APhO 2012, Експеримент — Задача 2. ELECTROMAGNETIC INDUCTION

13th Asian Physics Olympiad — India — Experimental Competition — Wednesday, 2nd May 2012 · 2 май 2012 г.

Please first read the following instructions carefully:
  1. The time available is 2½ hours for each of the two experimental questions.
  2. Use only the pen and equipment provided.
  3. In the boxes at the top of each sheet of paper write down your student code.
  4. There is a set of answersheets in which you have to enter your data and results. In addition blank sheets are provided which you can use.
  5. Use only pencil to draw graphs on the graph sheets.
  6. If you add any blank sheets to your main answersheets, write down the progressive number of each sheet (page number) on these additional sheets. You should enter the number of the question (Question No.) on such used blank sheets.
  7. If you use some blank writing sheets for notes that you do not wish to be marked, put a large ‘X’ across the entire sheet and do not include it in your numbering.
  8. Numerical results must be written with as many digits as appropriate; don’t forget the units.
  9. You should use mainly equations, numbers, symbols, graphs, figures and as little text as possible in your answers.
  10. At the end of the experiment arrange all sheets in the following order:
    1. Main Answersheet
    2. Used Graph sheets
    3. Used writing sheets
    4. The sheets which are marked with ‘X’
    5. Unused writing sheets and graph sheets
    6. The printed question paper
  11. Put all the papers inside the envelope and leave the envelope on your desk.
  12. You are not allowed to take any sheet of paper or any material used in the experiment out of the room.

Условие

The modern technique of eddy current testing employed for detecting defects under the surfaces of metallic objects is based on the principles of electromagnetic induction. The circulating currents induced in conducting bodies due to changing magnetic flux in the region where they are located are called eddy currents. The defects are detected by observing the changes in the resistance and inductance of a coil carrying alternating current when held near the surface of the object.

Unless the core of the solenoid has some ferromagnetic material the magnetic flux ϕ\phi is proportional to the current ii .The constant of proportionality is called self inductance of the coil and is represented by letter LL. The self-induced emf in a coil with inductance LL is, therefore, given by −Ldidt-L\dfrac{di}{dt}.

In a coupled system of circuits with current i1i_1 in one circuit the magnetic flux linked with it is given by

ϕ=L1i1+M21i2\phi = L_1 i_1 + M_{21} i_2

Where L1L_1 is the self inductance of that circuit and M21M_{21} is the mutual inductance of the coupled system.

Similar equation holds for the reverse, with M12=M21M_{12}=M_{21}

L-R Circuit

A sinusoidal alternating current with angular frequency ω\omega flowing through a series combination of resistance RR and inductance LL produces a voltage drop across the combination.

If we represent current by i=I0sin⁡ωti = I_0 \sin \omega t, the voltage drop across the resistance is equal to I0Rsin⁡ωtI_0 R \sin \omega t and that across the inductance it is I0ωLcos⁡ωtI_0 \omega L \cos \omega t. We can combine these to get the voltage across the RR-LL combination. The quantity ωL\omega L is called inductive reactance and is represented by symbol XX. One can readily show that the voltage across the coil is equal to I0Zsin⁡(ωt+θ)I_0 Z \sin(\omega t+\theta), where

Z=R2+X2(1)Z=\sqrt{R^2+X^2} \qquad (1)

and

θ=tan⁡−1(XR)(2)\theta=\tan^{-1}\left(\frac{X}{R}\right) \qquad (2)

Alternating voltage as well as current vary continuously in both magnitude and polarity during the course of time but the rms values of these quantities calculated over a cycle are independent of time and the relation V=IZV = IZ where both VV and II represent the rms values, is analogous to Ohm’s law. From this we see that

V2=(IR)2+(IX)2(3)V^2=(IR)^2+(IX)^2 \qquad (3)

(Note:- The concept of resistance is basically related to dissipation of electrical energy and the value of resistance of a coil in ac circuit can be different from its value determined by applying Ohm’s law with dc currents.)

When there are additional resistances and/or inductances in series with the coil, we can still consider the voltage drop across the combination as equal to the square root of the sum of the squares of voltage drops across the total resistance and across the total inductance.

Measurement of Inductance and Resistance of a Coil

For measuring alternating current and voltage, generally the rms values are noted. From Equations (2) & (3) we get

Vcos⁡θ=IR(4)V\cos\theta = IR \qquad (4)

and

Vsin⁡θ=IX(5)V\sin\theta = IX \qquad (5)

To obtain the values of RR and LL of the coil we can use the above equations. Voltage VV and current II can be measured. But there being three unknown quantities θ\theta, RR and XX we need one more equation.

Circuit: an AC source (~) in series with a resistor R′ and a coil drawn as a dashed box containing a resistor R₁ and an inductor L₁. Dashed arrows mark V_A across the whole combination, V_R′ across R′ and V across the coil.
Fig.1

If the applied voltage across the series combination consisting of a known resistor R′R' with the coil is VAV_\mathrm{A}, then an expression relating applied voltage VAV_\mathrm{A} to the voltage drops VR′V_{R'} across R′R', VV across the coil and the angle θ\theta is

VA2=VR′2+V2+2VR′Vcos⁡θ(6)V_A^2=V_{R'}^2+V^2+2V_{R'}V\cos\theta \qquad (6)

All quantities except θ\theta in Equation (6) are measurable.

Hence measuring the three voltages VAV_\mathrm{A}, VR′V_{\mathrm{R}'} and VV, and using Equations (4), (5) and (6), θ\theta, RR and XX can be determined. Knowing the frequency of the alternating current the value of LL can be calculated.

Alternatively, from Equations (4) and (6) we can express the value of RR in terms of the three measured voltages as

R=(R′2)[VA2−V2VR′2−1](7)R=\left(\frac{R'}{2}\right)\left[\frac{V_A^2-V^2}{V_{R'}^2}-1\right] \qquad (7)

The impedance ZZ of the coil can be calculated using the formula Z=VVR′R′Z=\dfrac{V}{V_{R'}}R' and the value of XX could be obtained from

X=Z2−R2(1A)X=\sqrt{Z^2-R^2} \qquad (1A)

Coupled Circuits

The energy supplied by the power source to the primary can be dissipated partially in the primary and the remainder in other mutually coupled secondaries. When no mechanical work is done, the energy dissipation is only in resistances. The inductances store energy in the magnetic field associated with them. With current II (rms value) the average stored energy in inductance LL is equal to 12LI2\dfrac{1}{2}LI^2.

When a current flows in the secondary the emf induced due to it in the primary brings about change in the primary current. Seen from the primary side the effect is a consequence of change in the effective resistance and reactance of the primary coil and there is no need to know the parameters in the secondary circuit. The total energy dissipated in primary as well as secondary circuits appears as if it is dissipated in the effective primary resistance when seen from the primary side.

The effective values of resistance RPER_\mathrm{PE} and inductance LPEL_\mathrm{PE} of primary can be related to a ‘reflected’ resistance RRR_\mathrm{R} and a ‘reflected’ inductance LRL_\mathrm{R} from the secondary side. The (average) power dissipated in the reflected resistance RRR_\mathrm{R} in the primary has to be equal to that in resistance RSR_\mathrm{S} in the secondary circuit. This gives

IP2RR=IS2(RS+RL)(8)I_P^2R_R=I_S^2\left(R_S+R_L\right) \qquad (8)

Similarly, we can relate the reflected inductance LRL_\mathrm{R} to the secondary inductance LSL_\mathrm{S} from

12LRIP2=12LSIS2(9)\frac{1}{2}L_RI_P^2=\frac{1}{2}L_SI_S^2 \qquad (9)

Considering the fact that the induced emf in the secondary due to an alternating primary current IPI_\mathrm{P} has magnitude equal to ωMIP\omega M I_\mathrm{P} we can write the equation corresponding to Kirchhoff’s loop rule for the secondary in terms of rms values of primary and secondary currents as

ωMIP=ISZS(10)\omega M I_P = I_S Z_S \qquad (10)

where ZSZ_\mathrm{S} is the impedance of the secondary circuit. When the secondary impedance is infinite the mutually induced emf appears as the voltage across the open ends of the secondary.

Photograph of red and black connecting wires with banana pins.
Connecting Wires
Photograph of a yellow digital multimeter with red and black probe leads.
Digital Multimeter
Photograph of a bobbin with two coaxial copper coils mounted on an acrylic board with blue terminals (1-START, 1-END) and green terminals (2-START, 2-END).
Coils wound one over the other

APPARATUS:

Photograph of a transparent acrylic board with a row of black and a row of red banana sockets.
Resistance Board
Photograph of the sine wave generator box with power switch, amplitude knob and output terminals.
Sine wave Generator
Photograph of a small acrylic resistance board with yellow sockets and resistors (the 300 ohm board).
Resistance Board
Photograph of a short cylindrical aluminium rod.
Aluminium Rod
Photograph of the complete set-up: generator, resistance board, coils, 300 ohm board, aluminium rod, wires and multimeter connected together.
Complete experimental set up

The apparatus provided for the experiment consists of the following.

  1. A sine wave generator with output of about 10 V (rms) at 1000 Hz frequency
  2. A digital multimeter (DMM) to be used as a voltmeter
  3. A pair of coaxial coils wound one around the other on a cylindrical hollow piece of non magnetic non conducting bobbin
  4. A piece of aluminum rod capable of fitting inside the bobbin
  5. Two series of ten resistors mounted on an acrylic board: one consisting of 100 ohms resistors and the other of 10 ohm resistors with banana pin sockets
  6. The required resistance for load RLR_\mathrm{L} or sampling resistance R′R' can be selected using this resistance board and connecting wire pieces. A separate acrylic board with resistance of 300 ohms is provided which can be used for sampling resistor when the other board is used for load resistance.
  7. A set of five red and five black wires with banana pins at their ends (the pair of longer red and black wires is meant for use as DMM probes).

The output of the sine wave generator of frequency 1000 Hz is to be used as the alternating source voltage. Use the 20 V range of the DMM to measure rms ac voltages.

The magnitudes of currents IPI_\mathrm{P} and ISI_\mathrm{S} when needed are to be calculated from the measured voltages across known values of R′R' in primary and RLR_\mathrm{L} in secondary respectively.

EXPERIMENT

PART 1: Determining Resistance and Inductance of a Coil with Air-core and Aluminium Core (3,4 т.)

Connect coil 1 (with blue terminals) in series with a resistor R′R' (to be selected from the resistance board) across the output terminals of the sine wave generator. The sine wave generator will be on before you start your experiment for stabilization of its output. Do not turn it off. Keep the output voltage amplitude maximum. (The DMM should show the output about 10 V).

The ac output of the generator may have some asymmetry. In that case, the readings of the DMM will show slightly different readings, when the input polarities of the DMM are interchanged between ‘V/Ω’ to ‘com’. To correct for the error due to asymmetry repeat each reading by interchanging the polarity of the probes of the DMM and take the mean of the two readings.

Choose the value of R´ to obtain VR′V_{\mathrm{R}'} and V approximately equal so that the systematic error in Z becomes negligible.

Circuit: AC source in series with R′ and coil 1 (dashed box with R₁ and L₁); arrows mark V_A, V_R′ and V. Below, coil 2 (dashed box with R₂ and L₂) is open, with its terminals marked by circles and the voltage V_O between them.
Fig.2

a) Measure the voltages VAV_\mathrm{A}, VR′V_{\mathrm{R}'} and VV as well as VOV_\mathrm{O} across the terminals of the other coil. Determine the resistance R1R_1 and inductance L1L_1 of coil 1 (with blue terminals) and estimate the uncertainties in the values determined. [0,9 т.]

b) Connect the other coil (coil 2 with green terminals), make the necessary measurements and determine R2R_2 and L2L_2. Evaluate the uncertainties in the values obtained. [0,9 т.]

c) Now insert the piece of aluminium rod at the core of the coils and repeat the procedure to obtain the values of inductances L1L_1* and resistance R1R_1* and the uncertainties in them for coil1. [0,8 т.]

d) Make necessary measurements and determine inductance L2L_2* and resistances, R2R_2* of coil 2 when it has an aluminium core. Estimate the uncertainties in the values. [0,8 т.]

Note: Uncertainty calculations need not be done for parts 2, 3 and 4.

PART 2: Mutual Inductance and Coupling Constant (3 т.)

Circuit: AC source in series with R′ and the primary coil (dashed box with R_P and L_P); arrows mark V_A, V_R′ and V_P. Below, the secondary coil (dashed box with R_S and L_S) is closed through a variable load resistor R_L, with the voltage V_O marked across it.
Fig.3

f) Mutual inductance MM can be obtained from the readings of VR′V_{\mathrm{R}'} and VOV_\mathrm{O} (recorded in Part 1). Find the average values for air core as well as aluminium core coils. The relation between the mutual inductance and the self inductances of the coupled coils is given by M=k (L1L2)1/2M = k\,(L_1L_2)^{1/2}. Determine the value of kk, the coupling coefficient in the two cases. [0,4 т.]

g) Select coil 1 (with blue terminals) as primary and coil 2 (with green terminals) as secondary. Connect the primary in series with the sampling resistor R′=300R' = 300 ohm across the output terminals of the generator. Connect across the secondary the variable load RLR_\mathrm{L}. The output voltage VOV_\mathrm{O} is to be measured across RLR_\mathrm{L}. Change RLR_\mathrm{L} and measure the voltages VAV_\mathrm{A}, VR′V_{\mathrm{R}'}, VV and VOV_\mathrm{O} corresponding to each value of RLR_\mathrm{L}. [0,8 т.]

h) A linear graph can be plotted combining various terms appearing in Equation (10) written in expanded form. Write the linear expression for plotting a graph whose slope can be used to obtain the value of MM and intercept to obtain the value of secondary reactance XSX_\mathrm{S}. [0,2 т.]

i) Calculate the necessary quantities using data from (g) to plot the graph corresponding to the expression obtained in the step (h) above. [0,9 т.]

j) Plot the graph and obtain the values of MM and XSX_\mathrm{S}. [0,7 т.]

PART 3: Relations between Effective Primary Impedance and the Reflected Quantities from Secondary (2,4 т.)

k) Use the data collected in Part 2 to determine the effective resistance RPER_\mathrm{PE} and the effective reactance XPEX_\mathrm{PE} of the primary corresponding to each value of RLR_\mathrm{L} in secondary. [0,6 т.]

l) Use the data of Part 2 to calculate the values of reflected resistance RRR_\mathrm{R} as defined in Equation (8) and of reflected reactance XRX_\mathrm{R} referring Equation (9) corresponding to the values of RLR_\mathrm{L}. [0,6 т.]

m) Plot graph of XPEX_\mathrm{PE} against XRX_\mathrm{R}. Taking into consideration the likely uncertainties of the quantities plotted write down the equation giving the relationship between the effective primary reactance and the reflected reactance. [0,6 т.]

n) Represent graphically the relation between RRR_\mathrm{R} and RLR_\mathrm{L} over the range of study and find the value of RLR_\mathrm{L} for which the reflected resistance attains a maximum. If needed, take some more observations to supplement the observations in Part 2 for locating the point with greater precision. [0,6 т.]

PART 4: Eddy Current Effects (1,2 т.)

o) A model based on the analysis of the data in Part 3 suggests how to estimate the ratio of inductance and resistance seen by the eddy currents set up in the core of a coil connected to a power source.

The analysis of the data of Part 2(i) and (j) should show that RPE=RP+RRR_\mathrm{PE} = R_\mathrm{P} + R_\mathrm{R}. The relation between XPEX_\mathrm{PE} and XRX_\mathrm{R} is obtained from Part 3(m).

Refer to the data collected in Part 1(c) and estimate the ratio of inductance and resistance as seen by eddy currents in the aluminium core when the power supply is connected to coil 1 and coil 2 respectively. [0,8 т.]

p) Connect the coils as in Part 2 Fig.3 and insert the piece of aluminium rod at the core of the coils. Write down the expression giving the power loss ΔP\Delta P in the aluminium core.

Set R′=300R' = 300 ohms and RL=1000R_\mathrm{L} = 1000 ohms. Adjust the magnitude of VAV_\mathrm{A} equal to 9.0 V. Make the necessary measurements and calculate the power dissipation due to the eddy currents in the aluminium core. [0,4 т.]

Отговори

Решение


Оригинал в Архива: exp2.pdf · официални решения: exp2_sol.pdf