Let's consider a wire located in a magnetic field and through which current flows (Fig. 12.6).
For each current carrier (electron), acts Lorentz force. Let us determine the force acting on a wire element of length d l
The last expression is called Ampere's law.
Ampere force modulus is calculated by the formula:
.
The Ampere force is directed perpendicular to the plane in which the vectors dl and B lie.
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Let's apply Ampere's law to calculate the force of interaction between two parallel infinitely long forward currents located in a vacuum (Fig. 12.7).
Distance between conductors - b. Let us assume that conductor I 1 creates a magnetic field by induction
According to Ampere's law, on the conductor I 2, from the side magnetic field, force acts
, taking into account that (sinα =1)
Therefore, per unit length (d l=1) conductor I 2, force acts
.
The direction of the Ampere force is determined by the left hand rule: if the palm of the left hand is positioned so that the magnetic induction lines enter it, and the four extended fingers are placed in the direction of the electric current in the conductor, then the extended thumb will indicate the direction of the force acting on the conductor from the field .
12.4. Circulation of the magnetic induction vector (total current law). Consequence.
A magnetic field, in contrast to an electrostatic one, is a non-potential field: the circulation of the vector In the magnetic induction of the field along a closed loop is not zero and depends on the choice of the loop. Such a field in vector analysis is called a vortex field.
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Let us consider as an example the magnetic field of a closed loop L of arbitrary shape, covering an infinitely long straight conductor with current l, located in a vacuum (Fig. 12.8).
The lines of magnetic induction of this field are circles, the planes of which are perpendicular to the conductor, and the centers lie on its axis (in Fig. 12.8 these lines are shown as dotted lines). At point A of contour L, vector B of the magnetic induction field of this current is perpendicular to the radius vector.
From the figure it is clear that
Where 
- length of the vector projection dl onto the vector direction IN. At the same time, a small segment dl 1 tangent to a circle of radius r can be replaced by a circular arc: , where dφ is the central angle at which the element is visible dl contour L from the center of the circle.
Then we obtain that the circulation of the induction vector
At all points of the line the magnetic induction vector is equal to
integrating along the entire closed contour, and taking into account that the angle varies from zero to 2π, we find the circulation
The following conclusions can be drawn from the formula:
1. The magnetic field of a rectilinear current is a vortex field and is not conservative, since there is vector circulation in it IN along the magnetic induction line is not zero;
2. vector circulation IN The magnetic induction of a closed loop covering the field of a straight-line current in a vacuum is the same along all lines of magnetic induction and is equal to the product of the magnetic constant and the current strength.
If a magnetic field is formed by several current-carrying conductors, then the circulation of the resulting field
This expression is called total current theorem.
The magnetic field has an orienting effect on the current-carrying frame. Consequently, the torque experienced by the frame is the result of the action of forces on its individual elements. Summarizing the results of a study of the effect of a magnetic field on various current-carrying conductors. Ampere established that the force d F, with which the magnetic field acts on the conductor element d l with current in a magnetic field is equal to where d l-vector, modulo equal to d l and coinciding in direction with the current, IN- vector of magnetic induction.
Direction of vector d F can be found, according to (111.1), by general rules vector product, which follows left hand rule: if the palm of the left hand is positioned so that the vector enters it IN, and place four extended fingers in the direction of the current in the conductor, then the bent thumb will show the direction of the force acting on the current.
Ampere force modulus (see (111.1)) is calculated by the formula
Where a-angle between vectors d l And IN.
Ampere's law is used to determine the strength of interaction between two currents. Consider two infinite rectilinear parallel current I 1 and I 2 ; (directions of currents are indicated in Fig. 167), the distance between them is R. Each of the conductors creates a magnetic field, which acts according to Ampere's law on the other conductor with current. Let's consider the strength with which the magnetic field of the current acts I 1 per element d l second conductor with current I 2 . Current I 1 creates a magnetic field around itself, the lines of magnetic induction of which are concentric circles. Vector direction B 1 is determined by the right screw rule, its module according to formula (110.5) is equal to
Direction of force d F 1, from which the field B 1 acts on section d l the second current is determined by the left-hand rule and is indicated in the figure. The force modulus, according to (111.2), taking into account the fact that the angle a between current elements I 2 and vector B 1 straight line, equal to
substituting the value for IN 1 ,
we get 
Arguing in a similar way, it can be shown that sapa d F 2 with which the magnetic field of the current I 2 acts on element d l first conductor with current I 1, directed in the opposite direction and modulo equal to
Comparison of expressions (111.3) and (111.4) shows that
i.e. two parallel currents of the same direction attract each other with by force
(111.5)
If currents have opposite directions, then, using the left-hand rule, we can show that between them there is repulsive force, defined by formula (111.5).
Biot-Savart-Laplace law.
An electric field acts on both stationary and moving electric charges in it. Key Feature magnetic field is that it acts only for moving ones There are electric charges in this field. Experience shows that the nature of the effect of a magnetic field on a current varies depending on the shape of the conductor through which the current flows, the location of the conductor and the direction of the current. Therefore, in order to characterize a magnetic field, it is necessary to consider its effect on a certain current. Biot-Savart-Laplace law for a current-carrying conductor I, element d l which creates at some point A(Fig. 164) field induction d B, is written in the form 
where d l- vector, modulo equal to length d l conductor element and coincident in direction with the current, r-radius vector drawn from element d l guide to the point A fields, r- radius vector module r. Direction d B perpendicular to d l And r, i.e. perpendicular to the plane in which they lie, and coincides with the tangent to the line of magnetic induction. This direction can be found by the rule for finding magnetic induction lines (right screw rule): the direction of rotation of the screw head gives the direction d B, If forward movement screw corresponds to the direction of current in the element.
Vector modulus d B is determined by the expression 
(110.2)where a is the angle between vectors d l And r.
For a magnetic field, as well as for an electric one, it is true superposition principle: the magnetic induction of the resulting field created by several currents or moving charges is equal to the vector sum of the magnetic induction of the added fields created by each current or moving charge separately:
Calculation of magnetic field characteristics ( IN And N) according to the given formulas is generally complex. However, if the current distribution has a certain symmetry, then the application of the Biot-Savart-Laplace law together with the superposition principle makes it possible to simply calculate specific fields. Let's look at two examples.
1. Direct current magnetic field- current flowing through a thin straight wire of infinite length (Fig. 165). At an arbitrary point A, remote from the axis of the conductor at a distance R, vectors d B from all elements of the current have the same direction, perpendicular to the plane drawing (“to you”). Therefore, the addition of vectors d B can be replaced by adding their modules. For the integration constant we choose the angle a(angle between vectors d l And r), expressing all other quantities through it. From Fig. 165 it follows that 
(arc radius CD due to the smallness of d l equals r, and angle FDC for the same reason it can be considered direct). Substituting these expressions into (110.2), we find that the magnetic induction created by one element of the conductor is equal to
(110.4)
Since the angle a for all forward current elements varies from 0 to p, then, according to (110.3) and (110.4),
Consequently, the magnetic induction of the forward current field
(110.5)
2. Magnetic field in the center of a circular conductor with current(Fig. 166). As follows from the figure, all elements of a circular conductor with current create magnetic fields in the center of the same direction - along the normal from the turn. Therefore, the addition of vectors d B can be replaced by adding their modules. Since all conductor elements are perpendicular to the radius vector (sin a=1) and the distance of all conductor elements to the center of the circular current is the same and equal R, then, according to (110.2),
Consequently, the magnetic induction of the field in the center of a circular conductor with current
The Ampere force is the force with which a magnetic field acts on a conductor carrying current placed in this field. The magnitude of this force can be determined using Ampere's law. This law defines an infinitesimal force for an infinitely small section of a conductor. This makes it possible to apply this law to conductors of various shapes.
Formula 1 - Ampere's Law
B induction of a magnetic field in which a current-carrying conductor is located
I current strength in the conductor
dl infinitesimal element of the length of a conductor carrying current
alpha the angle between the induction of the external magnetic field and the direction of the current in the conductor
The direction of Ampere's force is found according to the left-hand rule. The wording of this rule is as follows. When the left hand is positioned in such a way that the lines of magnetic induction of the external field enter the palm, and four extended fingers indicate the direction of current movement in the conductor, while the thumb bent at a right angle will indicate the direction of the force that acts on the conductor element.
Figure 1 - left hand rule
Some problems arise when using the left-hand rule if the angle between the field induction and the current is small. It is difficult to determine where the open palm should be. Therefore, to simplify the application of this rule, you can position your palm so that it includes not the magnetic induction vector itself, but its module.
From Ampere's law it follows that Ampere's force will be equal to zero if the angle between the line of magnetic induction of the field and the current is equal to zero. That is, the conductor will be located along such a line. And the Ampere power will have a maximum possible meaning for this system, if the angle is 90 degrees. That is, the current will be perpendicular to the magnetic induction line.
Using Ampere's law, you can find the force acting in a system of two conductors. Let's imagine two infinitely long conductors that are located at a distance from each other. Currents flow through these conductors. The force acting from the field created by conductor with current number one on conductor number two can be represented as:
Formula 2 - Ampere force for two parallel conductors.
The force exerted by conductor number one on the second conductor will have the same form. Moreover, if currents in conductors flow in one direction, then the conductor will be attracted. If in opposite directions, then they will repel each other. There is some confusion, because the currents flow in one direction, so how can they attract each other? After all, like poles and charges have always repelled. Or Amper decided that it was not worth imitating the others and came up with something new.
In fact, Ampere did not invent anything, since if you think about it, the fields created by parallel conductors are directed counter to each other. And why they are attracted, the question no longer arises. To determine in which direction the field created by the conductor is directed, you can use the right-hand screw rule.
Figure 2 - Parallel conductors with current
Using parallel conductors and the Ampere force expression for them, the unit of one Ampere can be determined. If identical currents of one ampere flow through infinitely long parallel conductors located at a distance of one meter, then the interaction force between them will be 2 * 10-7 Newton for each meter of length. Using this relationship, we can express what one Ampere will be equal to.
This video shows how a constant magnetic field created by a horseshoe magnet affects a current-carrying conductor. The role of a conductor carrying current in in this case performed by an aluminum cylinder. This cylinder rests on copper bars through which electric current is supplied to it. The force acting on a current-carrying conductor in a magnetic field is called the Ampere force. The direction of action of the Ampere force is determined using the left-hand rule.
The magnetic field has an orienting effect on the current-carrying frame. Consequently, the torque experienced by the frame is the result of the action of forces on its individual elements. Summarizing the results of a study of the effect of a magnetic field on various conductors with current, Ampere established that the force d with which the magnetic field acts on an element of a conductor d with current located in a magnetic field is directly proportional to the current strength I in the conductor and the vector product of an element of length d of the conductor for magnetic induction:
The direction of the vector d can be found, according to (3.3.1), using the general rules of the vector product, from which the rule of the left hand follows: if the palm of the left hand is positioned so that the vector enters it, and the four outstretched fingers are positioned in the direction of the current in the conductor, then the bent thumb will show the direction of the force acting on the current.
Ampere force modulus is calculated by the formula
Where a-angle between vectors d and .
Ampere's law is used to determine the strength of interaction between two currents. Let's consider two infinite rectilinear parallel currents I 1 and I 2 (the directions of the currents are indicated in Fig. 3.3.2), the distance between which is equal to R.
Each of the conductors creates a magnetic field, which acts according to Ampere's law on the other conductor with current. Let's consider the force with which the magnetic field of current I 1 acts on the element dl of the second conductor with current 1 2.
Current I 1 creates a magnetic field around itself, the lines of magnetic induction of which are concentric circles. The direction of the vector is given by the rule of the right screw, its module according to formula (3.3.2) is equal to
The direction of the force d 1 with which field 1 acts on the section dl of the second current is determined by the left-hand rule and is indicated in Fig. 3.3.1. Force module,
according to (3.3.2), taking into account the fact that the angle a between the current elements 1 2 and the vector
1 straight line is equal to
or, substituting the values for B 1, we get
Reasoning similarly, it can be shown that the force dF 2 with which the magnetic field of current I 2 acts on the element dl of the first conductor with current I 1 is directed in the opposite side and modulo equal to
The force of interaction between parallel currents. Ampere's law
If we take two conductors with electric currents, then they will attract each other if the currents in them are in the same direction and repel if the currents flow in opposite directions. The interaction force per unit length of the conductor, if they are parallel, can be expressed as:
where $I_1(,I)_2$ are the currents that flow in the conductors, $b$ is the distance between the conductors, $in the SI system (\mu )_0=4\pi \cdot (10)^(- 7)\frac(H)(m)\(Henry\per\meter)$ magnetic constant.
The law of interaction of currents was established in 1820 by Ampere. Based on Ampere's law, current units are established in the SI and SGSM systems. Since an ampere is equal to the strength of a direct current, which, when flowing through two parallel infinitely long straight conductors of an infinitely small circular cross-section, located at a distance of 1 m from each other in a vacuum, causes an interaction force of these conductors equal to $2\cdot (10)^(-7)N $ per meter of length.
Ampere's law for a conductor of arbitrary shape
If a current-carrying conductor is in a magnetic field, then each current carrier is acted upon by a force equal to:
where $\overrightarrow(v)$ is the speed of thermal movement of charges, $\overrightarrow(u)$ is the speed of their ordered movement. From the charge, this action is transferred to the conductor along which the charge moves. This means that a force acts on a current-carrying conductor that is in a magnetic field.
Let us choose a conductor element with a current of length $dl$. Let's find the force ($\overrightarrow(dF)$) with which the magnetic field acts on the selected element. Let us average expression (2) over the current carriers that are in the element:
where $\overrightarrow(B)$ is the magnetic induction vector at the location point of element $dl$. If n is the concentration of current carriers per unit volume, S is the cross-sectional area of the wire at a given location, then N is the number of moving charges in the element $dl$, equal to:
Let's multiply (3) by the number of current carriers, we get:
Knowing that:
where $\overrightarrow(j)$ is the current density vector, and $Sdl=dV$, we can write:
From (7) it follows that the force acting on a unit volume of the conductor is equal to the force density ($f$):
Formula (7) can be written as:
where $\overrightarrow(j)Sd\overrightarrow(l)=Id\overrightarrow(l).$
Formula (9) Ampere's law for a conductor of arbitrary shape. The Ampere force modulus from (9) is obviously equal to:
where $\alpha $ is the angle between the vectors $\overrightarrow(dl)$ and $\overrightarrow(B)$. The Ampere force is directed perpendicular to the plane in which the vectors $\overrightarrow(dl)$ and $\overrightarrow(B)$ lie. The force that acts on a wire of finite length can be found from (10) by integrating over the length of the conductor:
The forces that act on conductors carrying currents are called Ampere forces.
The direction of the Ampere force is determined by the rule of the left hand (The left hand must be positioned so that the field lines enter the palm, four fingers are directed along the current, then the thumb bent by 900 will indicate the direction of the Ampere force).
Example 1
Assignment: A straight conductor of mass m of length l is suspended horizontally on two light threads in a uniform magnetic field, the induction vector of this field has a horizontal direction perpendicular to the conductor (Fig. 1). Find the current strength and its direction that will break one of the threads of the suspension. Field induction B. Each thread will break under load N.
To solve the problem, let’s depict the forces that act on the conductor (Fig. 2). Let us consider the conductor to be homogeneous, then we can assume that the point of application of all forces is the middle of the conductor. In order for the Ampere force to be directed downwards, the current must flow in the direction from point A to point B (Fig. 2) (In Fig. 1 the magnetic field is shown directed towards us, perpendicular to the plane of the figure).
In this case, we write the equilibrium equation of forces applied to a conductor with current as:
\[\overrightarrow(mg)+\overrightarrow(F_A)+2\overrightarrow(N)=0\ \left(1.1\right),\]
where $\overrightarrow(mg)$ is the force of gravity, $\overrightarrow(F_A)$ is the Ampere force, $\overrightarrow(N)$ is the reaction of the thread (there are two of them).
Projecting (1.1) onto the X axis, we get:
The Ampere force module for a straight final conductor with current is equal to:
where $\alpha =0$ is the angle between the magnetic induction vectors and the direction of current flow.
Substitute (1.3) into (1.2) and express the current strength, we get:
Answer: $I=\frac(2N-mg)(Bl).$ From point A and point B.
Example 2
Task: A conductor in the form of half a ring of radius R flows D.C. force I. The conductor is in a uniform magnetic field, the induction of which is equal to B, the field is perpendicular to the plane in which the conductor lies. Find the Ampere force. Wires that carry current outside the field.
Let the conductor be in the plane of the drawing (Fig. 3), then the field lines are perpendicular to the plane of the drawing (from us). Let us select an infinitesimal current element dl on the semiring.
The current element is acted upon by an Ampere force equal to:
\\ \left(2.1\right).\]
The direction of force is determined by the left-hand rule. Let us select the coordinate axes (Fig. 3). Then the force element can be written through its projections ($(dF)_x,(dF)_y$) as:
where $\overrightarrow(i)$ and $\overrightarrow(j)$ are unit vectors. Then we find the force that acts on the conductor as an integral over the length of the wire L:
\[\overrightarrow(F)=\int\limits_L(d\overrightarrow(F)=)\overrightarrow(i)\int\limits_L(dF_x)+\overrightarrow(j)\int\limits_L((dF)_y)\ left(2.3\right).\]
Due to symmetry, the integral $\int\limits_L(dF_x)=0.$ Then
\[\overrightarrow(F)=\overrightarrow(j)\int\limits_L((dF)_y)\left(2.4\right).\]
Having examined Fig. 3, we write that:
\[(dF)_y=dFcos\alpha \left(2.5\right),\]
where, according to Ampere’s law for the current element, we write that
By condition $\overrightarrow(dl)\bot \overrightarrow(B)$. Let us express the length of the arc dl through the radius R angle $\alpha $, we obtain:
\[(dF)_y=IBRd\alpha cos\alpha \ \left(2.8\right).\]
Let us carry out integration (2.4) for $-\frac(\pi )(2)\le \alpha \le \frac(\pi )(2)\ $substituting (2.8), we obtain:
\[\overrightarrow(F)=\overrightarrow(j)\int\limits^(\frac(\pi )(2))_(-\frac(\pi )(2))(IBRcos\alpha d\alpha ) =\overrightarrow(j)IBR\int\limits^(\frac(\pi )(2))_(-\frac(\pi )(2))(cos\alpha d\alpha )=2IBR\overrightarrow(j ).\]
Answer: $\overrightarrow(F)=2IBR\overrightarrow(j).$