High School Physics Magnetic Field Questions, see Supplement

Mark the dots on the diagram and set the incoming magnetic field to point A.

1) The key is – the direction perpendicular to the y-axis enters the quadrant from n of y= -1 2r.

At the same time, understand that the velocity of p entering the electric field is not perpendicular and is at an angle, and is set to decompose to vx=vsin

vy=vcosθ

Coming out vertically at n, it means that the deceleration from the vertical direction to the n point is 0 (the horizontal direction is a constant velocity) and the equation a=eq m

vcosθ*t=r

vsinθ=at

2a*r/2=(vsinθ）²

Solution: tan = 1 = 45°

e=mv²/2qr

2) (3) Draw the approximate trajectory from A to P, and if the analyzed angle AOP=135°, the central angle corresponding to the trajectory of AP is also 135°

The time of AP is t=3t8=3 m 4bq

The time of MA is (2 2-1)r v

The total time is ( 2 2-1)r v + 3 m 4bq and the coordinates corresponding to m are [0,( 2 2+1)r].

Why is the question different from the picture? Ask for detailed 424489524

I'm going to go to the evening repair, and the next analysis is not understood if it is in contact.

Is the diagram drawn incorrectly, and the direction of the magnetic field is outward? If yes:

The motion of the particle in the electric field is a flat-like motion in turn, because we know the two points n and p, and according to the angle between the speed of the flat-like motion and the horizontal direction is twice the tangent between the displacement and the horizontal direction, we can know that the velocity of the point p is 135 degrees with the x-axis! Down.

Then the intersection point is obtained by making a perpendicular line (vertical velocity direction) between the point p and the point of the incoming magnetic field, because the intersection point should be the center of the circle, so the distance to the injector point and the point p is equal, and the motion radius of the injector point can be calculated by plane geometry (i.e., the trajectory of motion is determined).

Then find the velocity of motion within the magnetic field, and the rest is easy to say.

That's probably the case, you can check it yourself, right?

d First of all, if the particles are moving in a straight line, they need to be balanced by forces in the horizontal direction at least initially.

Hence the magnitude of the electric field force at the beginning is equal to the Lorentz force.

If you take one microelement along the direction of the particle's motion, the particle velocity is considered to be approximately constant, the Lorentz force is equal to the beginning, the direction is horizontal to the right, and the magnitude is unchanged.

At this point, the direction of the electric field force is already at an angle to the horizontal, and the electric field force becomes smaller as the distance from the origin increases.

It is then broken down to the horizontal direction, which is smaller and significantly smaller than the Lorentz force, so the horizontal direction produces the effect of the force.

It is no longer possible to do linear motion.

a b c are all achievable, and there are still questions welcome hi me, (hee-hee.

As you said, if you shoot in a direction of 45 degrees from the y axis, you must enter the ox at 45 degrees, and the angle of the center of the circle is 90 degrees, and you can't shoot from point A (on the left side of point A, the radius of this will be 2A). The central angle of the circle tangent to the top is 180 degrees, which is the largest central angle in all scenarios, and since all particles have the same period, the largest central angle is the longest.

First of all, from the point of view of the trajectory of motion, the resultant external force is not enough to provide the centripetal force, if the electric field force is vertically upward, then the resultant external force will exceed the centripetal force, so the electric field force is vertically downward.

Then at point p: qvb=mv2rSo v=qbr m, and the kinetic energy at this point is 1 2mv 2

The Lorentz force does not do work, only the electric field force does the work.

The work done by the electric field force is: fs=qer

1/2mv^2+qer=

qbr)^2+2qerm]/2m

1) The particle is centrifugal at point p, so the electric field should be downwards 2) The particle flies to p'The point electric field does negative work w=-eqr

From r=mv0 qb, v0 and initial kinetic energy ek are found in p'The kinetic energy of the point is ek-eqr

The direction of the electric field is vertically downward, (because from the motion at point p, the resultant external force is not enough to provide centripetal force, so the electric field force is to cancel the Lorentz force, so it is vertically downward) mv (bq)=r, get v=brq m,; Lorentz doesn't do work, so ek=1 2mv 2-eqr, just bring v in.

If you're still considering option d, you're not sure how the magnetic field force of the straight wire will be on the bow coil.

The straight wire is in the center of the circle, and at the beginning, its magnetic field is parallel to the direction of the current in the arc, and there is no force effect, but only the magnetic field force of the current in the chord part.

First make a schematic diagram of the magnetic field, and then use the left-hand rule to judge the force at both ends of A and B!

In addition, it is true that "the same direction current attracts the opposite direction and the opposite direction repels it". According to the force, the A end is facing the inside of the paper, and the B end is rotating outside the paper, and the direction of the current is gradually the same, so it is close.

1. When it starts to slide, the Loren magnetic force is vertically roded upward.

fn+qvb=mgcosθ

f=ufn=u(mgcosθ-qvb)

So mgsin -ufn=ma

mgsin -umgcos +uqvb=ma, doing an accelerated motion with increasing acceleration.

Because v is increasing, the magnetic force of Loren is increasing. When qvb=mgcos, it is no longer affected by friction, this time a is the maximum, a=gsinv=mgcosθ/qb

2. When qvb > mgcos, fn is reversed, that is, fn=qvb-mgcos so mgsin +mgcos -qvb=ma, do a deceleration motion with a gradual decrease in acceleration.

When A=0, V is maximum, and V=(GSIN +GCOS) qb

The direction of the Lorentz force is perpendicular to the rod, which affects the support force and thus the friction force.

The resultant force is the sliding component of gravity minus the frictional force. When f=0, a max=gsinc

At this point, qvb = mgcosc v=...Can.

When a=0, v is maximum. At this time ...

Because the particle q is positively charged, the direction of the Lorentz force is to the left, so the particle is deviated to the left, and the particle can only be shot from the three boundaries of ah, af, and fo, and because the radius is perpendicular to the velocity, the center of the circle can only be on the negative half axis of x. From r=mv0 qb, it can be seen that when the particle is emitted from point h, the radius is infinite, then b=0, and the direction of particle motion changes to 0 degrees. When the particle is emitted from the AH and AF edges, b>0, with the increase of b, the radius is getting smaller and smaller, when the particle is emitted from the FO boundary, the radius is the smallest, b is the largest, and the radius is, and b=mv0 can be found by the radius formula, and the direction of particle motion changes by 180 degrees. Give a good review and encouragement.