*Notes to a video lecture on http://www.unizor.com*

__Solenoid__

Solenoid is a wire spiral like on this picture

It can be considered as a combination of individual circular wire loops.

We will consider an ideal case when the wire is infinitesimally thin,

the length of a solenoid along its axis is infinite and there is a

direct electric current running through it, that is running through each

of its loops.

Our purpose, as in a case of a single loop, is to determine the magnetic field inside such a solenoid.

Position our infinite in length solenoid along the X-axis such that the

X-axis is the axis of a solenoid. Let the radius of a solenoid's loops

be

*and the density of the loops per unit of length along X-axis be*

**R***, that is, each, however small, segment*

**N***d*on the X-axis contains

**X***loops.*

**N·**d**X**Assume a direct electric current

*is running through all loops.*

**I**Since a solenoid is infinite, magnetic field created by an electric

current running through it should be the same at any point on its axis.

So, let's calculate the intensity of a magnetic field

*at the origin of coordinates. At any other point on the X-axis it will be the same.*

**B**The method of calculation the magnetic field is as follows.

**Firstly**, we will calculate the intensity of the magnetic field at

the origin of coordinates, generated by a single loop at the distance

*from this point.*

**X****Secondary**, we will take into account the density of the loops and

increase the calculated intensity of a single loop by a number of loops

on a segment from

*to*

**X***, assuming*

**X+**d**X***d*is infinitesimal increment.

**X**The

**final step**is to integrate the obtained intensity for all

*from*

**X***to*

**−∞***.*

**+∞**1.

*Magnetic field from a single loop at distance*

*from the origin of coordinates***X**Calculations of the magnetic field at any point on the axis going

through a wire loop is similar to but more complex than the calculation

of the magnetic field at the center of a loop we studied in the previous

lecture.

The picture below represents the general view of the wire loop with a point

*,*

**O**where we want to evaluate its magnetic field and, below this general

view, the same loop as viewed from a plane this loop belongs to

perpendicularly to the X-axis.

As we know, the intensity

*of the magnetic field at point*

**OB***, generated by an infinitesimal segment*

**O***d*of a wire loop located at point on a loop

**S***is perpendicular to both electric current*

**A***at point*

**I***and vector*

**A***from a point on a loop*

**AO***to a point of interest - the origin of coordinates*

**A***.*

**O**The direction of the electric current at point

*is perpendicular to the plane of a picture. Therefore, vector of intensity*

**A***is within a plane of a picture, as well as line*

**OB***. Now all lines we are interested in are on the plane of a picture and vector of magnetic field intensity*

**AO***at point of interest - the origin of coordinates*

**OB***is perpendicular to a line from point*

**O***on a loop to a point of interest*

**A***:*

**O**

**AO****⊥**.

*OB*Let

*be a projection of the endpoint of vector*

**B'***onto X-axis. From perpendicularity of*

**OB***to*

**AO***follows that*

**OB***and triangles Δ*

**∠AOP=∠OBB'***and Δ*

**AOP***are similar.*

**OBB'**As we know, the magnitude of magnetic field intensity at point

**O**depends on the distance from the loop segment with electric current,

the length of this segment, the magnitude of the electric current in it

and a sine of the angle between the direction of an electric current in a

segment and the line from a segment to a point of interest:

Since the direction of the electric current is perpendicular to a plane

of a picture, the angle between the direction of an electric current in a

segment and the line from a segment to a point of interest is 90°, so

its sine is equal to 1, and the formula for intensity of a magnetic

field at point

*produced by a segment*

**O***d*at point

**S***on the loop is*

**A***d*[

**B = μ**d_{0}·I·**S /***]*

**4π(R²+X²**We should take into consideration only the X-component of the vector of

magnetic field intensity because the Y-component will be canceled by the

corresponding Y-component of a vector of intensity produced by a

segment of a loop at diametrically opposite to point

*location.*

**A**Let

**k = sin(∠OBB').**Hence,

*.*

**k = sin(∠AOP) = R/√X²+R²**Then the X-component of vector

*is*

**OB***d*[

**B**d_{x}= μ_{0}·I·**S·k /***]*

**4π(R²+X²)***[*

**=**

= μd= μ

_{0}·I·**S·R /***]*

**4π(R²+X²)**^{3/2}To get a magnitude of the magnetic field intensity at point

*from an entire loop we have to integrate the above expression by*

**O***from*

**S***to*

**0***, and the result will be*

**2πR***[*

**B**_{x}= μ_{0}·I·2πR² /*]*

**4π(R²+X²)**^{3/2}Simplifying this gives

*[*

**B(X) = μ**_{0}·I·R² /*]*

**2(R²+X²)**^{3/2}Incidentally, if our loop is located at

*, the formula above gives*

**X=0***,*

**B**_{x}= μ_{0}·I / (2R)that corresponds to the results presented in the previous lecture for intensity of a magnetic field at the center of a loop.

2.

*Magnetic field from all loops located at distance from*

*to***X***from the origin of coordinates***X+**d**X**This is a simple multiplication of the above expression for

*by the number of loops located at distance from*

**B**_{x}*to*

**X***from the origin of coordinates, which is*

**X+**d**X***, where*

**N·**d**X***is a density of the loops in a solenoid.*

**N**The result is the intensity of a magnetic field at point

*produced by an infinitesimal segment of our infinitely long solenoid from*

**O***to*

**X**

**X+**d**X***[*

**μ**d_{0}·I·R²·N·**X /***]*

**2(R²+X²)**^{3/2}3.

*Finding the magnetic field at any point inside a solenoid by integrating the intensity produced by its segment from*−∞

**X**to**X+**d**X**by**X**for an entire length of a solenoid from*to*+∞

The above expression should be integrated by

*from*

**X***to*

**−∞***to get a total value of the intensity of the magnetic field produced by an entire infinitely long solenoid at point*

**+∞***(or any other point on its axis).*

**O***∫*

**B = γ·**_{[−∞;+∞]}

**d****X / (R²+X²)**^{3/2}where

**γ = μ**_{0}·I·N·R²/2Substitute

*in the integral with*

**X=x·R***d*.

**X=R·**d**x**Infinite limits for

*will be infinite limits for*

**X***.*

**x**Then integral will be

∫

_{[−∞;+∞]}

*∫*

**R·**d**x / (R²+R²·x²)**

= (1/R²)·^{3/2}== (1/R²)·

_{[−∞;+∞]}

**d****x / (1+x²)**^{3/2}The value of the integral can be easily calculated.

The indefinite integral is

**x/√1+x² + C**So, the value of the definite integral between infinite limits is equal to

*.*

**1 − (−1) = 2**Therefore,

**B = γ·2/R² = μ**_{0}·I·NIMPORTANT: The intensity of a magnetic field at any point on an axis of

an infinitely long solenoid does not depend on its radius.

Textbooks usually state that the intensity calculated above is the same

at all points inside an infinitely long solenoid, not only at the points

along its axis. The justification and the proof of this are beyond the

scope of this course.

## No comments:

Post a Comment