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Условие (A) Setup (B) Investigation of the magnetic braking force Отговори Решение

APhO 2005, Експеримент — Задача 2. MAGNETIC BRAKING ON AN INCLINED PLANE

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Всички материали от APhO 2005

EXPERIMENTAL COMPETITION · 28 април 2005 г.

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Thursday, April 28th, 2005

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Условие

INTRODUCTION

When a magnet moves near a non-magnetic conductor such as copper and aluminum it experiences a dissipative force called magnetic braking force. In this experiment we will investigate the nature of this force.

The magnetic braking force depends on:

the strength of the magnet, determined by its magnetic moment (μ);
the conductivity of the conductor ( $\sigma_C$ );
the size and geometry of both magnet and the conductor;
the distance between the magnet and conducting surface (d); and
the velocity of the magnet (v) relative to the conductor.

In this experiment we will investigate the magnetic braking force dependencies on the velocity (v) and the conductor-magnet distance (d). This force can be written empirically as:

F_{MB} = -k_0\, d^{\,p} v^{n} \qquad (1)

where

$k_0$ is an arbitrary constant that depends on μ, $\sigma_C$ and geometry of the conductor and magnet which is fixed in this experiment.
d is the distance between the center of magnet to the conductor surface
v is the velocity of the magnet
p and n are the power factors to be determined in this experiment

EXPERIMENT

In this experiment error analysis is required.

Drawing of the ring (doughnut) magnet with its thickness marked t_M and a red arrow through its axis pointing out of one flat face towards N, the opposite face marked S.

APPARATUS

(1) Doughnut-shaped Neodymium Iron Boron magnet.

Thickness: $t_M = (6.3\pm0.1)$ mm

Outer diameter: $d_M = (25.4\pm0.1)$ mm

The poles are on the flat faces as shown:

(2) Aluminum bar (2 pieces)

(3) Acrylic plate for the inclined plane with a linear track for the magnet to roll

(4) Plastic stand

(5) Digital stop watch

(6) Ruler

(7) Graphic papers (10 pieces)

Additional information:

Local gravitational acceleration: $\boldsymbol{g = 9.8\ \mathrm{m/s^2}}$

Mass of the magnet: $\boldsymbol{m = (21.5\pm0.5)\ \mathrm{g}}$

North-South direction is indicated on the table.

You can read the operation manual of the stopwatch

This problem is divided into two sections:

(A) Setup and introduction

(B) Investigation of the magnetic braking force

Remarks:

Make sure that the plane is clean before your experiment

QUESTIONS

Please provide sufficient diagrams in your answers so that your work can be understood clearly

(A) Setup

Roll down the magnet along the track as shown. Choose a reasonably small inclination angle so that it does not roll too fast.

Drawing of the inclined acrylic plane on a stand with a vertical rod at the top, a central linear track and the ring magnet standing on the track. — Figure 1. Inclined plane setup without aluminum bars

Drawing of the inclined plane with two aluminum bars placed along the track on either side of the ring magnet; an inset labelled FRONT VIEW shows the magnet between the two bars with the distances d from the magnet centre to each bar, labelled MAGNET and ALUMINUM BAR. — Figure 2. A complete setup with aluminum bars

[1] As the magnet is very strong, it may experience significant torque due to interaction with earth’s magnetic field. It will twist the magnet as it rolls down and may cause significant friction with the track. What will you do to minimize this torque? Explain it using diagram(s). [1 т.]

Place the two aluminum bars as shown in Figure 2 with distance approximately d=5mm. Remember that the distance d is to the center of the magnet as shown in the inset of Figure 2.

Again release the magnet and let it roll. You should observe that the magnet would roll down much slower compared to the previous observation due to magnetic braking force.

[2]. Provide diagram(s) of field lines and forces to describe the mechanism of magnetic braking. [1 т.]

(B) Investigation of the magnetic braking force

The experimental setup remains the same as shown in Figure 2 with the same magnet-conductor distance approximately $d = 5$ mm (about 2 mm gap between magnet and conductor on each side).

[1] Keeping the distance d fixed, investigate the dependence of magnetic braking force on velocity (v). Determine the exponent n of the speed dependence factor in Equation 1. Provide appropriate graph to explain your result. [4 т.]

Now vary the conductor-magnet distance (d) on both left and right. Choose a fixed and reasonably small inclination angle.

[2] Investigate the dependence of the magnetic braking force on conductor-magnet distance (d). Determine the exponent p of the distance dependence factor in Equation 1. Provide appropriate graph to explain your result. [4 т.]

Отговори

Решение

2. MAGNETIC BRAKING ON AN INCLINED PLANE

SUGGESTED SOLUTION

(A) Setup and Introduction

A1. To minimize the torque due to interaction of the magnet and the earth’s magnetic field we have to set the orientation of the inclined plane so that the magnet will roll down with the poles aligned to the North-South direction as shown.

Drawing of the inclined plane with the ring magnet on the track; red arrows show the magnet's poles (N towards the left, S towards the right) aligned with the North-South direction marked by large blue letters N and S. — Figure 1. Adjusting the orientation of the inclined plane

A2.

Drawing of the ring magnet moving in direction x along a plate next to an aluminum bar: blue field lines B pass from the magnet into the bar, a red current loop I is induced in the bar around point A, the force on the magnet F_C-M points backwards (green), and an inset at point A shows B, the current I and the force F_M-C. — Figure 2. Field and interactions in the magnetic braking effect

Answer with some vector analysis:

Consider a point A on the conductor. As the magnet moves, its magnetic field sweeps the conductor inducing electric field and causing current flow due to Faraday’s law, whose direction can be determined using Lenz’s law. Let’s choose an arbitrary loop as shown. At point A, the magnetic field and the current will cause Lorentz force $F_{M-C}$ pointing at x+ direction. This force is acting on the electrons in the conductor

On the other hand, due to Newtons’ Third law there is reaction force $F_{C-M}$ with the same magnitude but with opposite direction acting on the magnet, which is the magnetic braking force.

(B) Investigation of the magnetic braking force

B1. Determination of the power factor n: Dependence of the magnetic braking force with the velocity

In this experiment the student has to be aware that the magnet should reach the terminal velocity first before start the timing. From observation we can see that the magnet reaches terminal velocity almost immediately. To make sure we let the magnet travels first for about 5 cm before we start measuring the time. Here we use s = 250 mm from start to finish to obtain speed: $v = s/t$ .

The angle of inclination is varied to take several data. Given l = 425 mm, we measure h where $\sin\theta = h/l$ .

(a) The inclined plane with aluminum bars and the magnet; red markers show about 5 cm run-up (~ 5 cm) before START and the timed distance s from START to FINISH. (b) SIDE VIEW: plane of length l on a stand, raised by height h at the upper end, inclination angle θ, the magnet on the plane. — Figure 3. a. Measurement of the velocity b. Measurement of the plane inclination

Because the magnet-conductor distance is kept constant ( $d \approx 5$ mm), the magnetic braking force only depends on the velocity of the magnet, so we can simplify:

F_{MB} = -k_0\, d^{\,p} v^{n} = -k_1 v^{n}

where $k_1 = k_0 d^{\,p}$ is constant in this experiment.

The ring magnet on the inclined plane with the contact point C; a red arrow mg sinθ points down the slope and a blue arrow F_MB points up the slope. — Figure 4. Force diagram of the rolling magnet

When the magnet reaches the terminal velocity then the total torque should be zero. The equation of the motion at the contact point C will be:

\sum \tau_C = 0

mg\sin\theta\, R + F_{MB} R = 0

mg\sin\theta - k_1 v^{n} = 0

\sin\theta = \frac{k_1}{mg} v^{n}

To calculate the power factor n:

\ln\sin\theta = \ln\left(\frac{k_1}{mg}\right) + n\ln\left(v\right)

The experimental data:

H (mm)	$\bar{t}$ (s)	Sin $\theta$	$v$ mm/s	$\ln(v)$	$\ln(\sin\theta)$
23±0.5	22.98±0.005	0.054	10.88	2.39	-2.92
40	12.78	0.094	19.56	2.97	-2.36
50	10.17	0.118	24.58	3.20	-2.14
60	8.62	0.141	29.00	3.37	-1.96
70	6.96	0.165	35.92	3.58	-1.80
80	6.09	0.188	41.05	3.71	-1.67
91	5.48	0.214	45.62	3.82	-1.54
101	5.05	0.238	49.50	3.90	-1.44
111	4.57	0.261	54.70	4.00	-1.34
120	4.17	0.282	59.95	4.09	-1.26
130	3.72	0.306	67.20	4.21	-1.18
150	3.25	0.353	76.92	4.34	-1.04
170	2.81	0.400	88.97	4.49	-0.92

Table 1. Experimental data for power factor n determination

Note:

Column in bold are the data directly taken from the experiment.
Typical error for h measurement is shown in the first row: $h = \left(23 \pm 5\right) mm$ . Similar error applies for the rest of h data.
Data $\bar{t}$ are the average data taken from 3 to 5 measurement. Even though standard deviation error is quite small (±0.1s), the error should be dominated by response delay of the observer in pressing the stopwatch. Widely accepted value for human eye response is 0.25 sec, in this experiment we choose more conservative value (±0.5 s)

Plot of ln(sin θ) (vertical axis, -3.0 to -1.0) against ln(v) (horizontal axis, 2.5 to 4.5): thirteen data points on a straight red fitted line, the central point at about (3.58, -1.8) with a blue error bar. — Figure 5. Graph of $\ln(\sin\theta)$ vs $\ln(v)$ . Typical error bar is shown in the central data.

Using linear regression method or graphical method as shown in Fig. 5 one can determine n from the slope.

n = 0.96

Whose result is very close to the theoretical value of n = 1. From the data shown in Fig. 5 (as well as the coefficient of correlation r = 0.9995), it can be shown that this experiment is very good in demonstrating the linear velocity dependence of the magnetic braking force. This result has been repeated and verified by three independent persons and apparatus setups.

Error estimate of n:

Instead of laboring on detailed error propagation analysis that could be very time consuming, in olympiad context one can make the error estimate as follows:

The typical error of the data points in Fig 5 can be obtained from the central data:

\ln v = 3.58 \pm 0.075

\ln\left(\sin\theta\right) = -1.8 \pm 0.075

whose errors propagated from the uncertainties in h and t.

The power factor n can be obtained from the slope of Fig. 5: $n = \Delta y / \Delta x$ where $y = \ln\left(\sin\theta\right)$ and $x = \ln v$ .

From the data in Fig. 5. we have: $\Delta x = 2.1$ and $\Delta y = 2.0$ , and the typical errors: $\delta x = 0.075$ and $\delta y = 0.075$ .

So the error estimate for n:

\frac{\Delta n}{n} = \sqrt{\left(\frac{\delta x}{\Delta x}\right)^2 + \left(\frac{\delta y}{\Delta y}\right)^2} = \sqrt{\left(\frac{0.075}{2.1}\right)^2 + \left(\frac{0.075}{2.0}\right)^2} = 0.05

\Delta n = 0.05\, n = 0.048

So we can conclude the result of our experiment is:

\boxed{n = 0.96 \pm 0.05}

B2. Determination of the power factor p: Dependence of the magnetic braking force with the magnet-conductor distance

In this experiment we use one value of inclination angle, h=50 mm (l=425mm) so that $\theta = \arcsin s\left(h/l\right) = 6.8^{\circ}$ . Distance travelled remains: s = 250 mm, and the timing is done after the magnet travel first for about 5 cm as before.

The equation of motion, similar to previous section:

\sum \tau_C = 0

mg\sin\theta\, R + F_{MB} R = 0

mg\sin\theta - k_0\, d^{\,p} v^{n} = 0

v^{-n} = \frac{k_0}{mg\sin\theta}\, d^{\,p}

-n\ln v = \ln\left(\frac{k_0}{mg\sin\theta}\right) + p\ln d

Here again p can be obtained using linear regression or graphical method where we use the previously obtained value: $n = 0.96 \pm 0.05$ .

The experimental data:

d mm	$\bar{t}$ S	$V$ mm/s	$-n\ln v$	$\ln d$
4.5±0.5	13.53±0.005	18.48	-2.80	1.50
5.5	9.60	26.04	-3.13	1.70
6.5	6.70	37.31	-3.47	1.87
7.5	4.99	50.10	-3.76	2.01
8.5	3.47	72.05	-4.11	2.14
9.5	2.87	87.11	-4.29	2.25
10.5	2.14	116.82	-4.57	2.35
11.5	1.66	150.60	-4.81	2.44

Table 2. Experimental data for power factor p determination

Note:

Distance d is measured from the center of the magnet.

Plot of -n ln(v) (vertical axis, -5.0 to -2.5) against ln(d) (horizontal axis, 1.5 to 2.5): eight data points on a falling straight red fitted line, the central point at about (2.01, -3.76) with blue error bars. — Figure 6. Graph of $-n\ln(v)$ vs $\ln(d)$ . Typical error bar is shown in the central data.

From linear regression calculation we have:

p = -2.16

So the magnetic braking force is very sensitive with the magnet–conductor distance d in which the relationship is almost inversely quadratic. In brief, the further the magnet from the conductor the weaker the magnetic braking force becomes. This result has been repeated and verified by three independent persons and apparatus setups.

Error estimate of p:

Similar to previous section, we use the central data shown in Fig. 6.:

\ln d = 2.01 \pm 0.105

-n\ln\left(v\right) = -3.76 \pm 0.095

The power factor p can be obtained from the slope of line in di Fig. 6: $p = \Delta y / \Delta x$ where $y = -n\ln v$ and $x = \ln d$ .

For the data shown in Fig. 6 we obtain: $\Delta x = 0.94$ dan $\Delta y = 2.01$ , with typical error: $\delta x = 0.105$ and $\delta y = 0.095$

So the error estimate for p:

\frac{\Delta p}{p} = \sqrt{\left(\frac{\delta x}{\Delta x}\right)^2 + \left(\frac{\delta y}{\Delta y}\right)^2} = \sqrt{\left(\frac{0.105}{0.94}\right)^2 + \left(\frac{0.095}{2.01}\right)^2} = 0.12

\Delta p = 0.12\, p = 0.26

So we can conclude the result of our experiment is:

\boxed{p = -2.2 \pm 0.3}

[Marking Scheme] Experimental Question 2

Magnetic Braking on an inclined plane

(A.1) 1.0	1.0	Adjusting the orientation of the inclined plane track
(A.2) 1.0	1.0	Explanation using appropriate diagram of field and force lines
(B.1)	0.5	Recognizing v terminal
4.0	0.5	Equation of motion
	0.5	Obtain data >= 5 sets + corresponding calculations , 0.1 each data set.
	0.25	Units
	0.5	Graph
	0.25	Linear regression or graphic analysis
	0.5	Final result: $0.5 \le n \le 1.5$
	0.5	Final result: $0.9 \le n \le 1.1$
	0.5	Error analysis
(B.2)	0.5	Equation of motion
4.0	0.5	Obtain data >= 5 sets + corresponding calculations , 0.1 each data set.
	0.25	Units
	0.5	Graph
	0.25	Linear regression or graphic analysis
	0.5	Correct sign (-) for final result p
	0.5	Final result: $1.0 \le \lvert p \rvert \le 3.0$
	0.5	Final result: $1.5 \le \lvert p \rvert \le 2.5$
	0.5	Error analysis

Оригинал в Архива: exp_problems.pdf · официални решения: exp_solutions.pdf