6 On both sides, when the conductive liquid flows stably through the flowmeter, connect the upper and lower sides of the flowmeter to the two ends of the ammeter with series resistance R outside the tube, where I represents the measured current value, the known resistivity of the fluid, and what is the flow rate regardless of the internal resistance of the ammeter? When the analytical conductive fluid flows through the tube, the anions and cations in it will be deflected to the upper and lower surfaces of the tube by the magnetic field force. After the upper and lower surfaces are charged, on the one hand, anions and cations will be hindered by the electric field force and continue to deflect until the electric field force and magnetic field force reach a balance. On the other hand, for an external resistor, the upper and lower surfaces are equivalent to the power supply, so that the current in the resistor satisfies the ohm's law of closed circuit. Let the velocity v of the conductive fluid accumulate charges on the upper plate and the lower plate due to the deflection of positive and negative ions in the magnetic field, and form an electric field between the two electrodes. When the electric field force qE and Lorentz force qvB are balanced, E=Bv, electromotive force E'=Bcv, and internal resistance r= C/ab, the current in the circuit connected in series with R: I=Bcv/(R+r), v=I(R+? c/ab)/Bc； Fluid flow: Q=vbc=I(bR+? C/a)/B Problem-solving Review Because the electromagnetic flowmeter is a pipe, there is no structure in it that hinders the flow of fluid, so it can be used to measure the flow of highly viscous and corrosive fluid. This topic is a motion model which comprehensively applies Ohm's law of closed circuit and the motion knowledge of charged particles in electromagnetic field, also known as Hall effect, and is applied to many instruments and equipment, such as speed selector, magnetic fluid generator and so on. Example 6 is shown in the figure. The magnetic induction intensity of the uniform magnetic field is B, and the direction is perpendicular to the xOy plane. At a certain moment, a proton enters the magnetic field from the point (l0,0) along the negative direction of the Y axis. At the same time, an alpha particle enters the magnetic field from the point (-l0,0), and the velocity direction is in the xOy plane. Let the proton mass be m and the electric quantity be e, without considering the interaction between proton and α particle. (1) If a proton can pass through the coordinate origin O, what is its speed? (2) If the alpha particle can touch the proton when it reaches the origin for the first time, find the velocity of the alpha particle. The key to solve the problem of circular motion analysis of charged particles in magnetic field is its center and radius. If we can find these two quantities in the topic first, the problem-solving process will become concise, and the remaining work is to solve the problem by using the radius formula and the periodic formula. (1) If the proton can pass through the origin, the trajectory radius of the proton motion is R=L0/2. Another concept of this problem is that the center of the circle must be on a straight line perpendicular to the velocity, so the center of the proton trajectory must be on the X axis; (2) The last question is about the radius of circular motion, but the focus of this question is the period of circular motion. When two particles meet at the origin, their movement time must be the same, that is, tα=TH/2, and the trajectory of α particles moving to the origin is an arc, so let the corresponding central angle be? , there is tα=2? M/2Be, available? =? /2, the trajectory radius of α particle R=L0/2=4mv/B2e, the answer is v= eBL0/(4m), and the angle with the positive direction of X axis is? /4, right up; In fact, it is also possible that the alpha particle moves 3T/4 to reach the origin, and then the proton reaches the origin for the second time. In this case, the answer to velocity is the same, but the angle between the initial velocity direction of α particles and the positive direction of X axis is 3? /4, upper left; Looking back at similar problems, the focus is not on magnetic field force, but on the combination of mathematical knowledge and physical concepts. The key here is to use the vertical relationship between the linear velocity of circular motion and the radius of trajectory, the numerical relationship between chord length, arc length and circular radius, and the geometric relationship between central angle and arc to determine the center position and radius, period and motion time of arc. Of course, r=mv/Bq, T=2? The two formulas of m/Bq play a connecting role here. As shown in the figure, in Example 7, there are two metal balls A and B with the same diameter on the smooth insulating horizontal desktop, with the mass of Ma = 2m and MB = m respectively. Ball B has a positive charge of 2q and stands still in a uniform magnetic field with a magnetic induction intensity of B; The uncharged ball A enters the magnetic field at the speed v0 and collides with the ball B. If the pressure of the ball B on the desktop is exactly zero after the collision, what is the pressure of the ball A on the desktop? The physical knowledge related to solving this problem includes contact electrification, momentum conservation, Lorentz force, force balance and force analysis, and the most critical thing is the collision process, which is the turning point of all States and processes, and the selection and determination of physical quantities is also the starting point; After the collision, the electric quantity of ball B is Q, the electric quantity of ball A is Q, let the speed of ball B be vb and the speed of ball A be VA; Taking B as the research object, Bqvb = mbg can be VB = mg/BQ; Taking the collision process as the research object, there is momentum conservation, that is, mav0=mava+mbvb, and the known quantity can be substituted into VA = VO-MG/(2bq); Va is already included in this expression. If we analyze the force after a collision, we will find that N+Bqva=2mg, then we will get n = (5/2) mg-bqv0; Looking back at solving the problem, the focus of this problem is the combination of Lorentz force and momentum. In fact, you can also ask the size of the internal energy generated during the collision, and the energy problem is combined. Example 8. As shown in the figure, on the xOy plane, the coordinate of point A is (0, L), a boundary in the plane passes through the circular uniform magnetic field area of point A and the coordinate origin O, and the magnetic field direction is perpendicular to the paper. An electron (mass m, electric quantity e) starts from point A, and its initial velocity v0 is parallel to the positive direction of X axis, while it moves in the magnetic field, just from point B (not shown in the figure) Find the magnetic induction intensity of (1) magnetic field; (2) the coordinate o1of the magnetic field center; (3) the movement time of electrons in the magnetic field. Analytical electrons move in a uniform circular motion in a uniform magnetic field, shoot from point A to point B, and exit the magnetic field region. Therefore, there are points A, O and B in the circular magnetic field region, and the trajectory of electrons is shown by dotted lines in the figure, and its corresponding center is at point O2, so that aObOR22 becomes an angle. AOb260, as shown in the figure: RRLRRmvBe22 2 0 60? Substitute sin into the above formula to get RLBmveL 220, that is, the time for electrons to fly in the magnetic field; tTmBeLvL v 60360 162322300？ Because the circumferential angle ⊙O 1? AOb90, then the straight line segment of ab is the diameter of the circular magnetic field region, then y a v0 Ox 8 aORL 1 12, then the coordinate O 1, xaOL of the center of the magnetic field region? 16032 sinylal1602cos, that is, O 1 coordinate 32 12LL. The key to solve this problem is that the incident direction and the exit direction form a certain angle (600 in the problem). From the geometric relationship, it is determined that the arc of charged particles is 1/6 circle, and then the circumference of 1/6 circle is determined through the geometric relationship. As shown in the figure, Example 9 has a uniform electric field ENC20 104. /a uniform magnetic field b parallel to the y-axis in the I-quadrant region and perpendicular to the oxygen plane in the IV-quadrant region. Charged particle a, whose mass is mkg1121010? . , electric quantity qC 14 10 10? . From point A on the Y axis, the electric field is injected at a speed of vms 15 4 10/ parallel to the X axis, OAm? 4 102, find: (1) the position, speed and direction of particle a reaching the x axis; (2) At the same time that particle A injects electric field, particle B with the same mass and charge as A injects a uniform magnetic field from a point B on the Y axis at a speed v2 parallel to the X axis, and two particles A and B collide head-on on on the X axis (regardless of the gravity and coulomb force between the two particles). Try to determine the position of point B and the magnetic induction intensity of uniform magnetic field. The analytic particle A is positively charged, and after entering the electric field, the acceleration in the opposite direction along the Y axis is obtained under the action of electric field force, AEQMMSMS1010201020104412/KLOC. 220042102010127. (). So we can know the position of point P: OPvtmm? 401020108010572 ... (). The speed at which particle A reaches point P, vvatmst? 122 5 240 10./ ? The angle between vt and x axis:? 45 9 (2) According to the conclusion of (1), the time for B to make uniform circular motion in uniform magnetic field is also ts? 20 107., the trajectory radius ropm282102 obm12102, and the rotation angle of particle b is 34? , the exercise time is 3 8T 382? mBqt？ ? Bm qtT340 18？ Example 10 As shown in Figure 4, a small ring with a mass of 1g and a positive charge of 4× 10-4C is sleeved on a long straight insulating rod, and the dynamic friction coefficient between them is μ = 0.2. The rod is placed in a horizontal and mutually perpendicular uniform electric field and a uniform magnetic field. The plane of the rod is perpendicular to the magnetic field, and the angle between the rod and the electric field is 37. If E = 10N/C and B = 0.5t, the small ring starts from rest. Find: (1) When the acceleration of the small ring is maximum, the speed and acceleration of the ring; (2) When the speed of the small ring is maximum, the speed and acceleration of the ring. Analysis (1) After the small ring starts from rest, the force on the ring is shown in Figure 5. With the increase of speed, the Lorentz force perpendicular to the rod direction increases, so the positive pressure between the upper side of the ring and the rod decreases, the friction decreases and the acceleration increases. When the ring speed is v, the positive pressure is zero and the friction disappears. At this time, the ring has the maximum acceleration am. The direction parallel to the rod is: mgsin 37-qecos 37 = mam solution: am = 2.8m/S2；; In the direction perpendicular to the rod, there are: bqv = mgcos 37+qesin 37 solutions: v = 52m/s (2). After the above state, the Lorentz force is very small due to the increasing speed of the ring. Then the friction force F is generated again, and the acceleration A of the rod decreases. V↑BqV↑N↑f ↑a↓ The result of the above process is that A is reduced to zero, and the maximum speed Vm of the ring at this time. Parallel bars direction: mgsin 37 = eq cos 37+f=μN vertical bar direction: bqvm = mgcos 37+qesin 37+n and f=μN solution: VM = 122m/s at this time: a = 0 cases: 1 1 as shown in figure 7. A charged body A is negatively charged, and its electric quantity is q 1, which can just rest at point C in this space. The other charged body B is negatively charged, and its electric quantity is q2. It makes uniform circular motion with radius r in the vertical plane passing through point A. As a result, A and B collide and stick together at point C. Try to analyze the motion properties of A and B after sticking together. Analysis: Let the mass of A and B be m 1 and m2 respectively, the velocity of B is v, A is at rest, and at 10, there is a uniform circular motion of Q 1e = M 1gB in the vertical plane, which implies Eq2=m2g. At this time, the collision bonding process of A and B is M2 V+0 = (M 1+M2) V' A and B together, and the total electricity is Q65433. Therefore, they do uniform circular motion in the vertical plane at the speed v', so there is a solution: