Let's apply Ampere's law to calculate the force of interaction between two long straight conductors with currents I 1 and I 2 located at a distance d from each other (Fig. 6.26).
Rice. 6.26. Power interaction of rectilinear currents:
1 - parallel currents; 2 - antiparallel currents
Current carrying conductor I 1 creates a ring magnetic field, the magnitude of which at the location of the second conductor is equal to
This field is directed “away from us” orthogonally to the plane of the drawing. The element of the second conductor experiences the action of the Ampere force from the side of this field
Substituting (6.23) into (6.24), we get
|
|
With parallel currents the strength F 21 is directed towards the first conductor (attraction), with antiparallel - in reverse side(repulsion).
Similarly, conductor element 1 is affected by the magnetic field created by the current-carrying conductor I 2 at a point in space with an element with force F 12 . Reasoning in the same way, we find that F 12 = –F 21, that is, in this case Newton’s third law is satisfied.
So, the interaction force of two straight infinitely long parallel conductors, calculated per element of the length of the conductor, is proportional to the product of the current forces I 1 and I 2 flowing in these conductors, and is inversely proportional to the distance between them. In electrostatics, two long charged threads interact according to a similar law.
In Fig. 6.27 presents an experiment demonstrating attraction parallel currents and repulsion of antiparallel ones. For this purpose, two aluminum strips are used, suspended vertically next to each other in a slightly tensioned state. When parallel direct currents of about 10 A are passed through them, the ribbons are attracted. and when the direction of one of the currents changes to the opposite, they repel.
Rice. 6.27. Force interaction of long straight conductors with current
Based on formula (6.25), the unit of current is established - ampere, which is one of the basic units in SI.
Example. Along two thin wires, bent in the form of identical rings with a radius R= 10 cm, equal currents flow I= 10 A each. The planes of the rings are parallel, and the centers lie on a line orthogonal to them. The distance between centers is d= 1 mm. Find the forces of interaction between the rings.
Solution. In this problem it should not be confusing that we only know the law of interaction of long straight conductors. Since the distance between the rings is much less than their radius, the interacting elements of the rings “do not notice” their curvature. Therefore, the interaction force is given by expression (6.25), where we must substitute the circumference of the rings. We then obtain
From here it is not difficult to obtain an expression for the magnetic field induction of each of the straight conductors. The magnetic field of a straight conductor carrying current must have axial symmetry and, therefore, closed lines of magnetic induction can only be concentric circles located in planes perpendicular to the conductor. This means that vectors B1 and B2 of magnetic induction of parallel currents I 1 and I 2 lie in a plane perpendicular to both currents. Therefore, when calculating the Ampere forces acting on current-carrying conductors, in Ampere’s law one must put sin α = 1. From the law of magnetic interaction of parallel currents it follows that the induction modulus B magnetic field of a straight conductor carrying current I on distance R from it is expressed by the relation
| |
In order for parallel currents to attract and antiparallel currents to repel during magnetic interaction, the magnetic induction field lines of a straight conductor must be directed clockwise when viewed along the conductor in the direction of the current. To determine the direction of vector B of the magnetic field of a straight conductor, you can also use the gimlet rule: the direction of rotation of the gimlet handle coincides with the direction of vector B if, during rotation, the gimlet moves in the direction of the current. The magnetic interaction of parallel conductors with current is used in International system units (SI) to determine the unit of current - ampere:
Magnetic induction vector- this is the main force characteristic of the magnetic field (denoted B).
Lorentz force- the force acting on one charged particle is equal to
|
Under the influence of the Lorentz force, electric charges in a magnetic field move along curvilinear trajectories. Let us consider the most typical cases of motion of charged particles in a uniform magnetic field.
a) If a charged particle enters a magnetic field at an angle α = 0°, i.e. flies along the field induction lines, then F l= qvBsma = 0. Such a particle will continue its movement as if there were no magnetic field. The particle trajectory will be a straight line.
b) Particle with charge q enters a magnetic field so that the direction of its velocity v is perpendicular to the induction ^B magnetic field (Figure - 3.34). In this case, the Lorentz force provides centripetal acceleration a = v 2 /R and particle moves in a circle with radius R in a plane perpendicular to the magnetic field induction lines. under the influence of the Lorentz force : F n = qvB sinα, Taking into account that α = 90°, we write the equation of motion of such a particle: t v 2 /R= qvB. Here m- particle mass, R– radius of the circle along which the particle moves. Where can you find the relationship? e/m- called specific charge, which shows the charge per unit mass of the particle.
c) If a charged particle flies in at a speed v 0 into a magnetic field at any angle α, then this movement can be represented as complex and decomposed into two components. The trajectory of movement is a helical line, the axis of which coincides with the direction IN. The direction in which the trajectory twists depends on the sign of the particle's charge. If the charge is positive, the trajectory spins counterclockwise. The trajectory along which a negatively charged particle moves spins clockwise (it is assumed that we are looking at the trajectory along the direction IN; the particle flies away from us.
Let us determine the force with which conductors with currents I 1 and I 2 interact (attract or repel) (Fig. 3.19)
The interaction of currents occurs through a magnetic field. Each current creates a magnetic field that acts on another wire (current).
Let us assume that both currents I 1 and I 2 flow in the same direction. Current I 1 creates at the location of the second wire (with current I 2) a magnetic field with induction B 1 (see 3.61), which acts on I 2 with force F:
(3.66)
Using the left-hand rule (see Ampere's law), we can establish:
a) parallel currents of the same direction attract;
b) parallel currents of opposite directions repel;
c) non-parallel currents tend to become parallel.
Circuit with current in a magnetic field. Magnetic flux
Let there be a contour of area S in a magnetic field with induction B, the normal 
to which makes an angle α with the vector 
(Fig. 3.20). To calculate the magnetic flux Ф, we divide the surface S into infinitesimal elements so that within one element dS the field can be considered homogeneous. Then the elementary magnetic flux through an infinitely small area dS will be:
where B n is the projection of the vector 
to normal 
.
If the area dS is located perpendicular to the magnetic induction vector, then α = 1, cos α = 1 and dФ = BdS;
The magnetic flux through an arbitrary surface S is equal to:
If the field is uniform and the surface S is flat, then the value B n =const and:
(3.67)
For flat surface, located along a uniform field, α = π/2 and Ф = 0. The induction lines of any magnetic field are closed curves. If there is a closed surface, then the magnetic flux entering this surface and the magnetic flux leaving it are numerically equal and opposite in sign. Therefore, the magnetic flux through an arbitrary closed surface is zero:
(3.68)
Formula (3.68) is Gauss's theorem for the magnetic field, reflecting its vortex character.
Magnetic flux is measured in Webers (Wb): 1Wb = T m 2 .
The work of moving a conductor and a current-carrying circuit in a magnetic field
If a conductor or a closed circuit with a current I moves in a uniform magnetic field under the action of the Ampere force, then the magnetic field does work:
A=IΔФ, (3.69)
where ΔФ is the change in magnetic flux through the contour area or the area described by a straight conductor when moving.
If the field is non-uniform, then:
.
The phenomenon of electromagnetic induction. Faraday's law
The essence of the phenomenon electromagnetic induction is as follows: with any change in the magnetic flux through the area limited by a closed conducting loop, an E.M.F. arises in the latter. and, as a consequence, an inductive electric current.
Induction currents always counteract the process that causes them. This means that the magnetic field they create tends to compensate for the change in magnetic flux that this current caused.
It has been experimentally established that the value of E.M.F. induction ε i induced in the circuit depends not on the magnitude of the magnetic flux Ф, but on the rate of its change dФ/dt through the area of the circuit:
(3.70)
The minus sign in formula (3.70) is a mathematical expression Lenz's rules: the induced current in the circuit always has such a direction that the magnetic field it creates prevents the change in magnetic flux that causes this current.
Formula (3.70) is an expression of the basic law of electromagnetic induction.
Using formula (3.70), we can calculate the strength of the induction current I, knowing the circuit resistance R, and the amount of charge Q, passed during time t in the circuit:
If a segment of a straight conductor of length ℓ moves at a speed V in a uniform magnetic field, then the change in magnetic flux is taken into account through the area described by the segment during movement, i.e.
Faraday's law can be derived from the law of conservation of energy. If a current-carrying conductor is in a magnetic field, then the work of the current source εIdt for time dt will be spent on Lenz-Joule heat (see formula 3.48) and the work of moving the conductor in the field IdФ (see 3.69) can be determined:
εIdt=I 2 Rdt+IdФ (3.71)
Then 
,
Where 
and is the induced emf (3.70)
those. when Ф changes in the circuit, an additional emf ε i arises in accordance with the law of conservation of energy.
It can also be shown that ε i arises in a metal conductor due to the action of the Lorentz force on electrons. However, this force does not act on stationary charges. Then we have to assume that the alternating magnetic field creates an electric field, under the influence of which an induced current I i arises in a closed circuit.
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.
 |
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, a force acts on the conductor I 2 from the magnetic field
, 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 lines of magnetic induction enter into it, and the four outstretched fingers are positioned in the direction 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.
 |
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.
One of the manifestations of a magnetic field is its forceful effect on a current-carrying conductor placed in a magnetic field. Ampere established that a current-carrying conductor placed in a uniform magnetic field whose induction is acted upon by a force proportional to the current strength and the magnetic field induction:
F = IBℓsinα (15.22)
[α is the angle between the direction of the current in the conductor and the magnetic field induction].
This formula turns out to be valid for a straight conductor and a uniform field.
If the conductor has an arbitrary shape and the field is inhomogeneous, then expression (3.125) takes the form
dF = IBdℓsinα (15.23)
or in vector form
(15.24)
The product Idℓ is called the current element. Relations (15.23), (15.24) express Ampere's law.
To determine the direction of the force acting on a current-carrying conductor placed in a magnetic field, it is used left hand rule: if the left hand is positioned so that the lines of magnetic induction enter the palm, and the extended four fingers coincide with the direction of the current in the conductor, then the bent thumb will indicate the direction of the force acting on the current-carrying conductor placed in a magnetic field(Fig. 15.10) .
This force is always perpendicular to the plane in which the conductor and the vector lie. Knowing the direction and magnitude of the force acting on any section dℓ of the conductor, we can calculate the force acting on the entire conductor. To do this, you need to find the sum of the forces acting on all
conductor sections:
Using Ampere's law, consider interaction of parallel conductors with current (Fig. 15.11). Let us assume that in a homogeneous isotropic medium, the relative magnetic permeability of which is μ, two conductors are located at a distance d from each other. Let the current I 1 flow through one of them and the other - I 2 in the water direction.
Select element dℓ 2 on conductor 2. This element will be acted upon by the Ampere force
dF i = B 1 I 2 dℓ i
[ 
- induction of the magnetic field created by the first conductor at the location of the second conductor].
The vector is directed perpendicular to the direction of current I, therefore sinα=1. Taking this into account, we find
(15.25)
Using the left hand rule, we determine the direction of this force. To determine the force F 12, i.e. the force acting from conductor 1 on conductor 2, you need to sum up all the elementary forces dF i
The force with which two conductors interact is proportional to the product of the currents flowing through the conductors and inversely proportional to the distance between them.
If currents flow through conductors in the same directions, then conductors attract, and in opposite directions they repel.
Ampere's law is fundamental in the doctrine of magnetism and plays the same role as Coulomb's law in electrostatics.
15.5 Circuit with current in a magnetic field. Work on moving a conductor and a current-carrying circuit in a magnetic field
A current-carrying circuit with sides a and ℓ is placed in a magnetic field
(Fig. 15.12). An Ampere force acts on each side of the circuit. Forces act on the horizontal sides ℓ of the contour that stretch or compress) the contour without rotating it.
Each of the vertical sides a is acted upon by a force F = IBa. These forces create a couple of forces whose moment
M = Fℓcosφ (15.27)
[φ is the angle between the vector and the side of the contour ℓ.
The moment of force tends to rotate the contour so that the flow F penetrating the contour is maximum. Substituting the expression for force into formula (15.27), we have
М = IBaℓcosφ= ISBcosφ= p m Bcos(π/2-α)= = p m B sinα (15.28)
The quantity IS is called magnetic moment of the circuit p m.. Vector p m coincides with the direction of the positive normal to the contour plane.
Mechanical moment M, acting on a circuit with current in a uniform magnetic field is proportional to the magnetic moment p m of the circuit, the induction B of the magnetic field and the sine of the angle between the direction of the vectors p m (normal to the circuit) and .
In vector form, relation (15.28) has the form
M = (15.29)
Consider a conductor of length ℓ with current I, placed in a uniform external magnetic field, perpendicular to the plane contour and which can move freely in this field under the action of the Ampere force (Fig. 15.13).
Under the influence of this force, the conductor will move parallel to itself for a segment from position 1 to position 2. The work done by the magnetic field is equal to
dA=Fdx=IBℓdx=IBdS=IdФ, (15.30)
since ℓdx = dS is the area crossed by the conductor when it moves in a magnetic field, VdS = dФ is the flux of the magnetic induction vector penetrating this area. Thus,
dA= IdФ, (15.31)
those. the work done to move a current-carrying conductor in a magnetic field is equal to the product of the current strength and the magnetic flux crossed by the moving conductor.
Work on moving a conductor with current I from point 1 to point 2 is determined by the formula:
(15.32)
The work of moving a closed loop with current in a magnetic field is also determined by the formula. The formula remains valid for a circuit of any shape in an arbitrary magnetic field.
§ 15.5. Lorentz force. Motion of a particle in a magnetic field. Hall effect
Moving electric charges create a magnetic field around themselves, which propagates in a vacuum at the speed of light. When a charge moves in an external magnetic field, a force interaction of magnetic fields occurs, determined by Ampere’s law. The process of interaction of magnetic fields was studied by Lorentz, who derived a formula for calculating the force exerted by a magnetic field on a moving charged particle. Lorentz is the creator of classical electronic theory. His works in the fields of electrodynamics, thermodynamics, static mechanics, optics, radiation theory, and atomic physics are widely known. For his studies of the influence of magnetism on radiation processes, he was awarded the Nobel Prize in 1902.
The force exerted by a magnetic field on a moving charge is called Lorentz force And , is equal
F l = qυВ sinα (15.33)
where q is the particle charge; - particle speed; B is the magnetic field induction, α is the angle between the direction of the particle velocity and the magnetic induction vector .
This force is perpendicular to the vectors and.
The direction of the Lorentz force is determined according to the left hand rule: if you position your left palm so that the four outstretched fingers indicate the direction of movement of the positive charge, and the magnetic field vector enters the palm, then the outstretched thumb will show the direction of the Lorentz force acting on this charge.
With a change in the sign of the charge, the direction of the force changes to the opposite.
Analyzing expression (3.146), we can draw the following conclusions:
1. If charge speed =0; F l =0. A magnetic field does not act on a stationary particle.
2. If a particle flies into a magnetic field parallel to its lines of force. α=0°, sin0°=0; F l =0. A magnetic field does not act on a stationary charged particle; The particle will continue to move uniformly and in a straight line at the same speed that it had.
3. If the particle flies in perpendicular to the magnetic field lines ┴ . α=90°, sin90°=1; F l =qυV. The Lorentz force bends the trajectory of motion, acting as a centripetal force.
It is very important to use this phenomenon in the study of cosmic particles to determine the sign of the charge. The entry of a flying particle into a magnetic field causes a change in its trajectory depending on the sign of the charge (Fig. 3.59). In Fig. 3.59, the magnetic field induction vector is directed perpendicular to the plane of the drawing (away from us). The particle will move in a circle, the radius R of which can be determined from the equality of the centripetal force and the Lorentz force:
The greater the speed of a particle, the larger the radius of the circle along which it moves, but the period of revolution does not depend on either the speed or the radius of the circle.
(15.36)
4. If a particle moves at an angle β to the lines, then the trajectory of the particle will be a helical line (spiral) enclosing the magnetic field lines (Fig. 3.60).
The pitch h of the spiral is determined by υ t - the tangential component of the velocity υ of the particle. The radius of the spiral depends on υ n - the normal component of velocity υ.
In 1892, Lorentz obtained a formula for the force with which an electromagnetic field acts on any charged particle located in it:
(15.37)
This force is called electromagnetic Lorentz force , and this expression is one of the basic laws of classical electrodynamics.
When electric charge moves simultaneously in electric and magnetic fields, then the resulting force acting on the particle is equal to
F = qυВsinα+ qE (15.38)
In this case, the force has two components: from the influence of magnetic and electric fields. There is a fundamental difference between these components. An electric field changes the velocity and, consequently, the kinetic energy of a particle; a uniform magnetic field changes only the direction of its movement.
Hall effect
The American scientist E. Hall discovered that in a conductor placed in a magnetic field, a potential difference (transverse) arises in the direction perpendicular to the magnetic induction vector B and current I, due to the action of the Lorentz force on charges moving in this conductor (Fig. 3.62) .
Experience shows that the transverse potential difference is proportional to the current density j, magnetic induction and distance d between the electrodes:
Let us assume that electrons move with an ordered average speed υ and each electron is acted upon by a Lorentz force equal to eBυ. Under its action, the electrons are displaced so that one of the faces of the sample will be charged negatively, the other - positively, and an electric field will arise inside the sample, i.e. υ B = eE.
Therefore, the transverse potential difference is equal to
Average speed υ electrons can be expressed in terms of current density j, since j=ne υ , That's why
Equating this expression to formula (15.39), we obtain .
The Hall constant depends on the electron concentration.
Using the measured value of the Hall constant, you can: 1) determine the concentration of current carriers in the conductor (at famous character conductivity and charge of carriers); 2) judge the nature of the conductivity of semiconductors, since the sign of the Hall constant coincides with the sign of the charge of current carriers. Used for multiplying direct currents in analog computers, in measuring technology(Hall Sensor
Examples of problem solving
Example.A rectangular frame with sides a = 5 cm and b = 10 cm, consisting of N = 20 turns, is placed in an external uniform magnetic field with induction B = 0.2 Tesla. The normal to the frame makes an angle with the direction of the magnetic field. Determine the torque acting on the frame if a current I=2A flows through it.
Given: a= 5cm=0.05m; b=10cm=0.1m; N=20; B=0.2 T; . ; I=2A.
Find: M.
Solution.The mechanical torque acting on a current-carrying frame placed in a uniform magnetic field is
,
- magnetic moment of the frame with current. Module M=p m Bsinα.
Since the frame consists of N turns, then M=Np m Bsinα (1)
where is the magnetic moment of the frame with current
p m =IS=I a b. (2)
Substituting formula (2) into expression (1), we find the required torque
M=NIB a bsinα.
Answer: M=0.02 N∙m
Example.Current flows through a thin wire ring. Determine how many times the induction in the center of the circuit will change if the conductor is given the shape of a square without changing the current strength in the conductor.
Solution. The vector in the center of the circular current is directed at the selected direction of the current (see figure), according to the rule of the right screw, perpendicular to us in the drawing (in the figure this is indicated by a dot in a circle). Its module
where I is the current strength; R is the radius of the ring; μ 0 - magnetic constant; μ is the magnetic permeability of the medium.
The side of the square inscribed in the ring is equal to (the circumference of the ring is 2πR). The vector in the center of the square is also directed perpendicular to the drawing towards us. The magnetic induction at the center of the square is equal to the sum of the magnetic inductions created by each side of the square. Then the module, according to the Biot-Savart-Laplace law,
From formulas (1) and (2) we obtain the relation
Answer:
Example.Two infinitely long straight parallel conductors located in a vacuum at a distance of R = 30 cm carry identical currents in the same direction. Determine the magnetic induction B of the field created by the currents at point A, lying on the straight line connecting the conductors and lying at a distance r=20cm to the right of the right wire (see figure). The current strength in the conductors is 20A.
Given:μ=1; R=30cm=0.3m; r=20cm=0.2m; I 1 = I 2 =I=20 A.
Find:B.
Solution. Let the currents be directed perpendicular to the drawing plane from 
us, which is indicated by crosses in the figure. Magnetic induction lines are closed and surround conductors with currents. Their direction is determined by the right screw rule. The vector at each point is directed tangentially to the magnetic induction line (see figure).
According to the principle of superposition, the magnetic induction of the resulting field at point A
where and is the magnetic induction of fields at this point created by the first and second conductors. Vectors and and are codirectional, so the addition of vectors can be replaced by the addition of their modules
B=B 1 +B 2. (1)
Magnetic induction of fields created by infinitely long straight conductors with current I 1 and I 2,
, 
(2)
where μ 0 – magnetic constant; μ is the magnetic permeability of the medium.
Substituting expression (2) into formula (1) and taking into account that I 1 =I 2 =I and μ=1 (for vacuum), we obtain the desired expression for magnetic induction at point A:
Answer: V=28 µT.
Example.Through two infinitely long straight parallel conductors located in a vacuum, the distance between which is d=15cm, currents I 1 =70A and I 2 =50A flow in one direction. Determine the magnetic induction B of the field at point A, located at a distance of r 1 = 10 cm from the first and r 1 = 20 cm from the second conductor.
Given:μ=1; d=15cm=0.15 m; I 1 =70A; I 2 =50A; r 1 =10cm=0.1m; r 2 =20cm=0.2m.
Find:B.
Solution. Let the currents be directed perpendicular to the drawing plane towards us. The magnetic induction vectors are directed tangentially to the magnetic induction lines.
According to the principle of superposition, magnetic induction at point A (see figure)
where and are, respectively, the magnetic induction fields created by current-carrying conductors I 1 and I 2(directions of vectors and currents I 1 and I 2 shown in the figure). The modulus of the vector according to the cosine theorem,
.
Substituting these expressions into formula (1), we find the required B:
.
Answer: V=178 µT.
Example.In the same plane with an infinitely straight current-carrying conductor
I = 10 A there is a rectangular wire frame (side a = 25 cm, b = 10 cm), through which current I 1 = 2 A flows. The long sides of the frame are parallel to the forward current, and the nearest of them is located from the forward current at a distance of c = 10 cm and the current in it is codirectional to the current I. Determine the forces acting on each side of the frame.
Given:I=10A; a=25cm=0.25m; b=10 cm=0.10 m;; I 1 =2 A; c=10cm=0.1m.
Find: F 1 ; F2; F 3; F 4;
Solution.
A rectangular frame is in a non-uniform direct current field with induction
(we consider the case of vacuum), where r is the distance from the direct current to the point in question.
The force with which the direct current field acts can be found by summing up the elementary forces determined by Ampere's law,
The vector within the frame is directed perpendicular to its plane beyond the drawing, and within each side the angle is . This means that within one side the elementary forces are parallel to each other and the addition of vectors
Can be replaced by adding their modules:
(2)
where integration is carried out over the corresponding side of the frame
The short sides of the frame are located equally relative to the wire, and therefore the forces acting on them are numerically equal, but oppositely directed. Their direction, as well as the direction of other forces (see figure), is determined by the left-hand rule. Along each of the short sides of the rectangle, the magnetic induction changes [see. formula (1)]. Then, after performing the integration [taking into account (2)],
.
The long sides of the frame are parallel to the forward current, being at distances c and c+b from it, respectively. Then
;
,
where and 
.
Answer: F 1 =10 µN; F 2 =2.77 µN; F 3 =5 μN; F 4 =2.77 µN.
Example.An electron that has passed through an accelerating potential difference U=1 kV flies into a uniform magnetic field with induction B=3mT perpendicular to the magnetic induction lines. Determine: 1) the force acting on the electron; 2) the radius of the circle along which the electron moves; 3) the period of revolution of the electron.
Given: m=9.11∙10 -31 kg; e=1.6∙10 -19 C; U=1kV=1∙10 3 V; B=3mT=3∙10 -3 T; α=90º.
Find: 1)F; 2) R; 3) T.
Solution. When an electron moves in a magnetic field with a speed v, it is acted upon by the Lorentz force
F l =eυBsinα,
where α is the angle between vectors and (in our case α=90º). Then
When passing through an accelerating potential difference, the work of forces electrostatic field sends a message to the electron kinetic energy ,
Substituting expression (2) into formula (1), we find the desired force acting on the electron,
From mechanics it is known that constant force, perpendicular to the speed, and this is the Lorentz force (1), causes motion in a circle. It imparts normal acceleration to the electron, where R is the radius of the circle. According to Newton's second law, F=ma, where F=eυB. Then
where does the desired radius of the circle take into account (2)
Electron orbital period
Substituting expressions (3) and (2) into formula (4), we find the required electron circulation period
Answer: 1)F=9∙10 -15 N; 2) R=3.56 cm; 3) T=11.9 ns.
Example.A proton, having a speed of υ = 10 4 m/s, flies into a uniform magnetic field with induction B = 10 mT at an angle α = 60º to the direction of the magnetic induction lines. Determine the radius R and pitch h of the helix along which the proton will move.
Given: υ=10 4 m/s; e=1.6∙10 -19 C; m=1.67∙10 -27 kg; B=10mT=10∙10 -3 T; α=60º.
Find:R; h.
Solution. The movement of a proton in a uniform magnetic field with a speed directed at an angle α to the vector occurs along a helical line (see figure). To prove this, let us decompose the velocity vector into components parallel (υ x =υcosα) and perpendicular (υ y =υsinα) to the induction vector.
Movement in the direction of the field occurs with a uniform speed υ x, and in the direction perpendicular to the vector, under the influence of the Lorentz force - in a circle ( =const, υ x =const). As a result of the addition of two movements, the trajectory of the resulting proton movement is a helical line (spiral).
The Lorentz force imparts normal acceleration to the proton (R is the radius of the circle). According to Newton's second law, F=m a n, where F l =eυ y B – Lorentz force. Then
Where does the desired radius of the helical line along which the proton will move come from?
Helix pitch equal to the distance traversed by the proton along the x axis during one full revolution, i.e.
h=υ x T= υTcosα, (1)
where is the rotation period
(2)
Substituting formula (2) into expression (1), we find the desired pitch of the helix
Answer: R=9.04mm; h=3.28 cm.
Example.A uniform magnetic field with a strength of H=2kA/m is created between the plates of a flat capacitor located in a vacuum. The electron moves in the capacitor parallel to the capacitor plates and perpendicular to the direction of the magnetic field at a speed of υ=2 Mm/s. Determine the voltage U applied to the capacitor if the distance d between its plates is 1.99 cm.
Given: μ=1; N=2kA/m=2∙10 3 A/m; υ=2Mm/s=2∙10 6 m/s; d=1.99 cm=1.99∙10 -2 m).
Find:U.
Solution.
Let's assume that the magnetic field is directed perpendicular to the drawing away from us. What is indicated in the figure with crosses. An electron can move perpendicular to the direction of the magnetic field and parallel to the plates of the capacitor (with the selected direction of the magnetic field and charges on the plates) only as indicated in the figure. In this case, the Coulomb force (Y is the electric field strength) is balanced by the Lorentz force F l =eυB (its direction is determined by the left-hand rule). Then
Formula expressing the relationship between magnetic induction and magnetic field strength
For the case of vacuum (μ=1) it has the form B=μ 0 N. Substituting this formula into expression (1), we find the desired voltage on the capacitor plates
Answer: U=100B.
Example.A current I = 5 A is passed through a cross-section of a copper plate (copper density ρ = 8.93 g/cm 3) with a thickness of d = 0.1 mm. The plate with the current is placed in a uniform magnetic field with induction B = 0.5 T, perpendicular to the direction current and the edge of the plate. Determine the transverse (Hall) potential difference arising in the plate if the concentration of n free electrons is equal to the concentration of n" conductor atoms.
Given: ρ=8.93 g/cm 3 =8.93∙10 3 kg/m 3 ; d=0.1mm=1∙10 -4 m; I=5A; B=0.5 T; n = n" ; M=63.5∙10 -3 kg/mol.
Find:Δφ..
Solution. The figure shows a metal plate with current density in a magnetic field perpendicular (as in the problem statement). In this direction, the speed of current carriers in metals - electrons - is directed from right to left. Electrons experience the Lorentz force, which in this case directed upwards. At the upper edge of the plate there is an increased concentration of electrons (it will be charged negatively), and at the lower edge there is a lack of electrons (it will be charged positively). Therefore, an additional transverse electric field appears between the edges of the plate, directed from bottom to top.
In the case of a stationary distribution of charges in the transverse direction (the strength E B of the transverse field will reach such a value that its action on the charges will balance the Lorentz force)
or Δφ=υВα (1)
Where A– width of the plate; Δφ - transverse (Hall) potential difference.
Current strength
I=jS=neυS=neυ a d, (2)
where S is the cross-sectional area of the plate with thickness d; n - electron concentration; υ - average speed ordered movement of electrons.
Substituting (2) into (1), we get
According to the conditions of the problem, the concentration of free electrons is equal to the concentration of conductor atoms. Hence,
, (4)
where N A =6.02∙10 23 mol -1 – Avogadro’s constant; V m - molar volume copper; M – molar mass copper; ρ is its density.
Substituting formula (4) into expression (3), we find the desired
Example.Magnetic induction B on the axis of a toroid without a core (external diameter of the toroid d 1 = 60 cm, internal diameter d 2 = 40 cm), containing N = 200 turns, is 0.16 mT. Using the vector circulation theorem, determine the current strength in the toroid winding.
Given: d 1 =60 cm=0.6 m; d 2 =40 cm=0.4 m; N=200; B=0.16 mT=0.16∙10 -3 T.
Find: I.
Solution. Vector circulation
, (1)
those. equal to the algebraic sum of the currents covered by the circuit along which the circulation is calculated, multiplied by the magnetic constant. As a contour, we choose a circle located in the same way as the magnetic induction line, i.e. a circle of some radius r, the center of which lies on the axis
toroid. From the symmetry condition it follows that the modulus of the vector at all points of the magnetic induction line is the same, and therefore expression (1) can be written in the form
(2)
(take into account that the current strength in all turns is the same, and the circuit covers the number of currents, equal to the number turns of the toroid). For the center line of the toroid). For the centerline of a toroid. Substituting r into (2), we obtain the required current strength:
.
Answer: I=1 A
Example.In the same plane with an infinite straight wire through which a current I = 10A flows, there is a square frame with side a = 15 cm. Determine the magnetic flux Ф penetrating the frame if the two sides of the frame are parallel to the wire, and the distance d from the wire to the nearest side of the frame is 2 cm.
Given: I=10A; a=15 cm=0.15 m; d=2 cm=0.02m.
Find: F.
Solution.
Magnetic flux Ф through a surface area is calculated by the formula:
The square frame is in a non-uniform direct current field with induction
(we consider the case of vacuum), where x is the distance from the wire to the point in question.
The magnetic field is created by direct current (the direction is shown in the figure), and the vector is perpendicular to the plane of the frame (directed perpendicular to the drawing from us, which is shown in the figure by crosses), therefore for all points of the frame B n = B.
We divide the frame area into narrow elementary areas of width dx and area a dx (see figure), within which the magnetic induction can be considered constant. Then the flow through the elementary platform
. (1)
By integrating expression (1) in the range from to, we find the desired magnetic flux
.
Answer: Ф=0.25 µWb
Example.A circular conducting circuit with radius r=6cm and current I=2A is established in a magnetic field so that the plane of the circuit is perpendicular to the direction of a uniform magnetic field with induction B=10mT. Determine the work that must be done to slowly rotate the contour at an angle relative to the axis that coincides with the diameter of the contour.
Given: r=6 cm=0.06 m; I=2 A; B=10 mT=10∙10 -3 T; .
Find: A ext.
Solution. Work done by field forces to move a closed conductor with current I
A=I(F 2 -F 1), (1)
where Ф 1 and Ф 2 are the magnetic induction fluxes penetrating the circuits in the initial and final positions. We consider the current in the circuit to be constant, since when the circuit is slowly rotated in a magnetic field, induced currents can be neglected.
Flux of magnetic induction through a flat circuit of area S in a uniform magnetic field with induction B
where α is the angle between the normal vector to the contour surface and the magnetic induction vector.
In the initial position, Fig. a, circuit (the circuit is established freely), the flux of magnetic induction is maximum (α=0; cosα=1) and Ф 1 =BS (S-circuit area), and in final position, rice. b (; cosα=0), Ф 2 =0.
Then, substituting these expressions into formula (1), we find that
(take into account that the area of the circular contour is S=πr 2).
The work of external forces is directed against the field forces (equal to it in magnitude, but opposite in sign), therefore the required work
A in =πIBr 2 .
Answer: A vn=226 µJ.