A comprehensive question on high school physics electromagnetic induction

As shown in the figure below, the third section is the coil in the stage of leaving the magnetic field, at this time the AB edge is cutting the magnetic inductance line, if the right hand is used, the right hand should be placed on the AB edge, the magnetic field is perpendicular to the paper side outward, the palm should be facing down (let the magnetic inductance line penetrate the palm), the thumb points to the direction of speed (speed to the right), at this time the four fingers point to the direction of the current in the AB, the current direction in the AB is downward, and the current direction in the coil is counterclockwise, so it is correct.

According to Lenz's law, the direction of the magnetic field of the induced current in the coil should be the same as the direction of the original magnetic field, the direction of the original magnetic field is perpendicular to the paper side, and the direction of the magnetic field of the induced current should also be perpendicular to the paper side, and then use the spiral rule of the right hand, the thumb is perpendicular to the paper side outward (pointing to the direction of the magnetic field of the induced current), the four fingers refer to the direction of the induced current, and the current direction in the coil is counterclockwise. As shown in the figure below:

This one is very simple.,And isn't there an analysis next to it.。

Choose C, both methods are correct. **If something goes wrong, you can mention it.

The direction of gravity must be downward, the direction of the electric field force of the positively charged particles must be to the right, and the direction of the resultant force of the two forces is as shown in my figure, and the charged particles move in a straight line at a uniform speed, and the force is balanced, indicating that the Lorentz force must be the resultant force of gravity and the electric field force.

tana = qe/mg = 2×10-6 *10√3 / (2×10-6 * 10) =√3

a = 60 degrees.

The direction of motion of the charged particles is 60 degrees from the direction of the electric field to the right and upwards, f lo = qvb = qe sin60

v =e/bsin60 = 10√3 / 1*(√3/2) = 20m/s

The direction of the resultant force of gravity and the electric field force should be determined first, the direction of motion of the electrons and the direction of the resultant force should be perpendicular, and the force of the magnetic field should be balanced with the resultant force of the gravitational force and the electric field force. to determine the direction of motion and velocity of the particles. The electric field strength is missing from this question.

Not gravity downward.

Here the direction of the electric field force is determined, but since the velocity is uniform, the net force of the electric field force and the magnetic force of Loren is opposite to the gravitational force.

But in order for the velocity to be uniform, the electric field force must coincide with the magnetic force.

From the conservation of energy, fx=1 2mv +q, that is, the work done by the external constant force f is converted into heat energy and the kinetic energy of the rod;

From bvl=u, u r=i to obtain v=ri bl, and the above equation and data can be used to obtain the answer. We must know how to flexibly apply the law of conservation of energy.

BLV R=I0, FX-W Magnetism = 1 2*M*VV, Energy Consumed by Resistance = W Magnetism.

When the wires AB and CD move at a constant speed to the right along the direction of the guide rail at velocity V, these two wires are equivalent to the power supply, and the electromotive force of each wire is.

The electromotive force of the BC segment of the wire is EBC BLV and the internal resistance of this part of the wire is R 3

This part of the power supply and the external resistor form a closed loop, and the potential difference between BC is U [EBC (R R)] R

u＝blv*r / [(r / 3)＋r ]＝3blv / 4

The electromotive force of the AB segment is EAB b*l*v

The electromotive force of the cd segment is ECD b*l*v

There is no loop in the two stages of AB and CD, and there is no internal voltage in them, so the potential difference of AD is.

uad＝eab＋u＋ecd＝b*l*v＋（3blv / 4）＋b*l*v＝11*blv / 4

The wire cuts the magnetic field line, and the induced electromotive force e = segment resistance is r 3, then the current intensity of the closed circuit i = e r total = blv (r + r 3) = 3blv 4r

The potential difference between AD is the potential difference between BC, that is, the voltage on the resistor R, U=IR=3BLV 4

i.e. the potential difference is 3blv 4

The team will answer for you.

The potential difference between the two points of AD on the conductor is the end-of-road voltage.

According to the right-hand rule, the wire is to the right and the magnetic field is outward, so the current is determined by a d so d is the positive pole of the power supply and a is the negative pole of the power supply.

uad=ir

i=e/2r

uad=-e/2=-biv/2

It should be two points at BC asked.。。。 The potential difference between the two points of AB is 0, first judge the potential of the two points of BC, and use the wire ABCD as the power supply The right-hand rule knows that the potential of B is high, so it is positive.

Then look at the size.

Use e=blv to find the potential, because the series connection is r, the potential is divided directly by 2, i.e. e 2 = blv 2

This type of cutting is a type of rotary cutting of metal rods.

The formula for its resulting electromotive force is:

e = 1 2 b·l 2· = = 3V power supply internal resistance is 1

The parallel connection of the two semi-rings is 1

So the trunk current is i=e (r1+rcd+r)= 3 (1+1+1)= 1a

So the electric power of the ring is p=i2·rcd = 1 1 =1w

When the maximum is when A and D coincide, the induced electromotive force e=BLV

v=wr, where r is the midpoint value.

Hello, that's how it came.

The rod starts the source. If the velocity is V, then the instantaneous EMF is BVL. For a dry circuit, the resistance is r 2 (understand?) ), then the current in the trunk circuit is BVL (R 2), then the ampere force it receives is BIL, and the substitution is option A.

You know that the left and right resistors are the same, then q heat is generated on the AC resistor, and the resistance at the other end is also q, so the total heat is 2q, looking at the whole process, it is known by the conservation of energy, kinetic energy-heat = elastic potential energy, so it is option C.

I don't know how to ask questions.

At first, I thought that there should be no induced voltage, but after careful analysis, when the metal rod moves to the left, the closed part on the left side is getting larger and larger because of the magnetic flux, according to Faraday's law of electromagnetic induction, the induced electromotive force in the counterclockwise direction will be generated, and the size is evenly distributed along the closed part on the left, and the induced electromotive force reflected on the metal rod is pointed upward; In the closed part on the right, because the magnetic flux is getting smaller and smaller, a clockwise induced electromotive force is generated, and the induced electromotive force reflected on the metal rod is also pointed upward, so it does not cancel each other out.

On the contrary, because the rate of change of the magnetic flux on both sides is the same, the amount of induced electromotive force in the metal rod is also the same. Two induced electromotive forces of the same size are equivalent to a parallel relationship, so they do not affect each other.

Draw the x,y axis with the center of the circle as the origin, assuming that the angle between the contact between the metal rod and the ring and the x-axis at a certain position is , then with the movement of the metal rod, this angle changes between 0 180°. When at a certain angle, the effective cutting length of the rod l=2asin, the induced electromotive force at this time:

E=BVL=BV2ASIN, the unit length resistance of the metal rod is 3R2A, then the resistance at length L is R= 3R2A*(2ASIN)= 3RSIN

When in the position shown cos=1 2, so at this time:

3. Induced electromotive force e= 3bva

The effective resistance of the metal rod r=3r2

1) In the position shown in the figure, the equivalent resistance of the left part of the ring is R1=2R2 *(3)=R3, the equivalent resistance of the right part is R2=5R3, and the equivalent resistance R is filed after the two resistors are filed'=5/18r

So you can clearly draw an equivalent circuit diagram and find :

The magnitude of the current on the rod i = e (r+r')=√3bva/(3r/2+5/18r)=16√3bva/9r

The voltage at both ends of the rod umn=i*r'=16√3bva/9r*(5/18r)=40√3bva/91

2) Obviously, since the current in the metal rod is upward, the magnetic field force f should be to the left according to the left-hand rule. (In fact, there is no need to judge, it must be the opposite direction of movement).

f=bil=b*(16√3bva/9r)*(3a)=16b^2va^2/3r

First, find the effective length and effective resistance of the rod, and calculate the induced electromotive force and internal resistance r. The rod divides the ring into two parts, which are connected in parallel as the external resistance r, and the parallel resistance value is obtained, i=e (r+r) umn=i*r (m is on top).

The magnetic field force f=bia 3 on the metal rod is to the left.

I think the idea on the first floor is correct, as long as the problem of instantaneous induced electromotive force and average induced electromotive force is dealt with. This problem is basically solved. Otherwise, the landlord can also use the knowledge of the rate of change of magnetic flux and derivatives to solve the problem of transient and changing changes.

According to the distance from the rod to the diameter is a 2, the effective length of the cut can be found at this time, the resistance of the rod can be found according to the proportion, the ring is divided into two halves at this time, the resistance of each is calculated according to the arc length, the current on the rod can be calculated, and the voltage at both ends of the rod is the road terminal voltage at this time.

The ampere force can be solved according to f=bil.