Open Library is an open library of educational information. Magnetic field induction increases uniformly The phenomenon of electromagnetic induction

A1. The copper ring is in an external magnetic field so that the plane of the ring is perpendicular to the lines of magnetic induction. Induction magnetic field increases evenly. Induction current in the ring

3) equal to zero 4) constant

A2. In 3 seconds, the magnetic flux penetrating the wire frame steadily increased from 6 Wb to 9 Wb. What is the value of the EMF of induction in the frame in this case?

1) 1V 2) 2V 3) 3V 4) 0V

A3. The conductive loop moves at a constant speed in a constant uniform magnetic field so that the magnetic induction vector B is perpendicular to the plane of the loop. The contour velocity vector v is perpendicular to the vector B. In this case, over time, the induction EMF in the contour

1) increases 2) decreases

3) is constant and not equal to 0 4) is equal to 0

A4. The current in a coil with an inductance of 1 H has doubled. Magnetic flux through the coil

1) increased 2 times 2) increased 4 times

3) decreased by 2 times 4) decreased by 4 times

A5. The current energy in the coil is directly proportional

1) I 2 2) I3) I -1 4) I -2

A6. The figure shows a graph of the change in the current in the coil with inductance L = 6 H when the circuit is opened. Estimate the average value of the EMF of self-induction in the time interval s.

1) 36V 2) 18V

Part 2.

IN 1. In a uniform magnetic field with an induction of 0.04 T, a force of 24 mN acts on a conductor with a current. The length of the active part of the conductor is 20 cm, the current strength is 6A. At what angle is the conductor to the magnetic induction vector?

IN 2. In a uniform magnetic field with induction B = 5mT, a metal rod 50cm long is moving, perpendicular to the magnetic induction vector at a speed of 2m / s. What is the potential difference between the ends of the rod?

Part 3.

C1. EMF of self-induction arising in a circuit with an inductance of 0.2H, changes over time according to the law ε = 6t + 3 (V). By what law does the current in the circuit most likely change?

Option 2

A1. Changing over time, the magnetic field generates

1) vortex electric field 2) electrostatic field

3) constant magnetic field 4) gravitational field

A2. A thin copper ring with an area of 100 cm 2 is located in an external magnetic field so that the plane of the ring is perpendicular to the lines of magnetic induction. In 1 s, the magnetic induction increases uniformly from 1 mT to 2 mT. The EMF modulus of the induction arising in this case in the circuit is equal to

1) 10 -5 V 2) 10 -3 V 3) 10 -1 V 4) 0 V

A3. The inductance L of a closed conducting circuit is determined by the formula (I is the current in the circuit, Ф is the magnetic flux through the surface covered by the circuit) 1) L = Ф / I2) L = ФI3) L = I / Ф 4) L = ΔI / Ф.

A4. The dependence of the energy of the magnetic field in the coil inductance L on the current I in the coil is given by the graph

The figure shows the electrical circuit. In which lamp, after the key is closed, the current will last at its maximum value? 1) 1 2) 2 3) 3

4) in all at the same time

A5.

A6. What is the inductance of the wire frame if, at a current strength I = 3 A, a magnetic flux Ф = 6 Wb appears in the frame?

1) 0.5H 2) 2H 3) 18 H 4) there is no correct answer among the listed answers.

Option 3

1. The conducting loop moves at a constant speed in a constant uniform magnetic field so that the magnetic induction vector is perpendicular to the plane of the loop (Fig. 39). The contour velocity vector is perpendicular to the vector. In this case, over time, the EMF of induction in the circuit

A. increases; B. decreases;

V. constant and not equal to zero; G. is zero

2. What is the EMF of self-induction in a coil with inductance L = 3 H with a uniform decrease in current from 5 A to 1 A in 2 seconds?

A. 6V; B. 9V; V. 24V; G. 36 B.

3. Figure 40 shows a graph of the dependence of the magnetic flux through a conducting stationary loop from time to time. In what time interval is the induction EMF modulus in the circuit equal to zero?

A. 0 - 1 s; B. 1 - 3 s; V... 0 - 2 s; G. 3-4 p.

4. A coil with an inductance of 1 H is switched on to a voltage of 20 V. Determine the time during which the current in it reaches 30 A.

5. A conductor with an active length of 15 cm moves at a speed of 10 m / s perpendicular to the lines of induction of a uniform magnetic field with an induction of 2 T. What amperage occurs in a conductor if it is short-circuited? Circuit resistance 0.5 ohm.

Option 4

1. The magnetic flux of 1 Wb can be expressed in SI as

A. 1 Nm²; B. 1 T · m²; V. 1 T / s; G. 1 T / m²

2. The conducting circular loop moves translationally at a constant speed in the direction indicated in Figure 41, in the field of a straight conductor with current. It can be said about the induction current in the circuit that ...

A. it is directed clockwise;

B. it is directed counterclockwise;

V. it will not arise;

G. its direction depends on the modulus of the magnetic field induction.

A. 0.5 Gn; B. 2 G. V. 18 G.

4. What is the inductance of a turn of a wire, if at a current of 6 A a magnetic flux of 12 · 10 - 3 Wb is created? Does the inductance of a loop depend on the current in it?

5. What charge will pass through the cross-section of a loop, the resistance of which is 0.05 Ohm, with a decrease in the magnetic flux inside the loop by 15 mWb?

Option 5

1. The wire frame is in a uniform magnetic field.

a) The frame is turned around one of its sides.

b) The frame is moved across the magnetic induction lines.

v) The frame is moved along the lines of induction of the magnetic field.

Electric current arises

A. only in case a;B. only in case b;

V. only in case v;G. in all cases.

2. Figure 42 shows a graph of the change in the current in a coil with an inductance of 6 H when the circuit is opened. Estimate the average value of the EMF of self-induction in a time interval of 1 - 2 s.

A. 36V; B. 18V; V. 9V; G. 3 B.

3. What is the inductance of the wire frame if at a current strength of I = 3 A in the frame there is a magnetic flux Ф = 6 Wb?

A. 0.5 Gn; B. 2 G. V. 18 G. G. there is no correct answer among the listed answers.

4. What is the induction of the magnetic field if an EMF of 1.5 V is excited in a conductor with an active part length of 50 cm, moving at a speed of 10 m / s perpendicular to the induction vector?

5. The aluminum ring is located in a uniform magnetic field so that its plane is perpendicular to the magnetic induction vector. Ring diameter 25 cm, ring wire thickness 2 mm. Determine the rate of change of the magnetic induction over time, if in this case an induction current of 12 A appears in the ring. The specific resistance of aluminum is 2.8 · 10 -8 Ohm · m.

Option 6

1. Permanent straight magnet falls through the aluminum ring. Magnet Fall Acceleration Module

A. at the beginning of the span of the ring less than g, at the end more than g;

B. is equal to g; V. more g; G. less g.

2. Figure 43 shows the electrical circuit. In which lamp, after the key is closed, the current will last at its maximum value?

A. 1 B. 2 V. 3 G. All at the same time.

3. The inductance L of a closed conducting loop is determined by the formula

A. L = F / I B. L = Ф I

V. L = I / F G. L = ∆ I / F

4. Find the EMF of induction at the ends of the wings of an aircraft (wingspan 36.5 m) flying horizontally at a speed of 900 km / h, if the vertical component of the induction vector of the Earth's magnetic field is 5 · 10 - 3 T.

5. Two metal rods are arranged vertically and are closed at the top by a conductor. On these rods, without friction and breaking the contact, a jumper with a length of 0.5 cm and a mass of 1 g slides. The entire system is in a uniform magnetic field with an induction of 0.01 T, perpendicular to the plane of the frame. The steady-state speed is 1 m / s. Find the resistance of the jumper.

PRACTICAL WORK No. 5.
"Alternating current"

Option 1

1. What is the dependence of voltage on time t corresponds to harmonic vibrations?

A =? B =?

2. The graph (Fig. 44) shows the dependence of the current in the circuit on time. What is the current oscillation period?

A. 0.5s; B. 2 s; V. 1 s; G. 3 sec.

3. The period of free current oscillations in the electric circuit is equal to T. At some point, the energy of the electric field in the capacitor reaches a maximum. What is the minimum time after that the magnetic field energy in the coil will reach its maximum?

5. Write the equation of harmonic oscillations of the voltage at the terminals of the electric circuit, if the amplitude of oscillations is 150 V, the oscillation period is 0.01 s, and the initial phase is zero.

6. The current in the oscillatory circuit changes with time according to the law i= 0.01cos1000t. Find the inductance of the circuit, knowing that the capacity of its capacitor is 2 10 - 5 F.

Option 2

1. The oscillation period is 1 ms. The frequency of these vibrations is

A... 10 Hz; B. 1 kHz; V. 10 kHz; G. 1MHz

2. If the electrical capacity of a capacitor in an electric oscillatory circuit decreases 9 times, then the frequency of oscillations

A. will increase 9 times; B. will increase 3 times;

V. will decrease by 9 times; G. will decrease by 3 times.

3. A resistor, capacitor and coil are connected in series in the AC circuit. The amplitude of the voltage fluctuations across the resistor is 3 V, on the capacitor 5 V, on the coil 1 V. What is the amplitude of the fluctuations in the section of the circuit consisting of these three elements?

A. 3V; B. 5V; V. 5.7V; G. 9 B.

4. From the graph shown in Figure 45, determine the voltage amplitude and oscillation period. Write down the equation for the instantaneous voltage value.

7. In an oscillatory circuit, the dependence of the current strength on time is described by the equation i= 0.06sin10 6 πt. Determine the frequency electromagnetic waves and the inductance of the coil, if the maximum energy of the magnetic field is 1.8 · 10 - 4 J.

Option 3

1. Module the greatest value a quantity that changes according to a harmonic law is called

A. period; B. amplitude;

V. frequency; G. phase.

2. The change in the charge of the capacitor in the oscillatory circuit occurs according to the law q = 3cos5t (q is measured in microcoulomb, t - in seconds).

The amplitude of charge oscillations is

A. 3 μC; B. 5 μC;

V... 6 μC; G. 9 μC.

3. The graph (Fig. 46) shows the dependence of the current in the circuit on time. What is the effective value of the current strength?

4. The current value, measured in amperes, is given by the equation i= 0.28sin50πt, where t is in seconds. Determine the current amplitude, frequency and period.

5. The voltage across the capacitor plates in the oscillatory circuit changes according to the law u= 50cos10 4 πt. The capacitance of the capacitor is 0.9 μF. Find the inductance of the circuit and the law of variation with time of the current in the circuit.

Option 4

1. Which of the following expressions determines the inductive reactance of the inductor L in an alternating current circuit with a frequency ω ?

2. In a circuit consisting of a capacitor and a coil, free electromagnetic oscillations occur. If over time the initial charge imparted to the capacitor has halved, then the total energy stored in the capacitor is

A. decreased by half;

B. doubled;

V. decreased by 4 times;

G. has not changed.

3. The period of free oscillations in the circuit with an increase in electrical capacity

A. increases;

B. decreases;

V. does not change;

G. is always zero.

4. From the graph shown in Figure 47, determine the voltage amplitude, period and voltage value for the phase π / 3 rad.

5. The dependence of the current strength on time in the oscillatory circuit is determined by the equation i= 0.02sin500πt. Circuit inductance 0.1 G. Determine the period of electromagnetic oscillations, the capacity of the circuit, the maximum energy of the magnetic and electric fields.

Option 5

1. What expression determines the capacitance of a capacitor electrical capacity C in an alternating current circuit with a frequency ω ?

2. The ratio of the effective value of the harmonic alternating current to its amplitude is

A. 0; B. 1/; V. 2; G. 1/2.

3. The change in the charge of the capacitor in the oscillatory circuit occurs according to the law q = 10 - 4 cos10πt (C). What is the period of electromagnetic oscillations in the circuit (time is measured in seconds)?

A. 0.2 s; B.π / 5 s; V. 0.1π s; G. 0.1 sec.

4. A capacitor with a capacity of C = 5 μF is connected to an alternating current circuit with U m = 95.5 V and a frequency ν = 1 kHz (Fig. 48). What amperage will the ammeter, connected to the network, show? The resistance of the ammeter is negligible.

5. The charge on the capacitor plates of the oscillatory circuit changes according to the law q = 3 · 10 - 7 cos800πt. Circuit inductance 2 H. Neglecting the active resistance, find the capacitance of the capacitor and the maximum values of the energy of the electric field of the capacitor and the magnetic field of the inductor.

Option 6

1. What is the period of free oscillations in an electrical circuit from a capacitor with electrical capacity WITH and inductors L?

2. Find the maximum value alternating voltage if the effective value U = 100 V.

A. 70.7V; B. 141.4V; V. 200 V; G. 50 B.

A. Separates the modulating signal from the electromagnetic wave;

B. Amplifies the signal of one selected wave;

V. It selects from all electromagnetic waves the natural oscillations that coincide in frequency;

4. A coil with inductance L = 50 mH is connected to an alternating current generator with U m = 44.4 V and a frequency ν = 1 kHz. What amperage will the ammeter included in the circuit show?

5. The voltage across the capacitor plates in the oscillatory circuit changes according to the law u = 100cos10 4 πt. The electrical capacity of the capacitor is 0.9 μF (Fig. 49). Find the inductance of the circuit and the maximum value of the magnetic field energy of the coil.

PRACTICAL WORK №6.
"Radiation and reception of electromagnetic waves in the radio and microwave range"

Option 1

1. How far from the source intensity electromagnetic radiation depends on the distance to it?

A. In direct ratio;

B. Inversely;

V. Proportional to the square of the distance;

G. Inversely proportional to the square of the distance.

2. The frequency of infrared radiation is lower than the frequencies listed below, except ...

A. visible light;

B. radio waves;

V. ultraviolet radiation;

G. X-ray radiation.

3. The source of electromagnetic waves is ...

A. D.C;

B. stationary charge;

V. any accelerated moving particle;

G. any accelerated moving charged particle.

4. The electric field strength of a traveling electromagnetic wave in SI is given by the equation E= 5 · 10² sin (3 · 10 6 π ( x- 3 · 10 8 t X.

5. The height of the emitting antenna of the telecentre above the Earth's level is 300 m, and the height of the receiving antenna is 10 m. At what maximum distance from the transmitter can you receive?

Option 2

1. Which of the following waves are not shear waves?

A. Infrared;

B. Visible;

V. Sound;

G. Radio waves.

2. The frequency of emission of yellow light ν = 5.14 · 10 14 Hz. Find the wavelength of the yellow light.

A. 580 nm; B. 575 nm; V. 570 nm; G. 565 nm.

3. The field strength of a traveling electromagnetic wave in SI is given by the equation
E= 10²sin (4 10 6 π (2 10 8 t + x)). Find the amplitude, frequency of the wave and the speed of its propagation along the axis x.

4. The radar operates at 15 cm and emits pulses at 4 kHz. The duration of each pulse is 2 μs. What is the longest target detection range? How many vibrations are there in one impulse?

Option 3

1. Is there such a movement of an electric charge in which it does not emit electromagnetic waves?

A. There is no such movement.

B. There is, this is a uniform rectilinear movement.

V. There is this uniform movement around the circumference.

G. There is this rectilinear uniformly accelerated motion.

2. The flux density of electromagnetic radiation is 0.03 W / cm². In units of W / m2, it will be equal to

A. 0,0003; B. 3; V. 30; G. 300.

3. What is the function of the oscillatory circuit of a radio receiver?

A... Separates the modulating signal from the electromagnetic wave.

G. Accepts all electromagnetic waves.

i= 0,5sos 8 10 5 π t. Find the emitted wavelength.

5. What is the wavelength of the electromagnetic radiation of the oscillatory circuit, if the capacitor has a capacitance of 2 pF, the rate of change of the current in the inductor is 4 A / s, and the resulting EMF of the induction is 0.04 V?

Option 4

1. In what directions are the oscillations in the transverse wave?

A. In all directions.

B. Only in the direction of wave propagation.

V. Only perpendicular to the direction of wave propagation.

G. In the direction of wave propagation and perpendicular to this direction.

2. The radio receiver is tuned to a wavelength of 100 m. The natural frequency of the input oscillatory circuit is

A. 3 Hz; B. 300 kHz; V. 3 kHz; G. 3 MHz.

3. What is the function of the radio receiver antenna?

A. Separates the modulating signal from the electromagnetic wave.

B. Amplifies the signal of one selected wave.

V. It selects from all electromagnetic waves the natural oscillations that coincide in frequency.

G. Accepts all electromagnetic waves.

4. Electromagnetic waves propagate in a homogeneous medium at a speed of 2 · 10 8 m / s. What is the wavelength of electromagnetic oscillations in this environment, if their frequency in vacuum

6. When the current in the inductor changes by 1 A during 0.6 s, an EMF of 0.2 mV is induced in it. What is the length of the radio wave emitted by the generator, the oscillating circuit of which consists of this coil and a capacitor with a capacity of 14.1nF?

Option 5

1. When an electromagnetic wave propagates in a vacuum ...

A. only energy transfer occurs;

B. only momentum transfer occurs;

V. there is a transfer of both energy and momentum;

G. there is no transfer of energy or momentum.

2. How will the intensity of radiation of electromagnetic waves change with the same amplitude of their oscillations in the vibrator, if the frequency of oscillations is increased by 2 times?

A. Will not change.

B. Will increase 2 times.

V. Will increase 4 times.

G. Will increase 16 times.

3. Arrange the following types of electromagnetic waves in order of increasing wavelength:

A. visible light;

B. radio waves;

V. X-ray radiation;

G. infrared radiation.

4. The current in the open oscillatory circuit varies with time according to the law i= 0.8sin4 10 5 π t. Find the emitted wavelength.

5. How many electromagnetic oscillations with a wavelength of 375 m occur during one period of sound with a frequency of 500 Hz, pronounced in front of the tape recorder of the transmitting station?

Option 6

1. Consider two cases of electron motion in vacuum:

a) The electron moves uniformly and rectilinearly.

b) The electron moves uniformly and rectilinearly.

In what cases does the emission of electromagnetic waves occur?

A. a. B. b. V. a) and b). G. Neither a) nor b).

2. Which of the following devices is not required in the radio transmitter?

A. Antenna. B. Oscillatory circuit.

V. Detector. G. Continuous oscillation generator.

3. Among the waves of the long, short and ultrashort range, the waves have the highest propagation speed in vacuum ...

A. long range;

B. short range;

V. ultra-short range;

G. the propagation speeds of all waves are the same.

4. A radar station sends electromagnetic waves 10 cm long at a frequency of 2.25 GHz into an environment. What is the speed of waves in this medium and what will be the length of electromagnetic waves in a vacuum?

5. At what maximum distance can a target be detected on the sea surface by a ship's radar located at an altitude of 8 m above sea level? What should be the minimum time interval between adjacent pulses of such a locator?

In a uniform magnetic field, a straight conductor moves at a constant speed so that the velocity vector is perpendicular to the conductor. The magnetic induction vector is also perpendicular to the conductor and makes an angle with the vector α = 30 °. Then the same conductor begins to move at the same speed, in the same magnetic field, but in such a way that the angle α increases by 2 times. As a result of this, the following physical quantities will change: EMF modulus of induction arising in the conductor; the modulus of the electric field strength inside the conductor?

For each value, determine the corresponding change pattern:

1) will increase;

2) will decrease;

3) will not change.

Write down the numbers in the answer, arranging them in the order corresponding to the table:

Solution.

The EMF of induction for a conductor moving in a magnetic field perpendicular to the conductor is calculated by the formula: Therefore, with an increase in the angle between the speed and direction of the magnetic field, the EMF of induction in the conductor will also increase.

The modulus of the electric field strength inside the conductor is directly proportional to the induction EMF, therefore, the modulus of the electric field strength will also increase.

Answer: 11.

Julia Gorbacheva 14.04.2017 22:26

In the frame of reference of the conductor (where it is stationary), a constant electric field arises. If the conductor is in a constant electric field, then the magnitude of the electric field strength inside it is zero.

You can reason differently. If there is an electric field strength inside the conductor, then a force acts on the charge carriers in the conductor (for example, electrons). Under the action of this force, charge carriers move and an electric current exists in the conductor. Thus, the very statement that there is a nonzero electric field strength inside the conductor is equivalent to the statement that a constant current is maintained in the conductor.

Availability direct current in a conductor that does not form a closed circuit - this is an absurdity, contrary to the law of conservation of charge.

Anton

The charges in the conductor under consideration are acted upon by two forces balancing each other: the force from the side of the electric field created by the redistributed charges (during the transient process at the beginning of motion), and the Lorentz force from the side of the magnetic field. Without an electric field, the magnetic field would cause an electric current. During the transient process, this electric current leads to a redistribution of charges in the conductor.

At a nonzero electric field strength, a current arises in the conductor if there are no external forces that can increase or decrease this current, including completely compensate for the effect of the electric field.

15.1 The phenomenon of electromagnetic induction.

15.1.1 Discovery of the phenomenon of electromagnetic induction by M. Faraday.

The discovery of the magnetic action of current by H. K. Oersted in 1820 proved that electrical and magnetic phenomena are related. The theory of A.M. Ampere reduced the numerous magnetic phenomena he studied to the interaction of electric currents, that is, moving electric charges. After the discovery of Oersted and the works of Ampere, the English scientist Michael Faraday came to the idea of the opposite process - excitation electric current magnetism: if an electric current generates a magnetic field, then why can't a magnetic field excite an electric current? In 1822 in workbook M. Faraday, a record appears in which the task is formulated: "To turn magnetism into electricity." It took M. Faraday almost ten years of persistent and numerous experiments to solve the problem, which led to the discovery of the phenomenon of electromagnetic induction on August 29, 1831.

For a long time, M. Faraday carried a coil of wire and a permanent magnet in his pocket, trying at any free minute to come up with a new arrangement of the wire and magnet, which would lead to the appearance of an electric current. As has often happened in history, success came unexpectedly, however, it had to wait for almost ten years. To exclude the direct influence of the magnet on the device that records the current (galvanometer), M. Faraday placed magnets and conductors (usually coils) in one room, and the galvanometer in another. Having placed the coils and magnets once again, M. Faraday moved to another room to make sure once again that there was no electric current. Finally, one of the employees noticed that an electric current occurs only at the moment of the relative movement of the conductor and the magnet.

Now M. Faraday's experiments can be easily reproduced in school laboratory... Enough to connect wire reel to the galvanometer and insert a permanent magnet inside the coil. When the magnet is pushed into the coil, the galvanometer needle deflects, indicating the presence of current in the circuit (Fig. 104).

The current stops when the magnet is stationary. If you remove the magnet from the coil, then again the galvanometer registers the presence of current, only in the opposite direction. If you change the polarity of the magnet, then the direction of the current also changes. The magnitude of the current depends on the speed of movement of the magnet - the faster the magnet moves, the greater the strength of the resulting electric current. Similar results are obtained if the magnet is stationary and the coil is moving.

In other words, the result depends only on the relative movement of the coil and magnet.

Further, M. Faraday showed that an electric current appears in the circuit even when it is in a time-varying magnetic field. To demonstrate this phenomenon, it is possible in previous experiments to replace the permanent magnet with a coil connected to a direct current source (Fig. 105). The galvanometer registers the current only at the moments of switching on and off the current source. Please note that the coils are not connected to each other, the only connection between them is through the magnetic field.

Thus, in all cases, when the magnetic field changes, an electric current appears in a closed loop, which indicates the appearance of an electromotive force in it. M. Faraday connected his reasoning about electromagnetic phenomena with the properties of lines of force, which he perceived as quite real elastic threads and tubes. In such reasoning, an electric current arises when the lines of force of the magnetic field move and cross the circuit, due to which an EMF is induced (induced) in the circuit.

The phenomenon of the appearance of an electric current in a circuit when the magnetic field changes M. Faraday called the phenomenon of electromagnetic induction.

Further, we will not strictly follow the reasoning and experiments of M. Faraday, because in his time the nature of electrical and magnetic phenomena was absolutely unknown: even an electric current was not always associated with the movement of electric charges. Therefore, in our presentation we will use facts and ideas that became known much later.

15.1.2 Moving conductor in a magnetic field.

Today it is almost obvious that no configuration of a constant magnetic field can produce a constant electric current. To maintain the current in the electrical circuit, as we know, there must be a source of external forces that does the work to overcome the resistance forces. The magnetic field acts only on moving charges, and the forces acting on the charge (Lorentz force) are perpendicular to the particle's velocity vector, so it does not do work. Finally, if a stationary magnetic field could support an electric current, then this would be a direct path to the creation of a "perpetual motion machine", that is, to "free" energy production. Indeed, if the field is stationary, then its energy does not change, and the hypothetical electric current has energy and is capable of doing work. Therefore, for the EMF to appear in the circuit, there must be an external energy source. Energy can enter the circuit due to the work of external forces.

Consider a group of simple thought experiments that can be described theoretically. Let a cylindrical conductor move in a constant magnetic field, so that the velocity vector \ (~ \ vec \ upsilon \) is perpendicular to the cylinder axis, and the magnetic induction vector \ (~ \ vec B \) is perpendicular to both the conductor axis and its velocity ( Fig. 106). Free charges inside it move together with the conductor. From the side of the magnetic field, these charges will be acted upon by the Lorentz forces, directed, in accordance with the left-hand rule, along the axis of the conductor.

The most famous conductors are metals, where free charges are negatively charged particles - electrons. However, here and in what follows, we will consider the motion of positively charged particles, because the direction of the current is taken as the direction of positive particles.

As a rule, free charges move in a conductor in a chaotically equiprobable way in all directions, therefore, in a stationary conductor, the average value of the Lorentz force vector is zero. When the conductor moves, the directed movement of the conductor is superimposed on the chaotic thermal motion of free charges, due to which a nonzero resultant Lorentz force appears, which is the same for all particles. It is this constant force that leads to the emergence of an electric current - the directed movement of charged particles. This provides good reason to disregard the violent but chaotic heat movement.

Under the action of the Lorentz force, free charges will begin to shift towards the ends of the cylinder, where electric charges will be induced, described by surface densities ± σ ... In turn, these charges will begin to create an electric field, the action of which on charged particles will be directed in the direction opposite to the Lorentz force. At a constant speed of movement of the conductor, an equilibrium will be established at which the movement of charges will stop, but an electric field created by the induced charges will exist in the conductor. In the steady state, the Lorentz force \ (F_L = q \ upsilon B \) acting on the particle will be balanced by the force from the electric field \ (F_ (el) = q E \). Equating these forces, we determine the strength of the electric field in the conductor

\ (~ E = \ upsilon B \). (one)

Since the Lorentz force is the same at all points of the conductor, then the electric force must also be constant, that is, the resulting electric field is uniform. This electric field can also be characterized by the potential difference between the ends of the cylinder, which is equal to

\ (~ \ Delta \ varphi = E l = \ upsilon B l \), (2)

where l- conductor length.

The Lorentz force acting on free charges in a conductor can be an external force, that is, lead to the emergence of an electric current in a closed loop, if connected to a moving conductor.

Let the conductor in question AC can slide on two parallel rails (rails) connected to each other (Fig. 107). The whole system is placed in a uniform magnetic field, the induction vector of which \ (~ \ vec B \) is perpendicular to the plane of the tires. For simplicity, we will assume that the resistances of the buses and the moving conductor (jumper) are negligible compared to the resistance of the connecting resistor R... If an external force \ (~ \ vec F \) is applied to the moving conductor, as shown in the figure, then it will start moving. Under the action of the Lorentz force, free charges in the conductor will set in motion, creating excess charges at the ends. These charges will create an electric field in the entire loop formed by the jumper, busbars, and connecting resistor, so an electric current will flow in the loop. The Lorentz force acting on the charges of a moving conductor will play the role of an external, overcoming force acting from the electric field. The work of this force to move a unit charge (that is, EMF) is equal to the product of the Lorentz force by the distance between the tires

\ (~ \ varepsilon = \ frac (1) (q) F_L l = \ upsilon B l \). (3)

Despite the fact that this expression for the EMF completely coincides with the formula (2) for the potential difference, its meaning is fundamentally different. The potential difference is the possible work of the electric field forces; in the circuit under consideration, the direction of motion of charged particles is opposite to the direction of the force from the side of the electric field. The Lorentz force does work against the forces of the electric field, therefore it is external. The electric field does positive work by "pushing" charged particles along the busbars and the connecting resistor (which in this case form an external circuit).

According to Ohm's law, the strength of the electric current arising in the circuit is equal to

\ (~ I = \ frac (\ varepsilon) (R) = \ frac (\ upsilon B l) (R) \). (4)

Since an electric current flows through the conductor, an Ampere force acts on it from the side of the magnetic field, equal to

\ (~ F_A = I B l = \ frac (\ upsilon B ^ 2 l ^ 2) (R) \). (5)

The direction of this force is also determined by the "left hand rule", with the help of which it is easy to determine that this force is directed in the direction opposite to the velocity vector, therefore, formula (5) can be written in vector form

\ (~ \ vec F_A = - \ frac (B ^ 2 l ^ 2) (R) \ vec \ upsilon \). (6)

By its nature, this force completely coincides with the force of viscous friction (proportional to the speed and directed in the opposite direction), therefore it is often called the force of magnetic viscosity.

Thus, in addition to the constant external force \ (~ \ vec F \), the moving bridge is subject to the force of magnetic viscosity, which depends on the velocity. The equation of Newton's second law for the bulkhead has the form (in projection onto the direction of the velocity vector):

\ (~ ma = F - \ frac (B ^ 2 l ^ 2) (R) \ upsilon \). (7)

Under the action of these forces, the bulkhead will first move accelerated, and with an increase in speed, the acceleration modulus will decrease, and finally, the bulkhead will move at a constant speed, which is called steady motion speed\ (~ \ overline (\ upsilon) \). The magnitude of this speed can be found from the condition \ (F = F_A \), from which it follows

\ (~ \ overline (\ upsilon) = \ frac (FR) (B ^ 2 l ^ 2) \). (eight)

Let us now consider the conversion of energy in this system in a steady state of motion. For a time interval Δ t the jumper is displaced by a distance \ (\ Delta x = \ overline (\ upsilon) \ Delta t \), therefore, the external force does the work

\ (~ \ Delta A = F \ Delta x = F \ overline (\ upsilon) \ Delta t = \ frac (F ^ 2 R) (B ^ 2 l ^ 2) \ Delta t \). (9)

During the same time, an amount of heat will be released on the resistor equal to

\ (~ Q = I ^ 2 R \ Delta t = \ left (\ frac (\ upsilon B l) (R) \ right) ^ 2 R \ Delta t = \ frac (B ^ 2 l ^ 2) (R) \ left (\ frac (FR) (B ^ 2 l ^ 2) \ right) ^ 2 \ Delta t = \ frac (F ^ 2 R) (B ^ 2 l ^ 2) \ Delta t \). (10)

As expected, the amount of heat released is exactly equal to the work of the external force. Therefore, the source of energy for the electric current in the circuit is the device that moves the jumper (your hand can also be such a device). If the action of this force stops, then the current in the circuit will disappear.

Explain why when the magnetic field induction tends to zero, the speed of the jumper calculated by the formula (8) tends to infinity.
Explain why as the resistance of the resistor increases, the speed of the jumper increases.
Show that during acceleration, the work of the external force is equal to the sum of the change in the kinetic energy of the bulkhead and the amount of heat released on the bulkhead.

In this case, the magnetic field plays the role of a kind of mediator that helps to convert the energy of an external source (creating an external force) into the energy of an electric current, which is then converted into thermal energy. The very same external magnetic field does not change in this case.

The clause about the external field in this case is not accidental, the electric current induced in the circuit creates its own magnetic field \ (~ \ vec B "\). According to the thumb rule, this field is directed opposite to the external field \ (~ \ vec B \) (Fig. 108).

Let us now direct the direction of the external force to the opposite. This will change the direction of movement of the bridge, the Lorentz force, the electric current in the circuit and the induction of the magnetic field of this current (Fig. 109). That is, in this case, the direction of the induction vector \ (~ \ vec B "\) will coincide with the direction of the external field \ (~ \ vec B \). Thus, the direction of the induced field is determined not only by the direction of the external field, but also by the direction of movement of the bridge ...

We emphasize that the Ampère force, which plays the role of the viscosity force, is in this (and in all other) cases opposite to the speed of movement of the bridge.

Let's try to formulate general rule, which allows you to determine the direction of the induction current. In fig. 110 again depicts the schemes of the considered experiments, if you look at them from above. Regardless of the direction of movement of the bridge, the EMF of induction in the circuit in modulus is determined by formula (3), which we transform to the form

\ (~ \ varepsilon = \ upsilon B l = \ frac (B l \ Delta x) (\ Delta t) \), (11)

where Δ x = υ Δ t- the distance by which the jumper is displaced during the time interval Δ t... The expression in the numerator of this expression is equal to the change in magnetic flux through the contour BlΔ x = Δ Φ caused by a change in its area. Now let's pay attention to direction this EMF.

Of course, the electromotive force, as the work of external forces, is a scalar quantity, so it is not entirely correct to talk about its direction.

However, in this case, we are talking about the work of external forces along the contour, for which it is possible to determine the positive direction of the bypass. To do this, you must first choose the direction of the positive normal to the contour (obviously, the choice of this direction is arbitrary). As before, we will take the “counterclockwise” direction for the positive, if we look from the end of the positive normal vector, respectively, the “clockwise” direction will be considered negative (Fig. 111). In this sense, we can talk about the EMF sign: if, when walking in the positive direction (ie, "counterclockwise"), external forces do positive work, then the EMF value will be considered positive and vice versa.

In this case, the positive direction of the normal is compatible with the direction of the induction vector of the external field. Obviously, the direction of the induction current coincides with the direction of the EMF.

According to the accepted definition, in the case a) the induced EMF and the current in the circuit are negative, in the case of b) - are positive. It can be generalized: the sign of the EMF is opposite to the sign of the change in the magnetic flux through the circuit.

In this way, EMF of induction in the circuit is equal to the change in magnetic flux through the circuit, taken with the opposite sign:

\ (~ \ varepsilon = - \ frac (\ Delta \ Phi) (\ Delta t) \). (12)

The resulting rule can be interpreted somewhat differently. Let's pay attention to the direction of the magnetic field created by the induction current: with an increase in the magnetic flux through the circuit, this field is opposite to the induction of the external field, with a decrease in the magnetic flux, the field of the induction current is directed in the same way as the external field. That is, induction current field in the circuit prevents change magnetic flux through this circuit... This rule is universal for this phenomenon and is called Lenz rule .

This rule is closely related to the law of conservation of energy. Indeed, suppose the opposite: let the direction of the induction of the magnetic field created by the current in the circuit enhances change in magnetic flux through the circuit. In this case, we get a "self-accelerating" system: if the magnetic flux through the circuit accidentally increases, then this will lead to the appearance of an electric current, which will further increase the flux through the circuit, which will lead to an even greater increase in current, etc. Thus, it turns out that without an external source, the current in the circuit (and its energy) increases indefinitely, which contradicts the law of conservation of energy.

Please note that in this reasoning we take into account the magnetic flux not only of the external field, but also of the field created by the induced current. This field really needs to be taken into account: the Lorentz force acting on charged particles is determined by the total magnetic field at the location of the charge, regardless of the origin of this field. Thus, by means of a magnetic field, an electric current is able to act on itself - a changing current creates a changing magnetic field that affects the electric current. This phenomenon is called self-induction, we will get to know him in more detail later. Here, we note that in many cases this phenomenon can be neglected, since usually the induced fields are rather weak.

It can also be shown that the direction of the force of magnetic viscosity, which is always opposite to the speed of movement of a conductor in a magnetic field, is also associated with the Lenz rule.

The broadest generalization of Lenz's rule "for all occasions" sounds like this: the effect seeks to reduce the cause. Try to come up with examples from various branches of science on your own when this rule is true. It is more difficult (though possible) to come up with examples when this rule does not apply.

Let us consider another example of the occurrence of an EMF in a conducting circuit moving in a magnetic field. Let the field be created by a cylindrical permanent magnet, and the circular contour L moves with a speed \ (~ \ vec \ upsilon \) along the axis of this magnet, so that the plane of the contour remains perpendicular to the axis of the magnet all the time (Fig. 112).

In this case, the magnetic field is not uniform, but has axial symmetry. When the conductor moves in this field, the charged particles are affected by the Lorentz force directed along the conductor, constant in absolute value throughout the entire circuit. In this case, the Lorentz force again acts as an external force, leading to the appearance of an electric current in the circuit. The work of this force to move the charge along a closed loop is nonzero, so this force is not potential. Let's calculate the EMF of the induction arising in the circuit. A charged particle is acted upon by a force equal to

\ (~ F = q \ upsilon B_r \), (13)

where B r is the component of the induction vector, perpendicular to the velocity vector of the conductor, in this case it is directed radially. Since this force on the entire contour is directed tangentially to the contour and is constant in absolute value, its work on moving a unit charge, that is, EMF, is equal to

\ (~ \ varepsilon = \ frac (1) (q) F_L = \ upsilon B_r L \), (14)

where L- the length of the contour. To find an expression for the radial component of the induction vector, we use the magnetic flux theorem. As a closed surface, we choose a thin cylinder with thickness Δ z = υ Δ t, the axis of which coincides with the axis of the magnet, and the radius is equal to the radius of the contour (Fig. 113).

We represent the magnetic flux through this surface as the sum of fluxes through the lower base F 0, through the top base F 1 and through the side surface

\ (~ \ Phi_ (bok) = B_r L \ Delta z = B_r L \ upsilon \ Delta t \). (15)

The sum of these flows is zero

\ (~ \ Phi_0 + \ Phi_1 + \ Phi_ (bok) = 0 \). (sixteen)

Now let us correlate these surfaces with the contour in question.

The lateral surface of the cylinder is the surface that the contour in question sweeps out, so we linked the height of the cylinder to the speed of the contour. The lower base rests on the position of the contour at some point in time t... By convention, the external normal (shown in the figure) is considered a positive normal for a closed surface. When describing the magnetic flux through the contour, we agreed to consider the positive direction of the normal, the direction "along the field". That is, the flow through the circuit is opposite to the flow through a part of the closed surface. Therefore, in this case Φ 0 = −Φ (t), where Φ (t) is the flow through the circuit, at the moment of time t... The flow through the upper base is the flow through the circuit at the moment of time t + Δ t Φ 1 = Φ (t + Δ t). Another argument in favor of changing the sign in the flow through the bottom base is that if we calculate the change in flow, then we must keep the direction of the normal unchanged.

Now we rewrite relation (16) in the form

\ (~ - \ Phi (t) + \ Phi (t + \ Delta t) + B_r L \ upsilon \ Delta t = 0 \). (17)

From which we express the EMF of induction in the circuit (determined by formula (15))

\ (~ \ varepsilon = B_r L \ upsilon = - \ frac (\ Phi (t + \ Delta t) - \ Phi (t)) (\ Delta t) = - \ frac (\ Delta \ Phi) (\ Delta t ) \). (eighteen)

We got the same formula for the EMF of induction in the circuit as in the previous example.

In the considered example, the magnetic flux through the circuit decreases, since with increasing distance from the magnet, the field induction decreases. Therefore, in accordance with the obtained formula and Lenz's rule, the EMF of induction in the circuit is positive, in addition, the induction current creates a magnetic field directed in the same way as the field of a permanent magnet.

Please note that in the above conclusion, we did not make any assumptions about the dependence of the field induction vector on the coordinates. The only assumption was about the axial symmetry of the field. However, it can also be removed, for this, when calculating the EMF along the contour, it is simply necessary to break the latter into small sections, and then sum up the work of the Lorentz force over all sections.

Assignments for independent work.

Consider the direction of the field created by the induced current in the circuit in Fig. 112, show that Lenz's rule holds.
Show that in the circuit shown in fig. 112, the Ampere force acting on the circuit with the induced current is directed in the direction opposite to its speed.
Let an arbitrary contour in a short time interval have shifted from position 1 to position 2 in an arbitrary constant magnetic field. Using the expression for the Lorentz force and the magnetic flux theorem, prove in the general case the formula (18) for the EMF of induction in the circuit (Fig. 114).

Example 11.7.

Magnetic flux through a closed conductive loop with resistance R= 10 ohms changes over time t according to the law Ф =  t 2, where  = 10 Wb / s 2. Determine the amperage I in the loop at time t= 1 ms.

Solution.

The instantaneous value of the EMF induction, according to Faraday's law, is defined as

Then the current in the circuit according to Ohm's law is

mA.

The minus sign in the resulting expression indicates that the direction of the induction current is opposite to the direction of the positive bypass of the contour, which in turn is consistent with the direction of the normal vector to the surface stretched over the contour. The induction current is caused by a vortex electric field generated by a changing magnetic field if the circuit is stationary, and the Lorentz force if it moves in an inhomogeneous constant magnetic field.

Example 11.8.

On a long solenoid with a cross-sectional diameter d= 5 cm and containing n= 20 turns per 1 cm of length, tightly put on a circular turn of copper wire with a cross section s= 1 mm 2 (resistivity of copper
). Find the current in the turn if the current in the solenoid winding is increased at a constant rate
100 A / s. Disregard the magnetic field of the induction current.

Solution.

The magnetic field inside a long solenoid is uniform and equal

where n the number of turns per unit length, and I - instantaneous current value. Therefore, when choosing the direction of the normal to the coil surface along the direction of the field, the magnetic flux through this surface is

where
is the surface area of the loop.

With an increase in the current in the solenoid winding, the magnetic flux through the coil increases, and the resulting induction current is determined by the expression

where
, and the minus sign means that the induction current flows in the direction opposite to the direction of the positive bypass of the loop, matched with the direction of the normal.

Then, the value of the current through the loop at the moment of time t is equal to

mA.

Example 11 .9.

A flat contour (Fig. 13), which looks like two squares with sides a= 20 cm and b= 10 cm, is in a uniform magnetic field perpendicular to its plane. Field induction changes according to the law
, where B 0 = 10 mT and  = 100 s –1. Find the amplitude of the induction current in the circuit if the resistance is unit of its length
... Disregard the magnetic field of this current.

Solution.

The induction current in the frame is

Figure 14 shows the direction of the magnetic field, as well as the normals to the surface of each of the squares that make up the contour, matched by a single direction of positive traversal. Taking this into account, the total magnetic flux through the circuit is equal to

Considering that the loop resistance is
, we find the amplitude of the induction current

on the

Charge and change in magnetic flux

Example 11.10.

A square made of resistance wire R= 1 Ohm, placed in a uniform magnetic field, induction vector which is perpendicular to the plane of the square. Square side length a= 1 cm. The magnitude of the magnetic field induction is initially equal to B= 0.1 T and then it is reduced to zero. Find the value q charge, which as a result will move through the cross-section of the wire.

Solution.

The amount of electricity flowing through any cross-section of a circuit with resistance R when the magnetic flux through the circuit changes by an amount
, equals:

Note that the quantity q does not depend on the nature of the time dependence of the change in the magnetic flux, but is determined only by its initial and final values. Since the magnetic induction varies from to zero, the increment of the magnetic flux penetrating the circuit is equal to

The amount of charge that flows through the wire is determined by the expression

Cl.