There is an urgent need for 10 high school physics magnetic field exercises it is best to answer th

The main problem is that the working principle of ammeter and voltmeter is not understood, they are all modified from ammeters, and without current, the galvanometer has no indication. According to Faraday's law of electromagnetic induction, the magnetic flux in a closed circuit does not change, and no induced current is generated, so there is no indication in both meters.

There is no change in the magnetic flux in the closed loop, so there is no induced electromotive force and no current.

There is no change in the magnetic flux in the closed loop, and there is no induced current, but there is an induced electromotive force (dynamic electromotive force).

Then the base of the red isosceles triangle is (l +d ), r=[ (l +d ) 2] [d (l +d )]=(l +d ) 2d=mv bq

v=√（2qu/m）

q/m=2u/[b²（l²+d²）²/4d²]=8ud²/[b²（l²+d²）²

uq = mv 2 so v = 2uq m from the diagram we can see the circular motion l +(r-d) =r

bqv=mv²/r q/m=v/br

In this case, the radius of the circular motion is calculated r=mv qb, and it is obvious that the radius of the first one is 3 times the radius of the second one. Then draw a picture:

The key is to find the center of the circle of circular motion, draw the radius, and for the circular magnetic field, it is necessary to establish a triangle between the center of the circular magnetic field and the center of the circular motion, and grasp the symmetry. As long as you do this, it's basically fine.

I hope you can understand the essence of the method and hope it will be helpful to you.

From Newton's second law and the uniform circular motion: bqv=mvv r, t=2 r v, r=mv bq, t=2 m bq.

From the diagram, it can be seen that velocity v is injected into the magnetic field from point A in the direction of diameter AOB and through δt=t 6 and the magnetic field is emitted from point C.

Orbital radius r = ao 3 times the root number

When the velocity becomes v 3, the radius of motion is 3 times the root number of ao 3, and the geometric relation shows that the central angle of the circle that moves in the magnetic field is 120°, and the motion time is t 3, that is, 2δt.

Correct option: b

1), mv 2 2=qu v= (2qu m) and r=mv qb q m=v rb substituted (1) to obtain v=2u rb )

rA = l r B = l 2

So v A v B = r B r A = 1 2

2) by (1) (2) q m = 2u (br) 2t = 2 m qb = br 2 u

T A = T A 4 = BrA 2 4U

t B = t B 2 = brB 2 2u

tA:tB=:(rB2)=2:1

How are there three balls?

The answer is BCD, the velocity of the ball A is not charged, so the velocity that happens to pass through the highest point is when gravity happens to act as a centripetal force, that is, mg=mv 2 r, and the B ball is positively charged, so the velocity that happens to pass through the highest point is gravity + Lorentz force when it happens to act as a centripetal force, that is, mg+qvb=mv 2 r, the comparison of the two formulas, the velocity of the B ball is large, and b is correct; The mechanical energy from the release point to the highest point of the ring is conserved (Lorentz force does not do work), so the speed to the highest point of the ring is larger, and the falling height is larger, so the falling position of ball B is high, and c is correct; Because the mechanical energy is conserved, the mechanical energy of the three small balls all remains the same, d is correct.

The key to this problem is that when the highest point is just past the highest point, the force of the two balls is different, and the charged ball is subject to the Lorentz force, so the velocity is not the same, and the answer can be obtained by combining the conservation of energy or kinetic energy theorem.

First of all, when the block is placed in the electric field, it will be subjected to the electric field force qe=ma, and you can know that the acceleration of the object bridge block is a=qexm When the block moves, it will be subjected to the Lorentz force, and when the Lorentz force is equal to the gravity of the block, the block will leave the ground f Lo = g, qvb = mg, you can find that when the block leaves the ground, the velocity is v=mgxqb and then you chun can find vt 2-vo 2=2as, (mgxqb) 2=(2qexm)*s S=(m 3g 2)x(2q 3b 2e) can be found

From point O, the paper surface is shot into the square at a speed perpendicular to the CD edge, and the elapsed time t0 is just right out of the magnetic field from point C.

This known condition can be obtained with period t=2to (the central angle of the trajectory is ) This is understood?

If the incident and exit are at the same boundary of the magnetic field, then the angle of incidence and the angle of exit are equal (the angle between the chord) and the direction of velocity are deflected by 360°-60°, so the central angle of the circle is 300°

t=300°t/360°=5t/6=5to/3

t0 is the time taken to turn 180 degrees.

t=(2*pi*m) (q*b), independent of velocity, so the period is constant, so the magnitude of time and velocity in the magnetic field is irrelevant, while the direction is determined.

The time in the magnetic field illustrates the magnitude of the central angle corresponding to the trajectory, and it is not necessary to consider it according to the order in which the velocity changes from small to large and the radius from small to large.

The center of the circle is on the red line segment, and the angle of the center of the circle is always 300 degrees, which is shot out from the CD, and the corresponding time is 300 180*T0=5 3*to

The center of the circle is on the orange line segment, the angle of the center of the circle is from 180 degrees to 300 degrees, and it is shot from BC, and the corresponding time is t0 to 5 3*t0

The center of the circle is on the green line segment (very short on the figure), the angle of the center of the circle is from (120 degrees + gef) to 180 degrees, from BC, and the corresponding time is from T1 to T0, T1>2 3*T0

The orange and green parts can be considered together, from BC to ejection, the time is from T1 to 5 3*T0, T1 >2 3*T0).

The center of the circle is on the light blue line segment, the angle of the center of the circle is from 60 degrees to 120 degrees + gef, and it is shot out from AB, and the corresponding time is from 1 5 * T0 to T1

The center of the circle is on the pink ray, the center of the circle is from 0 degrees to 60 degrees, and it shoots out from AD, corresponding to the time from 0 to 1 3*t0 (t=0 corresponds to infinite velocity).