In the second year of high school, there are questions about the physics of electromagnetism

The voltage at both ends of AB is always the voltage at the end of the road.

When only the AB side enters the magnetic field, AB is the power supply, the internal resistance is R, the CD and EF are connected in parallel for the external circuit, and the parallel resistance of the external circuit is R 2.

u1=(1/3)blv

When CD enters the magnetic field, AB and CD are power supplies and two identical power supplies are connected in parallel, and EF is an external circuit.

u2=(2/3)blv=2u1

After EF also enters the magnetic field, AB, CD, and EF all cut the magnetic inductance wires, all of which are power supplies, and are connected in parallel for three identical power supplies, and the external circuit is open, and the current flowing through each power supply is 0, so the voltage at the end of the road is equal to the electromotive force, that is, U3=BLV

There is still electromotive force, because the conductor rod is cutting the magnetic inductance line, and you confuse the induced electromotive force with the induced current. When all are in, the induced current disappears, but the induced electromotive force remains.

The large circle and the small circle are made of the same material, so the resistance of the big circle is proportional to their circumference, and the circumference is 2 r

The resistance ratio is the radius ratio, so let the resistance of the large circle be 4r, and the small circle is r, then according to Lenz's law, the formula of induced electromotive force, at the beginning, the small circle is used as the power source, and its electric power Sun rents the potential k small circle area.

Then u (k small circle area) (4r 5r)1.

The second time, when the great circle is used as a power source and the small circle is a resistor, the great circle electromotive force k is the area of the great circle.

The second terminal voltage U2 (k large circle area) (r 5r)2. Divide 1 by 2 and find the ratio of their area by the ratio of their radius.

u2 4u, so d should be selected for this question

Solution:1. When the wire is parallel to the magnetic field strength, the force is zero, and the force is the largest at the angle of 90°. So the ampere force increases.

2。The direction of the ampere force is always perpendicular to the plane determined by both the wire and the magnetic field line (two intersecting lines determine a plane). In the process of rotation of the wire, it is unchanged from the plane determined by the magnetic induction wire. Therefore, the direction of the ampere force is always the same in the process.

Can you send a picture to see?

The direction of the current is decomposed vertically, and the force is applied only perpendicular to the direction of the magnetic field, and the direction is always perpendicular to the paper facing inward or outward (depending on the direction of the magnetic field).

Can you send a picture first?

As long as it's remembered.

Determine the direction of the ampere force experienced by a straight wire in a magnetic field.

Left-hand rule: Stretch out your left hand so that your thumb is perpendicular to the other four fingers and in a plane, let the magnetic lines flow in from the palm of your hand, the four fingers point in the direction of the current, and the thumb points in the ampere force direction (i.e., the direction of the conductor force).

The ampere force direction is always perpendicular to the magnetic field direction, because the magnetic field direction does not change, so the ampere force direction is always the same... I don't know if you can understand, I'm also a sophomore in high school, so I don't know if this analysis is right.

AD has to use the full reaction force (F+N), I can't explain it clearly, I'm sorry.

If the total anti-angle is greater than 30, select A

Less than 30,d

The analysis of recommended answers is a bit problematic.

b = μ*mg /il(cosθ+μsinθ)= mg /il(cosθ/μ+sinθ)=mg /ilsqrt(1+1/μ^2)sin(θ+arctan1/μ)

According to is known from 30° to 0°, if arctan1 <=60, i.e. >=sqrt(1 3), b monotonically becomes larger, a

else b first decreases and then increases, dConsistent with the analysis of total reactions.

Answer: ACD

Analysis: First of all, because the magnetic field is oblique to the upper left, then the direction of the ampere force on the metal rod is either oblique to the lower left or oblique to the upper right, and the metal rod is moving at a uniform speed to the right on the guide rail, indicating that the direction of the ampere force must be oblique to the upper right, (the force on the metal rod is: gravity, support force, friction force (to the left), ampere force).

Assuming that the angle between the ampere force and the horizontal plane is , then the variation range is 30° to 0°. This should be judged).

Since the metal rod is always in constant motion, then there is bil*cos +fn = mg bil*sin = *fn

It can be solved by two equations: b = *mg il(cos + sin) According to the 30° to 0° we know, cos increases, sin decreases, and cos increases at a smaller rate, while sin decreases at a larger rate (judging by the slope of the sine curve and the cosine curve).

Since the value of is not known, there are several cases:

When the comparison is small, the cos always increases faster than the *sin decreases, then the cos + sin always get bigger, then b always gets smaller, and b pairs.

When is not very small, the speed of cos increase is faster than that of *sin, and then because the speed of cos increases is getting smaller and smaller, and the speed of *sin is decreasing and the speed of cos is increasing, the speed of cos increasing is slower than that of *sin, then cos + sin first becomes larger and then smaller, then b first becomes smaller and then larger, d is right.

When it is larger, the speed of cos increase is slower than that of *sin, and because the speed of cos is getting smaller and smaller, and the speed of *sin is decreasing is getting bigger and bigger, so the speed of cos increasing cannot be faster than that of *sin, then cos + sin always becomes smaller, it is impossible to increase, then b always gets bigger, it is impossible to decrease, b is right and c is wrong.

It's not convenient to type, so I wrote down the whole process on paper and took a picture, click on it to see the big picture. (Hopefully not miscalculated).

Let the width be d and the letter 2 will be quadratic.

At equilibrium, v=2m s

fv+i2r=u*i

fv=b2d2v2/r

First list the basic formula, then look at the picture to find the special point, and then find the special position state in the mentions, and then set in the formula, generally you can do this kind of problem, it is generally not too difficult to find the special state.

The slope of the image is velocity, and then it is constant. The mechanical power of the electric motor w = is the traction force, and the traction force is equal to the gravitational force of the object plus the ampere force experienced by the pole.

a.When the metal frame enters the magnetic field, it can be obtained from Lenz's law: the magnetic flux in the wire frame increases, and a magnetic field perpendicular to the paper side will be generated, and the direction of the induced current can be induced by the right-hand rule: a d c b a a wrong.

b, when the metal wire frame leaves the magnetic field, it can be obtained from Lenz's law: the magnetic flux in the wire frame decreases, and a magnetic field perpendicular to the paper side will be generated, and the direction of the induced current can be induced by the right-hand rule: a b c d a b wrong.

c, this option can be seen from the energy point of view, when the DC edge of the metal wire frame enters the magnetic field and the AB edge leaves the magnetic field, the gravitational potential energy is equal, but in the process of the DC edge entering the magnetic field and the AB side leaving the magnetic field, electrical energy will be generated. From the law of conservation of energy, it can be concluded that the kinetic energy of the DC side entering the magnetic field is greater than the kinetic energy of the ab side leaving the magnetic field (this part of the kinetic energy is converted into electrical energy), so the speed of the DC side entering the magnetic field is always greater than the speed of the ab side leaving the magnetic field. C false.

d。Since d0"l, the metal wire frame will make a simple harmonic motion in the magnetic field d pairs.

If you don't understand, you can discuss it again

1.The force given by the electric field from the circumference of the uniform velocity is equal to the force of gravity, so the magnetic induction force can be considered, and the velocity of the circular motion can be obtained from the L and @ angles, and v can be obtained from the @ angle.

2.The vertical downward velocity can be obtained from the circular motion and the @ angle, and h can be obtained from the acceleration formula. This is a very typical question type, and the next time you encounter such a question type, it is usually the same pattern.

Answer: B

Analysis: Let the radius of the circular motion of the particle be r, find the center of the circle by the perpendicular line of the velocity through the incident point, and the declination angle of the velocity is equal to the angle of the center of the circle, make the presentation intention according to the topic, and find r=r according to the geometric relationship; The Lorentz force provides the centripetal force, then the velocity of the particle v=, so b pairs; A, C, D are wrong.

Let's draw a graph trajectory diagram. Well, connect the incident point and the exit point with the center of the magnetic field circle, and then connect the incident point and the exit point. It is easy to prove that the angle between the line of incidence and the center of the magnetic field and the vertical direction is 30 degrees, (use r 2 ah, draw a right triangle), the direction of velocity changes by 60 degrees, so the center angle corresponding to the trajectory is also 60 degrees (this is already a conclusion, you know) In this case, the two triangles you see are 60 degree equilateral triangles, and it's OK.

If only I could draw a picture, do you understand?

Have you seen the diagram of "Hook Difference Life", let me say again, it is not necessary to draw the same height, as long as it proves that the POM is 30 degrees. Of course, you can also go to the same high.

Solution: (1) The upward velocity is .

The electromotive force produced by v=d t is.

u=nblv

then the current is. i=u/r

The amount of charge flowing through the cross-section of each turn of the wire in the coil is.

q=it=nbld/r

2) During the lifting process, at least the gravitational potential energy of the coil and the electrical energy consumed by the coil should be supplied.

EP weight = mgd

EP = (U2 R)t

W person = EP weight + EP electricity = mgd + (nbld) 2 (rt).

Isn't it possible to find the amount of charge by finding the induced electromotive force from the change of magnetic flux, and then finding the induced current?

The second question is based on the conservation of energy.