Kelvin dripper

It's hard for people who haven't seen the mysterious Kelvin dripper works to understand that an original "symmetrical" device can develop into a very asymmetrical situation by itself. After all, there seems to be no difference between its left and right sides, but when you see the spark, there is a great potential difference between the water collected on the left and the water collected on the right. Looking back briefly, at least one thing has become clear. In order to make the potential difference between the two sides increase rapidly, the water falling into the left cup must be negatively charged relative to the water in the right cup. Once we accept (understand) this, as assumed, the water belt falling into the left cup is negatively charged and the water belt falling into the right cup is positively charged. We will soon understand that the negative charge of conductor (b, d) system is different from the positive charge of conductor (a, c) system. When the accumulated value reaches the critical discharge potential (about 3* 106V/m), electric sparks will appear. The remaining question is how and where water gets the charge. Why does the flow of water make the equipment negatively charged on one side and positively charged on the other?

The main activities are concentrated in the outlet where the water flows out. To illustrate what happened there, we assume that the sequence of events in the upper part of column A is as follows:

1. Due to some slight disturbance, cylinder A gets a tiny charge, which we assume is positive. If there is no water near the outlet, the upper electric field will look like (a) (note that they are mainly upward along the axis of the column mouth).

2. There are OH- ions and H+ ions in the water. If the water is pure, its PH value is 7, that is, every mole of water1107 will be decomposed into OH- ions and H+ ions. Due to the external field, H+ ions will rise and OH- will fall (as shown in Figure A). In this way, there is an induced charge just like an external field, and the distribution of the charge is shown in Figure (b). We only drew a small drop, but the separated charges in the water flow work in the same way.

3. Think of water as the sum of individual water drops. As can be seen from Figure (b), when water drops fall and break, some polarized charges (positive charges) in the upper part will remain at the outlet. In this way, the dispersed water droplets are negatively charged.

4. The next step is the automatic angular perturbation of the positive charge on the cylinder B (the falling negative charge reaches D through (b, d)), and the electric field line direction of the perturbation field at the B end is reversed (as shown in Figure A). The water droplets scattered and falling in cylinder B will be positively charged (unlike A).

The charge of falling water drops is opposite), and the charge left by the right exit is of course negative.

This is a self-running process under external control. The disturbance is getting bigger and bigger until the "dust collision" (here the air is broken down). This is an "active" example of successful use of physical instability (there are many perturbations of instability in the physical world: water mediation, successful fusion of metals by plasma instability in the laboratory, solar instability, sparks in the earth's core, etc. ).

Sometimes seemingly unstable growth will negate the law of conservation of energy, but this is definitely not the case here. Another form of energy is needed to provide instability through flow. In Kelvin water drops, its gravitational potential energy provides energy for the machine.

Two other points are used to perfect the story:

1. From giving A a tiny positive charge to ending the system (A, C) with a large number of positive charges, assuming that the initial perturbation is caused by negative charge, the above process will not change much except for the exchange of (A, C) and (B, D). Because of the randomness of the initial disturbance, we cannot predict which container is negatively charged. But once it starts to be negatively charged, it will always exist as long as there is water. Because sparks don't change the state of the system. However, when we started again the next day, A was positively charged.

2. What is the velocity of charge flowing out from under two glass tubes? Because their signs are opposite, the water in the conduit is not charged. But the water in the conduit has electricity. If the conduit is well insulated from the liquid, the machine will not work, and the residual charge at the outlet of the conduit will not move (no current passes through the liquid or conduit from one port of the conduit to another). Discrete charges in a liquid will quickly concentrate in one place (why). Therefore, both the catheter and the liquid must be conductive.

Question 4.2

Force on the current loop

Textbook 27- 12

As shown in the figure, the Z axis is established, and the positive direction of the Z axis points above the magnetic field. The magnetic force on a wire can be calculated by Giancoli Equation 27-4(69 1 page). Because the magnetic field is perpendicular to the wire, there are: the direction is as shown in the figure.

Because this is a problem in the horizontal plane, we can regard the wire loop as a whole and only get the force in the Z direction (any force in other directions will be offset by the component in the opposite direction). Therefore, to calculate the force acting on the loop, it is only necessary to integrate the components of each part in the z direction. As can be seen from the figure, the component of the wire in the Z direction points to the negative Z direction. Therefore, the resultant force on the wire is:

(The last step is to eliminate θ with geometric knowledge.)

Question 4.3

Lorentz force on electrons

Textbook 27-20

The force on charged particles in a magnetic field can be solved by Lorenz's law (Equation 27-5a on page 692 of Giancoli textbook). To calculate the cross product of components in Cartesian coordinates, there are:

Question 4.4

Spiral motion of electrons

Teaching materials 27-29

Like Giancoli's example 27-5 (page 694), we decompose the velocity into a component perpendicular to the magnetic field and a component parallel to the magnetic field. The component of velocity magnetic field in 45 direction is

⊥v makes the particle move in a circle (example 27-4 in the textbook (page 693)), so that we can find the radius of the moving circle:

Giancoli Equation 27-6 (page 694) gives the time required for a motion cycle:

The velocity of an electron in the direction parallel to the magnetic field is constant at v/, so its propagation path in time t is

Question 4.5

Torque of motor rotor coil

Teaching materials 27-33

Use Gancoli equation 27-12 (page 696) to find the potential energy of magnetic field (expressed by the physical quantity of magnetic field);

(If the current in a rectangular loop (area a*b) surrounded by n coils is I, then the magnetic moment μ=NIab).

(1)

(2)

Question 4.6

Mass spectrometer/spectrograph/spectroscope

Teaching materials 27-49

An ion with Q charge starts at rest, is accelerated by voltage V to obtain kinetic energy K=qV, and then enters magnetic field B. At this time, it is subjected to a force perpendicular to its velocity and makes a circular motion, but its velocity and kinetic energy remain unchanged. Newton's second law points out the relationship between the force on a particle in a magnetic field and its centripetal acceleration, so it can be obtained as follows: qVB=mv2/R, and when combined with K= 1/2mv2, it can be obtained as follows: K=q2B2R2/2m, from which it can be obtained as follows: m=qB2R2/2V.

Question 4.7

Accelerated motion of deuterium in cyclotron

Textbook 44- 10 (page 1 137)

Before solving this problem, I suggest you take a look at the contents of the electron cyclotron on Giancoli 1 16. )

(a) In question 4.6, we get the expression of kinetic energy of charged particles in cyclotron:

An ion has only one proton with a +e charge, and its mass can be found in Appendix D of Giancoli textbook (calculated by atomic mass):

Strictly speaking, we should subtract the mass of an electron from the mass of deuteron, but this has no effect on our work.

We hope that the kinetic energy of the particle at the exit radius 1.0 is10mev =1.6 *10/2j, so that we can solve the magnetic field strength:

(If K= 1/2mv2 is used to find the corresponding speed, cv 1.0≈ will be obtained, where c is the speed of light. So if particles can

The amount is very large, so the relationship between energy and speed in relativity should be applied in the course)

(b) The frequency of voltage change is the frequency of electron cyclotron (see Equation 44-2 on page 1 165438).

(c) Deuterium has an electron charge, so when it gets a voltage through 22KV, it gets an energy of 22KeV. It passes through the voltage gap twice in the process of one rotation, and then we can calculate its soft rotation number n from the given particle final energy 10MeV:

(d) The time required for each perturbation is T= 1/f(f is the frequency of the electron in (b)). Therefore, the total time required for deuterium to enter and run out is Δ t = nt = n/f = 4.6 *10-5s.

(e) The energy obtained after each complete deuterium revolution is Δ k = 2 * 22kev = 7.0 *10-15j. The time t after entering the cyclotron will become t/T=tf. Because the particle must make many turns in the cyclotron before leaving, we approximate its energy as a smooth function about time, not a function of gap jump. In this way, we can get the following energy relationship (where the velocity of deuterium is a function of time):

In order to find out the total distance of deuterium in cyclotron, we can integrate the velocity in the total time range of orbit:

Eliminate t and f(K/K=N) and get the following result δ δ.

Question 4.8

Force acting on conductor circuit

The textbook 28- 14 uses ampere's law (sections 28-4 and 28- 1).

(For convenience, we use A, B, C and I in the figure below to replace the physical quantities in the question. )

The force acting on the wire loop through which the current flows is determined by the magnetic field of the infinitely long straight wire above it. From pages 28- 1 0 of Giancoli7 10 and pages 28-4 of 7 12, we can know the magnitude of the magnetic field:

Where r is the vertical distance from a point to a straight line. Using the right-hand rule (see the chart on page 27-9 (b) of Giancoli689), we can know that the direction of the magnetic field inside the rectangular frame points to the inside of the paper (as shown in the figure), so that the force on the rectangular frame can be calculated:

(See Equation 27-4 on page Giancoli69 1)

For a small piece of wire on the upper edge of the wire, B is perpendicular to the paper and Idl is to the right, then the size of dF is IdlB. According to the right-hand rule, it points upward. Since the value of the upper edge part B of the conductor remains unchanged, the force on the upper edge is:

(The direction points to an infinite straight line)

For the bottom edge, Idl points to the inside of the paper, so the size of dF is IdlB, and the direction is downward. The value of b is, so the force on the lower line is:

(the direction points away from the infinite straight line)

Consider the left and right sideline together. Consider two independent Idl', one on the left line and the other on the right line, with the same length and the same distance to an infinite straight conductor. One advantage of the current loop is that the current direction of the left conductor is upward, while the current direction of the right conductor is downward (as viewed from the straight conductor), so the force on the left conductor (to the left) is cancelled by the force on the right conductor (to the right), and the resultant force dF' on the loop is zero. The total resultant force of the coils is

=2.6* 10-6N

Pointing straight wire

Question 4.9

Application of Biossa's Law (Giancoli28-30)

As shown in the figure, first consider a short segment of dl on an arc, and Biossa's law (Equation 28-5 on page Giancoli7 19) gives the magnetic field of this short segment at point C:

For two arcs (inner arc and outer arc), r? Are a short unit vector pointing to point C. For the inner arc r=R 1, Idl is perpendicular to R? Point to the left along the arc. So the contribution of this short segment on the arc to the magnetic field at point C is:

Pointing outward perpendicular to the paper (the direction is determined by the right-hand rule). The expression of is not expressed by the corresponding angular position of this small segment. In order to obtain the total contribution of the internal arc to the magnetic field at point C, we need to simply integrate dl (the total arc length is R 1θ) to obtain:

The direction is perpendicular to the paper and points outward.

The analysis of outer arc is similar to that of inner arc. R 1R2, Idl points to the right along the arc. Slightly change the above results: →

The direction is perpendicular to the paper and points to the inside.

Finally, consider a short section along this arbitrary straight line, and Idl is parallel to the displacement vector pointing to C, so. If this arc becomes a straight line, it does not contribute 0 to the magnetic field at point C? =×rIdl

Total magnetic field intensity at point C (remember1/r1>; 1/R2): Pointing outwards perpendicular to the paper.