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Hengyang Normal University 2007

Physics (II) Final Exam Question B (Answer Sheet)

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Multiple choice questions: (3*** 30 per question)

1. The vector of vacuum current element at point P represents the magnetic stress intensity at point P (B).

(1); (b) and: (c) and: (4).

2. The magnetic field with uniform magnetic stress takes a vertical closed surface, and then the magnetic flux passes through the closed surface: (d)

(1); (b) and: (c) and: (D) 0

3. The current I1= 4ai2 =1a. According to Ampere's law diagram, the closed curve C = (A).

(A)3μ0； (B)0；

(C)-3μ0； (D)5μ0

4. A straight cylinder with radius A carries current I, and the magnetic stress intensity at point p (a) is uniformly distributed (r>a) outside the cylinder.

(1); (b) and:

(c) and: (4)

5. The waveform at a certain moment is shown in Figure (b).

(a) potential energy at point a;

Potential energy at point b

(c) the potential energy of points A and C;

(d) potential energy at point b

6. The horizontal spring vibrates and pulls away from the equilibrium position by 5cm, and then carries out simple harmonic vibration through static release. If it is pulled to the axial direction and the vibration range is marked, the initial phase amplitude (b) of simple harmonic vibration is calculated.

(1); (b) and:

(c) and: (4)

7. The object is in simple harmonic vibration, the vibration range x=Acos(ωt+π/4)t=T/4(T period), and the acceleration of the object (d).

(1); (b) and: (c) and: (iv)

8. The simple resonance range is shown in the simple resonance displacement-time curve.

(A)x = 4 cos 2πt(m)； (3)

(B)x = 4 cos(πt-π)(m)；

(C)x = 4 cosπt(m)；

(D) x=4cos(2πt+π)(m)

9. Cosine wave propagates in the negative direction of X axis. It is known that the vibration range at X =- 1 m is y=Acos(ωt+). If the wave velocity is u, the wave path is (c).

(1); (b) and:

(c) and: (4)

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(a) The interference fringe spacing increases and moves in the O direction;

(b) The interference fringe spacing decreases and moves in the direction B;

(c) The interference fringe spacing decreases and moves in the O direction;

(d) The interference fringe spacing increases and moves in the O direction.

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Fill in the blanks: (3*** 18 per question)

1. Current around a straight wire I magnetic stress

2. Under the condition of coherence, coherent waves have the same vibration direction, the same frequency and constant phase difference.

3. The required interval T/4 (expressed in cycles) at the apogee of the resonance equilibrium position is half of the required interval T/ 12 (expressed in cycles).

4. Microscopically, the potential does not produce electrostatic Lorentz force.

5. When two resonance ranges x1= 0.03 cos ω tx2 = 0.04 cos (ω t+π/2) (Si), the combined amplitude is 0.05m m. ..

6. Describe the amplitude, angular frequency and initial phase of three characteristic quantities of harmonic transport.

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Three. Short answer: (6*** 12 per question)

1. Analysis of the influence of spring amplitude on physical quantities: vibration period, speed and acceleration vibration energy.

Reference solution: the spring vibration period T=2π 1 is only related to the internal properties and external factors of the system 1 and the amplitude 1.

Vmax =ωAA doubles VMAX doubles 1

Amax=ω2AA double amax triple 1

E = double Ka2A, quadruple E, 1

2. What are the characteristics of two coherent beams emitted by the same light source, and give an example?

Reference solution: the same light source emits two kinds of coherent light: wavefront amplitude 2 wavefront means that the original light source emits two parts of the same wavefront, and the Young's double-slit interference experiment of two light sources taking coherent light is 2; Amplitude refers to the common light source emitting light at the same point, and interfering with two coherent light films, such as reflection and refraction.

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IV. Calculation questions: (No.1 question 7, 8***3 1 for each question)

1. The ball connected with the light spring is simple harmonic motion with amplitude A along the X axis. The expression of this vibration is expressed by cosine function. If t=0, the motion state of the ball is different:

( 1)x0 =-A； (2) the equilibrium position moves in the X direction; (3) X = A/2 and negative to x to determine the corresponding initial phase.

Solution: (1) = π1; (2) =-π/2 1; (3) =π/3 1

Phasor diagram: Figure (1)1; Figure (2)1; Figure (3)2

2. The vibration amplitude of horizontal spring is A=2.0× 10-2m, and the period T=0.50st=0.

(1) The object x= 1.0× 10-2m is negative;

(2) The object X =- 1.0× 10-2m is transported.

Don't write two kinds of vibration expressions

Solution: The phasor diagram is known from the problem =4π2.

(1)φ 1= Its vibration expression x1= 2.0×10-2cos (4 π t+) (m) 3.

(2)φ2 = or- its vibration expression x1= 2.0×10-2cos (4 π t+) (m) 3.

Scheme 2: analysis (1) t = 0x0 =1.0×10-2m = a/2, v0.

From x0=Acosφ= knowing cosφ= then φ =

By v0 =-ω asin φ φ 0 φ =1.

The vibration expression is x1= 2.0×10-2cos (4 π t+) (m) 2.

(2)T = 0x 0 =- 1.0× 10-2m = A/2，v0 & gt0. 1

From x0 = acosφ=- knowing cosφ=- then φ =+(or)

From v0 =-ω asinφ >; 0 sinφ& lt； 0 φ = or-1

Its vibration expression

x 1 = 2.0× 10-2 cos(4πt+)(m)= 2.0× 10-2 cos(4πt-)(m)2

3. The coils shown in the figure are evenly and densely wound on the wood ring with complete cross section (the inner and outer radius of the wood ring is affected by the wood magnetic field cloth with the thickness of R 1R2) ***N turns to find the current. Is there less magnetic flux in the cross section of the magnetic field distribution tube inside and outside the I ring?

Solution: Appropriate Ampere Loop According to the Ampere Loop Theorem, the magnetic field of the outer ring of the circular ring is made into a circular Ampere Loop L perpendicular to the axis of the wooden ring.

Outside the circle =0, then from the ampere loop theorem B=0.

When the radius is r (r 1

2 B？ 2πr=μ0NI

By b = μ 0ni/(2π r) 2 in the ring

To require the flux of the annular pipe section, first consider the width dr, the height H, and the narrow strip area flux D φ = BHDR = DR2.

The magnetic flux through all sections of the pipe φ = 2.

4. Refractive index n 1= 1.52 Mirror coating n2= 1.38MgF2 antireflection film. Should the thickness of Apollo λ=550nm be less?

Solution: The anti-reflection film cancels the interference of reflected light and increases the intensity N of transmitted light.

2n2h = (2k+ 1), k = 0, 1, 2, … then h=(2k+ 1) 3.

K=0 1 antireflection film thickness

hmin = = = 9.96× 10-8(m)= 99.6 nm2

Solution 2: The interference of reflected light is cancelled in the antireflection film, so that the interference of transmitted light has a phase. Therefore, the thickness of the antireflection film is determined by the interference of the transmitted light, and the optical path difference 2n2h+λ/2 of the two transmitted lights when the surface passes through the secondary reflection (half-wave loss) 2 transmission mirror and the direct transmitted light passes through MgF2 is determined by the interference phase condition.

2n2h+ =kλ, k= 1, 2, 3, … then h = (k-) 3.

K= 1 1 antireflection film thickness hmin = = = 9.96×10-8 (m) = 99.6 nm2.

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Verb (abbreviation of verb) Title of evidence: (***9)

As shown in the figure, the current I flows through a straight wire, and the width of the other rectangular coil ***N turns, the aL speed v, is flat to the right. The test proves that the distance from the left side of the rectangular coil to the straight wire is d, and the coil should be at potential.

Solution: Use potential formula to solve.

Current I, magnetic field distribution of straight wire B=μ0I/2πx, which is perpendicular to the plane where the coil is located and to both sides, so the coil moves to the right, and the two sides of the coil generate potential (the two wires do not cut the magnetic field lines), and the left and right sides generate potential? The total potential of parallel coils is displayed on the same paper.

=? 1-? 2=N[ - ]3

=N[ ]

=N[ - ]= = 3

> 0? Where to? 1 in the same direction, i.e. clockwise 3.

Second, the distance d from the left side of the coil to the straight wire, the magnetic stress B 1=μ0I/2πd on the left side of the coil, and the speed v of the coil perpendicular to the paper surface carry the potential of the left wire.

1=N =N =NvB 1 =Nv L。

Clockwise magnetic stress B2=μ0I/2π(d+a) on the right side of the coil and the wire potential perpendicular to the paper surface on the right side of the coil.

2 =N =N =NvB2 =Nv L。

In response to the potential of the coil of the counter needle 3.

=? 1-? 2= Nv L-Nv L=

> 0 that is to say? Where to? 1 Same direction Clockwise 3

Three: by? = product path l is clockwise.

=N[ ]

=N[ ]=N()

=Nv L-Nv L= 6

> 0 that is to say? Clockwise direction 3, the same as the closed path L.

Solution 2: Solve by Lardy's law of electromagnetic response.

Non-uniform magnetic field of straight wire magnetic field B=μ0I/2πr The magnetic field in the coil plane is perpendicular to the coil plane, so the L-wide dr bin dS=Ldr is taken at the position far away from the straight wire R, and the magnetic flux of the bin passes through the needle direction.

dφ= = BDS cos 0 =

Plane magnetic flux of the connecting coil (let the distance from the left side of the coil to the straight line be x)

Φ= 3

The internal potential in the coil is determined by Lardy's law of electromagnetic response.

=-

Distance from left side of coil to straight line x=d Internal potential of coil

= 3

> 0? Direction and circuitous direction, namely clockwise direction 3.

According to the cold law, the direction of the potential is that the coil is flat to the right because the magnetic field is gradually weakening, and the magnetic field generated by the current will hinder the reduction of the original magnetic flux, that is, the current magnetic field should be in the same direction as the original magnetic field and the potential should be in the needle direction.

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