IPhO 1998, Експеримент — Задача 2. Part II. Magnetic flux linkage

29th International Physics Olympiad, Reykjavik, Iceland — Experimental competition · 6 юли 1998 г.

Условие

The response of two coils on a closed ferrite core to an external alternating voltage ($V_g$) from a sinusoidal signal generator is studied.

Theory

In the following basic theoretical discussion, and in the treatment of the data, it is assumed that the ohmic resistance in the two coils and hysteresis losses in the core have insignificant influence on the measured currents and voltages. Because of these simplifications in the treatment below, some deviations will occur between measured and calculated values.

Single coil

Let us first look at a core with a single coil, carrying a current $I$. The magnetic flux $\Phi$, that the current creates in the ferrite core inside the coil, is proportional to the current $I$ and to the number of windings $N$. The flux depends furthermore on a geometrical factor $g$, which is determined by the size and shape of the core, and the magnetic permeability $\mu = \mu_r\mu_0$, which describes the magnetic properties of the core material. The relative permeability is denoted $\mu_r$ and $\mu_0$ is the permeability of free space.

The magnetic flux $\Phi$ is thus given by

where $c=\mu g$. The induced voltage is given by Faraday's law of induction,

The conventional way to describe the relationship between current and voltage for a coil is through the self inductance of the coil $L$, defined by,

A sinusoidal signal generator connected to the coil will drive a current through it given by

where $\omega$ is the angular frequency and $I_0$ is the amplitude of the current. As follows from equation (3) this alternating current will induce a voltage across the coil given by

The current will be such that the induced voltage is equal to the signal generator voltage $V_g$. There is a 90 degree phase difference between the current and the voltage. If we only look at the magnitudes of the alternating voltage and current, allowing for this phase difference, we have

Two coils

Let us now assume that we have two coils on one core. Ferrite cores can be used to link magnetic flux between coils. In an ideal core the flux will be the same for all cross sections of the core. Due to flux leakage in real cores a second coil on the core will in general experience a reduced flux compared to the flux-generating coil. The flux $\Phi_B$ in the secondary coil B is therefore related to the flux $\Phi_A$ in the primary coil A through

Similarly a flux component $\Phi_B$ created by a current in coil B will create a flux $\Phi_A = k\Phi_B$ in coil A. The factor $k$, which is called the coupling factor, has a value less than one.

The ferrite core under study has two coils A and B in a transformer arrangement. Let us assume that coil A is the primary coil (connected to the signal generator). If no current flows in coil B ($I_B=0$), the induced voltage $\varepsilon_A$ due to $I_A$ is equal and opposite to $V_g$. The flux created by $I_A$ inside the secondary coil is determined by equation (8) and the induced voltage in coil B is

If a current $I_B$ flows in coil B, it will induce a voltage in coil A which is described by a similar expression. The total voltage across the coil A will then be given by

The current in the secondary coil thus induces an opposing voltage in the primary coil, leading to an increase in $I_A$. A similar equation can be written for $\varepsilon_B$. As can be verified by measurements, $k$ is independent of which coil is taken as the primary coil.

Experiment

Place the two U-cores together as shown in Figure 2, and fasten them with the rubber bands. Set the function generator to produce a 10 kHz, sine wave. Remember to set the multimeters to the most sensitive range suitable for each measurement. The numbers of the windings of the two coils, A and B, are: $N_A$ = 150 turns and $N_B$ = 100 turns (±1 turn on each coil).

1. Derive algebraic expressions for the self inductances $L_A$ and $L_B$, and the coupling factor $k$, in terms of measured and given quantities and write your results in field 1.a on the answer sheet. Draw circuit diagrams in field 1.b on the answer sheet, showing how these quantities are determined. Calculate the numerical values of $L_A$, $L_B$ and $k$ and write their values in field 1.c on the answer sheet.

2. When the secondary coil is short-circuited, the current $I_P$ in the primary coil will increase. Use the equations above to derive an expression giving $I_P$ explicitly and write your result in field 2.a on the answer sheet. Measure $I_P$ and write your value in field 2.b on the answer sheet.

3. Coils A and B can be connected in series in two different ways such that the two flux contributions are either added to or subtracted from each other.

3.1. Find the self inductance of the serially connected coils, $L_{A+B}$, from measured quantities in the case where the flux contributions produced by the current $I$ in the two coils add to (strengthen) each other and write your answer in field 3.1 on the answer sheet.

3.2. Measure the voltages $V_A$ and $V_B$ when the flux contributions of the two coils oppose each other. Write your values in field 3.2.a on the answer sheet and the ratio of the voltages in field 3.2.b. Derive an expression for the ratio of the voltages across the two coils and write it in field 3.2.c on the answer sheet.

4. Use the results obtained to verify that the self inductance of a coil is proportional to the square of the number of its windings and write your result in field 4 on the answer sheet.

5. Verify that it was justified to neglect the resistances of the coils and write your arguments as mathematical expressions in field 5 on the answer sheet.

6. Thin plastic spacers inserted between the two half cores (as shown in Figure 3) reduce the coil inductances drastically. Use this reduction to determine the relative permeability $\mu_r$ of the ferrite material, given Ampere's law and continuity of the magnetic field B across the ferrite - plastic interface.

Assume $\mu = \mu_0 = 4\pi\times 10^{-7}\ \mathrm{Ns^2/C^2}$ for the plastic spacers and a spacer thickness of 1.6 mm. The geometrical factor can be determined from Ampere's law

where $I_{total}$ is the total current flowing through a surface bounded by the integration path. Write your algebraic expression for $\mu_r$ in field 6.a on the answer sheet and your numerical value in field 6.b.

Решение

Оригинал в Архива: exp.pdf