current and conductors. Electricity- this is the movement of the smallest, invisible even in a microscope particles - electrons. Both water and electrons can only move if there is a forward and backward path for them. The water coming from the water pumping station flows through the water supply system to the tap (valve), then through the drain and sewerage it enters the irrigation fields, where it evaporates (Fig. 124). In the form of rain, it replenishes lakes and rivers, from which pumping stations again take water and, after cleaning, pump it into the water supply.

If you try to make the pump work with two closed pipes, as shown in Figure 125, then there will be no water flow, since there is no closed circuit for it. The same is true with electric current. It is taken from a generator (at a power plant), a battery (in a flashlight) or a battery (in a vehicle).

From one pole of the named current sources, it flows (Fig. 126) along the conductor to the consumer (lamps, radios, boilers, refrigerators, etc.), then from the consumer along the other conductor returns again to the current source. Electric current, just like water, can only flow if there is a closed circuit.

Just as the amount of water flowing through a pipe can be measured in cubic meters, the amount of electric current flowing through wires can be measured. For example, they say: a current of 1.5 amperes (1.5 a) flows in the network.

The amount of current, called current strength, depends on the conductor through which it flows.

Example. If we insert the plug of the heating device into the socket, then the electric current coming from the generator of the power plant through one of the cores of the cable or overhead wire laid underground will flow through the switchboard through the hidden wiring of the house to our apartment. Here, through a meter, a fuse, one of the two conductors leading to the outlet, it enters one of the terminals of the outlet. On one of the two pins of the plug, the current through the core of the cord will flow to the heating device. From here, having passed through the wire resistance of the device, it will flow in series through the second core of the cord, through the socket, apartment wiring, second fuse, meter, hidden house wiring and underground cable back to the power plant generator.

It has already been mentioned earlier that conductors conduct current well. Nevertheless, they offer some resistance to the flowing electric current, otherwise the current, as we will see later, would become infinitely large. Therefore, each electrical conductor can be considered as a resistance.

The wires that bring electric current to our apartment have a small amount of resistance.

The situation is different with wire resistance in a heating device. It also conducts current, but has a very large amount of resistance. The unit of resistance is ohm. This value is denoted by the Greek letter omega (Ω) and is written, for example, 10 Ω. The resistance value of a conductor depends on its material, length and cross section. Copper and aluminum have little resistance, so they conduct electricity very well and are used as materials for wires. For wire resistance in light bulbs and heating devices (boilers, heaters, pillows, irons, etc.), conductors with high resistance are required. The material for them is tungsten and some iron alloys. Long wires have a higher resistance than short ones. For example, a wire 2 m long has twice the resistance of a wire 1 m long. It goes without saying that both wires must be of the same material and have the same thickness.

Electric current flows in a closed circuit from generator 1 through fuses 2 of the whole house, meter 3, apartment fuses 4, plug 5, electric heater 6 and back to generator 1.

Thick wires have less resistance than thin wires. A wire with a cross section of 1.5 mm2 offers four times more resistance to the flowing current than a wire of the same material and the same length with a cross section of 6 mm2.

The greater the resistance of the wire, the lower the current that flows through this wire, all other things being equal.

Voltage. So, we know that the current flows from the current source through the circuit back to the source. Voltage creates current. It is measured in volts (V). Its value depends on the current source. So, for example, the voltage of one element, which is used in a flashlight or a transistor, is 1.5 e. In flat batteries, three elements are connected in series and give a voltage of 4.5 volts, respectively. Figure 127 shows a conditional image and a diagram of such a series connection of current sources.

When connected in series, the total voltage is obtained by adding the voltage of individual current sources, such as batteries. And vice versa, you can easily calculate the number of bulbs needed, for example for a Christmas tree, based on the mains voltage and the voltage of individual bulbs. The mains voltage is usually 220 V. Large power and heating installations use a voltage of 380 V.

The mains voltage can be easily found in each individual case by looking at the meter. This voltage is constant and does not depend on the current consumer.

Voltage, current and resistance are related to each other in a certain way: the voltage value is equal to the product of the current value and the resistance value: voltage \u003d current strength X resistance, or U \u003d IR.

For example, if the voltage in the outlet is 220 V and we connect a heating pad with a spiral resistance of 48.4 ohms to it, then the current in this network will be equal.

On the other hand, you can calculate the resistance of the wire if the current strength is known. Suppose that a current of 0.27 A flows through the spiral of an incandescent bulb connected to a network with a voltage of 220 V, then the resistance of the burning bulb is equal in this case.

How to find out the current strength in order to calculate the resistance will be shown in the next section.

Power. The electrical power that a device consumes is measured in watts or kilowatts. 1000 watts equals 1 kilowatt, or 1000 watts = 1 kilowatt. Power is indicated on the instruments. So, for example, on an incandescent light bulb, we can read 220 V / 40 W. This means that the light bulb must be connected to a network with a voltage of 220V and in this case the power consumption is 40 watts. We read on the boiler: 220 V / 750 W and on the heater - 220 V / 1000 W. This means that the boiler in a network with a voltage of 220 V consumes 750 watts, and the heater - 1000 watts, i.e. 1 kW.

Based on these two quantities - voltage and power, we can determine the strength of the current that flows through the resistances of the devices. The current strength is equal to the power divided by the voltage:

This current (Fig. 128) flows from the outlet through one of the cores of the cord, the coil of the heater and the other core back to the outlet. If, along with the heater, a 500-watt boiler is also connected to the same network, then an additional current will flow in it

Both of these currents are combined in the terminal block (Fig. 129) and flow together through the wire to the fuses through the meter Household wiring and cable. Thus, a current of 4.5 a + 2.3 a = 6.8 a flows through the wire to the terminal block.

However, other currents can branch off to the meter inside the apartment (in the kitchen, bathroom, etc.), which also need to be stacked together in order to calculate the total current. On each landing, in addition, it is also necessary to take into account a branch to other apartments, and the current in the cable consists of currents that flow in branches to individual houses (Fig. 130).

Inside the apartment, we can calculate the total current that flows through the fuses and the meter by adding up the power of all connected appliances (for example, light bulbs 40 watts, 41 watts, 40 watts, 100 watts, radio 50 watts, boiler 300 watts, stove 800 watts - total 1370 W) and divide the resulting amount by the voltage (1370 W: 220 V = 6.2 A).

The work of electric current. Each consumer of electrical energy has a meter, which is installed by the organization in charge of energy supply. This meter is installed so that all the current used in the household flows through it. At the same time, it also registers the mains voltage. The consumed electrical energy in the form of power is "counted" by the meter and must be paid. Electric Energy is the work of an electric current. It is equal to the power multiplied by the time during which the energy is consumed:

If the 1000 watt heater mentioned earlier is turned on for 2.5 hours, then it does the following work:

A boiler with a power of 500 W, when connected to the network for 0.5 hours, consumes:

0.5 kW X 0.5 h = 0.25 kWh.

Both appliances thus consume - each during their working time - 2.5 kWh + 0.25 kWh = 2.75 kWh. The meter will increase by this number and the consumed energy will be recorded. When determining the amount of payment for electricity, this amount is multiplied by the cost of one kilowatt-hour (for example, 4 kopecks).

Ministry of Education and Science Russian Federation

federal state budgetary educational institution

higher vocational education

"Tula State University"

Department of Physics

Semin V.A., Semina S.M.

METHODOLOGICAL INSTRUCTIONS

to practical exercises

by discipline

PHYSICS

Head Department of Physics _ D.M. Levin Methodological instructions were revised and approved at a meeting of the Department of Physics of the Faculty of Science, Protocol No. _ dated "" 200_.

Head Department of Physics _ D.M. Levin 1. Goals and objectives of practical classes:

a) Learning the basic physical phenomena and ideas, mastery of fundamental concepts, laws and theories of modern and classical physics, as well as methods of physical research.

b) Formation of scientific outlook and modern physical thinking.

c) Mastering the techniques and methods for solving specific problems from various areas physics.

The scope and timing of this type of work curricula full-time students of specialties 020000 natural Sciences, 090900 information security, 120,000 geodesy and land management, 130,000 geology, mineral exploration, 140,000 energy, power engineering and electrical engineering, 150,000 metallurgy, mechanical engineering and material processing, 160,000 aviation and rocket and space technology, 170,000 weapons and weapons systems, 190,000 200,000 instrument-making and optotechnics, 220,000 automation and control, 230,000 informatics and computer engineering, 240,000 chemical and biotechnology, 260,000 technology of food products and consumer goods, 270,000 construction and architecture, 280,000 life safety, environmental management and environmental protection 2. Lesson plan.

1. Analysis of students' questions on homework.

2. Solving typical problems on the board.

3. Independent solution by students of some tasks in the lesson and summing up.

4. Formulation of homework.

3. Topics of classes.

1. Calculation of tension electric field, created by discrete and distributed charges.

2. Calculation of the electric field potential created by discrete and distributed charges. Calculation of the electric field strength with a known function of the potential (x, y).

3. The charge that has passed through the cross section of the conductor. Joule-Lenz law. Ohm's laws and rules Kirchhoff.

4. Test work on topics 1–3.

5. Calculation of the current through the cross section of the conductor. Ohm's law in local and integral form. Theorem on the circulation of the magnetic induction vector.

6. Superposition of magnetic fields. A coil with current in a magnetic field. Lorentz force.

7. E.D.S. induction and self-induction. Electric damped and forced oscillations.

8. Maxwell's equations. Electromagnetic waves. Pointing vector.

9. Test work on topics 5–8.

10. Additional chapter. Using the Gauss theorem in differential and integral forms.

4. Electronic version http://physics.tsu.tula.ru/students/metodich_files/practich-elmag.pdf Lesson Calculation of electric field strength created by discrete and distributed charges.

A point charge q creates around itself an electric field with intensity kq er, (1.1) E r N m, r is the distance from the charge where k C to the point O, where the field is studied, er is a unit vector directed along the radius 1 vector r from point charge q to point O.

From (1.1) it follows that if the charge q is positive, then the electric field strength E is directed from the point O in the same direction as the vector er. If the charge q is negative, then the vector E is directed opposite to the vector er.

If two (or several) point electric charges are placed in space (see Fig. 1), then they will create an electric field at the point O, the strength of which Eres can be found using the principle of superposition of fields, that is, adding the strengths of the fields E1 and E2 vectorially , created by charges q1 and q2 at the point O independently of each other (parallelogram method). Thus, Figure 1 shows an example with a positive charge q1 and negative charge q2. At the point O, the charge q1 creates a field whose intensity modulus is equal to E1 21. Similarly, the charge q2 at the point O creates a field whose intensity modulus is equal to E2 22. Squaring the left and right parts of formula (1.2), we obtain the expression Erez E12 E22 2 E1 E2 cos, where is the angle between the vectors E1 and E2.

Thus, the modulus of the resulting field is equal to:

If there are three or more electric charges in space, then formula (1.2) is easiest to write in projections on the axes of the Cartesian coordinate system:

Using the Pythagorean theorem and formulas (1.4), one can find the modulus of the resulting field strength:

Charges q1 = 1 μC and q2 = 2 μC are located at the midpoints of adjacent sides of a square with side b = 1 m and create an electric field with a strength Eres at the point P located at the top of the square (see Fig.

rice. 2). Find the value of the horizontal and vertical projection of the vector Eres, as well as its module Eres You can find the cosine and sine of the angle:

We use formulas (3.4) and (3.5), and then also (3.7):

Erez Erez x Erez y 48.92 6, 432 49.3 kV/m The modulus of the vector Erez can be found using formula (3.3) without finding its projection:

Erez where dq – elementary charge, which can be distributed either over the volume dV, or over the surface dS, or on the section of the line dl.

In any of these cases, it is necessary to divide the charged region into small elements and express their charge in terms of density, for example, dq dV for the volume distribution (see Fig. 3). In this case, the application of the superposition principle (1.2) to find the electric field strength E in vector form causes great difficulties due to the infinite number of elementary charges dq distributed in space. In this case, it is necessary to use not the vector addition of the contributions of the fields dE, but the addition of their projections:

Problem example Determine the projection onto the x axis of the electric field strength created by this charge in the center of the semiring, if 0 = 1 μC/m.

As can be seen from Fig. 4, the projection onto the x-axis of the electric field strength created by the elementary charge dq dl at the point O is equal to:

Considering that dl Rd and cos d d sin, we get

1.1 The charge q1 = 1 µC is at the top of the square with side b = 1 m, and the charge q2 = 2 µC is in the center. Find the modulus of the electric field strength at the point P, located at another vertex of this square (see Fig.).

1.2 Charges q1 = 1 μC and q2 = - 2 μC are located at adjacent vertices of a square with side b = 50 cm.

Find the value of the horizontal projection of the electric field strength at the point P, located in the middle of the opposite side of the square (see Fig.).

Find the magnitude of the electric field strength at point A on the continuation of the rod at a distance a = 20 cm from its end (see Fig.). Answer: 180 kV / m A x 2, 0 x b, where x is the coordinate of a point on the rod, b \u003d 3 m is the rod length, A \u003d 2 μC / m3. What is the magnitude of the electric field strength created by this charge at the origin O, which coincides with the end of the rod?

Determine the projections of the electric field strength in the center of the ring on the x and y axes, drawn along two perpendicular diameters, if R = 2 m, 0 = 5 μC/m.

C is oriented towards...

a) 1; b) 2; at 3; d) 4; e) is equal to q1 and at a distance of 2a from q2, then the field strength vector at point C is oriented in the direction ...

1.8s. Charges q1 = 2 μC and q2 = 3 μC are located at adjacent vertices of a square with side b = 1.5 m.

Find the modulus of the electric field strength at the point P, located in the center of the square (see Fig.).

1.9s. The charge q1 = 4 µC is at the top of the square with side b = 60 cm, and the charge q2 = – 3 µC is in the middle of the side. Find the modulus of the electric field strength at the point P, located in the center of the square (see Fig.).

in the middle of the side. Find the value of the vertical projection of the electric field strength at the point P, located at the opposite vertex of the square (see Fig.).

Find the magnitude of the electric field strength at point A on the continuation of the rod at a distance a \u003d 10 cm from its end (see Fig.).

1.12s. A thin rod is unevenly charged. Electric charge distributed along it with the linear density of a point on the rod, b = 4 m is the rod length, А = 3 µC/m4. What is the magnitude of the electric field strength created by this charge at the origin O, which coincides with the end of the rod?

Determine the projection onto the x axis of the electric field strength created by this charge in the center of the semiring, if 0 = 400 nC.

Determine the value of the projection onto the x axis of the electric field strength created by this charge in the center of the ring, if, 0 = μC/m.

Calculation of the electric field potential created by discrete and distributed charges. Calculation of the electric field strength with a known function of the potential (x, y).

The electrostatic field of a point charge is characterized not only by the intensity vector E (see (1.1)), but also by the potential:

From (2.1) it can be seen that the potential is a scalar quantity, which can be either positive or negative, depending on the sign of the charge.

Using the principle of superposition of fields, one can find the potential of the resulting electric field in given pointО as an algebraic sum of field potentials created by each charge independently of each other (see Fig. 1):

Find the potential of the electric field at the point P, located at another vertex of this square (see Fig.).

data into formula (2.2):

To calculate the potential of an electric field created by a distributed charge with a known function of volume, surface or linear charge density, we apply the superposition principle (2.2) in the form - where r is the distance from a small element with a charge dq to point O (see Fig.

3), dq dV for volume distribution, dq dS for surface distribution or dq dl for thin line distribution.

0 = 1 µC/m. Determine the potential created by this charge in the center of the semiring.

We single out the element dl = Rd on the semicircle and, given that the distance from the element to the point O is equal to r R, using formula (2.3) we calculate the potential at the point O:

ordinate of a point on the rod, b = 1 m is the rod length, 0 = 1 µC/m.

What is the magnitude of the potential created by this charge at the origin O, which coincides with the end of the rod?

Let us single out an elementary charge dq on a rod of length dx at a distance x from the origin O (see Fig. 5). Taking into account that r = x and dq = dx, we find by formula (2.3) the potential at the point O:

Let us consider a test particle with an electric charge q0, located in an electrostatic field with strength E and having potential energy W. As you know, the electrostatic field is potential, therefore, the work of the field to move the particle is equal to the loss of potential energy:

From (2.4) we can draw conclusions about the projections of the force acting on the particle:

where W x ; Wy; W z - partial derivatives with respect to x, y, z.

Let's represent the force in vector form:

Gradient of interaction energy of particle with field gradW W, where – i j k – differential operator "nabla".

We divide equation (2.6) by q0 and, taking into account that E,a, we obtain the relationship between the intensity electrostatic field E and electric potential:

where grad is the vector pointing towards the fastest increasing potential.

An equipotential surface is a surface in a force field, at each point of which the potential is the same. Thus, if a particle q0 moves along an equipotential surface, then its potential energy does not change, and no work is done on the particle in this case. From (2.4) it follows that the force acting on the particle is perpendicular to the displacement, and hence to the equipotential surface.

From (2.7) we can conclude that the strength E is directed in the direction of the fastest decrease in the potential perpendicular to the equipotential surface.

The module of the vector E can be found by the formula:

The potential of the electrostatic field depends on the coordinates according to the law Ax10 y10. Find the magnitude of the electric field strength at the point P x0, y0, if A = 2 V/m20, x0 1 m, y0 2 m.

Using formula (2.8), we calculate the projections of the intensity vector E:

The potential of the electrostatic field depends on the coordinates according to the law Ax10 By15. Find the modulus of the electric field strength at the point P x0, y0, if A = 2 V/m10, B = 3 V/m15, x0 1 m, y0 2 m.

Substituting the coordinates x x0, y y0, we get:

We substitute the result in (2.9):

Tasks for work in a practical lesson.

middle side. Find the potential of the electric field at the point P, located in the middle of the opposite side of the square (see Fig.). Answer: 130 kV 2.2 The charge q1 = 2 μC is located at the top of the square with side b = 40 cm, and the charge q2 = –3 μC is in the middle of the side. Find the potential of the electric field at the point P located in the middle of the side of the square at a distance a = 60 cm from its end (see Fig.). Answer: 22.0 kV 2.4 The positive charge is distributed over a thin half-ring of radius R = 50 cm with linear 0 = 2 μC/m. Determine the potential created by this charge in the center of the semiring. Answer: 11.7 kV 2.5 The potential of the electrostatic field depends on the coordinates according to the law Ax 5 y 2, where A = 4 V/m7. Find the magnitude of the electric field strength at the point P x0 1 m, y0 2 m. Answer: 81.6 V / m.

2.6 The potential of the electrostatic field depends on the coordinates according to the law Ax 4 By 3, where A = 2 V/m4, B = 3 V/m3. Find the magnitude of the electric field strength at the point P x0 2 m, y0 3 m.

2.7 The potential of the electrostatic field depends on the coordinates according to the law sin Ax B sin Cy. Find the magnitude of the electric field strength at the point P x0, y0. A 2 rad/m, B 3 V, C 4 rad/m, 2.8e. The figure shows the equipotential lines of the system of charges and the potential values ​​on them. The electric field strength vector at point A is oriented in the direction ...

2.9e. In some region of space, an electrostatic field is created, the intensity vector of which at the point Р(x1,y1) is directed along the x axis. What dependence of the electric field potential on the coordinates x, y can correspond to such a direction of intensity?

2.10e. placed on a metal ball positive charge Q. The dependence of the electric field potential on the distance to the center of the ball will be described by a graph...

2.11e. Two infinite parallel plates are uniformly charged with surface charge densities equal in magnitude and opposite in sign. If the X axis is directed perpendicular to the plates, then the dependence of the electric field strength on x will be represented by a graph ...

2.12e. The potential of the electric field depends on the x-coordinate, as shown in the figure. Which figure correctly reflects the dependence of the projection of the electric field strength on the x coordinate?

adjacent vertices of a square with side b = 20 cm.

Find the potential of the electric field at the point P dividing the side of the square into two equal segments (see Fig.).

2.14s. The charge q1 = 4 μC is located at the top of the square with side b = 40 cm, and the charge q2 = – 5 μC is at middle side. Find the potential of the electric field at the point P, located at the opposite vertex of the square (see Fig.). Answer: - 37.0 kV point A on the continuation of the rod at a distance a = 20 cm from its end (see Fig.). Answer: 29.0 kV 2.16s A thin rod is unevenly charged. The electric charge is distributed over it with a linear density of 0 x 5, 0 x b, where x is the coordinate of a point on the rod, b = 2 m is the length of the rod. What is the magnitude of the potential created by this charge at the origin O, which coincides with the end of the rod, if 0 = 10 μC/m6?

the potential created by this charge in the center of the semiring, if 0 = 1 μC/m. Answer: 14.1 kV 2.18s. The potential of the electrostatic field depends on the coordinates according to the law 5 x 3 6 y 4 (B). Find the magnitude of the electric field strength at the point P x0, y0 x0 3 m, y0 2 m. Answer: 235 V/m 2.19s. The potential of the electrostatic field depends on the coordinates according to the law A exp Bx C cos Dy. Find the magnitude of the electric field strength at the point P x0, y0. A 1 V, B 2 m–1, C 3 V, 2.20 s. In a certain region of space, an electrostatic field is created, the intensity vector of which at the point Р(x1,y1) is directed at a certain angle to the x axis (see Fig.). What dependence of the electric field potential on the coordinates x, y can correspond to such a direction of intensity?

2.21s. An electron moves in the Coulomb field of a charged particle from point A to point B in one case along trajectory 1, in the other case along trajectory 2. How do the magnitudes of the work performed by the electric field on the electron correlate in these two cases?

A charge that has passed through the cross section of a conductor.

Joule-Lenz law. Ohm's laws and Kirchhoff's rules.

The current strength is defined as the charge flowing through the cross section of the wire per unit of time, i.e.

If the dependence of the current strength I t is known, then from (3.1) we can express the charge flowing in a short period of time:

and for any period of time where 2 1 is the potential difference.

Ohm's law uses another value - voltage or voltage drop: U 1 2. Thus (3.4) can be rewritten in a different form: A qU.

For a small time interval, using (3.2), we transform (3.4) as follows:

where P IU is the electric power.

Using Ohm's law for a homogeneous section of the chain U IR, and substituting it into (3.5), we obtain the Joule-Lenz law:

Formula (3.6) takes into account the fact that the work of the electric field done on electric charges does not lead to an increase in their kinetic energy, and is released in the form of heat dQ.

An alternating electric current flows through a wire with a resistance R1 \u003d 20 Ohm. The current strength changes according to the law I 5t10 (A). What is the amount of heat released in the wire and the amount of electricity that has passed through the cross section of the wire over a period of time from Let us substitute the function of current strength from time into formula (3.3) and (3.6):

An alternating electric current flows through a wire with a resistance R1 \u003d 30 Ohm. The current strength changes according to the law I A sin t, where A \u003d 4 A / s, rad / s. What is the amount of heat released in the wire and the amount of electricity that passed through the wire during the time interval from t0 0 to t1 = 0.5 s?

We substitute the function of the current strength from time into the formula (3.3) and (3.6):

many elements such as resistors, capacitors, current sources, inductors. These elements are connected. A contour is a closed line drawn along the connecting wires so that it does not intersect itself anywhere. Figure 6 shows two circuits I and II. The traversal along these contours is chosen here clockwise (in general, you can choose arbitrarily).

Usually, the characteristics of all elements included in the circuit are known, i.e. resistor resistances, E.D.S. current sources, etc.

Kirchhoff's rules can help with this.

or the algebraic sum of all the strengths of the currents converging in the node is 0.

The currents flowing into the node are taken with the "-" sign, and the currents flowing out - the algebraic sum of the voltage drops on each circuit element is equal to the algebraic sum of the emf. in this circuit.

The voltage drop across the resistance is considered positive if the direction of the current through this resistance coincides with the direction of bypassing the circuit, chosen arbitrarily.

E.D.S. is considered positive if, when bypassing the contour, a transition is made through the source from "–" (smaller segment) to "+" (larger segment).

We write formula (9.3) for two contours:

If some currents are known, then the calculation of the circuit is simplified, and you can sometimes get by with solving just one equation.

Let us write formula (9.3) for contour I (see Fig. 7).

And let's express E1 from here:

Tasks for work in a practical lesson.

3.1 An alternating electric current flows through a wire with a resistance R1 \u003d 2 ohms. The current strength changes according to the law I A t 3, where A \u003d 2 A / s3 / 2. What is the amount of heat released in the wire and the amount of electricity passed through the wire during the time t = 2 s?

3.2 An alternating electric current flows through a wire with a resistance R1 \u003d 3 ohms. The current strength changes according to the law I A exp Bt, where A = 5 A, B = 0.5 s–1. What is the amount of heat released in the wire during the time t1 \u003d 2 s, as well as the charge that passed through the wire during the same time?

3.3 An alternating electric current flows through a wire with a resistance R1 = 3 ohms. The current strength changes according to the harmonic law I A cos t, where A \u003d 4 A, \u003d / 3 s–1. What is the amount of heat released in the wire for half a period, as well as the charge that passed through the wire during the same time?

3.4. Through the resistance R \u003d 5 Ohm, a current begins to flow, increasing with time according to the law I At 2. What heat will be released on the resistance by the time t \u003d 5 s, if during this time through the resistance I 4 and I 5 flowing through the resistors R1, R4 and R5 respectively;

2) emf value. E2.

Ans.: I1 1 A (to the left); I 4 1.5 A (down); I 5 3.5 A (up); E2 = 1 V.

3.6e. The figure shows a part of the electrical circuit for which only I3 1 A is known. What is the current strength through the resistance R2?

a) 1.0 A; b) 0.6 A; c) 0.5 A; d) cannot be calculated, because not enough data 3.8e. The electric field strength in the conductor was increased by 2 times. How has the specific thermal power changed (heat released per unit time per unit volume)?

a) doubled b) increased by 4 times;

c) increased by 8 times; d) decreased by 2 times.

3.9e. The strength of the current flowing through the conductor varies with time, as shown in the figure. What charge will flow through the cross section of the conductor in the time interval a) 7 C; b) 12 C; c) 10.5 C; d) 1.5 C.

3.10e. A 10 ohm rheostat is connected to a current source with an internal resistance of 1 ohm, as If the rheostat slider is moved from the extreme right position to the left, then the current power in the rheostat will be ... a) first increase and then decrease b) first decrease and then increase c) increase continuously d) decrease continuously 3.11e. A 0.5 ohm rheostat is connected to a current source with an internal resistance of 1 ohm, as if the rheostat slider is moved from the extreme right position to the left, then the current power in the rheostat will be ... a) first increase and then decrease b) first decrease and then increase c) increase continuously d) decrease continuously 3.12s. An alternating electric current flows through a wire with a resistance R1 \u003d 25 ohms. The current strength changes according to the law I A sin t, where = 40 s–1. What is the amount of electricity that has passed in half the period if 5 J of heat has been released in the wire during this time?

3.13s. An alternating electric current flows through a wire with a resistance R1 \u003d 12 ohms. The current strength changes according to the law I A exp Bt, where B = 0.01 s–1. How much heat will be released in the wire in two seconds if a charge of 5 C passed through the wire during this time?

source E2. What is the emf E3?

Answers: 0.33 Ohm, I 3 \u003d 1 A (right), I 4 \u003d 0.75 A (down), 3.15e. The figure shows a part of the electrical circuit for which only some parameters are known: R1 4 ohms, R2 1 ohms, and the source is 1 5 V and has zero internal resistance. The potentials are 1 8 V, 2 2 V, and the current through the resistance R1 is I1 1 A.

Ohm's law in local and integral form.

Theorem on the circulation of the magnetic induction vector.

The current density is equal to the strength of the current flowing through a single area located perpendicular to the current lines:

Knowing the distribution of the current density in space, we can calculate the total current through an arbitrary surface S:

where the vector dS dS n, and n is the unit vector of the normal to the area dS ; is the angle between the vectors j and n.

If the conductor is made in the form of a thin strip, and the linear current density i is known, then where dx is the width of the strip along which the current dI flows. Ohm's law in local form states that the current density is proportional to the electric field strength E that creates this current:

where is the specific conductivity of the substance that conducts the current.

The reciprocal of conductivity is called resistivity:

From (2.8) in the case of a uniform electric field in Transforming (4.3) we can derive Ohm's law for a homogeneous section of the circuit:

is the resistance of a section of length l with a cross section S.

A current flows through an inhomogeneous cylindrical wire of radius R = 2 mm. Find the strength of the current flowing through the cross section of the conductor if the dependence of the current density on the distance r from the axis is given by the form Divide the cross section of the conductor (a circle of radius R) into rings of radius r and width dr (see Fig.

Fig. 8). The area of ​​such a ring is dS 2rdr, and the angle between j and dS is 0. Using formula (4.2), we find the total current flowing through the entire cross section of the conductor:

through the cross section of the conductor, if the current density depends on the distance x from one of the side faces according to the law j x j0, where j0 2 A/mm2; b = 5 mm.

(square b b) into narrow strips dx wide and b high (see Fig. 9). The area of ​​such a strip is equal to dS bdx and the angle between j and dS is equal to 0. Using formula (4.2), we find the total current flowing through the entire cross section of the conductor:

Task example Let us single out on the plane parallel to the midline at a distance x a narrow strip of width dx (see Fig. 10). Using formula (4.3) find the theorem on the circulation of the magnetic induction vector:

– circulation in a closed loop of the induction vector magnetic field equal to algebraic bounded contour multiplied by the magnetic constant 0 4 107 H/m. The current strength is considered positive if the direction of the current at the point of intersection with the surface S coincides with the direction of the positive normal to the surface at this point, and negative if the direction of the current is opposite to the direction of this normal. The positive normal is determined by the rule of the right screw with respect to the direction of bypass Г (see Fig.).

If one of the currents is covered by the circuit N times, then in formula (4.9) such a current will add up N times.

A current flows through a cylindrical conductor of radius R = 2 mm, the density of which varies with the distance r from the conductor axis according to the law j j0 exp Br 2, where B 1 mm–2 j0 = 3 A/mm2. Find the induction of the magnetic field at a point located at a distance r1 R 2 from the axis of the conductor.

line of induction of the magnetic field in the form of a circle of radius R 2, the axis of which coincides with the axis of the conductor. We divide the cross section of this circuit into strips of radius r, width dr and area dS 2rdr and find the total current flowing through this circuit:

Tasks for work in a practical lesson.

4.1. A current flows through an inhomogeneous cylindrical wire of radius R = 2 mm. Find the strength of the current flowing through the cross section of the conductor, if the current density depends on the distance r to the axis according to the law:

b) j r j0 exp Br 2, where j0 = 4 A/mm2, V= 0.01 mm–2.

section b b current flows. Find the strength of the current flowing through the cross section of the conductor, if the current density depends on the distance x of the current flowing over the entire strip, if the linear current density depends on the distance x up to 4.4 Oe. In two homogeneous cylinders made of the same material, flows D.C.. What can be said about the relationship between the current densities in cylinder A and in cylinder B?

a) Based on the figure, it is impossible to say for sure. You need to know the exact relationship between the length and area of ​​a cylinder.

constant voltage. What can be said about the relationship between the magnitudes of the tensions d) Based on the figure, it is impossible to say definitely. You need to know the exact relationship between the length and area of ​​a cylinder.

4.6e In some closed circuit there is a section consisting of two resistors connected in series. At the connection points of resistors A and B, the potentials 1 and 2 are known (see Fig.).

Potential 3 at point C is ... a) 6 V b) 0 V c) 7.5 V d) -1.5 V 4.7e In some closed circuit there is a section consisting of three resistors connected in series. At the connection points of resistors A and C, the potentials A and C are known (see Fig.). Thermal power is allocated in the NPP section equal to... a) 20 W b) 36 W c) 28 W d) 14 W 4.8. A current flows through a cylindrical conductor of radius R, the density of which varies with the distance r from the axis of the conductor according to the law j j0, where j0 = const. Find the ratio of magnetic field inductions at points located at distances r1 2 R and r2 R 2 from the axis of the conductor. Answer: B1 B2 2.

4.9. Long wires of different configurations carry different currents. I1 1 A, I 2 2 A, I 3 3 A, I 4 4 A, I 5 5 A. Find the circulation of the magnetic field induction vector created by these currents along the closed loop G.

4.10c. A current flows through an inhomogeneous cylindrical wire of radius R = 3 mm. Find the strength of the current flowing through the cross section of the conductor, if the current density depends on the distance r to the axis according to the law b) j r j0 sin Br 2, where j0 = 3 A/mm2, V= 0.01 mm–2.

flowing through the cross section of the conductor, if the current density depends on the distance x from one of the side faces according to the law:

4.12c. A current flows along the middle line of a conducting strip with a width of 2b = 8 mm. Find the strength of the current flowing through the entire strip if the linear current density depends on the distance x 4.13e. Two homogeneous cylinder of the same material are connected in parallel to a constant voltage source. What can be said about the ratio of thermal power PA and PB released in these cylinders?

d) Based on the figure, it is impossible to say for sure. You need to know the exact relationship between the length and area of ​​a cylinder.

the ratio of thermal power PA and PB released in these d) Based on the figure, it is impossible to say for sure. You need to know the exact relationship between the length and area of ​​a cylinder.

circuit there is a section consisting of three resistors connected in series. At the connection points of resistors A and C, the potentials A and C are known (see Fig.).

In the NPP section, a thermal power equal to ...

4.16c. Long wires of different configurations carry different currents. Find the circulation of the magnetic field induction vector created by these currents along the closed loop G.

a) 7.3; b) - 9.3; c) 9.3; d) - 11.3; e) 11.3. (μT m) A coil with current in a magnetic field. Lorentz force.

Consider a few simple examples of creating a magnetic field by electric currents of different configurations:

- induction of the magnetic field of a straight wire at a distance R from it.

is the magnetic field induction at the center of the coil with radius R.

- induction of the magnetic field created by a segment with current at point O at a distance a from the line on which The direction of the magnetic field B is determined by the right screw rule (see the drawings for formulas (5.1) - (5.3)).

The induction of a magnetic field created by a conductor of complex configuration is found according to the principle of superposition of fields:

where Bi is the induction of the field created by a part of a wire with a simple shape.

Electric current I = 1 A flows through a long wire, bent as shown in Fig.4. Find the induction of the magnetic field created by this current at point O if R = 1 m.

As can be seen from Fig. 11, the magnetic field at point O is created by a segment of length R and an arc of radius 2R with a turn angle of 135. An electric current directed to point O does not create a magnetic field in it.

We use formula (5.2) to find the induction of the magnetic field created by the arc:

We find the induction of the magnetic field created by the segment using formula (5.3), substituting the following data:

The resulting field is equal to the sum of these fields, since the induction vectors B of the arc and B of the segment are directed at the point O in one direction:

Cut in the back of the arc 0.237 µT Answer: 0.237 µT;

as shown in Fig.5. Find the induction of the magnetic field created by this current at the center of a circle of radius R = 1 m.

Fig.12 Let's find separate contributions to the magnetic field induction at the point O (the center of the circle), created by two semi-infinite straight conductors and a conductor in the form of an arc with a turn angle.

For a beam with a current flowing against the x-axis at a distance R from the point O, we use the half contribution from formula (5.1):

Similarly for the second beam with current flowing along the y-axis at a distance of 2R from the point O:

For a conductor in the form of an arc in 3 4 circles with radius R, we use formula (5.2):

The directions of vectors B1, B2 and B3 are different:

Using the superposition principle (5.4) and the Pythagorean theorem, we find the modulus of induction of the resulting magnetic field at the point O:

A small coil of area S with current I has a magnetic moment pm I S I S n, which is directed along the positive normal n, determined by the right screw rule with respect to the direction of current through this coil. Such a magnetic moment, interacting with an external magnetic field with induction B, has an interaction energy. In an effort to occupy the position in space with the lowest potential energy (5.5), the coil turns its magnetic moment along the field induction B. In an inhomogeneous magnetic field, such a coil is affected by a force that tends to pull the coil into an area with greater induction.

If a particle with an electric charge q and mass m flies with a speed v into a magnetic field c which is perpendicular to the speed of the particle v and induction B. This leads to a curvature of the trajectory without changing the speed of the particle (since the Lorentz force does not do work).

Consider the situation when a particle flies into a magnetic field perpendicular to the induction B. In this case, it will move in a circle with a constant speed, and the Lorentz force will be a centripetal force (see Fig. 13).

Find the radius of the circle using Newton's second law:

Ampere's force acts on the section of the conductor dl with current I in a magnetic field:

A positively charged particle with a charge q = 1 μC and a mass m flies in a circle about the time t, after which the particle velocity will be directed a) along the x axis; b) against the x-axis. Find the distance traveled during this time.

From the vector expression (7.1) it follows that the Lorentz force acting on the particle at the initial moment of time is directed along the x axis, so the particle will move as shown in Fig.14. It follows from this figure that after a quarter of a turn or after a time t T 4 the velocity of the particle will be directed parallel to the x axis, and after three quarters of the period (t 3T 4) it will be antiparallel to the x axis. Using the formula for the radius of the circle (5.8) and the period (5.9), we get the answer:

Answers: a) t = 0.157 ms; S = 1.57 m; b) t = 0.471 ms; S = 4.71 m.

Tasks for work in a practical lesson.

5.1. Electric current flows through a long wire bent as shown in the figure. Find the induction of the magnetic field created by this current at the center of the semicircle and at the center of the rectangle. I \u003d 1 A, R \u003d 1 m, a \u003d 1 m, b \u003d 2 m.

Answers: a) 0.314 µT; b) 0.414 μT; c) 0.214 μT; d) 0.894 µT 5.2. Electric current flows through a long wire bent as shown in the figure. Find the induction of the magnetic field created by this current at the center of the arc. I = 1 A, R = 1 m, 1200.

Answers: a) 0.209 μT; b) 0.109 μT; c) 0.309 μT;

Z. A coil of radius R with current I 2 is located parallel to the XY plane. The center of the coil lies on the Y axis at a distance of 2R from the origin. Find the induction of the magnetic field created by these currents at the center of the coil. I1 1 A, I 2 2 A, R = 1 m. Answer: 1.26 µT 5.4. Current I flows through a long wire bent as shown in the figure. Find the induction of the magnetic field created by this current at the center of a circle of radius R. I 1 A, R = 1 m.

Answers: a) 0.426 μT; b) 0.236 μT; c) 0.314 µT; d) 0.372 μT;

5.5. A small coil with current, having a magnetic moment pm 1 Am2, is kept in a non-uniform magnetic field on the x-axis at an angle = 60 to it. Determine the projection of the force Fx acting on the coil at the point with coordinate x0 = 1 m, if the magnitude of the magnetic field induction on the x axis changes according to the law B x Ax 3, where A 1 T/m3. Answer: 1.5 N;

A charged particle with a charge m/s flies in a circle. The magnetic field induction B = 1 μT and is directed along the y axis. At the initial moment of time, the particle velocity v was directed along the x axis. Find:

A) after what time t the velocity of the particle for the first time becomes directed along the z axis;

B) the path S traveled by the particle during this time;

C) the maximum removal of the particle from the x-axis;

D) the maximum distance from the z-axis, Answers: A) 0.236 s; B) 47.1 m; C) 20 m; D) 10 m.

5.7e. The magnetic field is created by two long parallel conductors with currents I and I2, located perpendicular to the plane of the drawing. If I1 = 2I2, then the induction vector B of the resulting field at point A is directed...

two straight long parallel conductors with oppositely directed currents, and I1 2 I 2. The induction B of the magnetic field is zero at some point in the section ...

1) a; 2) b; 3) c; 4) d; 5) there is no such point; 6) in the middle between the wires;

5.9e. A current-carrying loop with a magnetic moment pm is in a uniform magnetic field with induction B. Where is the moment of forces acting on the loop directed?

a) perpendicular to the drawing "from us";

b) perpendicular to the drawing "to us";

c) along the induction of the magnetic field;

d) against the induction of a magnetic field.

5.10e. The figure shows the trajectories of charged particles having the same speed 5.11 oersteds. In a magnetic field for a horizontal conducting rod hangs on two threads. Thread tension is zero. What is the relationship between the directions of the magnetic field and the current in the rod?

a) the current flows from L to M, the induction is directed away from us;

b) the current flows from L to M, the induction is directed to the right;

c) the current flows from M to L, the induction is directed away from us;

d) the current flows from M to L, the induction is directed upwards;

5.12s. Electric current flows through a long wire bent as shown in the figure. Find the induction of the magnetic field created by this current at the center of the circle and square, if I = 1 A, R = 1 m, a = 1 m.

Answers: a) 0.314 µT; b) 0.528 μT; c) 0.428 μT. d) 1.13 µT 5.13s. Current I 1 A flows through a long wire bent as shown in the figure. Find the induction of the magnetic field created by this current at the center of a circle of radius R= 1 m.

Answers: a) 0.537 μT; b) 0.537 μT; c) 0.384 μT; d) 0.481 μT;

5.14s. A small coil with current, having a magnetic moment pm 2 Am2, is kept in a non-uniform magnetic field on the x axis at a point with coordinate x0 = 0.5 m. The direction of the magnetic moment of the coil is opposite to the direction of the magnetic field induction. Determine the projection of the force Fx acting on the coil if the magnitude of the magnetic field induction on the x axis changes according to the law B x Ax 5, where A = 3 T/m5. Answer: - 1.875 N.

5.15e. An electron flies in a circle in a uniform magnetic field as shown in the figure. Where is the direction of the magnetic field induction vector?

currents I1 and I2, located perpendicular to the plane of the drawing. If I1 = 2I2, then the induction vector B of the resulting field at point A is directed...

a) 1; b) 2; at 3; d) 4; e) B 15.17e. Two coaxial turns carry the same current in the same direction. The distance between the centers of the turns is 2 cm. The upper turn creates a magnetic field with induction B = 1 μT at point A located on the axis at a distance of 1 cm from its center. What is the magnitude of the induction of the magnetic field created by two turns?

a) 2 μT; b) 0 μT; c) 2 μT; d) 4 μT.

moving in parallel lines at some distance from each other. The magnetic force acting on a right-handed charge has a direction...

Electric damped and forced oscillations Let's consider a closed contour G of arbitrary shape in a non-uniform magnetic field, which limits a certain surface S (see Fig. 15). where is the angle between the vector B and the normal n to the area of ​​the surface dS through which the magnetic field penetrates.

When the flow F changes in time, an E.D.S. occurs in the G circuit.

induction - electromotive force, equal to the rate of change of the magnetic flux (law electromagnetic induction Faraday):

If the circuit were made of a conductive substance, then an electric current would flow through it.

The flow F can change for the following reasons.

1) The induction of the magnetic field B changes.

2) The geometric dimensions of the contour change, i.e. the area S changes.

3) The orientation of the contour in space changes, i.e. angle changes.

In case 1), a vortex electric field Evortex arises in space, which acts on the free electrons of the conducting circuit.

In cases 2) and 3), due to the movement of the conductor in a magnetic field, the Lorentz force acts on free electrons in it.

where the proportionality factor L is called the loop inductance. If the current in the circuit begins to change, then an E.D.S. will appear in it. self-induction:

The sign "-" in formulas (6.2) and (6.4) means that when the magnetic flux changes through a closed circuit, such an EMF arises in it, which tends to reduce the flux change. This is Lenz's rule. As a result of the increase in current strength in Fig. 16, and hence the induction B, there is a vortex electric field directed against the current I in the circuit.

penetrates a uniform magnetic field at an angle Find the emf modulus. induction in the circuit at time t = 1 s, if We determine the dependence of the magnetic flux on time:

According to formula (6.2), we determine the module of E.D.S. induction:

A current I flows through a conducting circuit with inductance L. Find the emf module. self-induction in the circuit at the time t \u003d 1 s, if both the current and the inductance change with time according to the laws We use the formula (6.4):

Fig. 17), consisting of a series-connected resistor with resistance R, a capacitor with capacitance C and where 0 is the initial phase of oscillations, is the cyclic frequency of natural damped oscillations.

The damping coefficient and the cyclic frequency of natural undamped oscillations 0 are determined as follows:

The logarithmic damping decrement and the relaxation time (the time during which the amplitude decreases by a factor of e = 2.72) are defined as follows:

The amplitude of oscillations in the circuit decreases with time according to the law:

where A0 is the initial amplitude. Since the energy of undamped and weakly damped oscillations is proportional to the square of the amplitude W A2, then, using (6.9), we obtain:

Free weakly damped oscillations occur in the circuit, in which the charge on the capacitor changes with time according to the law q q0 exp at sin bt. Estimate the time after which the energy of the circuit will decrease by a factor. q0 1 μC; a 0.05 s–1; b 10 s–1. What will be the attenuation coefficient if:

a) increase the resistance R in the circuit by 2 times?

b) increase the inductance L in the circuit by 2 times?

c) increase the capacitance C in the circuit by 2 times?

The energy of the circuit W is proportional to the square of the oscillation amplitude, therefore, using formula (6.9) and taking into account that = a = 0.05 s–1, we obtain: W A2 q0 e 2 at From the ratio of the energies of the circuit at the initial moment of time and at the moment of time t from formula (6.7) it follows that a) if the resistance in the circuit is doubled, then the attenuation coefficient will also increase twice: b) when the inductance changes twice, the attenuation coefficient c) when the capacitance changes twice, the attenuation coefficient does not will change, since it does not depend on the capacitance C (see formula (6.7)).

Answers: t = 6.93 s; a) = 0.1 s–1; b) = 0.025 s–1; c) = 0.05 s–1.

in and amplitude q0. The dependence of the charge on the capacitor on time will look like this:

In order to find an expression for the current strength in the circuit, we differentiate (6.11) with respect to time:

where I 0 q0 в - current amplitude To calculate the voltage drop across the inductor, the expression for the self-induction EMF is used, but with the opposite sign U L L dI dt. Substituting expression (6.12) here, we get:

where U 0 L q0 L2 is the amplitude value of the voltage on the inductor.

Now you can analyze the phases of voltage fluctuations on the circuit elements: on the capacitor, inductor and resistor.

The voltage across the capacitor can be found from (6.11):

where U 0C q0 C is the voltage amplitude across the capacitor. From (6.14) and (6.13) it can be seen that the voltages on the capacitor and on the inductor oscillate in antiphase.

We find the voltage across the resistor from Ohm's law and (6.12):

where U 0 R q0 in R is the voltage amplitude across the resistor. From (6.15) and (6.14) it can be seen that the voltage across the resistor is ahead in phase by the voltage across the capacitor.

Since the circuit elements are connected in series (see Fig. 18), the voltage at the source terminals is the sum of the voltages across the capacitor, coil and resistor. But it is necessary to add such voltages taking into account the phases, that is, use the phase diagram.

Substituting into (6.16) expressions for stress amplitudes from (6.13), (6.14) and (6.15), we obtain the expression Fig.19. Phase If (6.16) is divided by the amplitude of the current I 0 from (6.12), then you can find the circuit impedance or impedance:

where X L U 0 L I 0 Lв - reactive inductive resistance;

X C U 0C I 0 1 in C - reactive capacitance;

R U 0 R I 0 – active resistance resistor.

The expression X X C X L is called the total reactance of the circuit.

From (6.19) you can find an expression called the amplitude-frequency characteristic for the current:

Analyzing the amplitude-frequency characteristics (6.17) and (6.20) for the charge and current, one can find the resonant frequencies at which the amplitudes q0 and I reach a maximum:

From (6.21) it can be seen that the resonant frequency for the charge on the capacitor is less than for the current. But if the attenuation is weak, i.e. 0, then these frequencies can be approximately considered equal.

Tasks for work in a practical lesson.

6.1. A circular conducting coil of radius R = 1 m penetrates a uniform magnetic field at an angle = 60 to the normal coil. The magnetic field induction changes with time according to the law B t At 5, where A = 3 T/s5. Find the emf module. induction in the circuit at time t \u003d 2 s Answer: 376.8 At an angle \u003d 30 to the plane of the circuit. The magnetic field induction changes with time according to the law B t At 3, where A = 4 T/s3. Find the emf module. induction in the circuit at the moment 6.3. A current I flows through a conducting circuit with inductance L. Find the emf module. self-induction in the circuit at the time t \u003d 2 s, if both the current and the inductance change with time according to the laws of the figure. Find the modulus of the average value of the EMF of self-induction in the time interval from t1 = 0 to t2 = 20 s.

perpendicular to the magnetic field lines. The magnitude of the induction varies depending on time according to the law B 2 5t 2 102 T. What is equal to magnetic flux through the frame?

6.5. Free oscillations occur in the circuit, in which the charge on the capacitor changes with time according to the law q q0 exp 4t sin 3t, where q0 1 μC. Find the logarithmic decrement of the contour damping. What will be the period of oscillation if the resistance R is reduced to zero?

Answers: \u003d 8.37, T \u003d 1.256 s.

6.6. Free oscillations occur in the circuit, in which the charge on the capacitor changes with time according to the law q q0 exp 5t sin 4 3t, where q0 3 μC. What will be the relaxation time of oscillations if:

a) remove one resistance R from the circuit?

b) add one more resistance R in series?

6.7. Free weakly damped oscillations occur in the circuit, in which the charge on the capacitor changes with time according to the law q q0 e 0.1t sin 3t, where q0 5 mC.

How many times will the circuit energy decrease in t \u003d 1 s?

6.8e. The figure shows a graph of damped oscillations of an electric charge on a capacitor, described by the equation for the current in the inductor of an oscillatory circuit, consisting of a capacitor with a capacitance C, a coil with an inductance L and a resistor with a resistance of a) 40 H; b) 5 H; c) 2.5 H; d) not enough data capacitor connected in series and connected to the source alternating current, changing according to the law I 0.1cos 3.14t (A). The figure shows the phase diagram of voltage drops on the indicated elements. The amplitude values ​​of the voltages are respectively equal: on the resistance U R V, on the inductor U L 5 V, on the capacitor U C 2 V. Establish a correspondence between the resistance and its numerical value.

3. The total resistance permeates the uniform magnetic field at an angle = 30 to the normal of the coil. Induction The magnetic field changes with time according to the law B t At 4, where A = T/s4. Find the emf module. induction in the circuit at the moment t = 2 s.

Answer: 1740 B 6.13s. Free oscillations occur in the circuit, in which the charge on the capacitor changes with time according to the law q q0 exp 4t sin bt, where q0 3 μC.

Find the cyclic oscillation frequency if the logarithmic damping decrement of the circuit is = 2. Answer: 12.56 s– 6.14 s. Free oscillations occur in the circuit, in which the charge on the capacitor changes with time according to the law q q0 exp 4t sin 3t, where q0 2 μC.

What will be the relaxation time of oscillations if:

a) add another resistance R in parallel?

b) remove one resistance R? Answers: a) 0.375 s; b) 0.125 s;

6.15s. Free oscillations occur in the circuit, in which the charge on the capacitor changes with time according to q q0 exp 4t sin 3t, where q0 4 μC.

b) add another resistance R in parallel?

and II can correspond to the frequency response of the following values:

a) I - charge on the capacitor; II - current in the coil;

b) I - charge on the capacitor; II- voltage on the capacitor;

c) I - current in the coil; II - charge on the capacitor;

and the capacitor are connected in series and connected to an alternating voltage source, the elements. Establish a correspondence between the amplitude values ​​of the voltages on these elements and the amplitude value of the source voltage.

Answer: 1 - a); 2 – b) 6.18e. In a uniform magnin field, a conducting jumper moves with a uniformly increasing speed (see Fig.). If the resistance of the jumper and guides can be neglected, then the dependence induction current from time to time can be represented by a graph ...

Electromagnetic waves. Pointing vector.

An alternating magnetic field with induction B generates in space a vortex electric field with strength E, for which the circulation theorem in integral and differential form is written as follows:

A vortex magnetic field with strength H is generated in space by conduction currents with density j and an alternating electric field with induction D. The theorem on the circulation of the vector H in integral and differential form looks like this:

jcm is called the displacement current density.

where If to the equations (7.1) and (7.2), which are the theorems on the circulation of the vectors E and H, add the Gauss theorems in the integral and differential form for the vectors D and B where is the volume density of external charges.

then we get the system of Maxwell equations (7.1) - (7.4), which is supplemented by constitutive equations that are valid in an isotropic non-ferromagnetic substance in weak fields. Here, is the dielectric constant of the medium, the magnetic permeability of the medium, is the specific conductivity of the medium.

0 = 8.85 1012 F/m is the electrical constant.

0 = 4 107 H/m is the magnetic constant.

A uniform electric field is created between the plates of a flat air capacitor, the magnetic field strength (or displacement current density) inside the capacitor at the time t = 0.5 s, if E0 1 kV/m.

There are no conduction currents between the capacitor plates, i.e. 0 j

8.85 1012 103 2 3.14 1 55.6 10 9 A/m A uniform magnetic field is created between the poles of the magnet, the induction of which depends on time according to the law B B0 cos. Find the modulus of the electric field strength between the poles at a distance r = 5 cm from the axis of the magnet at time t s, if B0 2 T.

Using formula (7.1), in integral form, we find the circulation E along a closed contour in the form of a circle of radius r with an axis coinciding with the axis of the magnet, and express the intensity modulus:

The equation for a plane electromagnetic wave propagating along the x axis is written for both the vector E and H:

where is the magnetic permeability of the medium (=1 for vacuum and air), is the permittivity of the medium.

A particular case of solution (7.6) is the equation of a plane electromagnetic wave:

where is the cyclic frequency of oscillations of the vectors E and H, k is the wavenumber.

From (7.7) it follows that the oscillations of the electric and magnetic vectors occur in one phase with amplitudes E0 and H 0. The values ​​of these amplitudes are related by the relation:

whence follows the equality of the volumetric energy densities of the magnetic and electric fields in the wave:

The Poynting vector (energy flux density of an electromagnetic wave) is directed along the wave velocity v em (see Fig. 20).

The Poynting vector can be expressed in terms of the volumetric density of electromagnetic energy wem wel wmagn:

The energy carried by an electromagnetic wave through an arbitrary surface S in time is found as an electromagnetic wave incident at an angle to the normal of the surface and partially reflected by it, exerts pressure on it:

where r is the reflection coefficient.

Tasks for work in a practical lesson.

7.1. Between the plates of a flat air capacitor, an alternating homogeneous electrical field. Find the displacement current density inside the capacitor at the moment t = 0.5 s, if the electric field strength changes with time according to the law Answers: a) 70.8 nA/m2; b) 29.5 nA/m 7.2. An alternating uniform magnetic field is created between the poles of a magnet. Find a module electrical force, acting on a charged particle with a charge q = 4 μC, located in a magnetic field depends on time according to the law Answers: a) 7.84 1010 N; b) 2, 7 108 N 7.3. A plane electromagnetic wave propagating in a dielectric is described by the wave function u u 0cos(ax bt), where a = 0.04 m –1, b 6 106 s 1. Find permittivity dielectric. The speed of light in vacuum is c = 3108 m/s.

7.4e. The equation of a plane wave propagating along the OX axis has the form 0.01e equals...

electromagnetic wave. Energy flux density vector electromagnetic field oriented towards...

7.6e. In an electromagnetic wave, the intensity vectors of the electric and magnetic fields oscillate so that the phase difference of their oscillations is ...

7.7e. Light falls on a black plate. If the volume density of the electromagnetic energy of the wave is doubled, and the area of ​​the plate is doubled, then the light pressure on the plate...

a) will decrease by 2 times; b) will increase by 2 times;

c) will increase by 4 times; d) 4 times decrease; d) will not change.

7.8e. A parallel beam of light fell on a blackened flat surface at an angle of 45 to the normal and produced a pressure p on it. What pressure will produce the same beam of light, falling normally on a mirrored flat surface?

7.9e. The following system of Maxwell equations:

always true for an alternating magnetic field...

7.10e. A magnet is inserted into the dielectric ring. In this case, in a dielectric...

b) nothing happens

symmetric electric field changes with What energy will cross the cylindrical surface of radius r = 1 cm and length b = 1 m for the time interval 0 t 1 s?

7.12c. An alternating uniform electric field is created between the plates of a flat air capacitor. Find the modulus of the magnetic field in c, if the electric field strength changes with time according to the law Answers: a) 8.13 1017 T; b) 3.08 1014 T uniform magnetic field. Find the modulus of the electric force acting on a charged particle with a charge q = 5 μC located at a distance of 2 cm from the magnet axis at the moment t = 2 s. The magnetic field induction depends on time according to the law Answers: a) 3.6 μN; b) 1.96 µN 7.14e. The following system of Maxwell equations:

L S L S S S

a) in the presence of charged bodies and conduction currents;

b) in the absence of charged bodies and conduction currents;

c) in the absence of charged bodies;

d) in the absence of conduction currents;

7.15e. There is a magnet in the metal ring. In this case, in the ring...

a) a vortex electric field is generated;

b) nothing happens

c) an electrostatic field is generated;

7.17e. A parallel beam of light fell on a mirrored flat surface at an angle of 45 to the normal and exerted pressure p on it.

What pressure will be produced by the same beam of light, falling normally on a blackened flat surface?

Using Gauss' Theorem in Differential An electric field can be represented graphically by drawing lines of force. Line of force - the electric field strength E is directed tangentially. Therefore, if we place a charged particle at rest in an electric graphic image fields can be defined as the density lines of force, i.e. the number of lines intersecting a unit transverse area:

Then the number of field lines crossing the site can be found as follows:

where the vector dS is equal in absolute value to the area dS and is directed along the normal to this area. The value dФE in formula (8.2) is called the flow of the electric field strength vector E through the area dS.

One can prove the Gauss theorem for the electric field strength in vacuum:

is the flow of the electric field strength vector E through an arbitrary closed surface, is equal to the sum charges inside this surface divided by 0, where, 0 8.85 1012 F/m is the electrical constant, is the charge density.

Using the Ostrogradsky theorem for the electric field strength vector, one can obtain the Ostrogradsky-Gauss theorem in differential form for the electric field strength in vacuum:

Along with the Gauss theorem for the electric field strength, the Gauss theorem for the electric induction vector D is often used, which is included in the Maxwell equations (7.3) So:

The strength of the electrostatic field is given by the formula E i Ax3 y 4 j By 2 x 5, where A = 3 V/m8, B = 4 V/m8. Using the Gauss theorem in differential form, find the volume charge density at the point P x0, y0, where x0 1 m, y0 2 m.

It can be seen from the vector expression for E that By formula (8.7) we calculate div E:

div E From formula (8.6) we calculate 0 div E 8.85 1012 160 1.42 109 C/m3.

a figure, along the axis of symmetry of which a uniformly charged segment of length l = 6 cm with a linear charge density = 2 μC / m is placed. The middle of one of the cones.

In the general case, the calculation of the electrical displacement flux through the shaded area of ​​the cone using the DdS formula causes great difficulties. But the charged rod is located on the axis of the cone symmetrically with respect to the plane of the base of the cone. Thus, we can conclude that the flow through the shaded area is equal to half the flow through the entire surface of the figure in Fig.22.

The flow of the vector D through a closed surface can be calculated according to the Ostrogradsky-Gauss law using the formula (7.3):

Where does the answer come from: ФD \u003d 60 nC m. Find the flow of the electric field strength vector through a small surface area. along the normal to the surface of the sphere. The angle between the vector E and any area on the sphere dS is equal to 0. The modulus of tension on the surface of the sphere is equal to E 2. The flux of the vector E can be easily calculated using the formula (8.3):

Tasks for work in a practical lesson.

8.1. The strength of the electrostatic field is given by the formula Using the Gauss theorem in differential form, find the volume charge density at the point P x0, y0, where x0 1 m, y0 2 m.

Answers: a) 0.354 nC/m3; b) 0.266 nC/m3.

8.2 The strength of the electrostatic field is given by the formula Using the Gauss theorem in differential form, find the volume charge density at the point P x0, y0.

Answers: a) - 0.11 nC/m3; b) 2, 4 1012 C/m Find the flux of the electric intensity vector 8.4 The charge q is placed in the center of the upper face of a cube with side a. Find the electrical displacement vector flow through all other faces.

8.5 The charge q1 is placed at the center of the sphere, and the charge q2 is placed at a distance R 2 from the center. Find the flow of the electric field strength vector through the surface of the sphere. q1 5 nC, q2 3 nC, R 3 m.

8.6e. A point charge +q is at the center of a spherical surface. If we add a charge +q outside the sphere, then the flow of the electrostatic field strength vector E through the surface of the sphere ... will increase by 2 times; b) will decrease by 2 times; c) will not change 8.7. A uniformly charged ring of radius r and linear charge density is placed inside a sphere of radius R. The center of the ring coincides with the center of the sphere. Find the flow of the electric field strength vector through the surface of the sphere.

8.8. Above an infinite flat surface, uniformly charged with a surface charge density, is located at a distance h. The planes of the plate and the surface are at an angle. Find the flow of the electric field strength vector through the surface of the plate.

a uniformly charged plane with a surface charge density. An electrical displacement was placed on the plane through the surface of a quarter of a sphere.

charge density. At a large distance r there is a round plate of radius R. The angle between the plane of the plate and the perpendicular to the thread passing through the center of the plate is equal.

Find the flux of the electric displacement vector through the surface of the plate. 1 μC/m, 300, R 1 cm, r 12 m, h 5 m.

In this case, the magnitude of the flow of the electric field strength vector through side surface cone...

a) increased b) decreased c) did not change d) there is not enough data on the ratio of the height of the cone and its radius of the electrostatic field strength is a) S1 ; b) S2 ; c) S3; d) S1 and S3; e) there is no such surface 8.13e. Electric charge q is distributed uniformly inside a sphere of radius R1.

The radius of the sphere was increased to R2 = 2R1, and the charge was evenly distributed over the new volume. How many times the flow of the electric field intensity vector through the spherical surface of radius R1 has decreased.

the charge q2 is at a distance b from the center. Find the flow of the electric field strength vector through the surface of the sphere.

Find the flow of the electric tension vector 8.16s. Above an infinite plane, uniformly charged with a surface charge density, in a parallel plane at a distance h is a small circle of radius R.

Find the flow of the electric field strength vector through the surface of the circle. 1 nC/m2, R 3 cm, h 1 m. Answer: 160 mVm 8.17s. The electric field is created by an infinite straight uniformly charged filament with a linear charge density. At a large distance, r runs parallel to the plane of the plate. Find the flux of the electric displacement vector through the surface of the plate. 2 µC/m, R 1 cm, r 5 m. Answer: 20 nC of the top of the cone outside (Fig. a). It was moved along the axis of the cone to a point near the top, but inside (Fig.b). At the same time, the magnitude of the flow of the electric intensity vector a) increased b) decreased c) did not change d) there is not enough data on the ratio of the height of the cone and its radius 8.19 Oe. The electric charge q is distributed uniformly inside the square parallelepiped b1b1 and height h.

The square rib was increased to b = 3b1, leaving the height unchanged, and the charge was evenly distributed over the new volume. How many times did the flow of the electric field strength vector through the surface of the parallelepiped decrease with square section b1b1.

1) 3 times 2) 9 times 3) 27 times 4) did not change 8.20e. Given system point charges in vacuum and closed surfaces S1, S2 and S3. The flow of the electrostatic field strength vector is equal to zero through...

a) S1; b) S2 ; c) S3; d) S1 and S3; e) there is no such surface 8.21c. The strength of the electrostatic field is given by the form E i A cos Bx j C exp Dy ;

Loy Using the Gauss theorem in differential form, find the volume charge density at the point P x0, y0.

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