.0 \(\times\) 10\(^{-8}\) G. Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop 16 cm (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

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

(a) How large a current would a very long, straight wire have to carry so that the magnetic field 2.00 cm from the wire is equal to 1.00 G (comparable to the earth's northward-pointing magnetic field)? (b) If the wire is horizontal with the current running from east to west, at what locations would the magnetic field of the wire point in the same direction as the horizontal component of the earth's magnetic field? (c) Repeat part (b) except the wire is vertical with the current going upward.

A closed curve encircles several conductors. The line integral \(\oint\overrightarrow{B}\) \(\cdot\) \(\overrightarrow{dl}\) around this curve is 3.83 \(\times\) 10\(^{-4}\) T \(\cdot\) m. (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

To use a larger sample, the experimenters construct a solenoid that has the same length, type of wire, and loop spacing but twice the diameter of the original. How does the maximum possible magnetic torque on a bacterium in this new solenoid compare with the torque the bacterium would have experienced in the original solenoid? Assume that the currents in the solenoids are the same. The maximum torque in the new solenoid is (a) twice that in the original one; (b) half that in the original one; (c) the same as that in the original one; (d) one-quarter that in the original one.
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