Physics, Problems with Magnetic Fields, Physics Solving Problems with Electromagnetic Fields.

Left electric right power generation.

The magnetic field of the induced current hinders the change of the original magnetic flux.

Lenz's law - d current is the maximum, and in d, the magnetic field of the induced current is the same as the original magnetic field, which is strengthened; c is not closed, and the increase is small;

b current is second, and in b, the magnetic field of the induced current is the same as the original magnetic field, which is strengthened; A is not closed, and the increase is less.

d>b>c>a

I don't understand it too much, but I just understand that the vertical conductor rod moves at v0 and 2v0, what is the problem?

Teach you a good way, in electromagnetism, the force related to the left hand, the rest of the right hand.

It can be proved that in a uniform magnetic field, the magnetic field force of the energized wire placed perpendicular to the direction of the magnetic field is equivalent to the magnetic field force of the energized wire connected at both ends of the wire. As shown in the figure below, the magnetic field force experienced by DBCE is equal to the magnetic field force experienced by the conductor DE.

Therefore, at this time, the magnetic field force experienced by the entire wire is: f=fde=bil 2Direction downward.

When the magnetic field is above the dashed line, the magnetic field force experienced by the wire is equivalent to the magnetic field force experienced by DE, except that the current direction is opposite, as shown in the figure below.

So the magnetic field force experienced by the wireframe at this time is: f'=fde=bil 2, direction up.

You should be able to understand the other equilibrium equations.

I think A, B, and D should be chosen

If it is known that it is a circular motion with a uniform velocity, then the resultant force must be perpendicular to the direction of v and the magnitude is exactly equal to the centripetal force. Therefore, the gravitational force and the electric field force are balanced, and the direction of the electric field force is upward. Radius of motion r=mv qb

Again. mg=qe

v=sqr(2gh). b right.

Since the electrostatic field is a conservative position, w = -δep

The electric field force does negative work.

The electric field energy can be increased because the gravitational force and the electric field force are equal in magnitude and opposite directions, and the work done is only zero (or from the energy point of view).

e=ep+ek=c

Constant. Unchanged.

And the ek has not changed.

Therefore, EP remains unchanged).

The first 3 questions are easy to do, and the 4th question is what this process refers to, probably using calculus, or using energy to calculate.

Answer: A. When AB moves at a constant speed to the right, the magnetic field lines are cut uniformly to produce a stable and constant induced electromotive force, and the induced current in the right circuit is clockwise (from A to B).

Since it is a stable induced current, this induced current does not produce an induced current on the left solenoid l, because the magnetic flux of the magnetic field induced in the left solenoid l is constant.

If there is no induced current in the CD, there is no magnetic field force, so the CD does not move.

b. When AB accelerates to the right, the induced current in the right coil increases with time.

l induced current, from Lenz's law, it can be seen that the induced current flows from C to D.

The cd is subjected to ampere force to the left and moves to the right.

When C. Cd accelerates to the left, the induced current on the right flows from B to A. The induced current on the left flows from D to C, and Cd moves to the left.

D. When Cd decelerates to the left, the induced current on the left flows from C to D, and Cd moves to the right.

So, B and D are both correct.

If you don't understand something, please hi me. ₁

Needless to say, if a is incorrect, there is no magnetic flux change is impossible to have induced current, analysis b, first of all, you have to understand that the energized rod is the left-handed rule, the energized spiral is the right-handed rule, these understand, the right magnetic field is from the inside out, the rod moves to the right, the left-hand rule, judge the current pointing to a, right, the spiral right-hand rule, thumb down, prove that the magnetic flux is downward, because it is an accelerated motion, the magnetic flux is increasing, then L2 must be up, L2 is right-handed rule, thumb up, The current is C pointing to D, the magnetic field on the left is from the outside to the inside, the left hand rule, the four fingers point to D, then the thumb must point to the right, B is correct, on the contrary, C is definitely not right, then D, the same way, it's just that the magnetic flux of L1 is reduced, then the direction of L2 must be the same as L1.

In fact, it is the weakening principle of the induced magnetic field, which is very troublesome to explain, you think that the magnetic flux of L1 is increasing, then L2 will not let you increase, then the n level they form is on the same side, repulsion, you want to decrease, it does not let you decrease, then their magnetic poles, n, s are on the same side, attract.

AB When moving at a constant speed, according to the right-hand rule, the induced current in AB is constant, the magnetic flux in L1 is unchanged, the magnetic flux passing through L2 does not change, no induced current is generated in L2, Cd remains stationary, and A is incorrect; When AB accelerates to the right, according to the right-hand rule, the direction of the current flowing through AB is A pointing to B, the magnetic flux in L1 increases upward, the magnetic flux in L2 decreases (the obstacle increases), increases, the direction of the current through CD goes downward, CD moves to the right, and B is correct; In the same way, C is incorrect and D is correct. Choose b and d

Physics questions on the motion of charged particles in the electromagnetic field are hot topics in the college entrance examination, and almost all of them have such questions in the college entrance examination papers every year, and most of them are large-scale calculation questions and comprehensive questions.

Why is the physics question on the motion of charged particles in an electromagnetic field a hot topic in the college entrance examination? Because in order to test so many knowledge points of high school physics in a limited period of time with limited topics, it is necessary to test several knowledge points through a topic, that is to say, the topic should be comprehensive, and the physics questions with the motion of charged particles in the electromagnetic field as the theme can examine the electric field, electric field force, acceleration, velocity, displacement, uniform motion, uniform acceleration motion and magnetic field, Lorentz force, circular motion, centripetal force, centripetal acceleration, linear velocity, angular velocity, geometric drawing, mathematical calculus and many other knowledge points, so, Physics problems on the motion of charged particles in an electromagnetic field are easy to make comprehensive problems.

What you have to do is clear your mind. In fact, the steps of solving each big problem are similar, and you can form a fixed template. Which step to do first, which step to do later.

You have to know this. Moreover, there are only so many knowledge points about electromagnetic fields. I'm sure you know more or less about the classic question types in this area.

This is the physics information you found in the third year of high school, and there are questions like this!

Regardless of gravity, the particle motion is circular, and its radius is r=mv qb=so you can make the trajectory of the particle from the graph. I won't do a tula here, and then use the knowledge of analytic geometry to easily determine that the trajectory swept in the magnetic field corresponds to the radian angle of pi 3, so the running time is t=pi 3*m qb=.

I hope you can understand it, but if you can't understand it, it's because you're learning geometry too poorly. There's no way I'm going to help you face-to-face.

If an iron rod is magnetic, then the two ends are the most magnetic, called the "poles", and there is almost no magnetism in the middle. Therefore, if A is close to the middle of B, the attraction can only be caused by the magnetism of A, indicating that A is magnetic; Conversely, when B approaches the middle of A, there is no attraction, indicating that B is not magnetic. So A is right.

a If an iron rod is magnetic, it has the strongest magnetism at both ends (poles) and the weakest magnetism in the middle.

This is the use of the law of the second time, because you didn't shoot the arrow in the positive direction of the coil, so you can only say that the answer is one of b and d, and the question should have a diagram of the n pole and the s pole at both ends of the magnetic rod, see which one is closer to the coil, if the n pole is close, as n is extremely close, the magnetic inductance line through the coil becomes more, to resist this change, that is, to produce a counterclockwise (from the magnetic rod insertion) current, through the coil after the magnetic inductance line becomes less, to resist this change, so the current direction becomes clockwise.

It should be one of the AC, there is no diagram, I can't tell which one to choose. In the process of approaching-passing-moving away, the direction of the induced current is constant according to Lenz's law.

The answer must be one of AC, first of all, you have to know that the effect of Lenz's law is essentially achieved through ampere force, according to the same principle of magnetic flux increase and inverse decrease [a generalization of Lenz's law], when the magnetic rod is close to and far away, the ampere force on the coil is in one direction, so the current is also in the same direction...

Then the answer to the four-quadrant in AC is determined according to taking ** as the positive direction...