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Hengyang Normal University in the next semester of 2007.

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

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1. Multiple choice questions: (3 points for each small question, * * * 30 points)

1. If the potential vector of the current element to point P in vacuum is, the magnetic induction intensity generated at point P is (b).

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

2. In a uniform magnetic field with a magnetic induction intensity of, take a cubic closed surface with a side length of, and the magnitude of magnetic flux passing through the closed surface is: (d)

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

3. As shown in the figure, the current I 1=4 A and I2= 1 A in the two wires. According to Ampere's law, the closed curve C shown in the figure has = (A).

(A)3μ0； (B)0；

(C)-3μ0； (D)5μ0 .

4. a long straight cylinder with radius a carries current I, and the current I is evenly distributed on the cross section, then the outer side of the cylinder (r >；; A) the magnitude of magnetic induction intensity at point p is (a)

(1); (b) and:

(c) and: (D) 10.

5. The waveform diagram at a certain moment is shown in the figure, and the following statement is correct (B)

(a) Point A has the largest potential energy and the smallest kinetic energy;

(b) The potential energy and kinetic energy at point B are the largest.

(C) A and C have the largest potential energy and kinetic energy;

(d) Point B has the maximum kinetic energy and the minimum potential energy.

6. Pull the horizontal spring vibrator 5cm away from the equilibrium position, release it from the static state, do simple harmonic vibration and start timing. If the pulling direction is selected as the positive axis direction and expressed by the vibration equation, the initial phase and amplitude of this simple harmonic vibration are (b).

(1); (B)，；

(C)，； (D).

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

(1); (b) and: (c) and: (D) 10.

8. The displacement-time curve of simple harmonic vibration is shown in the figure, and the vibration equation of this simple harmonic vibration is as follows

(A)x = 4 cos 2πt(m)； (3)

(B)x = 4 cos(πt-π)(m)；

(C)x = 4 cosπt(m)；

x=4cos(2πt+π)(m).

9. Cosine waves propagate along the negative direction of the X axis. It is known that the vibration equation at X =- 1 m is y=Acos(ωt+), and if the wave velocity is u, the wave equation is (c).

(1); (b) and:

(c) and: (4)

10. As shown in the figure, two flat glass plates OA and OB form a airlight wedge, and a beam of flat monochromatic light is vertically incident on the optical wedge. When the included angle θ between plate A and plate B increases, the interference pattern will be (c).

(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 points for each small question, *** 18 points)

1. The magnetic induction around a long straight wire with current I is.

2. The coherent conditions of coherent waves are that the vibration direction is the same, the frequency is the same and the phase difference is constant.

3. The time required for the harmonic oscillator to move from the equilibrium position to the farthest point is T/4 (expressed in cycles), and the time required for completing half the distance is T/ 12 (expressed in cycles).

4. From the microscopic point of view, the non-electrostatic force that produces electromotive force is Lorentz force.

5. When the two harmonic dynamics equations are x 1=0.03cosωt and X2 = 0.04 cos (ω t+π/2) (Si), their combined amplitude is 0.05m.. ..

6. The three characteristic quantities describing simple harmonic vibration are amplitude, angular frequency and initial phase.

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Three. Short answer: (6 points for each small question, *** 12 points)

1. When the amplitude of the spring oscillator is doubled, try to analyze how the following physical quantities will be affected: vibration period, maximum speed, maximum acceleration and vibration energy.

Reference solution: The period of the spring oscillator is T=2π 1 min, which is only related to the internal properties of the system and has nothing to do with external factors, so it has nothing to do with the amplitude. 1 point

Vmax =ωA, and when a doubles, Vmax doubles. 1 point

Amax=ω2A。 When a doubles, amax doubles. 1 point

E= kA2。 When A doubles, E quadruples. 1 point

2. What are the ways to divide the light emitted by the same light source into two parts to become coherent light? What are the characteristics of these methods and give examples?

Reference solution: There are two methods to split the light emitted by the same light source into coherent light: wavefront splitting and amplitude splitting. Wavefront splitting method refers to the method of obtaining coherent light by using two parts of the same wavefront emitted by the original light source as two sub-light sources, such as Young's double-slit interference experiment. 2 points; Fractional amplitude method refers to the method of "splitting" the light emitted by ordinary light sources at the same point by reflection and refraction, so as to obtain coherent light, such as thin film interference.

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Fourth, the calculation problem: (1 problem 7 points, cross-question 8 points, ***3 1 point)

1. There is a small ball connected to a light spring, which makes a simple harmonic motion with an amplitude of A along the X axis. The expression of vibration is expressed by cosine function. If t=0, the motion state of the ball is as follows:

( 1)x0 =-A； (2) The overbalanced position moves to the positive direction of X; (3) Crossing X = A/2, moving in the negative direction of X ... trying to determine the corresponding initial phase.

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

The phasor diagram is as follows: Figure (1)1; Figure (2)1; Figure (3)2 points

2. A horizontal spring oscillator with amplitude A=2.0× 10-2m and period t=0.50s ... When t = 0,

(1) The object passes through x= 1.0× 10-2m and moves in the negative direction;

(2) When the object passes through X =- 1.0× 10-2m, it moves forward.

Write the vibration expressions in the above two cases respectively.

Scheme 1: phasor diagram method. From the topic =4π2 points

(1)φ 1=, and its vibration expression is x1= 2.0×10-2cos (4 π t+) (m) 3 points.

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

Solution 2: Analysis method. (1) because when T=0, x0 = 1.0× 10-2m = a/2, v0.

From x0=Acosφ=, we know that cosφ=, φ =+-,

By v0 =-ω asinφ < 0, where sin φ >; 0, so φ = 1 min.

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

(2) because T=0, x0 =-1.0×10-2m = a/2, and v0 > 0. 1 point

From x0 = acos φ =-, we know that cos φ =-, φ =+(or,),

From v0 =-ω asinφ >; 0, with sin φ

Its vibration expression

X1= 2.0×10-2cos (4π t+) (m) = 2.0×10-2cos (4π t-) (m) 2 points.

3. As shown in the figure, the coils are evenly and densely wound on a wooden ring with a rectangular cross section (the inner and outer diameters of the wooden ring are R 1 and R2, respectively, and the thickness is h, so the wood has no influence on the magnetic field distribution). * * * has n cycles. Find the distribution of magnetic field inside and outside the ring after applying current I. What is the magnetic flux through the cross section of the pipe?

Solution: Select the Ampere Loop appropriately, and then discuss the magnetic field outside the loop and inside the loop in two cases according to the Ampere Loop Theorem. The circle perpendicular to the central axis of the wooden ring and centered on the central axis is the ampere loop L.

If the perimeter is outside the ring, because =0, it can be obtained from Ampere's loop theorem, and B=0 outside the ring.

If the circumference is inside the ring, the radius is r (r 1

Two o'clock, b? 2πr=μ0NI

Therefore, point b = μ 0ni/(2π r) 2 is in the ring.

In order to find the magnetic flux through the cross section of the ring pipe, we can first consider that the magnetic flux through the narrow strip with the cross section width of dr and the height of H in the ring pipe is d φ = bhdr = dr2 for 2 minutes.

The magnetic flux through the entire cross section of the tube is φ = 2 minutes.

4. The lens with refractive index of n 1= 1.52 is coated with a layer of MgF2 antireflection film with n2= 1.38. If this film is suitable for light with wavelength λ=550nm, what is the minimum thickness of the film?

Solution 1: The antireflection film is used to counteract the interference of reflected light, thus increasing the intensity of transmitted light. Because n is empty

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

When k=0 1, the minimum thickness of antireflection film can be obtained.

hmin = = = 9.96× 10-8(m)= 99.6 nm2。

Scheme 2: For the antireflection film, the interference of reflected light is offset, that is, the interference of transmitted light is constructive. Therefore, the thickness of the antireflection film can also be obtained by interference enhancement of transmitted light. When the light is reflected twice (with half-wave loss) on the upper and lower surfaces of MgF2 for 2 minutes and then transmitted to the lens and meets the transmitted light directly transmitted through MgF2, the optical path difference between the two transmitted lights is 2n2h+λ/2. By interfering with constructive conditions, there are

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

When k= 1 1, the minimum thickness of antireflection film hmin = = = 9.96×10-8 (m) = 99.6 nm2.

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Verb (abbreviation of verb) Proof: (***9 points)

As shown in the figure, there is a current I in the long straight wire, and another rectangular coil ***N turns, with a width of A and a length of L, moves to the right at a speed of V, which proves that when the distance from the left side of the rectangular coil to the long straight wire is D, the induced electromotive force in the coil is 0.

Solution 1: Use electromotive force formula to solve.

Method 1: The magnetic field distribution of a long straight wire with current I is B=μ0I/2πx, and the direction is inward perpendicular to the plane of the coil. For the upper and lower sides of the coil, because the direction of the coil is perpendicular to the direction of the coil, when the coil is translated to the right, the upper and lower sides of the coil will not generate induced electromotive force (the upper and lower wires do not cut the magnetic field lines), but only the left and right sides will generate electromotive force. And electromotive force is generated on the left and right sides? In the same direction, they are all parallel to the paper and can be regarded as parallel, so the total electromotive force in the coil is

=? 1-? 2 = n [-] 3 points

=N[ ]

= n [-] = = 3 points

> 0, then? Direction and? The direction of 1 is the same, that is, clockwise for 3 minutes.

Method 2: When the distance from the left side of the coil to the long straight wire is d, the magnetic induction intensity of the left side of the coil is B 1=μ0I/2πd, and the direction is perpendicular to the paper surface and inward. When the coil moves at speed v, the electromotive force in the left conductor is

1=N =N =NvB 1 =Nv L。

Clockwise for 3 minutes. The magnetic induction intensity B2 on the right side of the coil = μ 0i/2π (d+a), and the direction is perpendicular to the paper surface and inward. When the coil moves, the electromotive force in the right conductor is

2 =N =N =NvB2 =Nv L。

The direction is counterclockwise for 3 minutes. So the induced electromotive force in the coil is

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

> 0, that is? Direction and? The direction of 1 is the same, clockwise for 3 minutes.

Method 3: by? =, the integral path L is clockwise, and there are

=N[ ]

=N[ ]=N()

= NV L-NV L = 6 points

> 0, that is? In the same direction as the closed path L, clockwise for 3 minutes.

Solution 2: Solve by Faraday's law of electromagnetic induction.

Because the magnetic field of a long straight wire is an uneven magnetic field B=μ0I/2πr, the magnetic field direction is perpendicular to the coil plane and inward. Therefore, take a small slot DS = Ldr with length L and width DR at the distance from the long straight conductor R, and the winding direction of the return path is clockwise, then the magnetic flux passes through the slot.

dφ= = BDS cos 0 =

The magnetic flux passing through the whole coil plane (when the distance from the left side of the coil to the long straight wire is x) is

φ = 3 points

The induced electromotive force in the coil is given by Faraday's law of electromagnetic induction as follows

=-

When the distance from the left side of the coil to the long straight conductor is x=d, the induced electromotive force in the coil is

= 3 points

Because? > 0, so? The direction is the same as the detour direction, that is, 3 minutes clockwise.

The direction of induced electromotive force can also be judged by Lenz's law: when the coil moves to the right, the magnetic field gradually weakens and the magnetic flux passing through the coil decreases, so the magnetic field generated by induced current will hinder the reduction of the original magnetic flux, that is, the magnetic field of induced current should be the same as the original magnetic field, so the direction of electromotive force is clockwise.