Lorentz force. Solving problems and exercises according to the model

Dutch physicist H. A. Lorenz at the end of the 19th century. established that the force exerted by a magnetic field on a moving charged particle is always perpendicular to the direction of motion of the particle and the lines of force of the magnetic field in which this particle moves. The direction of the Lorentz force can be determined using the left-hand rule. If you position the palm of your left hand so that the four extended fingers indicate the direction of movement of the charge, and the vector of the magnetic induction field enters the outstretched thumb, it will indicate the direction of the Lorentz force acting on the positive charge.

If the charge of the particle is negative, then the Lorentz force will be directed in the opposite direction.

The modulus of the Lorentz force is easily determined from Ampere's law and is:

F = | q| vB sin?,

Where q- particle charge, v- the speed of its movement, ? - the angle between the vectors of speed and magnetic field induction.

If, in addition to the magnetic field, there is also an electric field, which acts on the charge with a force , then the total force acting on the charge is equal to:

.

Often this force is called the Lorentz force, and the force expressed by the formula ( F = | q| vB sin?) are called magnetic part of the Lorentz force.

Since the Lorentz force is perpendicular to the direction of motion of the particle, it cannot change its speed (it does not do work), but can only change the direction of its motion, i.e. bend the trajectory.

Such a curvature of the trajectory of electrons in a TV picture tube is easy to observe if you bring a permanent magnet to its screen - the image will be distorted.

Motion of a charged particle in a uniform magnetic field. Let a charged particle fly in at a speed v into a uniform magnetic field perpendicular to the tension lines.

The force exerted by the magnetic field on the particle will cause it to rotate uniformly in a circle of radius r, which is easy to find using Newton's second law, the expression for purposeful acceleration and the formula ( F = | q| vB sin?):

.

From here we get

.

Where m- particle mass.

Application of the Lorentz force.

The action of a magnetic field on moving charges is used, for example, in mass spectrographs, which make it possible to separate charged particles by their specific charges, i.e., by the ratio of the charge of a particle to its mass, and from the results obtained to accurately determine the masses of the particles.

The vacuum chamber of the device is placed in the field (the induction vector is perpendicular to the figure). Charged particles (electrons or ions) accelerated by an electric field, having described an arc, fall on the photographic plate, where they leave a trace that allows the radius of the trajectory to be measured with great accuracy r. This radius determines the specific charge of the ion. Knowing the charge of an ion, you can easily calculate its mass.

DEFINITION

Lorentz force– the force acting on a point charged particle moving in a magnetic field.

It is equal to the product of the charge, the modulus of the particle velocity, the modulus of the magnetic field induction vector and the sine of the angle between the magnetic field vector and the particle velocity.

Here is the Lorentz force, is the particle charge, is the magnitude of the magnetic field induction vector, is the particle velocity, is the angle between the magnetic field induction vector and the direction of motion.

Unit of force – N (newton).

The Lorentz force is a vector quantity. The Lorentz force takes its greatest value when the induction vectors and direction of the particle velocity are perpendicular ().

The direction of the Lorentz force is determined by the left-hand rule:

If the magnetic induction vector enters the palm of the left hand and four fingers are extended towards the direction of the current movement vector, then the thumb bent to the side shows the direction of the Lorentz force.

In a uniform magnetic field, the particle will move in a circle, and the Lorentz force will be a centripetal force. In this case, no work will be done.

Examples of solving problems on the topic “Lorentz force”

EXAMPLE 1

EXAMPLE 2

Exercise Under the influence of the Lorentz force, a particle of mass m with charge q moves in a circle. The magnetic field is uniform, its strength is equal to B. Find the centripetal acceleration of the particle.

Solution Let us recall the Lorentz force formula:

In addition, according to Newton's 2nd law:

In this case, the Lorentz force is directed towards the center of the circle and the acceleration created by it is directed there, that is, this is centripetal acceleration. Means:

1. Calculate the Lorentz force acting on a proton moving at a speed of 106 m/s in a uniform magnetic field with an induction of 0.3 Tesla perpendicular to the induction lines.
2. In a uniform magnetic field with an induction of 0.8 T, a force of 1.5 N acts on a conductor with a current of 30 A, the length of the active part of which is 10 cm. At what angle to the magnetic induction vector is the conductor placed?
3. Which of the electron beam particles
deviate by a larger angle in the same magnetic field - fast or slow? (Why?)
4. Accelerated in an electric field by a potential difference of 1.5 105 V, a proton flies into a uniform magnetic field perpendicular to the lines of magnetic induction and moves uniformly in a circle with a radius of 0.6 m. Determine the speed of the proton, the magnitude of the magnetic induction vector and the force with which the magnetic field acts on a proton.

Literature: -

Internet resources.

-

Topic No. 10 Electromagnetic oscillations.

Solving problems and exercises according to the model.

Read the theoretical material by choosing one of the sources listed in the bibliography.

Find formulas for solving problems.

Write “Given” to the problem statement.

Problem 1. In an oscillating circuit, the inductance of the coil is 0.2 H. Current amplitude is 40 mA. Find the energy of the magnetic field of the coil and the energy of the electric field of the capacitor at the moment when the instantaneous value of the current is 2 times less than the amplitude. Neglect the loop resistance.

Problem 2. A frame with an area of ​​400 cm 2 has 100 turns. It rotates in a uniform magnetic field with an induction of 0.01 Tesla, and the rotation period of the frame is 0.1 s. Write the dependence of the emf on time that occurs in the frame if the axis of rotation is perpendicular to the lines of magnetic induction.

Task 3. A voltage of 220V is supplied to the primary winding of the transformer. What voltage can be removed from the secondary winding of this transformer if the transformation ratio is 10? Will it draw power from the network if its secondary winding is open?

Literature: - G.Ya. Myakishev B.B. Bukhovtsev Physics. Textbook for 11th grade. – M., 2014.

Internet resources.

- Landsberg G.S. Elementary physics textbook - M. Higher School 1975.

Yavorsky B.M. Seleznev Yu.A. Reference Guide to Physics - M.Nauka, 1984.

Solving problems for calculating the parameters of an oscillatory circuit.

Read the theoretical material by choosing one of the sources listed in the bibliography.

Find formulas for solving problems.

Write “Given” to the problem statement.

1. What kind of capacitance is needed in the oscillatory circuit so that with an inductance of 250 mH it can be tuned to an audio frequency of 500 Hz .

2. Find the inductance of the coil if the voltage amplitude is 160 V, the current amplitude is 10 A, and the frequency is 50 Hz .

3. The capacitor is connected to an alternating current circuit of standard frequency with a voltage of 220V. What is the capacitance of the capacitor if the current in the circuit is 2.5 A .

4. In one box there is a resistor, in another there is a capacitor, in the third there is an inductor. The leads are connected to external terminals. How can you find out what is in each of them without opening the boxes? (Given sources of direct and alternating voltage of the same size and a light bulb.)

Literature: - G.Ya. Myakishev B.B. Bukhovtsev Physics. Textbook for 11th grade. – M., 2014.

Internet resources.

- Landsberg G.S. Elementary physics textbook - M. Higher School 1975.

Yavorsky B.M. Seleznev Yu.A. Reference Guide to Physics - M.Nauka, 1984.

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  • movement of a charged particle in a uniform magnetic field;
  • application of the Lorentz force.
Depending on the planning of the material, from 1 to 3 lessons can be devoted to studying this topic, including lessons in problem solving.

Lesson Objectives

Study the motion of a charged particle in a uniform magnetic field, practice solving problems on the topic “The effect of a magnetic field on a moving charge. Lorentz force."

New material in this lesson is studied while students simultaneously work with a computer model. Students must obtain answers to the worksheet questions using the capabilities of this model.

No. Lesson steps Time, min Techniques and methods
1 Organizing time 2
2 Repetition of the material studied on the topic “Lorentz force” 10 Frontal conversation
3 Studying new material using a computer model “Motion of a charged particle in a uniform magnetic field” 30 Working with the worksheet and model
4 Homework explanation 3

Homework: § 6, No. 849 (Collection of problems. 10–11 grades. A.P. Rymkevich - Moscow Bustard, 2001).

Worksheet for the lesson

Sample answers
Model “Charge Movement in a Magnetic Field”

Full name, class ___________________________________________________

1.

under what conditions does a particle move in a circle?

Answer: a particle moves in a circle if the velocity vector is perpendicular to the magnetic field induction vector.

2.

Provided that the particle moves in a circle, set the maximum values ​​for the particle speed and the magnitude of the magnetic field induction. What is the radius of the circle along which the particle moves?

Answer: R = 22.76 cm.

3.

Reduce the particle speed by 2 times. Do not change the magnetic field. What is the radius of the circle along which the particle moves?

Answer: R = 11.38 cm.

4.

Reduce the particle speed by 2 times again. Do not change the magnetic field. What is the radius of the circle along which the particle moves?

Answer: R = 5.69 cm.

5.

How does the radius of the circle along which the particle moves depend on the magnitude of the particle's velocity vector?

Answer: The radius of the circle along which the particle moves is directly proportional to the magnitude of the particle's velocity vector.

6. Reset the maximum values ​​for the speed and magnitude of the magnetic field induction (the particle moves in a circle).
7.

Reduce the magnetic induction value by 2 times. Don't change the speed of the particle. What is the radius of the circle along which the particle moves?

Answer: R = 45.51 cm.

8.

Reduce the magnetic induction value by 2 times again. Don't change the speed of the particle. What is the radius of the circle along which the particle moves?

Answer: R = 91.03 cm.

9.

How does the radius of the circle along which the particle moves depend on the magnitude of the magnetic field induction?

Answer: The radius of the circle along which the particle moves is inversely proportional to the magnitude of the magnetic induction of the field.

10.

Using the formula for the radius of a circle along which a charged particle moves in a magnetic field (formula 1.6 in the textbook), calculate the specific charge of the particle (the ratio of the particle’s charge to its mass).


11.

Compare the specific charge of the particle with the specific charge of the electron. Draw a conclusion.

Answer: the result obtained corresponds to the tabulated value of the specific charge of the electron.

12.

Using the left-hand rule, determine the sign of the particle charge in a computer experiment. Draw a conclusion.

Answer: analysis of the trajectory of a particle according to the left-hand rule allows us to say that it is a negatively charged particle. Taking into account the previously obtained result of the equality of the specific charges of the particle under study and the electron, we can conclude that the particle represented in the model is an electron.

13. Perform the following experiments under this condition: υ x = 5∙10 7 m/s, υ z = 0 m/s, B = 2 mT. 14.

Calculate the Lorentz force acting on the charge.


15.

Calculate the acceleration that this force imparts to this charge (according to Newton’s second law).

F L = 1.6∙10 –14 N,

m = 9.1∙10 –31 kg.

____________________

a – ?

Answer: the acceleration of the charge is 1.76∙10 16 m/s 2.

16.

Calculate the radius of the circle in which the particle is moving using the centripetal acceleration formula.

υ = 5∙10 7 m/s,

a = 1.76∙10 16 m/s 2.

____________________

R – ?

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