The classification of any electrical wire includes the main parameters represented by conductivity, cross-sectional area or diameter, the materials from which the conductor is made, typical features of insulating protection, the level of flexibility, as well as indicators of thermal resistance.

The area or cross-section of a conductor is one of the most important criteria for choosing a wire.

The most widely used wire brands are PUNP and PUGNP, as well as VPP, PNCB and PKGM, which have the following main technical characteristics that are very important for obtaining a safe connection:

• PUNP- a flat wire product of an installation or so-called installation type, with single-wire copper conductors in PVC insulation. This variety is distinguished by the number of cores, as well as the rated voltage within 250 V with a frequency of 50 Hz and the operating temperature from minus 15 ° C to plus 50 ° C;
• PUGNP- flexible variety with stranded conductors. The main indicators, which are represented by the nominal voltage level, frequency and temperature operating conditions, do not differ from similar PUNP data;
• APB- aluminum single-core variety, round wire with protective PVC insulation and a single-wire or multi-wire core. The difference of this type is resistance to mechanical damage, vibrations and chemical compounds. The operating temperature ranges from minus 50 °C to plus 70 °C;
• PBC- stranded copper variety with PBX insulation, giving the wire high density and traditional rounded shape. The heat-resistant conductor is designed for a nominal level of 380 V at a frequency of 50 Hz;
• PKGM- power mounting variety, represented by a single-core copper wire with organosilicon rubber or fiberglass insulation impregnated with a heat-resistant composition. The operating temperature ranges from minus 60 °C to plus 180 °C;
• PCB- heating single-core version in the form of a single-wire wire based on galvanized or blued steel. The operating temperature ranges from minus 50 °C to plus 80 °C;
• WFP- a single-core copper variety with a multi-wire core and insulation based on PBX or polyethylene. The operating temperature ranges from minus 40 °C to plus 80 °C.

In conditions of low power, a copper wire ShBBP with protective external PBX insulation is used. The stranded type of core has excellent flexibility, and the wired product itself is rated for a maximum of 380 V, at a frequency within 50 Hz.

Wire products of the most common types are sold in bays, and most often have a white color of the insulation.

Conductor cross-sectional area

In recent years, there has been a noticeable decrease in the quality characteristics of manufactured cable products, as a result of which the resistance indicators - the wire cross section - suffer. The diameter of any conductor must necessarily comply with all parameters declared by the manufacturer.

Any deviation, even 15-20%, can cause significant overheating of electrical wiring or melting of the insulating material, so the choice of area or thickness of the conductor should be given increased attention not only in practice, but also from the point of view of theory.

Cross section of conductors

The parameters most important for the correct selection of the conductor section are reflected in the following recommendations:

• conductor thickness - sufficient for unhindered passage of electric current, with the maximum possible heating of the wire within 60 ° C;
• conductor cross-section - sufficient for a sharp decrease in voltage, not exceeding the permissible values, which is especially important for very long wiring and significant currents.

Particular attention must be paid to the maximum operating temperature, above which the conductor and protective insulation become unusable.

The cross section of the conductor used and its protective insulation must necessarily ensure full mechanical strength and reliability of electrical wiring.

Conductor Cross Section Formula

As a rule, wires have a circular cross section, but the current rating must be calculated according to the cross-sectional area. In order to independently determine the cross-sectional area in a single-core or stranded wire, the sheath, which is insulation, is carefully opened, after which the diameter is measured in a single-core conductor.

The area is determined in accordance with the well-known physical formula even for schoolchildren:

S = π x D²/4 or S = 0.8 x D², where:

• S is the cross-sectional area in mm 2 ;
• π is the number of π, the standard value is 3.14;
• D is the diameter in mm.

Conductor

Measurements of a stranded wire will require its preliminary fluffing, as well as the subsequent counting of the number of all veins inside the bundle. Then the diameter of one constituent element is measured and the cross-sectional area is calculated in accordance with the standard formula above. At the final stage of measurements, the areas of the veins are summarized in order to determine the indicators of their total cross section.

In order to determine the diameter of the wire core, a micrometer or caliper is used, but if necessary, you can use a standard student ruler or centimeter. The measured vein of the wire must be wound as tightly as possible on the stick with two dozen turns. Using a ruler or centimeter, it is required to measure the winding distance in mm, after which the indicators are used in the formula:

D = l/n,

• l is represented by the winding distance of the vein in mm;
• n is the number of turns.

It should be noted that a larger cross-section of the wire allows you to provide a margin in terms of current, as a result of which the level of load on the wiring can be slightly exceeded.

In order to independently determine the wire section of a monolithic core, it is required to measure the diameter of the inner part of the cable without protective insulation using a conventional caliper or micrometer.

Correspondence table for wire diameters and their cross-sectional area

The determination of the cable or wire section according to the standard physical formula is one of the rather laborious and complex processes that do not guarantee the most accurate results, therefore it is advisable to use special, ready-made tabular data for this purpose.

Cable core diameter	Section indicators	Conductors with stranded copper type
		Power in network conditions 220 V	Current	Power in network conditions 380 V
1.12 mm	1.0 mm2	3.0 kW	14 A	5.3 kW
1.38 mm	1.5 mm2	3.3 kW	15 A	5.7 kW
1.59 mm	2.0 mm2	4.1 kW	19 A	7.2 kW
1.78 mm	2.5 mm2	4.6 kW	21 A	7.9 kW
2.26 mm	4.0 mm2	5.9 kW	27 A	10.0 kW
2.76 mm	6.0 mm2	7.7 kW	34 A	12.0 kW
3.57mm	10.0 mm2	11.0 kW	50 A	19.0 kW
4.51 mm	16.0 mm2	17.0 kW	80 A	30.0 kW
5.64 mm	25.0 mm2	22.0 kW	100 A	38.0 kW
6.68 mm	35.0 mm2	29.0 kW	135 A	51.0 kW

How to determine the cross section of a stranded wire?

Stranded wires are also known as stranded or flexible cables, which are single-core type wires tightly twisted into one bundle.

In order to independently correctly calculate the cross section or area of ​​stranded wires, it is necessary to initially calculate the cross section of each wire in the bundle, after which the result should be multiplied by their total number.

Current strength

( if ).

current density

where S- cross-sectional area of ​​the conductor.

Current density in a conductor

where is the speed of the ordered movement of charges in the conductor, n- concentration of charges, e- elementary charge.

Dependence of resistance on conductor parameters

where l- conductor length, S- cross-sectional area of ​​the conductor, - specific resistance, - specific conductivity.

Resistivity versus temperature

,

where is the temperature coefficient of resistance, is the resistivity at .

Resistance in series (a) and parallel (b) connection of conductors

where is the resistance of the conductor, n is the number of conductors.

Ohm's law:

for a homogeneous section of the chain

,

for an inhomogeneous section of the chain

,

for closed circuit

where U- voltage on a homogeneous section of the circuit, - potential difference at the ends of the circuit section, - EMF of the source, r- internal resistance of the current source.

Short circuit current

Current work over time t

Current power

Joule-Lenz law (the amount of heat released when current passes through a conductor)

Current source power

Current source efficiency

.

Kirchhoff rules

1) - for nodes;

2) - for contours,

where is the algebraic sum of the strengths of the currents converging in the node, is the algebraic sum of the EMF in the circuit.

2.1. A voltage of 1 V is maintained at the ends of a copper wire 5 m long. Determine the current density in the wire (copper resistivity ).

BUT. b.

C.D.

2.2. A 5 ohm resistor, a voltmeter and a current source are connected in parallel. The voltmeter shows a voltage of 10 V. If you replace the resistor with another one with a resistance of 12 ohms, then the voltmeter will show a voltage of 12 V. Determine the EMF and the internal resistance of the current source. Ignore the current through the voltmeter.

A.B.

C.D.

2.3. Determine the current strength in a circuit consisting of two elements with an EMF equal to 1.6 V and 1.2 V and internal resistances of 0.6 Ohm and 0.4 Ohm, respectively, connected by the same poles.

A.B.C.D.

2.4. The galvanic cell gives an external resistance of 0.5 ohms with a current of 0.2 A. If the external resistance is replaced by 0.8 ohms, then the current in the circuit is 0.15 A. Determine the strength of the short circuit current.

A.B.C.D.

2.5. A load is connected to a current source with an EMF of 12 V. The voltage at the source terminals is 8 V. Determine the efficiency of the current source.

A.B.C.D.

2.6. The external current source circuit consumes 0.75 W of power. Determine the current strength in the circuit if the source EMF is 2V and the internal resistance is 1 ohm.

A.B.C.D.

2.7. A current source with an EMF of 12 V and an internal resistance of 1 ohm is connected to a load with a resistance of 9 ohms. Find: 1) the current strength in the circuit, 2) the power released in the external part of the circuit, 3) the power lost in the current source, 4) the total power of the current source, 5) the efficiency of the current source.

2.8. The winding of the electric boiler has two sections. If one section is turned on, the water boils after 10 minutes, if the other, then after 20 minutes. After how many minutes will water boil if both sections are turned on: a) sequentially; b) in parallel? The voltage at the terminals of the boiler and the efficiency of the installation should be considered the same in all cases.

A. [a) 30 min; b) 6.67 min] C. [a) 6.67 min; b) 30 min]

C. [a) 10 min; b) 20 min] D. [a) 20 min; b) 10 min]

2.9. An ammeter with a resistance of 0.18 ohms is designed to measure current strength up to 10 A. What resistance should be taken and how to turn it on so that this ammeter can measure current strength up to 100 A?

A.V.

C.D.

2.10. A voltmeter with a resistance of 2000 ohms is designed to measure voltages up to 30 V. What resistance should I take and how to turn it on so that this voltmeter can measure voltages up to 75 V?

A.V.

C.D.

2.11 .* The current in a conductor with a resistance of 100 ohms increases uniformly from 0 to 10 A within 30 s. What is the amount of heat released during this time in the conductor?

A.B.C.D.

2.12.* The current in a conductor with a resistance of 12 ohms decreases uniformly from 5 A to 0 within 10 s. How much heat is released in the conductor during this time?

A.B.C.D.

2.13.* A 3 ohm conductor carries a uniformly increasing current. The amount of heat released in the conductor in 8 s is 200 J. Determine the charge that has flowed through the conductor during this time. At the initial moment of time, the current was equal to zero.

A.B.C.D.

2.14.* The current in a conductor with a resistance of 15 ohms increases uniformly from 0 to a certain maximum within 5 s. During this time, an amount of heat of 10 kJ was released in the conductor. Find the average value of the current in the conductor for this period of time.

A.B.C.D.

2.15.* The current in the conductor increases uniformly from 0 to some maximum value within 10 s. During this time, an amount of heat of 1 kJ was released in the conductor. Determine the rate of current rise in the conductor if its resistance is 3 ohms.

A.B.C.D.

2.16. On fig. 2.1 = =, R 1 = 48 Ohm, R 2 = 24 Ohm, the voltage drop U 2 across the resistance R 2 is 12 V. Neglecting the internal resistance of the elements, determine the current strength in all sections of the circuit and the resistance R 3.

	R4

Rice. 2.1 Fig. 2.2 Fig. 2.3

2.17. On fig. 2.2 = 2V, R 1 = 60 ohms, R 2 = 40 ohms, R 3 = R 4 = 20 ohms, R G = 100 ohms. Determine the current strength I G through the galvanometer.

2.18. Find the current strength in the individual branches of the Wheatstone bridge (Fig. 2.2), provided that the current flowing through the galvanometer is zero. EMF source 2V, R 1 \u003d 30 Ohm, R 2 \u003d 45 Ohm, R 3 \u003d 200 Ohm. Ignore the internal resistance of the source.

2.19. On fig. 2.3 = 10 V, = 20 V, = 40 V, and resistance R 1 = R 2 = R 3 = 10 ohms. Determine the strength of the currents through the resistances ( I) and through sources (). Ignore the internal resistance of the sources. [ I 1=1A, I 2=3A, I 3=2A,=2A,=0,=3A]

2.20. On fig. 2.4 \u003d 2.1 V, \u003d 1.9 V, R 1 \u003d 45 Ohms, R 2 \u003d 10 Ohms, R 3 \u003d 10 Ohms. Find the current in all parts of the circuit. Ignore the internal resistance of the elements.

Rice. 2.4 Fig. 2.5 Fig. 2.6

2.21. On fig. 2.5 resistances of voltmeters are equal to R 1 = 3000 ohms and R 2 = 2000 ohms; R 3 \u003d 3000 Ohm, R 4 \u003d 2000 Ohm; \u003d 200 V. Find the readings of voltmeters in the following cases: a) key To open, b) key To closed. Ignore the internal resistance of the source. [a) U 1 \u003d 120 V, U 2 \u003d 80 V, b) U 1 \u003d U 2 \u003d 100 V]

2.22. On fig. 2.6 \u003d \u003d 1.5 V, internal resistances of sources r 1 \u003d r 2 \u003d 0.5 Ohm, R 1 \u003d R 2 \u003d 2 Ohm, R 3 \u003d 1 Ohm. The resistance of the milliammeter is 3 ohms. Find the milliammeter reading.

2.23. On fig. 2.7 = = 110 V, R 1 = R 2 = 200 Ohm, voltmeter resistance 1000 V. Find the voltmeter reading. Ignore the internal resistance of the sources.

Rice. 2.7 Fig. 2.8 Fig. 2.9

2.24. On fig. 2.8 \u003d \u003d 2V, the internal resistances of the sources are 0.5 Ohm, R 1 \u003d 0.5 Ohm, R 2 \u003d 1.5 Ohm. Find the current in all parts of the circuit.

2.25. On fig. 2.9 = = 100 V, R 1 = 20 ohms, R 2 = 10 ohms, R 3 = 40 ohms, R 4 = 30 ohms. Find the ammeter reading. Ignore the internal resistance of the sources and the ammeter.

2.26. What current is shown by the ammeter in fig. 2.10, whose resistance R A \u003d 500 Ohm, if \u003d 1 V, \u003d 2 V, R 3 \u003d 1500 Ohm and the voltage drop across the resistance R 2 is 1 V. Neglect the internal resistance of the sources.

2.27. On fig. 2.11 \u003d 1.5 V, \u003d 1.6 V, R 1 \u003d 1 kOhm, R 2 \u003d 2 kOhm. Determine the readings of the voltmeter if its resistance R V \u003d 2 kOhm. Ignore the source resistance.

		V
	BUT
		V

Rice. 2.10 Fig. 2.11 Fig. 2.12

2.28. On fig. 2.12 resistance R 1 \u003d 5 Ohm, R 2 \u003d 6 Ohm, R 3 \u003d 3 Ohm. Find the ammeter reading if the voltmeter shows 2.1 V. Neglect the resistance of the source and the ammeter.

2.29 . Determine the EMF of the source in the circuit in fig. 2.13, if the strength of the current flowing through it is 0.9 A, the internal resistance of the source is 0.4 ohms. R 1 \u003d 30 Ohm, R 2 \u003d 24 Ohm, R 3 \u003d 50 Ohm, R 4 \u003d 40 Ohm, R 5 \u003d 60 Ohm.

2.30. Find the ammeter readings in the circuit in fig. 2.14, if the EMF is 19.8 V, the internal resistance is 0.4 Ohm, R 1 \u003d 30 Ohm, R 2 \u003d 24 Ohm, R 3 \u003d 50 Ohm, R 4 \u003d 40 Ohm, R 5 \u003d 60 Ohm.

Rice. 2.13 Fig. 2.14 Fig. 2.15

2.31 . Find the values ​​of all resistances in the circuit in fig. 2.15, if a current of 0.4 μA flows through the resistance R 1, a current of 0.7 μA through the resistance R 2, 1.1 μA through the resistance R 3, no current flows through the resistance R 4. Ignore the internal resistance of the elements. E 1 \u003d 1.5 V; E 2 \u003d 1.8 V.

Rice. 2.16 Fig. 2.17 Fig. 2.18

2.32. Determine E 1 and E 2 in the circuit in fig. 2.16, if R 1 \u003d R 4 \u003d 2 Ohms, R 2 \u003d R 3 \u003d 4 Ohms. The current flowing through the resistance R 3 is 1A, and no current flows through the resistance R 2. Internal resistances of elements r 1 =r 2 =0.5 Ohm.

2.33. Determine the current strength in all sections of the circuit in the circuit in fig. 2.17, if E 1 \u003d 11 V, E 2 \u003d 4 V, E 3 \u003d 6 V, R 1 \u003d 5 Ohm, R 2 \u003d 10 Ohm, R 3 \u003d 2 Ohm. Internal resistance sources r 1 =r 2 =r 3 =0.5 Ohm.

2.34. In the diagram in fig. 2.18 R 1 \u003d 1 Ohm, R 2 \u003d 2 Ohm, R 3 \u003d 3 Ohm, the current through the source is 2A, the potential difference between the points 1 and 2 equal to 2 V. Find the resistance R 4 .

Electromagnetism

Basic formulas

Magnetic induction is related to the magnetic field strength by the relation

where - magnetic constant,

Magnetic permeability of an isotropic medium.

The principle of superposition of magnetic fields

where is the magnetic induction created by each current or moving charge separately.

The magnetic induction of the field created by an infinitely long straight current-carrying conductor,

where is the distance from the conductor with current to the point at which the magnetic induction is determined.

Magnetic induction of the field created by a rectilinear conductor with a current of finite length

,

where are the angles between the current element and the radius vector drawn from the point under consideration to the ends of the conductor.

Magnetic field induction in the center of a circular conductor with current

where is the radius of the circle.

Magnetic field induction on the axis of a circular conductor with current

,

where is the radius of a circular coil, is the distance from the center of the coil to the point at which the magnetic induction is determined.

Magnetic field induction inside a toroid and an infinitely long solenoid

where is the number of turns per unit length of the solenoid (toroid).

Magnetic field induction on the axis of a solenoid of finite length

,

where are the angles between the axis of the coil and the radius vector drawn from a given point to the ends of the coil.

Ampere force acting on a conductor element with current in a magnetic field,

where is the angle between the directions of the current and the magnetic induction of the field.

Magnetic moment of the circuit with current

where is the area of ​​the contour,

The unit normal vector (positive) to the contour plane.

The torque acting on a current-carrying circuit placed in a uniform magnetic field is

,

where is the angle between the direction of the normal to the contour plane and the magnetic field induction.

The force of interaction between two straight parallel conductors with currents and

,

where is the length of the conductor, is the distance between them.

Magnetic flux through the pad

where , is the angle between the direction of the magnetic induction vector and the normal to the site.

Magnetic flux of an inhomogeneous field through an arbitrary surface

where integration is carried out over the entire surface.

Magnetic flux of a uniform field through a flat surface

The work of moving a current-carrying conductor in a magnetic field

where is the flux of magnetic induction crossed by the conductor during its movement.

The Lorentz force acting on a moving charged particle in a magnetic field is

where is the charge of the particle, is the velocity of the particle, is the angle between the directions of the particle velocity and the magnetic field induction.

E.D.S. induction

The potential difference at the ends of a conductor moving in a magnetic field is

where is the speed of the conductor, is the length of the conductor, is the angle between the directions of the speed of the conductor and the magnetic induction of the field.

E.D.S. self-induction

where is the inductance of the circuit.

Solenoid Inductance

,

where is the cross-sectional area of ​​the solenoid, is the length of the solenoid, is the total number of turns.

The energy of the magnetic field of the circuit with current

Volumetric magnetic field energy density

.

3.1. On fig. 3.1 shows a section of two rectilinear infinitely long current-carrying conductors. The AC distance between the conductors is 10 cm, I 1 \u003d 20 A, I 2 \u003d 30 A. Find the magnetic induction of the field caused by currents I 1 and I 2 at points M 1, M 2 and M 3. Distances M 1 A \u003d 2 cm, AM 2 \u003d 4 cm and CM 3 \u003d 3 cm.

A.V.

C.D.

3.2. Solve the previous problem, provided that the currents flow in one

direction.

A.V.

C.D.

3.3. Two rectilinear infinitely long conductors are located perpendicular to each other and are in the same plane (Fig. 3.2). Find the magnetic induction of the field at points M 1 and M 2 if I 1 \u003d 2 A and I 2 \u003d 3 A. Distances AM 1 \u003d AM 2 \u003d 1 cm, DM 1 \u003d CM 2 \u003d 2 cm.

Rice. 3.2 Fig. 3.3

A.V.

C.D.

3.4. Two rectilinear infinitely long conductors are located perpendicular to each other and are in mutually perpendicular planes (Fig. 3.3). Find the magnetic induction of the field at points M 1 and M 2 if I 1 \u003d 2 A and I 2 \u003d 3 A. Distances AM 1 \u003d AM 2 \u003d 1 cm and AC \u003d 2 cm.

A.V.

C.D.

3.5. On fig. 3.4 shows a section of three rectilinear infinitely long current-carrying conductors. Distances AC=CD=5 cm; I 1 =I 2 =I; I 3 \u003d 2I. Find a point on the straight line AD at which the magnetic field induction caused by currents I 1 , I 2 , I 3 is equal to zero.

A.B.

C.D.

3.6. Solve the previous problem under the condition that all currents flow in the same direction.

A.B.

C.D.

3.7. Two circular turns with a radius of 4 cm each are located in parallel planes at a distance of 0.1 m from each other. Currents I 1 \u003d I 2 \u003d 2 A flow through the turns. Find the magnetic field induction on the axis of the turns at a point located at an equal distance from them. The currents in the coils flow in the same direction.

A.B.C.D.

3.8. Solve the previous problem under the condition that the currents flow in opposite directions.

A.B.C.D.

3.9. A current of 2A flows through a long conductor bent at an angle. Find the magnetic induction of the field at a point lying on the bisector of this angle and at a distance of 10 cm from the vertex of the angle.

A.B.C.D.

3.10. Along a conductor bent in the form of a rectangle with sides a= 8 cm and in\u003d 12 cm, current flows with force I\u003d 50 A. Determine the strength and magnetic induction of the field at the intersection of the diagonals of the rectangle.

A.V.

C.D.

3.11. A current of force I = 2 A flows through a wire frame, which has the shape of a regular hexagon. In this case, a magnetic field B = 41.4 μT is formed in the center of the frame. Find the length of the wire from which the frame is made.

A.B.C.D.

3.12. A conductor bent in the form of a circle carries a current. The magnetic field at the center of the circle is B = 6.28 μT. Without changing the strength of the current in the conductor, he was given the shape of a square. Determine the magnetic induction of the field at the point of intersection of the diagonals of this square.

A.B.D.

3.13. The solenoid winding contains two layers of tightly adjacent turns of wire with a diameter of d = 0.2 mm. Determine the magnetic induction of the field on the axis of the solenoid if the current flows through the wire I = 0.5 A.

A.B.C.D.

3.14. A thin ring with a mass of 15 g and a radius of 12 cm carries a charge uniformly distributed with a linear density of 10 nC/m. The ring rotates uniformly with a frequency of 8 s -1 about an axis perpendicular to the plane of the ring and passing through its center. Determine the ratio of the magnetic moment of the circular current created by the ring to its angular momentum.

A.B.C.D.

3.15. In two infinitely long straight parallel conductors, the distance between which is 25 cm, currents of 20 and 30 A flow in opposite directions. Determine the magnetic induction of the field at a point at a distance of 30 cm from the first and 40 cm from the second conductor.

A.B.C.D. [27.0 μT]

3.16. Determine the magnetic induction of the field on the axis of a thin wire ring with a radius of 10 cm, through which a current of 10 A flows, at a point located at a distance of 15 cm from the center of the ring.

A.B.C.D.

3.17. A wire bent in the form of a square with a side equal to 60 cm flows a direct current of 3 A. Determine the magnetic induction of the field in the center of the square.

A.B.C.D.

3.18. The current flowing through a wire ring of copper wire with a cross section of 1.0 mm 2 creates a magnetic field induction of 0.224 mT in the center of the ring. The potential difference applied to the ends of the wire forming the ring is 0.12 V. What current flows through the ring?

A. V. S. [2 A] D.

3.19. A current of 2 A, flowing through a coil 30 cm long, creates a magnetic field induction of 8.38 mT inside it. How many turns does the coil have? Consider the diameter of the coil to be small compared to its length.

A.B.C.D.

3.20. An infinitely long wire forms a circular loop tangent to the wire. The radius of the loop is 8 cm. A current of 5A flows through the wire. Find the induction of the magnetic field at the center of the loop.

A.B.C.D.

3.21*. Find the distribution of the magnetic induction of the field along the axis of a circular coil with a diameter of 10 cm, through which a current of 10A flows. Make a table of values ​​for values ​​in the interval 0-10 cm every 2 cm and build a graph with a scale. [ ] .

3.22*. Determine, using the vector circulation theorem, the magnetic field induction on the axis of a toroid without a core, through the winding of which, containing 300 turns, a current of 1A flows. The outer diameter of the toroid is 60 cm, the inner diameter is 40 cm.

3.23. Two infinite rectilinear parallel conductors with the same currents flowing in the same direction are at a distance R from each other. To move them apart to a distance of 3R, 220 nJ work is expended for each centimeter of the conductor length. Determine the current strength in the conductors.

A.B.C.D.

3.24. A straight conductor 20 cm long, through which a current of 40A flows, is in a uniform magnetic field with an induction of 0.5 T. What work is done by the field forces, moving the conductor by 20 cm, if the direction of movement is perpendicular to the lines of magnetic induction and the conductor.

A.B.C.D.

3.25. In a uniform magnetic field, the induction of which is 0.5 T, the conductor moves uniformly at a speed of 20 cm / s perpendicular to the field. The length of the conductor is 10 cm. A current of 2A flows through the conductor. Find the power required to move the conductor.

A.B.C.D.

3.26. Magnetic induction of a homogeneous field 0.4 T. In this field, a conductor 1 m long moves uniformly at a speed of 15 cm / s so that the angle between the conductor and the field induction is equal to . A current of 1A flows through the conductor. Find the work of moving the conductor in 10 s of movement.

A.B.C.D.

3.27. A conductor 1 m long is located perpendicular to a uniform magnetic field with an induction of 1.3 T. Determine the current in the conductor if, when it moves at a speed of 10 cm / s in a direction perpendicular to

field and conductor, for 4 s the energy of 10 J is spent on moving the conductor.

A.B.C.D.

3.28. In a uniform magnetic field with an induction of 18 μT in a plane perpendicular to the lines of induction, there is a flat circular frame, consisting of 10 turns with an area of ​​100 cm 2 each. A current of 3A flows in the frame winding. What should be the direction of the current in the loop so that when it is rotated around one of the diameters, the field forces do positive work? What is the magnitude of this work?

A.B.C.D.

3.29. A square circuit with a side of 20 cm, through which a current of 20 A flows, is freely established in a uniform magnetic field with an induction of 10 mT. Determine the change in the potential energy of the contour when turning around an axis lying in the plane of the contour by an angle.

A.B.C.D.

3.30. A current of 10A flows through a circular coil with a radius of 15 cm. The coil is located in a uniform magnetic field with an induction of 40 mT so that the normal to the contour plane makes an angle with the magnetic induction vector. Determine the change in the potential energy of the circuit when it is rotated through an angle in the direction of increasing angle.

A.B.C.D.

3.31. A round frame with a current area of ​​20 cm 2 is fixed parallel to the magnetic field with an induction of 0.2 T, and a torque of 0.6 mN·m acts on it. When the frame was released, it turned on and its angular velocity became 20 s -1 . Determine the strength of the current flowing in the frame.

A. B. C. D. [15 A]

3.32. Two long horizontal conductors are parallel to each other at a distance of 8 mm. The upper conductor is fixed motionless, while the lower one hangs freely under it. What current must be passed through the upper wire so that the lower one can hang without falling? A current of 1A flows along the lower one and the mass of each centimeter of the conductor length is 2.55 mg.

A.B.C.D.

3.33 . The flux of magnetic induction through the cross-sectional area of ​​the solenoid (without core) is 5 μWb. The length of the solenoid is 35 cm. Determine the magnetic moment of this solenoid.

A.B.C.D.

3.34. A circular contour is placed in a uniform magnetic field so that the plane of the contour is perpendicular to the field lines of force. Magnetic field induction 0.2 T. A current of 2A flows through the circuit. The radius of the contour is 2 cm. What work is done when the contour is rotated by ?

A.B.C.D.

3.35*. Next to a long straight wire carrying a current of 30A, there is a square frame with a current of 2A. The frame and the wire lie in the same plane. The axis of the frame passing through the midpoints of the opposite sides is parallel to the wire and spaced from it at a distance of 30 mm. Frame side 20 mm. Find the work to be done to rotate the frame around its axis by . .

3.36*. Two straight long conductors are at a distance of 10 cm from each other. The conductors carry currents of 20A and 30A. What work per unit length of the conductors must be done to push these conductors to a distance of 20 cm? .

3.37. A proton accelerated by a potential difference of 0.5 kV, flying into a uniform magnetic field with an induction of 0.1 T, moves in a circle. Determine the radius of this circle.

A.B.C.D.

3.38. An alpha particle with a speed of 2 Mm/s flies into a magnetic field with an induction of 1 T at an angle . Determine the radius of the turn of the helix that the alpha particle will describe?

A.B.C.D.

3.39. A magnetic field with an induction of 126 μT is directed perpendicular to the electric field, the intensity of which is 10 V/m. An ion flying at some speed flies into these crossed fields. At what speed will it move in a straight line?

A.B.C.D.

3.40. An electron accelerated by a potential difference of 6 kV flies into a uniform magnetic field at an angle to the direction of the field and begins to move along a helix. The magnetic induction of the field is 130 mT. Find the pitch of the helix.

A.V.S. [1.1 cm] D.

3.41. The proton flew into a uniform magnetic field at an angle to the direction of the field lines and moves in a spiral, the radius of which is 2.5 cm. The magnetic field induction is 0.05 T. Find the kinetic energy of the proton.

A.V.

C.D.

3.42. Determine the frequency of an electron in a circular orbit in a magnetic field with an induction of 1 T. How will the frequency of revolution change if an alpha particle rotates instead of an electron?

3.43. A proton and an alpha particle, accelerated by the same potential difference, fly into a uniform magnetic field. How many times is the radius of curvature of the proton trajectory less than the radius of curvature of the alpha particle trajectory?

A.B.C.D.

3.44. A particle carrying one elementary charge flew into a uniform magnetic field with an induction of 0.05 T. Determine the angular momentum that the particle had when moving in a magnetic field if its trajectory was an arc of a circle with a radius of 0.2 mm.

A.V.

C.D.

3.45. An electron moves in a circle in a uniform magnetic field with an induction of 31.4 mT. Determine the period of revolution of the electron.

A.B.C.D.

3.46. Find the ratio q / m for a charged particle if, flying at a speed of 10 8 cm / s into a uniform magnetic field with a strength of 2 10 5 A / m, it moves along an arc of a circle with a radius of 8.3 cm. The direction of the particle velocity is perpendicular to the direction magnetic field.

A.B.C.D.

3.47. An electron accelerated by a potential difference of 3 kV flies into the magnetic field of the solenoid at an angle to its axis. The number of ampere-turns of the solenoid is 5000. The length of the solenoid is 26 cm. Find the step of the helical trajectory of an electron in the magnetic field of the solenoid.

A.B.C.D.

3.48. A charged particle moves in a magnetic field in a circle with a speed of 1 Mm/s. The magnetic induction of the field is 0.3 T. The radius of the circle is 4 cm. Find the charge of the particle if its kinetic energy is known to be 12 keV.

A.V.

C.D.

3.49*. The Serpukhov proton accelerator accelerates these particles to an energy of 76 GeV. If we disregard the presence of accelerating gaps, then we can assume that accelerated protons move along a circle with a radius of 236 m and are kept on it by a magnetic field perpendicular to the plane of the orbit. Find the required magnetic field. .

3.50*. The charged particle passed through an accelerating potential difference of 104 V and flew into electric (E = 100 V/m) and magnetic (B = 0.1 T) fields crossed at a right angle. Determine the ratio of the charge of a particle to its mass if, moving perpendicular to both fields, the particle does not experience deviations from a rectilinear trajectory. .

3.51. In a uniform magnetic field with an induction of 0.1 T, a frame containing 1000 turns rotates uniformly. Frame area 150 cm 2 . The frame does 10 rpm. Determine the maximum emf. framed induction. The axis of rotation lies in the frame plane and is perpendicular to the field direction.

A.B.C.D.

3.52. The wire coil is located perpendicular to the magnetic field, the induction of which varies according to the law B = B o (1 + e to t), where B o = 0.5 T, k = 1 s -1. Find the value of the emf induced in the coil at a time equal to 2.3 s. The coil area is 0.04 m 2 .

A.B.C.D.

3.53. A square frame made of copper wire is placed in a magnetic field with an induction of 0.1 T. The cross-sectional area of ​​the wire is 1 mm 2 , the area of ​​the frame is 25 cm 2 . The normal to the frame plane is parallel to the field lines. What charge will pass through the frame when the magnetic field disappears? The specific resistance of copper is 17 nOhm m.

A.B.C.D.

3.54. A ring of aluminum wire is placed in a magnetic field perpendicular to the lines of magnetic induction. Ring diameter 20 cm, wire diameter 1 mm. Determine the rate of change of the magnetic field if the strength of the induction current in the ring is 0.5A. The specific resistance of aluminum is 26 nOhm m.

A.B.C.D.

3.55. In a magnetic field with an induction of 0.25 T, a rod 1 m long rotates with a constant angular velocity of 20 rad/s. The axis of rotation passes through the end of the rod parallel to the field lines. Find e.m.f. induction at the ends of the rod.

A.B.C.D.

3.56. A ring of wire with a resistance of 1 mΩ is placed in a uniform magnetic field with an induction of 0.4 T. The plane of the ring makes an angle with the lines of induction. Determine the charge that will flow through the ring if it is pulled out of the field. The area of ​​the ring is 10 cm2.

A.B.C.D.

3.57. A coil containing 10 turns, each with an area of ​​4 cm 2 , is placed in a uniform magnetic field. The axis of the coil is parallel to the field induction lines. The coil is connected to a ballistic galvanometer with a resistance of 1000 ohms, the coil resistance can be neglected. When the coil was pulled out of the field, 2 μC flowed through the galvanometer. Determine the field induction.

A.B.C.D.

3.58. On a rod made of non-magnetic material 50 cm long and 2 cm 2 in cross section, a wire is wound in one layer so that there are 20 turns per centimeter of the rod length. Determine the energy of the magnetic field of the solenoid if the current in the winding is 0.5A.

A.B.C.D.

3.59. Find the potential difference at the ends of the axis of the car, which occurs when it moves horizontally at a speed of 120 km / h, if the axis length is 1.5 m and the vertical component of the earth's magnetic field strength is 40 A / m.

A.B.C.D.

3.60. A coil of wire is put on a solenoid with a length of 20 cm and a cross-sectional area of ​​30 cm 2 . The solenoid winding has 320 turns and a current of 3A flows through it. What emf is induced in the turn put on the solenoid, when the current in the solenoid disappears within 0.001 s?

A.V.C. [0.18 V] D.

3.61. A coil with a diameter of 10 cm, having 500 turns, is in a magnetic field. The axis of the coil is parallel to the lines of magnetic field induction. What is the average value of emf. induction in the coil, if the magnetic field induction increases within 0.1 s from zero to 2 T?

A.B.C.D.

3.62*. A flywheel with a diameter of 3 m rotates around a horizontal axis at a speed of 3000 rpm. Determine the emf induced between the rim and the wheel axle if the plane of the wheel makes an angle with the plane of the magnetic meridian. The horizontal component of the earth's magnetic field is 20 μT. .

3.63*. A copper hoop with a mass of 5 kg is located in the plane of the magnetic meridian. What charge is induced in it if it is rotated about the vertical axis by ? The horizontal component of the earth's magnetic field is 20 μT. The density of copper is 8900 kg / m 3, the resistivity of copper is 17 nOhm m. .

3.64*. In a uniform magnetic field, the induction of which is 0.5 T, a coil containing 200 turns tightly adjacent to each other rotates uniformly at a frequency of 300 min -1. The cross-sectional area of ​​the coil is 100 cm 2 . The axis of rotation is perpendicular to the axis of the coil and the direction of the magnetic field. Determine the maximum emf induced in the coil. .

The copper conductor has a length of 500 m and a cross-sectional area of ​​0.5 mm2. A) what is the current strength in the conductor when the voltage at its ends is 12V? Resistivity of copper 1.7 times 10 -8 powers of ohms times m b) Determine the speed of the orderly movement of electrons. The concentration of free movement for copper equal to 8.5 times 10 to the 28th degree of meters to minus 3 degrees, and the electron charge modulus equal to 1.6 times 10 to the minus 19th degree C c) A second copper conductor of twice the diameter was connected in series to the first conductor . What will be the speed of the ordered movement of electrons in the second conductor?

Solution for question a)
What do we know about current, voltage and resistance?

I=U/R, U=I*R
I - current in Amperes,
U - voltage in Volts
R - resistance in ohms
What is a current of 1 ampere?
This is such a current at which in 1 second a charge of 1 Coulomb passes through the conductor.
1 A = 1 C/s(1 ampere equals 1 coulomb per second)
What do we know from the conditions?
U = 12 V - voltage
p \u003d 1.7 * 10e-8 Ohm * m - specific resistance "ro" (the resistance value of a conductor with a cross section of 1 square meter and a length of 1 meter).
Our conductor has a section S = 0.5 mm ^ 2 or 0.0000005 m ^ 2 or 0.5 * 10e-6 m ^ 2 (in one square meter 1000000 square millimeters - 1000 * 1000) and length L = 500m
We get the resistance of the conductor
R=p*L/S\u003d 1.7 * 10e-8 * 500 / 0.5 * 10e-6 \u003d 0.000000017 * 500 / 0.0000005 \u003d 17 ohms
The current will then be:
I=U/R\u003d 12 / 17 A (0.706. Ampere)
Solution for question b)
The current I is also expressed in terms of the following quantities:
I=e*n*S*Vav
e - electron charge, C
n - electron concentration, pcs/m^3 (pieces per cubic meter)
S - sectional area, m^2
Vav - average speed of the ordered movement of electrons, m/s
That's why
Vav=I/(e*n*S)= (12/17) / (1.6*10e-19 * 8.5*10e+28 * 0.5*10e-6) = 11.657*10e-3 m/s (or 11.657 mm/s)
Solution for question c)
We argue similarly to solutions a) and b)
First you need to find the total current (total resistance).
Because Condition c) refers to the diameter, we conclude that all wires are round.
The length of the second wire is not specified. Let's say it is also 500m.
The area of ​​a circle is determined by the ratio:
S=(pi*D^2)/4,
where D is the diameter of the circle,
pi = 3.1415926.
Thus, when the diameter is doubled, the cross-sectional area of ​​the wire is quadrupled,
when the diameter is tripled, the cross-sectional area of ​​the wire is increased nine times, etc.
Total S2=S1*4\u003d 0.5 * 10e-6 * 4 \u003d 2 * 10e-6 M^2
If the cross-sectional area of ​​a wire is quadrupled, then, with the same length, its resistance will decrease by a factor of four.
Total R2=R1/4= 17/4 ohms = 4.25 ohms
The total resistance in series connection adds up, so
I=U/R=U/(R1+R2)\u003d 12 / (17 + 17/4) \u003d 48/85 \u003d 0.5647. A
The ordered electron velocity for the second conductor would then be:
Vav=I/(e*n*S2)= (48/85)/(1.6*10e-19 * 8.5*10e+28 * 2*10e-6) = 0.02076*10e-3 m/s (or 0.02076 mm/s)

"Conductors and dielectrics" - The electrical characteristics of a medium are determined by the mobility of charged particles in it. Dielectrics. Free charges are charged particles of the same sign that can move under the influence of an electric field. Dielectrics - gases, distilled water, benzene, oils, porcelain, glass, mica, etc. External electric field.

"Golden Section" - Pokrovsky Cathedral (St. Basil's Cathedral). Admiralty. Protection of the Virgin on the Nerl. Painting in the foyer of the second floor. Research objectives: Golden section - proportion. St. Basil's Cathedral. The purpose of the study: To derive the law of the beauty of the world from the point of view of mathematics. Golden section in architecture. Made by 10th grade student Yulia Smetanina.

"Sections of a parallelepiped" - 1. Introductory speech of the teacher - 3 min 2. Activation of students' knowledge. Rectangle CKK'C' - section ABCDA'B'C'D'. Homework. The cutting plane intersects the faces along segments. ? MNK- section of the parallelepiped ABCDA'B'C'D'. Task: build a section through the edge of the parallelepiped and point K. Independent work of students.

"Proportions of the golden section" - The division of the segment by the "golden section". "Golden Pentagon". Euclid, Leonardo da Vinci, Luca Pacioli. "Golden Rectangle". Inanimate nature. For example, the ratio of land and water on the Earth's surface is in the golden ratio. The harmony of the universe is based on numbers. "Golden section" in nature, art and architecture.

“Construction of sections” - If the section is rendered, then an open line is drawn, two thickened strokes. Section designation. Some dimensions of part elements are more convenient to show on sections. Sections in the drawings are divided into extended and superimposed. Sections are made to the same scale as the image to which they refer.

"Conductor in an electrical circuit" - Solve the problem. Connection of conductors. Electric bulbs in a Christmas tree garland are connected in series. Determine the resistance of the circuit The resistance of each resistor is 3 ohms. 1. Two conductors with a resistance of 4 ohms and 2 ohms are connected in series. Series connection I = I1 = I2 U = U1 + U2 R = R1 + R2 For identical conductors R = nR1.

When charged particles move, an electric charge is transferred from one place to another. However, if charged particles perform random thermal motion, as, for example, free electrons in a metal, then there will be no charge transfer (Fig. 143). An electric charge moves through the cross section of the conductor only if, along with the chaotic movement, the electrons participate in an ordered movement (Fig. 144). In this case, it is said that an electric current is established in the conductor.

From the 7th grade physics course, you know that the ordered (directed) movement of charged particles is called electric current. An electric current arises from the orderly movement of free electrons in a metal or ions in electrolytes.

However, if you move a neutral body as a whole, then, despite the orderly movement of a huge number of electrons and atomic nuclei, an electric current does not arise. Total chargetransferred through any section of the conductor will be equal to zero, since charges of different signs move with the same average speed. The current in the conductor will arise only if, when the charges move in one direction, the positive charge transferred through the section is not equal in absolute value to the negative one.

Electric current has a certain direction. The direction of movement of positively charged particles is taken as the direction of the current. If the current is formed by the movement of negatively charged particles, then the direction of the current is considered opposite to the direction of movement of the particles.

Current action. We do not directly observe the motion of particles in a conductor. However, the presence of an electric current can be judged by the actions or phenomena with which it is accompanied.

First, the conductor through which the current flows heats up.

Secondly, the electric current can change the chemical composition of the conductor, for example, release its chemical constituents (copper from a solution of copper sulphate, etc.). Of such kind

processes are observed not in all conductors, but only in solutions (or melts) of electrolytes.

Thirdly, the current has a magnetic effect. So, the magnetic needle near the current-carrying conductor rotates. The magnetic effect of the current, in contrast to the chemical and thermal ones, is the main one, since it manifests itself in all conductors without exception. The chemical effect of the current is observed only in electrolytes, and heating is absent in superconductors (see § 60).

Current strength. If an electric current is established in the circuit, then this means that an electric charge is transferred all the time through the cross section of the conductor. The charge transferred per unit of time serves as the main quantitative characteristic of the current, called the current strength. If a charge is transferred through the cross section of the conductor in time, then the current strength is equal to:

Thus, the current strength is equal to the ratio of the charge transferred through the cross section of the conductor over a time interval to this time interval. If the current strength does not change with time, then the current is called constant.

Current strength, like charge, is a scalar quantity. It can be both positive and negative. The sign of the current strength depends on which of the directions along the conductor is taken as positive. Current strength if the direction of the current coincides with the conditionally chosen positive direction along the conductor. Otherwise

The strength of the current depends on the charge carried by each particle, the concentration of particles, the speed of their directed movement and the cross-sectional area of ​​the conductor. Let's show it.

Let the conductor have a cross section of area 5. For the positive direction in the conductor, we take the direction from left to right. The charge of each particle is equal. The volume of the conductor, limited by sections and 2, contains particles, where is the concentration of particles (Fig. 145). Their total charge If the particles move from left to right with an average speed, then in time all the particles enclosed in the volume under consideration will pass through section 2. Therefore, the current strength is equal.

Does electric current have power? Yes, just imagine... And what is strength for? Well, how for what, in order to do useful work, or maybe not useful :-), The main thing is to do something. Our body also has power. Someone has such strength that he can gouge a brick to smithereens with one blow, while the other cannot even lift a spoon :-). So, my dear readers, electric current also has power.

Imagine the hose you use to water your garden.

Let the hose be a wire, and the water in it is an electric current. We opened the faucet a little and water ran through the hose. Slowly, but still, she ran. The power of the jet is very weak. We can't even spray someone with a hose like that. And now let's open the faucet to the fullest! And we have such a jet that it’s even enough to water the neighbor’s plot :-).

Now imagine that you are filling a bucket. Will you fill it faster with a pressure from a hose or from a faucet? The diameter of the hose and the faucet are the same

Of course, the pressure from the yellow hose! But why is this happening? The thing is that the volume of water for an equal period of time comes out of the faucet and the yellow hose is also different. Or in other words, more water molecules run out of the hose than out of the faucet in the same time.

It's the same story with wires. That is, for an equal period of time, the number of electrons running through the wire can be completely different. Now we can define the strength of the current.

So, the current strength is the number of electrons passing through the cross-sectional area of ​​the conductor per unit of time, let's say, per second. Below in the figure, this same cross-sectional area of ​​\u200b\u200bthe wire through which the electric current runs is shaded with green lines.

• for direct current -

where I - direct current strength;

• for non-direct current - in two ways:

1) according to the formula -

Q = 〈 I 〉 Δ t ,

where 〈 I 〉 - average current strength;

2) graphically - as the area of ​​a curvilinear trapezoid (Fig. 8.1).

In the International System of Units, charge is measured in coulombs (1 C).

The current strength is determined by the speed, concentration and charge of the current carriers, as well as the cross-sectional area of ​​the conductor:

where q is the charge modulus of the current carrier (if the current carriers are electrons, then q = 1.6 ⋅ 10 −19 C); n is the concentration of current carriers, n = = N /V; N - the number of current carriers that have passed through the cross section of the conductor (located perpendicular to the speed of movement of the current carriers) during the time Δt, or the number of current carriers in the volume V = Sv Δt (Fig. 8.2); S is the cross-sectional area of ​​the conductor; v is the modulus of the speed of movement of current carriers.

The current density is determined by the strength of the current passing through the unit area of ​​the cross section of the conductor, located perpendicular to the direction of the current:

where I is the current strength; S is the cross-sectional area of ​​the conductor (located perpendicular to the speed of movement of the current carriers).

The current density is vector quantity.

The direction of the current density j → coincides with the direction of the velocity of positive current carriers:

j → = q n v → ,

where q is the charge modulus of the current carrier (if the current carriers are electrons, then q = 1.6 ⋅ 10 −19 C); v → - speed of movement of current carriers; n is the concentration of current carriers, n = N /V; N - the number of current carriers that have passed through the cross section of the conductor (located perpendicular to the speed of movement of the current carriers) during the time Δt, or the number of current carriers in the volume V = Sv Δt (Fig. 8.2); v is the modulus of the speed of movement of current carriers; S is the cross-sectional area of ​​the conductor.

In the International System of Units, current density is measured in amperes per square meter (1 A / m 2).

The current strength in gases (the electric current in gases is caused by the movement of ions) is determined by the formula

I = N t ⋅ | q | ,

where N /t is the number of ions that pass through the cross section of the vessel every second (every second); |q | - ion charge module:

• for a singly charged ion -

|q | = 1.6 ⋅ 10 −19 C,

• for a doubly charged ion -

|q | = 3.2 ⋅ 10 −19 C

Example 1. The number of free electrons in 1.0 m 3 of copper is 1.0 ⋅ 10 28 . Find the value of the speed of the directed movement of electrons in a copper wire with a cross-sectional area of ​​​​4.0 mm 2, through which a current of 32 A flows.

Solution. The speed of the directed movement of current carriers (electrons) is related to the current strength in the conductor by the formula

where q is the charge modulus of the current carrier (electron); n is the concentration of current carriers; S is the cross-sectional area of ​​the conductor; v is the modulus of the speed of the directed movement of current carriers in the conductor.

We express from this formula the desired value - the speed of current carriers -

v = I q n S .

To calculate the speed, we use the following values ​​of the quantities included in the formula:

• the magnitude of the current and the cross-sectional area of ​​the conductor are given in the condition of the problem: I = 32 A, S = 4.0 mm 2 = 4.0 ⋅ 10 −6 m 2;
• the value of the elementary charge (equal to the modulus of the electron charge) is a fundamental constant (constant value): q = 1.6 ⋅ 10 −19 C;
• concentration of current carriers - the number of current carriers per unit volume of the conductor -

n = N V = 1.0 ⋅ 10 28 1 = 1.0 ⋅ 10 28 m −3 .

Let's do the calculation:

v = 32 1.6 ⋅ 10 − 19 ⋅ 1.0 ⋅ 10 28 ⋅ 4.0 ⋅ 10 − 6 = 5.0 ⋅ 10 − 3 m/s = 5.0 mm/s.

The speed of the directed movement of electrons in the specified conductor is 5.0 mm/s.

Example 2. The current strength in the conductor increases uniformly from 10 to 12 A in 12 s. What charge passes through the cross section of the conductor in the specified time interval?

Solution. The current in a conductor changes over time. Therefore, the charge transferred by the current carriers through the cross section of the conductor, located perpendicular to the speed of the current carriers, for a certain period of time, can be calculated in two ways.

1. The desired charge can be calculated using the formula

Q = 〈 I 〉 Δ t ,

where 〈 I 〉 - average current strength; ∆t - time interval, ∆t = 12 s.

The current strength increases uniformly in the conductor; therefore, the average current strength is given by

〈 I 〉 = I 1 + I 2 2 ,

where I 1 - the value of the current at the initial time, I 1 = 10 A; I 2 - the value of the current at the final moment of time, I 2 \u003d 12 A.

Substituting the expression for the average current strength into the formula for calculating the charge, we get

Q \u003d (I 1 + I 2) Δ t 2.

Calculation gives value

Q \u003d (10 + 12) ⋅ 12 2 \u003d 132 C \u003d 0.13 kC.

The figure shows the dependence I (t) specified in the condition of the problem.

The charge transferred by the current carriers through the cross section of the conductor, located perpendicular to the speed of the current carriers, for a specified period of time, is numerically equal to the area of ​​the trapezoid bounded by four lines:

• straight line I (t);
• perpendicular to the time axis, restored from the point t 1 ;
• perpendicular to the time axis, restored from the point t 2 ;
• time axis t .

We will calculate using the formula for the area of ​​a trapezoid:

Q \u003d 12 + 10 2 ⋅ 12 \u003d 132 C \u003d 0.13 kC.

Both methods of calculating the charge transferred by current carriers over a specified period of time give the same result.

The idea of ​​electric current can be approached from different positions. One of them is macroscopic, the other is based on the analysis of the conduction mechanism. For example, the flow of liquid through pipes can be considered as a continuous movement of matter, but it can also be analyzed in terms of the movement of fluid particles.

The first concept of electric current arose at that stage in the development of physics, when the mechanism of conduction was not yet known. It was then that the physical quantity arose - current strength, which shows how much electric charge passes through the cross section of the conductor per unit time. Current strength. The unit of current strength is ampere (A): .

Two features of this quantity follow from the definition of the current strength. One of them is the independence of the current strength from the cross section of the conductor through which the current flows. The second is the independence of the current strength from the spatial arrangement of the circuit elements, which you could see more than once: no matter how you move the conductors, this does not affect the current strength. The current is called permanent if the current does not change over time.

Thus, the idea of ​​an electric current, its strength arose when it was not yet clear what it was.

The study of the electrical conductivity of various substances showed that in different substances various charged free particles move under the influence of an electric field in the process of current flow. For example, in metals they are electrons, in liquids they are positive and negative ions, in semiconductors they are electrons and "holes". Not only the types of particles are different, but also the nature of their interaction with the substance in which the current flows. So, free electrons in metals move freely between the nodes of the crystal lattice for some time, then collide with ions located at the nodes. In electrolytes, ions interact with each other and with the atoms of the liquid.

But for all substances there is: particles in the absence of a field move randomly, when a field arises, a very small amount of speed is added to the speed of chaotic movement either in the direction of the field (for positive particles) or in the direction opposite to the field (for negative particles). This extra speed is called drift speed. The average speed of chaotic motion is hundreds of meters per second, the drift speed is several millimeters per second. However, it is this small addition that explains all the actions of the current.

For any substances, you can get a formula for calculating the current strength: , where is the concentration of charged particles, is the charge of one particle, is the cross-sectional area.

In this way, electricity is the orderly movement of charged particles.

It may seem that this formula contradicts the assertion that the current strength is independent of the cross-sectional area of ​​the conductor. But this independence is an experimental fact. It can be explained by the fact that the drift velocity is greater where the cross section is smaller, and particles drift more slowly through a larger cross section.

The experimental fact is that when applied to a conductor permanent potential difference goes through it D.C.. This fact contradicts, at first glance, the formula . Indeed, at a constant potential difference in a substance, a field with a constant field strength is created. Consequently, a constant force acts on free particles and their speed must increase. It turns out that at a constant voltage, the current strength should increase in proportion to time. This does not happen because when a current flows in a substance, electrical resistance. It is it that ensures the constancy of the current strength at a constant potential difference.

To measure the resistance, it is necessary to investigate the dependence of the current on the voltage. This dependency graph is called current-voltage characteristic. Three types of current-voltage characteristics are possible (Fig. 40).