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A Physics Problem Based on an Actual Experiment

Yohtaro UENO and Toshiaki SHIBATA
Department of Physics, Tokyo Institute of Technology
Oh-okayama, Meguro, Tokyo Japan

Most problems on physics of the university entrance examinations in Japan are certainly appropriate for evaluating calculating ability when only limited understanding of physics is needed. However, it is difficult to evaluate to what extent applicants understand basic physical ideas through these exams. We have confirmed this via the 1999 Spring entrance examination by including in one of such problems as above a question that asks why there occurs a physical process for which you have calculated.

In recent days most university teachers of natural sciences have seen a strong tendency of their students to show less academic improvement than in the past; students are deficient in the ability to think logically by themselves. In view of the strong influence of entrance examinations on high school and junior high school education in Japan, the fact that the conventional type of problems have persisted dominantly for a long time is without doubt a major factor that has brought about such serious matters.

In order to get over this difficulty we have also tried to pose problems based on or strongly related to experiments and observations. These kinds of problems involve questions that require understanding of basic physical ideas and are favorable to those who have experimental and observational experience.

The problem given below is the only one that we made by doing an experiment. We hope that this type of problems would encourage high school teachers and students who have spent time in doing a lot of experiments. Finally, it is worth noting that one can complement to some extent the shortage of the conventional type of problems by adding questions that require reasoning and explanations.

Problem:

In order to study the force between two electric currents running in parallel lead wires, an experiment was performed using the apparatus as shown in Fig. 1. Although each of the lead wires 1 (abcdef) and 2 (gh) is a part of a different closed circuit, the remaining part of the circuit is abbreviated in Fig. 1. Lead wire 2 (hereafter abbreviated as LW2) is positioned horizontally, and its length is equal to segment cd of LW1. LW1 is attached to a Y-shaped insulator between a and b between e and f. These connected parts are supported by two insulated fulcrums at the midpoint of segments ab and ef, so that the system is balanced. Segments ab and ef are co-linear and serve as the axis of the balance. (Fig. 2 shows the apparatus viewed along the axis). When the balance is in balance, the plane of LW1 is horizontal and cd lies directly under gh. The force exerted on segment cd of LW1 can be measured by the adjusting the weights on the fine-adjustment weight pan. The abbreviated parts of the circuits in Fig. 1 are designed so that their influence on measuring the force be negligible. The quadrilateral bcde is a rectangle where bc is 0.1[m] and cd is 0.3[m].

We want to pass electric currents of up to 5[A] through LW1 and LW2. What is the most important thing we have to know for the purpose, in addition to the electromotive force V[V] of the battery and the electric resistance R[W ] of the lead wire? Write also the equation which shows how it affects the electric currents.

Firstly, we sent electric current I₁=5[A] in LW1 and no current in LW2, and balance the system by adjusting the weight. Next, keeping 1₁=5[A], we sent electric current I₂[A] in the direction opposite to 1₁. Varying the value of I₂[A], we measured the force exerted on segments cd and gh which was caused by the electric current 1₂[A]. The distance between segments cd and gh was kept at 0.01[m] all the time. The force was repulsive and we obtained the result shown in Table 1. Plot this data on the graph on the examination answer sheet, choosing appropriate coordinates.

Next, by moving LW2 upwards vertically while keeping it horizontal, we varied distance r[m] between segments cd and gh. Following the same steps as in (b), we measured the force exerted on segments cd due to electric current I₂[A], while keeping I₁=I₂=5[A]. The results are shown in Table 2. Plot this data on the graph on the examination answer sheet, choosing appropriate coordinates.

Let us denote by F[N] the force which acts between segment cd in WL1 and segment gh in WL2. Taking into account the principle of action and reaction of a force and making use of plots obtained in (b) and (c), derive an equation which expresses F[N] in terms of electric currents I₁[A] I₂[A] and distance r[m]. Note that the derivation should be logically done step by step. What is the value of the proportionality constant in the equation?

The proportional constant obtained in (d) turned out to bw smaller than the value obtained by the formula applicable to (d). let us consider the qualitative influence of each one of the following items on the experiment. If it makes the constant smaller, write S on the answer sheet. F it makes the constant larger n other cases, write SL. If I has no influence, write N.

The electric current is passing through segments ab and ef.

Segments cd and gh are not sufficiently long.

Geomagnetic fields are acting on segment cd and gh.

Although segments cd and gh are parallel, they may not be completely in a vertical plane.

The force acting between the lead wires, observed in (b) and (c), can be regarded as working via the magnetic flux density existing around one of the lead wires. In (d), the equation for the force which acts between the lead wires with r=0.3[m] was derived. Use that equation to derive an equation which expresses the magnitude of magnetic flux density B[T] at the position of LW2, generated by I₁[A] running ins segment cd with r[m]. Using the derived equation, obtain the value of B[T] when I₁=5[A] and r=0.01[m].

Table 1

Table 2

Electric Current I₂[A]

Force [N]

Distance r [m]

Force =[N]

0.0

0.0

0.010

10 x 10^-5

0.5

1.2 x 10^-5

0.015

7.0 x 10^-5

1.5

3.6 x 10^-5

0.020

5.0 x 10^-5

2.0

3.8x 10^-5

0.025

4.2 x 10^-5

2.5

5.6 x 10^-5

0.30

3.7 x 10^-5