High School Physics Electromagnetic Induction, Electromagnetic Induction High School Physics?

The magnetic flux has not changed, and there is no induced current.

AD and BC do the cutting magnetic inductance line movement, the electromotive force is the same, UA=ub, UC=UD, and the electromotive force is judged to be high at the A end according to the right hand. The C-terminal is low potential, so UA>UC

So choose a CD

Reverse thinking.

Combined with Lenz's law, the clockwise induced current generated in b indicates that the magnetic field in b decreases from the vertical paper face inward, or increases from the vertical paper face outward.

1) If the magnetic field in B decreases inwards on the perpendicular paper side, it means that the current in A is clockwise and decreases, so that C can be judged to be decelerating to the left.

1) If the magnetic field in B increases outwards on the perpendicular paper side, it means that the current in A is counterclockwise and increases, so that C can be judged to accelerate to the right.

To sum up, c decelerates to the left or accelerates to the right.

Let's work backwards: if you want b to generate a clockwise induced current, then the magnetic field generated by the induced current according to Lenz's law is perpendicular to the inward direction of your computer screen (hehe, how can you generate a magnetic field inward from the vertical screen?). There are two scenarios:

1. The magnetic field generated by the current in the coil is perpendicular to the screen and gradually increases. In order to make the magnetic field generated by A outward, the current in A must be counterclockwise, then C must move to the right (right-hand rule), and in order to make the outward magnetic field generated by A gradually increase, the current must gradually increase, that is, the induced electromotive force generated by C gradually increases, and the rate of change of the magnetic flux must gradually increase (the law of electromagnetic induction), that is, C must accelerate the motion. Therefore, c must accelerate to the right;

2. (In fact, as long as you do more questions, you can guess that the other situation is: deceleration to the left, let's push it), the other situation is that the magnetic field generated by the current in the coil A is perpendicular to the screen inward, and gradually decreases. In order to make the magnetic field generated by the current in A inward, the current in A must be in a clockwise direction, then C must move to the left (right-hand rule), and in order to make the inward magnetic field generated by A gradually decrease, the current must gradually decrease, that is, C produces a depth-induced electromotive force that gradually decreases, and to make the induced electromotive force generated by C gradually decrease, the rate of change of the magnetic flux must gradually decrease (the law of electromagnetic induction), so C must slow down.

Therefore, c must slow down the movement to the left.

Draw a diagram to give a detailed physical context.

The top edge, the left and right sides are not cut, and the bottom edge is cut. The left and right sides are subjected to ampere forces, but the equal and large reverses cancel each other out. The lower edge is subjected to an upward ampere force that does negative work on the coil, causing the loss of mechanical energy to be converted into electrical energy.

The upper edge does not enter the magnetic field and is not subject to ampere force.

Any process of doing work will inevitably lead to energy transformation, and the positive and negative aspects of work determine the direction of energy transformation.

In an electric motor, ampere force does positive work, and electrical energy is converted into mechanical energy.

In a generator, ampere force does negative work, and mechanical energy is converted into electrical energy.

Solution: The coil of the circuit in term D A is equivalent to a bar magnet, and the upper end is the n pole. When the subbar magnet moves to the left, the magnetic flux through the B coil decreases, and according to Lenz's law, the induced current generated by the reduction should be from top to bottom (the lower end of the magnetic field generated by the induced current according to the right-hand rule is the n pole).

When choosing a at the beginning, the CD rod is stationary, due to the uniform movement of the AB rod to the left, cutting the magnetic inductance line, and the induced current is generated on the circuit I, at that moment the CD rod is subjected to the ampere force to the left, and there is a tendency to move to the left, due to the tendency to move, it must be subjected to the static friction force like the right, at this time there are two situations:

In the first case, if the ampere force is greater than the static friction force, the CD rod will make a variable acceleration motion with decreasing acceleration until the velocity is v"=v-(mg r) (bbll); bbll is b squared multiplied by l squared, forgive me for not being familiar with the computer].

Second case: If the ampere force is less than the static friction force, the CD rod remains stationary.

I will give you a suggestion: this kind of topic must understand the principle, the physics exam questions or comprehensive questions involve a lot, if you don't understand my explanation or ask other masters, this topic is very important, I hope you have been improving!

AB moves to the left to cut the magnetic inductance line to produce an induced current from C to the generated current, and the magnetic field gives CD a force to the left, and the magnitude of this force depends on the speed v of AB. The process is as follows:

Induced electromotive force: E=BLV, L is the length of AB and CD.

Let the resistance of cd be r and the mass be m.

Then the current in cd: i=blv r.

Lorentz force on cd: f=bil=b*b*l*l*v r.

Total force on cd: f=f- mg.

When V is larger, the force on CD is greater, and when the force is greater than the maximum static friction force on CD, CD moves.

So the answer is probably moving to the left.

When the AB rod moves to the left, the magnetic flux increases, an induced current is generated in the loop, and the CD rod receives ampere force, and the F=BiL direction is to the left. When f is greater than the frictional force, the cd moves to the left, and when it is less than that, it does not move.

It is not rigorous to rely on a change in magnetic flux in general to determine whether an electric current is generated.

In fact, the method of generating induced electromotive force is divided into induction and kinetics, the essence of induction is that the changing magnetic field excites the vortex electric field to orient the electrons, and the kinetics are because the free electrons in the conductor are moved by the Lorentz force.

When a square closed coil cuts the magnetic field and moves forward, if the whole coil is in the magnetic field, and the magnetic flux does not change, no current is generated at this time, but in fact, the conductors in the front and back directions are induced electromotive force, but their magnitude is equal and the direction is opposite, so they cancel each other out (it can be understood that part of the electrons accumulate to one side and form an electric field so that the remaining electrons no longer accumulate, then a continuous current cannot be generated).

The apparent phenomenon is that if the magnetic flux does not change, no current is generated. If the coil accelerates, the electrons will accumulate in one direction, and there will be an electric current in the conductors on the left and right sides.

In this diagram, the disc cuts the magnetic inductance line, generating an electromotive force, so there is an electric current.

It's not in the book, my teacher said it.

You think of a copper disc as 1 with a straight conductor, the conductor enters the magnetic field with magnetic flux, and the magnetic field comes out, without magnetic flux, and each conductor reciprocates like this, generating a continuous current.

Because although the total magnetic flux in the copper plate does not change, due to the rotation of the copper plate, the magnetic flux of each part of the copper plate is changing.

The difference between kinetic electromotive force and induced electromotive force. It's in the textbook.

Glad to answer for you :

The answer to this question is judged to be on **, I can dig Li Yi also talk about the idea: you can now see the left part as the power supply and then analyze the electric circuit, and finally disconnect S, C can only form a loop with R2, so all the Q on C flows through R and finally positive and negative neutralization.