According to a solenoid 0.25 m long, having a number of turns of 500

According to a solenoid 0.25 m long, having a number of turns of 500

We present to your attention a product - an electronic textbook on the theory of electromagnetism. In this textbook you will find a detailed description of the principles of operation of solenoids and much more that will help you better understand this fascinating science.

One of the examples you will study is a 0.25 m long solenoid that has 500 turns. With our tutorial, you will learn how to calculate the magnetic field energy of such a solenoid and understand why it is an important element in various electronic devices.

Our e-textbook is presented in an easy-to-read format that allows you to quickly find the information you need. It also contains many illustrations and examples to help you understand the material better.

By purchasing our electronic textbook, you get a unique opportunity to expand your knowledge in the field of electromagnetism and learn a lot about solenoids.

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The 0.25 meter long solenoid with 500 turns is an electromagnetic device that can convert electric current into a magnetic field. The solenoid can be used in various applications such as electromechanical locks, automatic doors, automobile equipment and many others. Due to their compactness and high performance, solenoids are widely used in various fields of industry and science.

This product is a solenoid, that is, a coil consisting of a wire that is wound around a cylindrical body. The length of the solenoid is 0.25 meters, and the number of turns is 500.

The solenoid carries a current of 1 Ampere and has a cross-sectional area of ​​15 square centimeters. There is an iron core inside the solenoid.

To solve problem 31033 it is necessary to use the formula for calculating the energy of the magnetic field of the solenoid:

W = (L*I^2)/2,

where W is the energy of the magnetic field of the solenoid, L is the inductance of the solenoid, I is the current strength passing through the solenoid.

The solenoid inductance can be calculated using the formula:

L = (μ0 * N^2 * A) / l,

where μ0 is the magnetic constant, N is the number of turns, A is the cross-sectional area, l is the length of the solenoid.

Thus, substituting the known values, we get:

L = (4π * 10^-7 * 500^2 * 0.0015) / 0.25 = 0.0472 Henry

Then, substituting the obtained values ​​into the formula for calculating the magnetic field energy, we obtain:

W = (0.0472 * 1^2) / 2 = 0.0236 J

Thus, the magnetic field energy of the solenoid is 0.0236 J.

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