Physical Electromagnetism, University Physics Electromagnetism

Flat throwing motion:

1.The kinetic energy theorem derives the initial velocity v entering the composite field

2.Uniform acceleration motion from the vertical direction: y=1 2*a*t 2a=(qe) mv=(qu) (mdv).

3.Horizontal: x=vt

It's too general to do circular motions in general.

Force analysis, use centripetal force test, f=qvb and other formulas to do.

A composite field of gravity and magnetism.

The effects of gravity and magnetic fields on charged particles are to be considered.

It generally moves in a curvilinear motion, but in most cases it is not a circular motion but a flat throw.

The composite field of electric field force and magnetic force is generally also a curvilinear motion, which belongs to the flat throwing, and the electric field force and magnetic field force should be considered.

If there is a Lorentz force involved in the composite field, and the sum of the work algebra of the other forces except the Lorentz force is 0, then the particle moves in a uniform circular motion.

The most important thing is force analysis.

Generally, it is a combination of electric and magnetic fields! The magnetic field provides the Lorentz force, the formula f=qvb, and the direction of the force is determined by the left-hand rule (the magnetic field line passes through the palm, the finger direction is the current direction, and the thumb direction is the force direction)! For the most part, the Lorentz force provides the centripetal force!

However, in the recombination length, the particle is also subjected to the electric field force, and the direction of the electric field force is judged by the direction of the electric field strength and the chargeability of the particle! In most of the cases we are doing now, the electric field and the magnetic field are perpendicular to each other, the particles enter the composite field with two perpendicular fields, the vector of the electric field force and the Lorentz force and the centripetal force that provides the particles to do circular motion, however, sometimes there are also situations where the two forces are balanced and the particles move in a uniform straight line! It is good to start the analysis mainly from the resultant force of the particles!

I am a senior student in the physics class of a key middle school in Guangdong, and I hope I can help you!

[Analysis].

When the switch is at S1.

The voltmeter measures the voltage at both ends of the resistor 1, and from i=u r, it can be seen that the total resistance of i= circuit is 12 ohms.

Therefore the electromotive force of the power supply e=

When the switch is located at S2, the voltmeter measures the voltage at both ends of the resistor 2, and the total resistance of the circuit is 10 ohms at this time, so the current is the indication of the voltmeter at this time.

Let the electromotive force be e

s at 1 o'clock:

u1=i1*r1=e*r1/(r+r1)

2=10e/(2+10)=10e/12...1>

S at 2 o'clock:

u2=i2*r2=e*r2 (r+r2)=8e (2+8)=<2> the simultaneous solution gives u2=

Solution: From the figure, it can be seen that the voltmeter measures the voltage of the resistance r1 or r2, because the circuit in the figure is connected in series, so the voltage is proportional to the resistance, and u is the power supply voltage r1 r=u1 u r2 r=u2 u, so u2=

The current is the electromotive force resistance.

EMF = (n=1, the Greek letter is too hard to beat, understand, it should be called fai t) fai = bs

Suppose the radius is r

The ratio of the area of the circle to the square r 2 2r 2= 2 b is equal for the circle and the square.

Resistance is related to length, the ratio of circumference 2 r (4 2r) = (2 2) i circle: i square = 1: 2

The ratio of electromotive force is the ratio of area.

e=δ δt=δbs circle δt

Let the radius of the circle rs circle = r 2

s square = 2r 2

The resistance ratio is the wire length ratio.

Circumference 2 r

The circumference of the square is 4r 2

i=e ri circle i square = 2 1

Option A does not rotate because there is no torque, and the distance between the currents in the same direction is closer (equal in size), so they attract each other

See the figure of renting a round ruler and a scale.

1) m w*w*r/2=qe

w*r=v can be recited to get r=mv*v (2e).

Question: midfield strength e=ker r, what does it mean, but I don't understand.

I don't have a diagram, but I know the content of the diagram (I've seen this issue).

Analysis: If the velocity of the wire when it just leaves the mercury surface is v, then there is.

v 2 2gh (in the process of air, the mechanical energy is conserved, or with the vertical upward throwing law, etc.) v root number (2gh).

In the very short time t of energizing, the amount of electricity passing through the wire is q, then the current is i q t, and the magnitude of the magnetic field force experienced by the wire is f b* i * l bq l t

In this case, the magnetic force is much greater than the gravitational force of the wire, so the effect of gravity can be ignored.

Then the momentum theorem gives f* t m v (during energizing), i.e. (bq l t)* t m v

So q m v (bl) m * root number (2gh)] (bl).

Bend into the shape shown in the figure.

Bend into the shape shown in the figure.

shown in the figure.

The algorithm is correct, but the formula is wrong, and the denominator is r, not r squared.

He's using the wrong formula...The formula denominator of the point charge potential is r and not the square of r.