Please send a finale question about high school physics electromagnetism. Thank you. It s a little h

As shown in the figure, a smooth metal guide rail cdefg, oh cd fg, def=600, is fixed on the smooth inclined plane with an inclination angle of 300. A conductor rod AB with mass m is pulled by the motor, moves from the bottom of the inclined plane along the OH direction at a constant speed v0, slides on the guide rail and reaches the top of the inclined plane, AB OH. The resistance of the CD and FG sections of the metal guide rail is not counted, the material and cross-sectional area of the DEF section and the AB rod are the same, the resistance per unit length is R, O is the midpoint of the AB bar, and the whole inclined plane is in the uniform magnetic field with the vertical inclined surface and the magnetic induction intensity is B.

Find: 1) the magnitude of the current in the circuit when the conductor rod slides on the guide rail;

2) The voltage at both ends of AB when the conductor rod moves to the DF position;

3) Pull the conductor rod from the bottom end to the top end of the motor to do the work done externally.

Solution: (1) Let the equivalent cutting length of the AB rod be L, then.

The electromotive force is = (1 point).

The loop resistance is =3l r (1 point).

The loop current is i= (1 minute).

Score (1 point).

2) When the AB stick slides to DF.

1 point) 2 points.

2 points. Score (1 point).

3) The work done by the motor is 1 point.

1 point) is the electric heat generated when the AB rod slides on the def, which is numerically equal to the work done to overcome the ampere force, so (2 points).

It is the electric heat generated when the AB rod slides on the CDFG rail, the current is constant, and the resistance is constant.

1 point) (1 point).

As shown in the figure, the metal ball A with a mass of M1 and a charge of +Q and a metal ball B with a mass of M2 M1 and a charge of +Q are hung on the ceiling with an insulating lightweight thin wire of equal length, they are just in contact when they are stationary, and an insulating paper is pasted at the AB contact place, so that there is no charge transfer during the AB collision, there is a uniform magnetic field with a magnetic induction intensity of B on the left side of the vertical paper facing the vertical paper, and there is a uniform electric field with a vertical downward direction on the right side of PQ, and the field strength is small for E. Now pull the ball B to the position that the thin line is 53 degrees from the vertical direction (the thin line is just straightened) and release it freely, and the hem will collide with the ball A elastically at the lowest point. Due to the electromagnetic damping effect, ball A will stop at the lowest point before collision again, find out how many times after the collision the thin line of suspension B deviates from the vertical direction of the angle is less than 37°?

Q is released at rest from the point p, and the charge accelerates in a uniform linear motion v 2 = (2eq m)|po|①。The charge is shot perpendicular to cm, and the charge moves in the magnetic field r=cm=a=mv qb, i.e., v=aqb mBy |po|=(ab)^2q\2me ⑵∵po|The maximum charge trajectory in the magnetic field is tangent to cm, and the plot of r=(a r) 2, i.e., r=a3 v=qb 3m, and v 2=(2eq m)|po|∴|po|=qb^2\18me

(1) Because the particle passes perpendicular to the cm, then we can know the radius of the particle's circumferential motion r=a qvb=mvv r The velocity of the particle entering the magnetic field v=qba m So the kinetic energy of the incoming magnetic field e=mvv 2 So the electric field force work qed=mvv 2 So d=qbbaa 2me (2) According to the title, it can be known that at this time r=a 3 In the same way, we can know that d=qbbaa 18me period t=2 pie m qb So the motion time t=piem qb (3) Because Loren's magnetic force does not work, So if you want to move three times, the second time is tangent to cm. So at this point the radius r=a 5 in the same way we know d=qbbaa 25me

Isosceles triangle CDM below the x-axis

Alas, I miss myself in high school. The college entrance examination physics questions are all correct, and the multiple-choice questions ...... wrong

Now it's all forgotten, and it can't help you. good luck！~

The answer is 20 m s Let's consider gravity and electric field force first, assume that the electric field force is horizontal to the right, of course, gravity is vertically downward, and the resultant force of these two is 4*10 -5n From the relationship between the action force and the reaction force, we know that the magnetic field force is equal to its magnitude and the direction is opposite qvb=4*10 -5n Find v=20 m s, and then judge the direction from the left-hand rule as the plane perpendicular to the gravitational force and the electric field force inward.

You're asking about the direction of the magnetic field, and the direction of velocity isn't up or down.

From the geometric relationship, it can be seen that the motion time of a positive charge in a magnetic field is t 3, and the motion time of a negative charge in a magnetic field is t 6. Both of them are equal in mass and in charge, so the period is equal. Therefore, t positive:t negative = 2:1

Select the CD and go back to give you the steps.

<> the drawings I made for you with CAD, isn't that enough?

1.It can be obtained according to the kinetic energy theorem.

The work done by the induced current is wf: wg+wf=0-0

wf=-mgd

Because the velocity from the coil entering the magnetic field to leaving the magnetic field coil is constant, the work done by gravity in this process is: wg=mgd

So a pair. 2.When the coil enters the magnetic field, the ampere force f must be greater than mg, because only in this way can the coil decelerate from just entering the magnetic field to fully entering the magnetic field.

When the coil is completely in the magnetic field to the ab side, just leaving the magnetic field, because the induced current is 0, the coil accelerates with an acceleration of g.

When the ab edge of the coil has just left the magnetic field, the velocity is likely to be v again

Based on the above analysis, it can be seen that the velocity is minimal when the coil is just fully entered into the magnetic field.

Let the minimum speed be vmin

According to the kinetic energy theorem: (apply the kinetic energy theorem to the coil from the beginning to the moment when it is fully entered into the magnetic field, the initial velocity is 0, and the final velocity is vmin).

mg(h+l)+wf=1/2m*vmin^2-0

Note that in this process, the induced current does work wf=-mgd (when the coil completely enters the magnetic field and just leaves the magnetic field at the AB side, because the induced current is 0, the coil accelerates to accelerate the movement of g, and the induced current does not do work in this process).

Solve vmin = 2g under the root number (h+l-d).

So d right. 3.As for option C, I think it might be that you made a mistake.

The minimum velocity of the coil may be the square of the MGR B*L squared [(mgr) (bl)2].

This is to think that when the coil just enters the magnetic field, the ampere force f is equal to mg, and the coil moves at a uniform speed.

So: f=bil=b 2*l 2*v r=mg

Solve v=[(mgr) (bl) 2].

But this is not possible. Because according to the above analysis, the coil must decelerate as soon as it enters the magnetic field, and it is impossible to have a uniform velocity.

So C is wrong. Therefore: the answer AD