A very tangled electromagnetic induction problem, ask for help, physics master

Ad right. The original question was analyzed incorrectly.

Correct analysis: at the moment of S closing, L generates self-induced electromotive force, which is added at both ends of P, and P is bright; As the self-induced electromotive force decreases, p gradually dims until it goes out.

At the moment when S closes, the voltage at both ends of C is 0, and Q is not bright; As C is charged, the voltage at both ends of C increases, and Q gradually brightens.

At the moment of S closure, the current in P flows back to the negative pole of the power supply through capacitor C, without passing through Q (equivalent to Q being short-circuited by C).

A is true and B is false. When the circuit is disconnected after stabilization, the analysis of the original question is correct. d right.

This, I kind of forgot about the capacitors and coils.

However, I remember that capacitor charging is done instantaneously, it is just that there are electrons running over at both ends, and the time can be ignored, so your eyes can't feel the moment Q is short-circuited. Next, so the capacitor is charged instantly, at the speed of light! Then no electrons pass through.

So the Q light is not gradually on!

I mean, when it's closed, the rest, you seem to be analyzing?

No, let's ask it again, this question is only D right.

Yours is wrong!

Hurry up and ask the teacher!

I'm right about what A is wrong!

If you don't believe it, you'll have trouble doing the questions in the future.

What does it mean to gradually become brighter!

There is no such thing, C Shunma is charged! Velocity of light! You can't see such a short interval of time!

It just has an electromotive force at both ends, and it forms instantly!

Immediately it was broken.

Don't haha, I'm anxious for you.

The answer is correct, you heard it wrong in class, the capacitor is equivalent to an open circuit, the inductance is equivalent to a short circuit, and then combined with the answer, there is no problem. Hope.

Closing the transient process, L starts to be regarded as an open circuit, C is regarded as a short circuit, the answer is wrong, I have learned a bunch of circuits...

It should be wrong. I agree with you. But I'm still in eighth grade?!!

Let the outer diameter of the solenoid r, the number of turns and the width of the bridge rubber disturbance be n, the length is l, and the energizing current is i

1) The single-turn magnetic induction intensity b·dl=(u0*i) is integrated to obtain b=u0*i (2pi*r).

r r , 2) the magnetic induction intensity of a single turn b·dl = (u0*i*r ) r, and the integration obtains, b=u0*i*r (2pi*r )

2.The intensity of the n-turn magnetic induction is b·dx=(y l)*n*u0*i, and the integral obtains b=(n*u0*i) l

where y selects the length of the solenoid with partial turns (y l), and dx is the length of the microelement along the direction of the current.

There are two electromotive forces here, one is the kinetic induced electromotive force, e1 = blv, and the other is the induced induced electromotive force e2 = δ t, the e1 direction is upward, and the e2 direction is counterclockwise, the two directions are the same, so the electromotive force is equal to e1 + e2, so ab is chosen.

Strictly following Faraday's law of electromagnetic induction, you write out the form of a function of magnetic flux, and when you look at the derivative of time, you will understand.

。。。I'm poor at physics.,,I can't do it.。。

i=blv r, which is true, but when the velocity is v 2, the effective cutting length becomes smaller, so it is not half of the original current (the resistance is constant r).

What you have to know is that in this process of change, the velocity does not change uniformly, and by calculating you will know that the acceleration is decreasing, so it is doing the deceleration motion of decreasing acceleration, so you will know that when v=1 2v0, the conductor rod is not in the middle position, but should be on the left. So that you bring in the formula and you understand

According to the second law of bovine, bil-b l v r=ma is obtained because a is a constant quantity, so the derivatives of both sides are obtained.

So a=dv, dt=10m s

The left side of the equation under the initial condition is 1, so m=

Thank you!

1. The question should be "probably yes", then the answer should be ACD

When entering the magnetic field, it just does a uniform motion, f amp = bil = b (blv) l = b 2 * l 2 * v, again, f a = g

Then when the magnetic field is out, because b is greater than a, there is a period of time in the middle that the closed conductor frame is not subject to ampere force, but only by gravity, and accelerates.

When it comes out, F A = Bil = B (BLV) L = B 2 * L 2 * V, it must be greater than G, if it slows down to still cutting, but F A = G, then it will be at a constant speed.

2. The answer should be b

If you follow your statement, there should be problems with your diagram, one is that the rectangular wireframe should be attached to the magnetic field boundary.

The second is that d is greater than l

Then, we only need to look at the right side, when the right side just comes out of the right side of the magnetic field, there is no current induced, until the left side enters the magnetic field, at this time, the distance that the right side has left the magnetic field is (d-l), divided by the velocity to get the time.

Answer D At the beginning, AB cuts the magnetic inductance line, AB is the power supply, the current is counterclockwise, UAB is a negative value, and it is the road-end voltage, that is, the voltage of the three-segment wire, so the potential difference is large. The induced electromotive force then gradually decreases to zero.

In the same way, when the CD is cut, the potential difference is still negative, and the UAB is only a section of wire potential difference, so the maximum value is smaller than before.

First of all, the magnetic flux increases, and the current direction is counterclockwise, so the point B potential is high, and later, the magnetic flux decreases, and the current direction is clockwise, so the point B potential is high, so the correct one is d

Correct answer d

From b = blv and the right hand, it can be determined that the magnetic field enters the wireframe processab is the power supply, b is the positive pole of the power supply, uab "error, the magnetic field leaves the wireframe, cd is the power supply, d is the positive pole of the power supply, ab is the external circuit c error, correct answer d

Yes c pair. Analysis, in the process of pulling out the magnetic field in the metal frame, there is only one side that is still in the magnetic field and carefully cutting the magnetic inductance line has electromotive force e blv, and the part of the circuit outside the magnetic field is the external circuit, obviously, with the pull out of the frame, the resistance of the outer circuit gradually increases, and the resistance of the inner circuit gradually decreases, and the voltage measured by the voltmeter is the road end voltage (the voltage at both ends of the outer circuit is Xiao Annihilation) Kaiming, although the electromotive force remains unchanged, but the process of increasing the external resistance and decreasing the internal resistance must be the process of increasing the voltage at the road end, so C is right.

See my give**.

Sorry, this answer is wrong. I asked my physics teacher this afternoon. In the case of resistance of the rod, the terminal voltage is not equal to the electromotive force, and it cannot be calculated as such. To use differential equations, I won't. Let's wait for someone else.

But the second question can only be calculated with q=uc, which is clear, thank you!