Find the magnetic field strength at the center of the arc

Description of a digital product - Find the magnetic field strength at the center of the arc

This digital product is an electronic document containing information and formulas for calculating the magnetic field strength in the center of an arc with a radius of 0.2 m, closing two parallel semi-infinite conductors, at a given current strength in the circuit.

The document is written in Russian and is intended for a wide range of users interested in electromagnetism and physics.

The document is designed in a beautiful and understandable html format, which makes it easy to navigate the contents and quickly find the necessary information. The document presents basic formulas and examples of calculations that will help the user cope with the task of finding the magnetic field strength at the center of the arc.

By purchasing this digital product, you gain access to useful information that can be used for educational purposes, as well as in professional activities in the field of electrical engineering and physics.

Don't miss the opportunity to purchase this digital product and expand your knowledge in the field of electromagnetism!

This digital product is an electronic document containing a detailed solution to the problem of finding the magnetic field strength in the center of an arc with a radius of 0.2 m, closing two parallel semi-infinite conductors, at a given current in the circuit, which is 14 A. The document presents formulas and laws , used in solving the problem, as well as the derivation of the calculation formula and the answer.

The document is written in Russian and is intended for a wide range of users interested in electromagnetism and physics. The document is designed in a beautiful and understandable html format, which makes it easy to navigate the contents and quickly find the necessary information.

By purchasing this digital product, you gain access to useful information that can be used for educational purposes, as well as in professional activities in the field of electrical engineering and physics. Don't miss the opportunity to purchase this digital product and expand your knowledge in the field of electromagnetism!

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To solve this problem, it is necessary to use the Biot-Savart-Laplace law, which allows one to calculate the magnetic field created by a current at an arbitrary point in space. The formula of the law is as follows:

dH = (μ0/4π) * Idl x r / r^3

where dH is the element of the magnetic field created by the element of the length of the conductor dl, I is the strength of the current passing through the conductor, r is the distance from the element of length to the point at which the magnetic field is calculated, μ0 is the magnetic constant.

To find the magnetic field at the center of the arc, it is necessary to integrate the elements of the magnetic field dH over the entire arc. Since the arc consists of two semi-infinite conductors, the magnetic field at the center of the arc will be equal to the sum of the magnetic fields created by each of the conductors.

Let us first calculate the magnetic field created by one semi-infinite conductor. The conductor is located at a distance r from the center of the arc, so the formula of the Biot-Savart-Laplace law takes the following form:

dH = (μ0/4π) * Idl / r^2

where dl is the conductor length element.

Since the conductors are parallel, the elements of the magnetic field dH will be directed in one direction and will add up. Therefore, the magnetic field at the center of the arc will be equal to twice the magnetic field created by one conductor.

Thus, the magnetic field at the center of the arc will be equal to:

H = 2 * (μ0/4π) * I / r

where I is the current strength passing through each of the conductors, r is the radius of the arc.

Let's substitute the known values ​​into this formula:

H = 2 * (4π*10^-7 N/A^2) * 14 A / 0.2 m H = 4.4 * 10^-5 T

Thus, the magnetic intensity at the center of an arc with a radius of 0.2 m, closing two parallel semi-infinite conductors, if they lie in a plane perpendicular to the plane of the arc, with a current strength in the circuit of 14 A, is equal to 4.4 * 10^-5 T.

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