1. In this question we will limit ourselves to considering the movement of a charged particle in homogeneous constants fields.
IN magnetic field the Lorentz force will have only one magnetic component
which is always perpendicular to the trajectory of motion and therefore does not do any work, but only bends the trajectory without changing the magnitude of the speed. This kind of force is called gyroscopic.
In the general case, the particle velocity makes an angle with the vector (Fig. 3) and can be decomposed into two vectors (parallel and perpendicular to the vector)
where , , and the particle motion itself can be represented as a superposition of two motions with these velocities.
Let us first consider the movement of a particle with a speed parallel to the magnetic induction vector. In this case, and the particle moves along the magnetic field line.
In the second motion with speed, the Lorentz force does not change in magnitude and creates normal acceleration in a plane perpendicular to the vector. Therefore, the trajectory of such movement is a circle of radius r in this plane. The condition for circular motion, written on the basis of Newton's second law,
allows you to find the radius of the circle and the angular velocity of rotation of the particle
which are called cyclotron radius and cyclotron frequency.
The cyclotron radius is proportional to the momentum of the particle and inversely proportional to the magnitude of its charge and magnetic induction. The cyclotron frequency is inversely proportional to the mass of the particle and proportional to its charge and magnetic induction.
The directions of rotation of particles with positive and negative charges are mutually opposite due to the difference in the directions of the Lorentz force (Fig. 2). In vector form, the cyclotron frequency can be written as
For a positively charged particle, the direction of the angular velocity is opposite to the direction of the vector, for a negatively charged particle it coincides with the vector.
2. In the general case, when a particle is involved in rotational motion around the direction of the vector and in translational motion parallel to the line of force, the resulting motion of the particle will occur along a helical line. For positively charged particles, the helix corresponds to the left screw, for negatively charged particles – to the right (Fig. 4). If the vectors are directed opposite to each other, then vice versa.
This movement is used in systems that focus the electron beam in cathode ray tubes. The fact is that the pitch of the helix, determined by the product and the period of revolution,
for electrons escaping from the electron gun at different angles to the beam axis, does not depend on the angle due to its smallness ().
Therefore, all the electrons emitted from the electron gun at small but different angles will converge at one point after the revolution period. The pitch of the helix can be changed by varying the magnitude of the magnetic induction, which allows the electron beam to be focused on the screen of the cathode ray tube.
Conclusions.
1) The force acting on a charged particle from the magnetic field does no work. It causes rotational motion of particles around the direction of the magnetic induction vector with angular velocity.
2) In the general case, a charged particle moves along a helical line.
3. Magnetic field of a moving charge
1. Let a charged particle move with speed relative to the laboratory frame of reference K. In a system that moves with a particle, there is no magnetic field (), and the electric field is described by the formula
This is the usual electrostatic field of a stationary point charge.
In a stationary reference frame, in accordance with transformations (5), (6), we find
It follows that with slow movements a charged particle creates in the surrounding space an electric field the same as a stationary and magnetic one with induction
In this case, the radius vector is drawn from the charge to the observation point.
Let's analyze this expression. The magnitude of the magnetic induction vector
depends inversely proportional to the square of the distance from the charge to the field point in question, directly proportional to the size of the charge and its speed. But the spatial distribution of magnetic induction around a charge is more complicated than for an electric field.
The formula for magnetic induction includes the sine of the angle between the directions of velocity and the radius vector drawn from the charge to the observation point (Fig. 5).
Magnetic induction vanishes on a line passing through the charge parallel to the velocity vector (), and is maximum in a plane passing through the charge perpendicular to the vector ().
The direction of the magnetic induction vector is perpendicular to the velocity vector and radius vector (Fig. 5).
If, maintaining the angle a and the length of the vector, rotate the radius vector around the velocity vector, then its end will describe a circle. At each point of this circle the vector will be directed tangent to it. Consequently, such a circle will be a vector line (magnetic field line).
Experience shows that the principle of field superposition is satisfied for a magnetic field
The magnetic induction of the resulting field at a certain point is equal to the vector sum of the magnetic induction of the fields created by various sources at this point.
2. Let us now consider a magnetic field created at an arbitrary point by an infinitesimal segment of a thin conductor of length , through which a current of force flows I.
The quantity is called the current element. The direction of the vector coincides with the direction of the current. Since the current strength by definition, where S is the cross-sectional area of the conductor, then the current element can be expressed in terms of current density, where is the volume of the selected section of the conductor. Here it is taken into account that the vectors and coincide in direction.
All charge carriers located in this current element move in an orderly manner at an average speed and create the same magnetic induction at a given point in space. Therefore, we can obtain the resulting magnetic induction created by all charge carriers at an arbitrary point by multiplying the number of carriers in the current element, where n– concentration of charge carriers in a conductor, per magnetic induction created by one carrier at this point
Here the current density is expressed in terms of the average speed of the ordered movement of charge carriers. The radius vector is drawn from the current element to the observation point.
The resulting expression is called the Biot-Savart-Laplace law. It allows you to calculate the magnetic field of any system of conductors using the principle of superposition
The shaded variables refer to the integration point.
A comparison of formulas (8) and (9) shows that the configuration and distribution in space of the magnetic fields of the current element and the moving charge are identical (Fig. 6). The magnitude of the magnetic induction vector created by the current element is proportional to the magnitude of the current element, the sine of the angle between the direction of the current and the direction to the observation point and inversely proportional to the square of the distance from the source to the observation point
The current element creates maximum magnetic induction in a plane perpendicular to the current element, and does not create on a straight line passing through the current element, parallel to the vector. The lines of the tension vector are the essence of a circle around this straight line.
Conclusions.
1) The magnetic field of a moving charge is a consequence of the movement of a charged particle and its electric field.
2) The magnetic field of a current element and a moving charge have the same distribution of force characteristics in space. This is due to the fact that electric current represents the ordered movement of charged particles.
3) The current element and the moving charge create maximum magnetic induction in a plane perpendicular to the direction of movement of the charges. The lines of force in both cases are circles perpendicular to the tangent to the trajectory of motion. The magnetic field is not created on a straight line tangent to the trajectory of the charges.
4) Magnetic induction is inversely proportional to the square of the distance from the charge to the observation point. This is due to the distribution of the electric field of a charged particle in space and its transformation into a magnetic field during movement.
We strengthen our skills in solving and visualizing differential equations using the example of one of the most common evolutionary equations, remember the good old Scilab and try to understand whether we need it... Pictures under the cut (700 kilobytes)
Let's make sure the software is fresh
julia>] (v1.0) pkg>update #will you have time to make tea (v1.0) pkg> status Status `C:\Users\Igor\.julia\environments\v1.0\Project.toml` AbstractPlotting v0.9.0 Blink v0.8.1 Cairo v0.5.6 Colors v0.9.5 Conda v1.1.1 DifferentialEquations v5.3.1 Electron v0.3.0 FileIO v1.0.2 GMT v0.5.0 GR v0.35.0 Gadfly v1.0.0+ #master (https://github.com /GiovineItalia/Gadfly.jl.git) Gtk v0.16.4 Hexagons v0.2.0 IJulia v1.14.1+ [`C:\Users\Igor\.julia\dev\IJulia`] ImageMagick v0.7.1 Interact v0.9.0 LaTeXStrings v1. 0.3 Makie v0.9.0+ #master (https://github.com/JuliaPlots/Makie.jl.git) MeshIO v0.3.1 ORCA v0.2.0 Plotly v0.2.0 PlotlyJS v0.12.0+ #master (https://github .com/sglyon/PlotlyJS.jl.git) Plots v0.21.0 PyCall v1.18.5 PyPlot v2.6.3 Rsvg v0.2.2 StatPlots v0.8.1 UnicodePlots v0.3.1 WebIO v0.4.2 ZMQ v1.0.0
and let's start setting the problem
Movement of charged particles in an electromagnetic field
A charged particle with a charge moving in an EMF with speed is acted upon by the Lorentz force: . This formula is valid with a number of simplifications. Neglecting corrections for the theory of relativity, we assume the mass of the particle to be constant, so that the equation of motion has the form:
Let us direct the Y axis along the electric field, the Z axis along the magnetic field, and assume for simplicity that the initial velocity of the particle lies in the XY plane. In this case, the entire trajectory of the particle will also lie in this plane. The equations of motion will take the form:
Let's make it dimensionless: . Asterisks indicate dimensional quantities, and - the characteristic size of the physical system under consideration. We obtain a dimensionless system of equations of motion of a charged particle in a magnetic field:
Let's lower the order:
As the initial configuration of the model, we will choose: T, V/m, m/s. For a numerical solution we will use the package DifferentialEquations:
Code and graphs
using DifferentialEquations, Plots pyplot() M = 9.11e-31 # kg q = 1.6e-19 # C C = 3e8 # m/s λ = 1e-3 # m function model solver(Bo = 2., Eo = 5e4, vel = 7e4) B = Bo*q*λ / (M*C) E = Eo*q*λ / (M*C*C) vel /= C A = syst(u,p,t) = A * u + # ODE system u0 = # start cond-ns tspan = (0.0, 6pi) # time period prob = ODEProblem(syst, u0, tspan) # problem to solve sol = solve(prob, Euler(), dt = 1e-4, save_idxs = , timeseries_steps = 1000) end Solut = modelsolver() plot(Solut)
Here the Euler method is used, for which the number of steps is specified. Also, not the entire solution of the system is stored in the answer matrix, but only the 1st and 2nd indices, that is, the x and y coordinates (we don’t need velocities).
X = for i in eachindex(Solut.u)] Y = for i in eachindex(Solut.u)] plot(X, Y, xaxis=("X"), background_color=RGB(0.1, 0.1, 0.1)) title !("Particle trajectory") yaxis!("Y") savefig("XY1.png")#save the graph to the project folder
Let's check the result. Let's introduce instead X new variable. Thus, a transition is made to a new coordinate system, moving relative to the original one at a speed u in the direction of the axis X:
If we select and denote , the system will be simplified:
The electric field has disappeared from the last equations, and they represent the equations of motion of a particle under the influence of a uniform magnetic field. Thus, the particle in the new coordinate system (x, y) should move in a circle. Since this new coordinate system itself moves relative to the original one with speed , the resulting motion of the particle will consist of uniform motion along the axis X and rotation around a circle in a plane XY. As is known, the trajectory resulting from the addition of such two movements is, in the general case, trochoid. In particular, if the initial speed is zero, the simplest case of motion of this kind is realized - by cycloid.
Let's make sure that the drift speed is really equal E/B. For this:
- let’s spoil the response matrix by replacing the first element (maximum) with an obviously smaller value
- let's find the number of the maximum element in the second column of the response matrix, which is plotted along the ordinate
- Let's calculate the dimensionless drift speed by dividing the abscissa value at the maximum by the corresponding time value
Out: 8.334546850446588e-5
B = 2*q*λ / (M*C) E = 5e4*q*λ / (M*C*C) E/B
Out: 8.333333333333332e-5
With an accuracy of seventh order!
For convenience, we will define a function that accepts model parameters and a graph signature, which will also serve as the file name png, created in the project folder (works in Juno/Atom and Jupyter). Unlike Gadfly, where the graphs were created in layers, and then were output by the function plot(), in Plots, in order to create different graphs in one frame, the first of them is created by the function plot(), and subsequent ones are added using plot!(). In Julia, the names of functions that change accepted objects usually end with an exclamation point.
function plotter(ttle = "qwerty", Bo = 2, Eo = 4e4, vel = 7e4) Ans = modelsolver(Bo, Eo, vel) X = for i in eachindex(Ans.u)] Y = for i in eachindex( Ans.u)] plot!(X, Y) p = title!(ttle) savefig(p, ttle * ".png") end
At zero initial speed, as expected, we obtain cycloid:
plot() plotter("Zero start velocity", 2, 4e4, 7e4)
We obtain the trajectory of the particle when the induction and voltage are zero and when the sign of the charge changes. Let me remind you that the dot means sequential execution of the function with all elements of the array
Hidden away
plot() plotter.("B is zeroed E varies", 0, )
plot() plotter.("E is zero B varies", , 0)
q = -1.6e-19 # C plot() plotter.("Negative charge")
And let’s see how a change in initial velocity affects the trajectory of a particle:
plot() plotter.("Variation of speed", 2, 5e4, )
A little about Scilab
There is already enough information on Habré about Sailab, for example, so we will limit ourselves to links to Wikipedia and the home page.
On my own behalf, I’ll add about the availability of a convenient interface with checkboxes, buttons and graph output, and a rather interesting visual modeling tool, Xcos. The latter can be used, for example, to simulate a signal in electrical engineering:
Actually, our problem can be solved in Scilab:
Code and pictures
clear function du = syst(t, u, A, E) du = A * u + // ODE system endfunction function = model solver(Bo, Eo, vel) B = Bo*q*lambda / (M*C) E = Eo*q*lambda / (M*C*C) vel = vel / C u0 = // start cond-ns t0 = 0.0 tspan = t0:0.1:6*%pi // time period A = U = ode(" rk", u0, t0, tspan, list(syst, A, E)) endfunction M = 9.11e-31 // kg q = 1.6e-19 // C C = 3e8 // m/s lambda = 1e-3 / / m = modelsolver(2, 5e4, 7e4) plot(cron, Ans1) xtitle("Dimensionless coordinates and velocities","t","x, y, dx/dt, dy/dt"); legend("x", "y", "Ux", "Uy"); scf(1)//creating a new graphic window plot(Ans1(1, :), Ans1(2, :)) xtitle ("Particle trajectory","x","y"); xs2png(0,"graf1");// you can save graphs in different formats xs2jpg(1,"graf2");// however, it works every now and then
Information on the function for solving difurs ode. Basically this begs the question
Why do we need Julia?
... if there are already such wonderful things as Scilab, Octave and Numpy, Scipy?
I won’t say anything about the last two - I haven’t tried them. And in general, the question is complex, so let’s think offhand:
Scilab
On a hard drive it will take a little more than 500 MB, it starts quickly and the difuro calculation, graphics and everything else are immediately available. Good for beginners: excellent guide (mostly localized), there are many books in Russian. Internal errors have already been mentioned and, and since the product is very niche, the community is sluggish, and additional modules are very scarce.
Julia
As packages are added (especially any Python stuff a la Jupyter and Mathplotlib), it grows from 376 MB to quite more than six gigabytes. It doesn’t spare the RAM either: at the start it’s 132 MB and after you draw up graphs in Jupiter, it will easily reach 1 GB. If you work in Juno, then everything is almost like in Scilab: You can execute code directly in the interpreter, you can type in the built-in notepad and save as a file, there is a variable browser, a command log and online help. Personally, I am outraged by the absence of clear() , i.e. I ran the code, then started correcting and renaming it, but the old variables remained (there is no variable browser in Jupiter).
But all this is not critical. Scilab is quite suitable for the first couples; making a lab, a course, or calculating something in between is a very handy tool. Although there is also support for parallel computing and calling C/Fortran functions, it cannot be seriously used for anything. Large arrays plunge him into horror; in order to define multidimensional ones, he has to deal with all sorts of obscurantism, and calculations outside the framework of classical problems may well throw everything away along with the operating system.
And after all these pains and disappointments, you can safely move on to Julia, to rake even here. We will continue to study, fortunately the community is very responsive, problems are resolved quickly, and Julia has many more interesting features that will turn the learning process into an exciting journey!
1.5. Movement of charged particles in an electric field
1.5.1. An electron with a zero initial velocity enters a uniform electric field of strength E = 200 kV/m. How far will an electron fly if left to its own devices in time?t = 1 ns? What speed will he reach?
Solution
1. Let us determine the magnitude of the Coulomb force acting on an electron of mass m @ 1×10 -30 kg when it enters an electric field of intensity E
where e @ 1.6×10 -19 C is the charge of the electron.
2. For an electron, which is considered a particle within the framework of classical concepts, Newton’s second law is valid, through which the acceleration of the particle can be found
The impressive magnitude of the acceleration is due to the very small mass of the electron and the relatively large value of the force.
3. We will find the path traveled by the electron in a given period of time using kinematic relations
. (3)
4. The speed of the electron at the end of a given period of time is determined from the law of conservation of momentum
. (4)
1.5.2. The proton and electron must be accelerated to speed v = 30 mm/s. What potential difference must they pass through?
Solution
1. The work of moving a charge in an electric field in accordance with the theorem on the change in kinetic energy is equal to
, (1)
where v1 and v2 are the initial and final speed of the particle, m is the mass of the particle. If we assume that the acceleration of particles begins from a state of rest, then equation (1) can be simplified
. (1)
2. Potential difference required to accelerate an electron with mass me @ kg and charge e @ 1.6 × 10 -19 kg
. (2)
3. Potential difference required to accelerate a proton to a given speed, having mass mp @ 1.67×1kg and charge p @ 1.6×1C
. (3)
1.5.3. The potential difference between the cathode and anode isU = 90 V, distance isr = 1×10- 3 m. With what acceleration does the electron move from the cathode to the anode? How long does it take him to cover the distance?r. What is the speed of the electron v at the moment of impact with the anode surface? In what timet does the electron fly the distance from the cathode to the anode?
Solution
1. Using equation (1) of the previous problem, we determine the final speed of the electron before hitting the anode
2. Let us write down the kinematic equations of electron motion and determine the flight time of the electron from the cathode to the anode
3. We determine the acceleration of the electron from the upper equation of the system of equations (2)
. (3)
1.5.4. A speck of dust weighingm = 1×10- 12 kg, carrying an electric charge of five electrons, passed through an accelerating potential difference in a vacuumU = 3 10 6 V. What is the speed and kinetic energy of the dust grain?
Solution
1. The change in the energy of a dust grain, in accordance with the theorem on the change in kinetic energy, is equal to the work of the electric field forces
2. Let us express the energy of a dust grain in electron volts
. (2)
3. Determine the speed of a speck of dust
. (3)
1.5.5. A charged particle passing through an accelerating potential differenceU = 0.6 MV, acquired speed v = 5.4 Mm/s. Determine the specific charge of the particle (ratio of charge to mass).
Solution
1. Let’s write a theorem about the change in the kinetic energy of a particle and determine the specific charge
1.5.6. A proton whose initial speed was equal to v0 = 100 km/s, having passed through an accelerating electric field with a intensity E = 300 V/cm, it doubled its speed. How far did the proton travel if its velocity vector coincided in direction with the intensity vector?
Solution
1. Let us determine the magnitude of the Coulomb force acting on a proton with mass m = 1.67×1 kg and charge e = +1.6 -19 C
2. Let us write a theorem about the change in the kinetic energy of a proton when it passes through an electric field
, (2)
and determine the path s traversed by the proton
. (3)
1.5.7. An infinite plane is negatively charged with surface densitys = 35.4 nC/m2. The electron moves in the direction of the field line created by the plane. On distancey0 = 5×10- A 2 m electron had a kinetic energy K = 80 eV. What is the minimum distanceymin electron can approach the plane?
Solution
2. Braking force acting on the electron from the electric field
3. The electron will stop its movement at the moment of time when the work of the Coulomb force, braking its movement, becomes equal in value to the initial kinetic energy, the electron will travel a certain distance y
(4)
4. The distance to the plate at the moment the electron stops will be determined as
. (5)
1.5.8. An electron flying horizontally at a speed v0 = 1.6 Mm/s, flew into a uniform electric field with a intensity E = 90 V/cm, directed vertically. Determine the electron velocity vectorvthrought = 1 ns?
|
Solution
1. In a vertical electric field, the electron will be acted upon by the Coulomb force, which provides acceleration directed along the y axis
, (1)
where e @ 1.6×1C is the charge of the electron, m @ 1×1kg is the mass of the electron.
2. Since the projection of acceleration onto the x axis is zero, the horizontal motion of the electron will proceed with an initial speed v0, i.e. vx = v0, and the vertical component of the speed will be determined by the equation
in the case under consideration, taking into account equation (1):
3. Thus, after time t the modulus of the electron velocity will be equal to
. (4)
. (5)
|
1.5.9. An electron flies into a flat capacitor with a speed v0 = 2 Mm/s, directed perpendicular to the electric field strength vector. At what distanceh will the electron move to the lower plate of the capacitor during the flight of the capacitor plates? The length of the plates is x = 5 cm, the distance between the platesd = 2 cm, potential difference between platesU = 2 V.
Solution
1. Let us write down the kinematic equations of electron motion under the influence of the constant Coulomb force F = eE= eU/d
2. Since the electron moves along the horizontal axis at a constant speed, the time of flight of the capacitor can be determined as
3. The vertical displacement of an electron can thus be represented by the following equation
4. The vertical acceleration of the electron a is determined by Newton’s second law
. (4)
5. Substitute the acceleration value from equation (4) into equation (3)
. (5)
1.5.10. Proton anda- a particle from a state of rest passes an accelerating electric field. In what ratio will their speeds be?
Solution
1. As is known, an a-particle consists of two protons and two neutrons, therefore the charge of an a-particle is twice the charge of a proton, i.e. qa = 2qp, and the mass is four times greater, i.e. ma = 4 mp.
2. When particles pass through the same potential difference Dj by field forces, work will be done and they will acquire the corresponding kinetic energy
. (1)
1.6. Electric dipole. Properties of dielectrics
1.6.2. Dipolewith electric torque p = 0.12 nC×m is formed by two point chargesq. Determine the electric field strength E and potentialj at points A and B located at a distancer = 8 cm from the center of the dipole.
Solution
1. Let us determine the electric field strength of the dipole at point A, which is located at a distance of r = 0.08 m from the center of the dipole, and the radius vector r makes an angle with the dipole axis a = p/2, i.e. cosa = 0
2. Potential at point A
. (2)
https://pandia.ru/text/78/367/images/image042_0.gif" width="269" height="40">. (4)
1.6.3. Determine tension E and potentialj electric dipole with moment p = 4 pC×m at a distancer = 0.1 m from the center of the dipole in the directiona = 600 with electric torque vector.
|
Solution
, (1)
https://pandia.ru/text/78/367/images/image046_0.gif" width="267" height="44">. (2)
1.6.4. Dipole with electric moment p = 1 pC×m rotates uniformly with frequencyn = 10 3 s- 1 relative to the axis passing through the center of the dipole perpendicular to its arm. Obtain the law of potential change over time for a certain point located at a distance from the center of the dipoler = 1 cm and lying in the dipole plane. At the initial moment of time the potential is zeroj(0) = 0.
|
Solution
1. In this case, the electric torque vector, while remaining constant in magnitude, changes its position over time, in other words, the angle a = f(t). These circumstances lead to the fact that the magnitude of the potential will also become dependent on time
. (1)
2. Determine the amplitude value of the potential that will occur at t = 0, when cos(2pn0) = 1
. (2)
, (3)
and write the potential equation as a function of time
Thus, the potential at point A, located in the plane of rotation of the dipole, changes over time according to the cosine law.
1.6.5. Electric dipole with moment p = 0.1 nCIt is attached to an elastic thread. When an electric field of intensity E = 3 kV/m was created in the space where the dipole is located, perpendicular to the torque vector, the dipole rotated at an anglea = 300. Determine the torsion constant of the threadz, equal to the torque of the twisting force per 1 rad.
1.6.6. Perpendicular to the dipole arm with electric moment p = 12 pC×m a uniform electric field of intensity E = 300 kV/m is excited. Under the influence of the field, the dipole begins to rotate about an axis passing through its center. Determine the angular velocity of the dipolew at the moment it passes through the equilibrium position. The moment of inertia of the dipole relative to the axis perpendicular to the arm and passing through the center of the dipole is equal toJ=2×10- 9 kg×m2.
Solution
1. Kinetic energy of rotational motion is defined as
2. The mechanical moment acting on the dipole in an electric field is equal to
. (2)
3. The total work done when rotating a dipole in an electric field depends on the electric torque and the angle of rotation
. (3)
4. When passing through the equilibrium state, a1 = 0, a2 = p/2, since the electric field in the initial position of the dipole is perpendicular to its axis
1.6.7. A projection lamp bulb filled with krypton under pressure p = 20 MPa at a temperature T = 400 K is placed in an electric field of strength E = 2 MV/m. Find the dielectric constant of krypton and its polarization P. Take the polarizability of krypton equal toa = 4.5×10- 29 m3.
Solution
1. Dielectric constant e is included in the Clausius-Mossotti equation
where n is the concentration of krypton atoms.
2. We determine the concentration of gas molecules from the equation of molecular kinetic theory
, (2)
where kB = 1.4×1J/K is Boltzmann’s constant.
3. Substitute the value of n from equation (2) into equation (1)
4. Let us resolve equation (3) with respect to the dielectric constant e
. (3)
5. The magnitude of polarization is determined in general form by the equation
where pi is the dipole moment induced in the i-th atom, k is the number of atoms in volume DV. When atoms are in a uniform electric field, all atoms have dipole moments that coincide in direction and magnitude, this makes it possible to move from geometric addition in equation (4) to algebraic
6. On the other hand, the ratio of the total number of atoms to the volume they occupy is equal to the concentration k/DV = n
8. Let us express the local field strength E* through the external field strength E, and the concentration through pressure and temperature
(8)
1.6.8. Near the atom at a distance of r = 1 nm there isa- a particle that is a doubly ionized helium atom with a charge of 2|e|. Electric fielda- particles induce an electric moment of the atom p = 1×10- 32 Kl×m. Find the polarizability of this atom.
Solution
1. Polarizability a is proportional to the induced electric torque p and inversely proportional to the local magnetic field strength
2. In this case, the external electric field is created by the a - particle, which, in fact, will also be local
3. Let's combine equations (1) and (2)
. (3)
1.6.9. Water has a densityr = 103 kg/m3 and refractive indexn = 1.33. Determine Electronic Polarizabilityae water molecules.
Solution
1. The electronic polarizability of molecules is determined by the Lorentz - Lorentz formula
, (1)
where m = 18×10 - 3 kg/mol, NA = 6×1023 mol - 1.
2. Let us resolve equation (1) with respect to aе
. (2)
. (3)
1.7. Electrical capacity. Capacitors
1.7.1. Determine the electrical capacitance C of a solitary conducting sphere with radiusR = 1 m, immersed in transformer oil.
Solution
1. Dielectric constant of kerosene e = 2, the capacity of the ball is determined by the equation
1.7.2. Find the electrical capacitance C of a conducting sphere immersed in water. The radius of the sphere isR = 2 cm.
Solution
1. Let’s use the equation of the previous problem, taking into account the dielectric constant of water e = 80
1.7.3. Determine the electrical capacity of the Earth, taking it to be a sphere of radiusR@ 6,4 ×105 m.
Solution
1. Let's use the equation for the electrical capacity of the ball
1.7.4. Two metal balls with radiiR1 = 2 cm andR2 = 6 cm is connected by a conductor with a negligible capacitance and an electric charge is impartedQ = 1 nC. Determine the surface charge density.
Solution
1. Let’s write down the equations for the electric capacity of the balls
2. The electric capacity of the ball is determined, as is known, by the amount of charge placed on it and the potential C = Q/j. Since the balls are connected by a capacitive conductor, the potential of both balls will be the same, but the electrical capacitances will be different
where Q1, Q2 and C1, C2 are the charges and electrical capacities of the balls, respectively.
3. In accordance with the law of conservation of charge
4. We form a system of equations from which we can find the charge of each ball
, (4)
, (5)
. (7)
https://pandia.ru/text/78/367/images/image086.gif" width="343" height="43">. (9)
1.7.5. Ball radiusR1 = 6 cm charged to potentialj1 = 300 V, and the ball has a radiusR2 = 4 cm to a potential of 500 V. Find the potential of the balls after they are connected by a capacitive conductor.
Solution
1. Let's write down the equations that determine the electric capacity of the balls
2. Total capacity of the balls after connection
3. Since the potentials of the balls are known before they connect, their charges can be determined
4. Electric charge of the balls after they are connected by a capacitive conductor
5. Potential of the balls after their connection
1.7.6. Copper cannonball, massm = 10 kg due to friction during flight with the air acquired an electric charge equivalentN = 1010 uncompensated elementary charges. Determine the electrical capacity of the nucleus and its potential.
Solution
1. To determine the electrical capacity of a spherical cannonball, it is necessary to know its radius, which can be found from the known mass m and copper density r = 8.9 × 103 kg/m3
. (1)
2. Electrical capacity of a copper cannonball
3. Electric potential of the nucleus
. (3)
1.7.7. Charged conducting body of spherical shape with radiusR = 2 cm has electrical energyW = 1 J. Determine the potential of this body.
Solution
1. Electrical energy and the potential of a charged body are related by the following equation
1.7.8. Find the electric capacitance C of a flat capacitor with the area of the platess = 100 cm2 and the distance between themd = 0.1 mm filled with mica with dielectric constante = 7.
Solution
1. The electrical capacitance of a flat capacitor is determined by the equation
. (1)
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1.7.9. Between the plates of a flat capacitor charged to a potential differenceU = 600 V, there are two layers of dielectrics: thick glassd1 = 7 mm and ebonite thicknessd2 = 3 mm. Area of each plates = 200 cm2. Determine the electrical capacitance of capacitor C, displacementD, voltage E and potential drop on each dielectric layer.
Solution
1. Let us take the dielectric constant of glass e1 = 7, the permeability of ebonite e2 = 3. The design proposed in the problem can be considered as two series-connected capacitors, and
, (1)
, (2)
2. Determine the charge of the capacitor
3. Surface electric charge density s, which coincides in value with the displacement value D
. (5)
4. Since C1 = C2 = 180 pF, then j1 = j2 = 300 V, and for the field strength E the following relations can be written
. (6)
. (7)
1.7.10. Distance between the plates of a parallel-plate capacitord = 1.3 mm, plate area iss = 20 cm2. In the space between the capacitor plates there are two layers of dielectrics: mica thickd1 = 0.7 mm and ebonite thicknessd2 = 0.3 mm. Determine the electrical capacitance of such a capacitor.
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Solution
1. This design of an electrical capacitance can be considered as three capacitors connected in series: one with a mica dielectric, the second with an ebonite dielectric, and the third with an air dielectric.
2. Dielectric constant of mica e1 = 7, dielectric constant of ebonite e2 = 3, dielectric constant of air e3 = 1.
3. Three capacitors connected in series have a common capacitance determined by the equation
. (1)
4. The capacitances of individual capacitors are respectively equal
5. Compare equations (1) and (2)
. (3)
6. Let's transform the last equation to a simpler form
, (4)
, (5)
. (6)
7. Divide the numerator and denominator of equation (6) by the product of dielectric constants (e1e2e3)
, (7)
, (8)
8. Substitute into equation (8) the tasks and reference data specified by the condition
1.7.11. An electric charge with a density ofs = 0.2 µC/m2. Distance between platesd = 1 mm. How much will the potential difference on the capacitor plates change if the distance between the plates is tripled?
Solution
1. The potential difference on the plates of the capacitor and its charge are related by the following relationship
2. As the distance between the plates increases, the capacitance of the capacitor and the potential difference between the plates changes, in other words
3. Determine the potential difference when changing the distance between the plates
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1.7.12. Two cubes of electrical capacity C1 and C2 are charged to potentialsj1 andj2 respectively. Determine the capacity of a rectangular prism made up of these cubes.
Solution
1. Since the connected bodies do not represent capacitors in the classical sense, it is not possible to use series or parallel connection formulas to find the total capacitance. In this case, the laws of conservation of charge and energy apply.
2. Let's write down the laws of conservation of charge and energy
. (1)
3. Let us rewrite the system of equations (1) taking into account the values of the charges of the cubes and their total energy Wo
. (2)
4. Let us combine the equations of system (2)
, (3)
. (4)
1.7.13. On a flat capacitor with a paraffin dielectric (e = 2) voltage appliedU = 4000 V. Distance between platesd = 2 mm. Determine surface charge densitys on the covers.
Solution
1. Let us express the electrical capacitance of the capacitor in terms of its electrical and geometric parameters
2. Substitute the given values into equation (1)
. (2)
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1.7.14. A flat capacitor consists of two round conducting plates with a radiusr = 1 cm, the space between which is filled with vinyl plastic with dielectric constante = 3. What is the maximum chargeQmax should be on the plates so that at an electric field strength E = 45 kV/mm, electrical breakdown of the dielectric occurs?
Solution
1. To solve the problem, we use equation (1) of the previous problem
. (3)
2. Let us resolve equation (3) with respect to the charge Q
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1.7.14. Electrostatic balances are a device in which the force of gravity is compensated by the force of attraction between differently charged plates located at a distanced = 1 mm. What additional weight must be placed on the scale so that the distance between the plates is maintained when charging the capacitor with voltageU = 1 kV? The area of the plates iss = 5×10- 3 m2.
Solution
1. Determine the Coulomb force acting on a positively charged plate
2. On the other hand, the charge of a capacitor can be expressed through its capacitance and the potential difference between the plates
3. Substitute the charge value from equation (2) into equation (1)
4. Determine the mass overload m for balancing the scales
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1.7.15. Electrostatic scales are designed in such a way that one of the capacitor plates is fixedly fixed, and the second is connected to a spring with a stiffness coefficientk. The area of the capacitor plates iss. Determine the elongation of the springDl when imparting charges of equal magnitude and opposite sign to the platesQ.
Solution
1. Using equation (1) of the previous problem, we determine the magnitude of the force that arises during the interaction of oppositely charged plates
2. The attraction of the plates will be accompanied by an elongation of the spring by the amount Dl and the appearance of an elastic force Fу = k×Dl, in other words
. (2)
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1.7.16. In a parallel-plate variable capacitor, the capacitance changes by increasing the distance between the plates. What work is done by the current source to which the plates are connected if the capacitance changes from C1 to C2, and the charge of the capacitor remains equalQ?
Solution
1. As was shown in previous problems, oppositely charged plates attract with force
2. With an elementary change in the distance between the plates by dу, the capacitance of the capacitor changes by dC, and elementary work is performed by an external energy source, such as a battery
the total work done when the distance changes from d1 to d2 is determined as
3. Let’s solve equations (2) regarding the charges and determine their difference
, (3)
1.7.18. The plates of a flat-plate air capacitor carry charges + 3Q and –Q. Determine the potential difference between the plates if the distance between themd, and their area- s.
Solution
1. We will assume that the electric field strength between two parallel charged plates is determined by the equation
where https://pandia.ru/text/78/367/images/image151.gif" width="165" height="40">. (2)
, (3)
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1.7.19. A flat-plate air capacitor is immersed in a liquid dielectric with a dielectric constante2 in two ways shown in the figure. How many times does the capacitance of the capacitor change?
Solution
1. When half the area of both plates is immersed in a liquid dielectric, then such a complex capacitor can be considered as two electrical capacitances connected in parallel
where e1 = 1 is the dielectric constant of air, e2 is the dielectric constant of liquid dielectric.
2. The change in capacity for the case considered above will be
where C0 = e0e1s/d is the electrical capacitance of the air capacitor.
3. When one plate is immersed in a dielectric, a complex capacitance is formed, which can be represented as two series-connected capacitors C2.1 and C2.2
, (3)
4. The ratio of capacities in this case is determined by the equation
. (6)
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1.7.20. In the absence of gravity, a flat air capacitor with plates of areas and the distance between themd1 is connected to a source with electromotive forcee. A conducting plate with a mass is tightly pressed to the bottom platem and thicknessd. At what speed will the plate hit the top plate if it is released?
Solution
1. An electric field is induced on a conductive plate pressed to the bottom plate, and negative charges will be concentrated on the side of the bottom plate, and positive charges on the opposite side. Since the plate is pressed tightly to the plate and its location is asymmetrical, some of the electrons from the plate will go to the plate, the charge of which can be defined as
, (1)
where C = e0s/(d1 – d2) is the capacity of the air capacitor formed by the metal plate and the top plate, e* is the emf of the current source.
2. The negatively charged metal plate will be attracted to the top positively charged plate of the capacitor. Due to Newton's second law, the presence of a force acting on a mass must inevitably lead to its movement. The motion of the plate is described by the law of conservation of energy, in particular, by the theorem on the change in kinetic energy. The work done by the electric field forces is equal to the change in the kinetic energy of the plate. Taking into account the immobility of the plate at the initial moment of time, the above can be represented as follows
, (2)
whence the speed of the plate at the moment it reaches the upper plate is determined by the equation
. (3)
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1.7.21. How many times will the capacitance of a flat air capacitor with plates of area change?s1 and the distance between themd1, if parallel to the plates a paraffin plate with an area ofs2 =s1/2 and thicknessd2 =d1/2?
Solution
1. In this case, when adding a plate, the capacitance can be represented as three capacitors, with serial and parallel connection. The electrical capacity of a capacitor formed by plates and an air gap is determined as
2. When introducing a plate with an air gap above it, it represents two series-connected capacitors C2.1 and C2.2, and a parallel capacitance C2.3
1.7.23. Determine the capacitance of the capacitor connection, if C1 = C2 = C3 = C4 = C5 = 1 µF
Solution
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1. Since all the capacitors involved in the circuit under consideration are the same, the potentials of points 2 and 4 will also be the same, which means that when the battery is connected to a current source, capacitor C5 will not be charged. In this regard, the above diagram can be simplified.
2. In the absence of capacitor C5, the circuit is a combination of series and parallel connection
. (2)
3. According to the conditions of the problem, all containers are the same in size, so let’s introduce the notation C1 = C2 = C3 = C4 = C, then
If a particle with charge e moves in space where there is an electric field with intensity E, then it is acted upon by a force eE. If, in addition to the electric field, there is a magnetic field, then the Lorentz force equal to e also acts on the particle, where u is the speed of the particle relative to the field, B is the magnetic induction. Therefore, according to Newton’s second law, the equation of particle motion has the form:
The written vector equation breaks down into three scalar equations, each of which describes movement along the corresponding coordinate axis.
In what follows we will be interested only in some special cases of motion. Let us assume that charged particles, initially moving along the X axis with speed, enter the electric field of a flat capacitor.
If the gap between the plates is small compared to their length, then edge effects can be neglected and the electric field between the plates can be considered uniform. By directing the Y axis parallel to the field, we have: . Since there is no magnetic field, then 
. In the case under consideration, the charged particles are only affected by the force from the electric field, which, for the chosen direction of the coordinate axes, is entirely directed along the Y axis. Therefore, the trajectory of the particles lies in the XY plane and the equations of motion take the form:
The movement of particles in this case occurs under the influence of a constant force and is similar to the movement of a horizontally thrown body in a gravitational field. Therefore, it is clear without further calculations that the particles will move along parabolas.
Let us calculate the angle by which the particle beam will deviate after passing through the capacitor. Integrating the first of equations (3.2), we find:
Integrating the second equation gives:
Since at t=0 (the moment the particle enters the capacitor) u(y)=0, then c=0, and therefore
From here we get for the deflection angle:
We see that the beam deflection significantly depends on the specific particle charge e/m
§ 72. Motion of a charged particle in a uniform magnetic field
Let's imagine a charge moving in a uniform magnetic field with a speed v perpendicular to V. The magnetic force imparts to the charge an acceleration perpendicular to the speed
(see formula (43.3); the angle between v and B is a straight line). This acceleration only changes the direction of the speed, but the magnitude of the speed remains unchanged. Consequently, the acceleration (72.1) will be constant in magnitude. Under these conditions, a charged particle moves uniformly in a circle, the radius of which is determined by the relation. Substituting here the value (72.1) for and solving the resulting equation for R, we obtain
So, in the case when a charged particle moves in a uniform magnetic field perpendicular to the plane in which the movement occurs, the trajectory of the particle is a circle. The radius of this circle depends on the speed of the particle, the magnetic induction of the field and the ratio of the particle's charge to its mass. The ratio is called specific charge.
Let us find the time T spent by the particle on one revolution. To do this, divide the circumference by the velocity of the particle v. As a result we get
From (72.3) it follows that the period of revolution of a particle does not depend on its speed; it is determined only by the specific charge of the particle and the magnetic induction of the field.
Let us find out the nature of the motion of a charged particle in the case when its speed forms an angle a other than a straight line with the direction of a uniform magnetic field. Let us decompose the vector v into two components; - perpendicular to B and - parallel to B (Fig. 72.1). The modules of these components are equal
Magnetic force has a modulus
and lies in a plane perpendicular to B. The acceleration created by this force is normal for the component.
The component of the magnetic force in direction B is zero; therefore, this force cannot affect the value. Thus, the movement of a particle can be represented as the superposition of two movements: 1) movement along direction B with a constant speed and 2) uniform movement in a circle in a plane perpendicular to vector B. The radius of the circle is determined by formula (72.2) with v replaced by . The trajectory of motion is a helix whose axis coincides with direction B (Fig. 72.2). The line step can be found by multiplying the rotation period T determined by formula (72.3):
The direction in which the trajectory twists depends on the sign of the particle's charge. If the charge is positive, the trajectory spins counterclockwise. The trajectory along which a negatively charged particle moves twists clockwise (it is assumed that we are looking at the trajectory along direction B; the particle flies away from us, if, and towards us, if).
16. Movement of charged particles in an electromagnetic field. Application of electron beams in science and technology: electron and ion optics, electron microscope. Charged particle accelerators.
Let's introduce the conceptelementary particle as an object, the mechanical state of which is completely described by specifying three coordinates and three components of the speed of its movement as a whole. Studyinteractions of elementary particles with em.m. Let us preface the field with some general considerations related to the concept of “particle” in relativistic mechanics.
Particle interaction with each other is described (and was described before the theory of relativity) using the concept of a force field. Each particle creates a field around itself. Every other particle in this field is subject to a force. This applies to both charged particles interacting with em. field, and massive particles that do not have a charge and are in a gravitational field.
In classical mechanics, the field was only a way of describing the interaction of particles as a physical phenomenon. The situation is changing significantly in the theory of relativity due to the finite speed of field propagation. The forces currently acting on a particle are determined by their location at the previous time. A change in the position of one of the particles is reflected in other particles only after a certain period of time. The field becomes physical reality through which the interaction of particles occurs. We cannot talk about the direct interaction of particles located at a distance from each other. Interaction can occur at any moment only between neighboring points in space (short-range interaction). That's why we can talk about the interaction of a particle with a field and the subsequent interaction of the field with another particle .
In classical mechanics, you can introduce the concept of an absolutely rigid body, which under no circumstances can be deformed. However, in the impossibility of existence absolutely rigid body can be easily verified using the following reasoning based on theory of relativity.
Let a rigid body be set in motion at any one point by an external influence. If there was a body absolutely solid, then all its points would have to move simultaneously with the one that was affected. (Otherwise the body would have to deform). The theory of relativity, however, makes this impossible, since the impact from a given point is transmitted to others at a finite speed, and therefore all points of the body cannot simultaneously begin to move. Therefore, under absolutely solid body we should mean a body, all dimensions of which remain unchanged in the frame of reference where it is at rest.
From the above, certain conclusions regarding the consideration of elementary particles . It is obvious that in relativistic mechanics particles, which we consider as elementary , cannot be assigned finite dimensions. In other words, within the strict special theory of relativityelementary particles should not have finite dimensions and, therefore, should be considered as point ones.
17. Own electromagnetic oscillations. Differential equation of natural electromagnetic oscillations and its solution.
Electromagnetic vibrations are called periodic changes in tension E and induction B.
Electromagnetic waves include radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, x-rays, and gamma rays.
In unlimited space or in systems with energy losses (dissipative), eigenelectric circuits with a continuous frequency spectrum are possible.
18. Damped electromagnetic oscillations. Differential equation of damped electromagnetic oscillations and its solution. Attenuation coefficient. Logarithmic damping decrement. Good quality.
electromagnetic damped oscillations arise in e electromagnetic oscillatory system, called LCR - circuit (Figure 3.3).
Figure 3.3.
Differential equation
we obtain using Kirchhoff’s second law for a closed LCR circuit: the sum of the voltage drops across the active resistance (R) and capacitor (C) is equal to the induced emf developed in the circuit circuit: 
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attenuation coefficient |
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This is a differential equation that describes the fluctuations in the charge of a capacitor. Let us introduce the following notation: The value β, as in the case of mechanical vibrations, is called attenuation coefficient, and ω 0 – natural cyclic frequency hesitation. With the introduced notation, equation (3.45) takes the form
Equation (3.47) completely coincides with the differential equation of a harmonic oscillator with viscous friction (formula (4.19) from the section “Physical foundations of mechanics”). The solution to this equation describes damped oscillations of the form q(t) = q 0 e -bt cos(wt + j) (3.48) where q 0 is the initial charge of the capacitor, ω = is the cyclic frequency of oscillations, φ is the initial phase of oscillations. In Fig. Figure 3.17 shows the form of the function q(t). The dependence of the voltage on the capacitor on time has the same form, since U C = q/C. |
(3.47)DECREMENT DECREMENT
(from Latin decrementum - decrease, decrease) (logarithmic attenuation decrement) - a quantitative characteristic of the rate of attenuation of oscillations in a linear system; represents the natural logarithm of the ratio of two subsequent maximum deviations of a fluctuating quantity in the same direction. Because in a linear system, the oscillating value changes according to the law (where the constant value is the damping coefficient) and the two subsequent maximum. deviations in one direction X 1 and X 2 (conventionally called “amplitudes” of oscillations) are separated by a period of time (conventionally called “period” of oscillations), then 
, and D. z..
So, for example, for mechanical oscillate system consisting of mass T, held in the equilibrium position by a spring with a coefficient. elasticity k and frictional force F T , proportional speed v(F T =-bv, Where b- coefficient proportionality), D. z.
At low attenuation. Likewise for electric. circuit consisting of inductance L, active resistance R and containers WITH, D. z.
.
At low attenuation.
For nonlinear systems, the law of damping of oscillations is different from the law, i.e., the ratio of two subsequent “amplitudes” (and the logarithm of this ratio) does not remain constant; therefore D. z. does not have such a definition. meaning, as for linear systems.
Good quality- a parameter of the oscillatory system that determines the width of the resonance and characterizes how many times the energy reserves in the system are greater than the energy losses during one oscillation period. Indicated by the symbol from English. quality factor.
The quality factor is inversely proportional to the rate of decay of natural oscillations in the system. That is, the higher the quality factor of the oscillatory system, the less energy loss for each period and the slower the oscillations decay.
19. Forced electromagnetic oscillations. Differential equation of forced electromagnetic oscillations and its solution. Resonance.
Forced electromagnetic oscillations are called periodic changes in current and voltage in an electrical circuit that occur under the influence of an alternating emf from an external source. An external source of EMF in electrical circuits are alternating current generators operating at power plants.
In order to carry out undamped oscillations in a real oscillatory system, it is necessary to compensate for the energy loss in some way. Such compensation is possible if we use any periodically acting factor X(t), which changes according to a harmonic law: When considering mechanical vibrations, the role of X(t) is played by an external driving force (1) Taking into account (1) the law of motion for a spring pendulum (formula (9) of the previous section) will be written as Using the formula for the cyclic frequency of free undamped oscillations of the pressure pendulum and (10) of the previous section, we obtain equation (2) When considering an electric oscillatory circuit, the role of X(t) is played by the external one supplied to the circuit accordingly emf periodically changing according to the harmonic law. or alternating voltage (3) Then the differential equation of oscillations of charge Q in the simplest circuit, using (3), can be written as Knowing the formula for the cyclic frequency of free oscillations of the oscillatory circuit and the formula of the previous section (11), we arrive at the differential equation (4) Oscillations that arise under the influence of an external periodically changing force or an external periodically changing emf, are called respectively forced mechanical And forced electromagnetic oscillations. Equations (2) and (4) will be reduced to a linear inhomogeneous differential equation (5) and further we will apply its solution for forced vibrations depending on the specific case (x 0 if mechanical vibrations is equal to F 0 /m, in the case of electromagnetic vibrations - U m/L). The solution to equation (5) will be equal (as is known from the course on differential equations) to the sum of the general solution (5) of the homogeneous equation (1) and the particular solution of the inhomogeneous equation. We are looking for a particular solution in complex form. Let's replace the right-hand side of equation (5) with the complex variable x 0 e iωt: (6) We will look for a particular solution to this equation in the form Substituting the expression for s and its derivatives (and) into expression (6), we will find (7) Since this equality should be true for all times, then time t must be excluded from it. This means η=ω. Taking this into account, from formula (7) we find the value s 0 and multiply its numerator and denominator by (ω 0 2 - ω 2 - 2iδω) We represent this complex number in exponential form: where (8) 
(9) This means that the solution to equation (6) in complex form will have the form Its real part, which is the solution to equation (5), is equal to (10) where A and φ are determined by formulas (8) and (9), respectively. Consequently, a particular solution to the inhomogeneous equation (5) is equal to (11) The solution to equation (5) is the sum of the general solution to the homogeneous equation (12) and the particular solution to equation (11). Term (12) plays a significant role only in the initial stage of the process (when oscillations are established) until the amplitude of forced oscillations reaches the value determined by equality (8). Graphically forced oscillations are shown in Fig. 1. This means that in a steady state, forced oscillations occur with a frequency ω and are harmonic; the amplitude and phase of the oscillations, which are determined by equations (8) and (9), also depend on ω.
Fig.1
Let us write down expressions (10), (8) and (9) for electromagnetic oscillations, taking into account that ω 0 2 = 1/(LC) and δ = R/(2L) : 
(13) Differentiating Q=Q m cos(ωt–α) with respect to t, we obtain the current strength in the circuit during steady oscillations: (14) where (15) Equation (14) can be written as where φ = α – π/2 - phase shift between current and applied voltage (see (3)). In accordance with equation (13) (16) From (16) it follows that the current lags in phase with the voltage (φ>0) if ωL>1/(ωС), and leads the voltage (φ<0), если
ωL<1/(ωС).
Выражения
(15) и (16) можно также вывести с помощью
векторной диаграммы. Это будет осуществлено
далее для переменных токов.
Resonance(fr. resonance, from lat. resono“I respond”) is the phenomenon of a sharp increase in the amplitude of forced oscillations, which occurs when the frequency of natural oscillations coincides with the oscillation frequency of the driving force. An increase in amplitude is only a consequence of resonance, and the reason is the coincidence of the external (exciting) frequency with some other frequency determined from the parameters of the oscillatory system, such as the internal (natural) frequency, viscosity coefficient, etc. Usually the resonant frequency is not much different from own normal, but not in all cases we can talk about their coincidence.
20. Electromagnetic waves. Electromagnetic wave energy. Energy flux density. Umov-Poynting vector. Wave intensity.
ELECTROMAGNETIC WAVES, electromagnetic oscillations propagating in space at a finite speed, depending on the properties of the medium. An electromagnetic wave is a propagating electromagnetic field ( cm. ELECTROMAGNETIC FIELD).
Let a particle of mass m and charge e fly with speed v into the electric field of a flat capacitor. The length of the capacitor is x, the field strength is equal to E. Shifting upward in the electric field, the electron will fly through the capacitor along a curved path and fly out of it, deviating from the original direction by y. Under the influence of the field force, F=eE=ma, the particle moves accelerated vertically, therefore
The time of movement of a particle along the x axis at a constant speed. Then . And this is the equation of a parabola. That. a charged particle moves in an electric field along a parabola.
3. Particle in a magnetic field Let's consider the movement of a charged particle in a magnetic field of strength N. The field lines are depicted by dots and are directed perpendicular to the plane of the drawing (towards us).
A moving charged particle represents an electric current. Therefore, the magnetic field deflects the particle upward from its original direction of motion (the direction of motion of the electron is opposite to the direction of the current)
According to Ampere's formula, the force deflecting a particle at any part of the trajectory is equal to
Current, where t is the time during which charge e passes through section l. That's why
Considering that , we get
The force F is called the Lorentz force. The directions F, v and H are mutually perpendicular. The direction of F can be determined by the left hand rule.
Being perpendicular to the velocity, the Lorentz force changes only the direction of the particle's velocity, without changing the magnitude of this velocity. It follows that:
1. The work done by the Lorentz force is zero, i.e. a constant magnetic field does not do work on a charged particle moving in it (does not change the kinetic energy of the particle)
Let us recall that, unlike a magnetic field, an electric field changes the energy and speed of a moving particle.
2. The trajectory of a particle is a circle on which the particle is held by the Lorentz force, which plays the role of a centripetal force.
We determine the radius r of this circle by equating the Lorentz and centripetal forces:
That. The radius of the circle along which the particle moves is proportional to the speed of the particle and inversely proportional to the magnetic field strength.
The period of revolution of a particle T is equal to the ratio of the circumference S to the particle speed v:6
Taking into account the expression for r, we obtain Therefore, the period of revolution of a particle in a magnetic field does not depend on its speed.
If a magnetic field is created in the space where a charged particle is moving, directed at an angle to its speed, then the further movement of the particle will be the geometric sum of two simultaneous movements: rotation in a circle with a speed in a plane perpendicular to the lines of force, and movement along the field with a speed . Obviously, the resulting particle trajectory will be a helical line
4. Electromagnetic blood speed meters
The operating principle of an electromagnetic meter is based on the movement of electric charges in a magnetic field. There is a significant amount of electrical charges in the blood in the form of ions.
Let us assume that a certain number of singly charged ions move inside the artery at a speed of . If an artery is placed between the poles of a magnet, the ions will move in the magnetic field.
For directions and B shown in Fig. 1., the magnetic force acting on positively charged ions is directed upward, and the force acting on negatively charged ions is directed downward. Under the influence of these forces, the ions move to the opposite walls of the artery. This polarization of arterial ions creates a field E (Fig. 2), equivalent to the uniform field of a parallel-plate capacitor. Then the potential difference in the artery U (whose diameter d) is related to E by the formula
