Чтобы найти сопротивления элементов $a$, $b$, $c$ и $d$:
Решая систему из полученных уравнений $(1)-(4)$, получим:
Выберем, к примеру, элемент $a$.
Метод 1
То же, что и в B1, с различными значениями тока $I$ в круглой катушке.
Метод 2
Соберём мостовую схему, подключив датчик с источнику питания, как показано на рисунке ниже:
Пусть при $I=0$ мост сбалансирован, так что $\Delta U=0$. Если установить в катушке ток $I$, сопротивление элемента $a$ станет равным $R+\Delta R$, и:\[\Delta U=\frac{\mathcal E\cdot R}{R+R+\Delta R}-\frac{\mathcal E}{2}\implies \delta(B)=\frac{\Delta R}{R}\approx -\frac{\Delta U}{\mathcal E/4}.\]Если же при $I=0$ мост не балансирован, так что $\Delta U=\Delta U_0$, то:\[\frac{\Delta R}{R}=-\frac{\Delta U-\Delta U_0}{\mathcal E/4}\implies\delta (B)=\frac{\Delta R}{R}\approx -\frac{\Delta U-\Delta U_0}{\mathcal E/4}.\]
Максимальное значение $\Delta R/R$ в эксперименте составляет около $10\,\%$. Погрешность результата при использовании мостовой схемы – порядка $1\,\%$.
$\mathcal E=6300~мВ$
$I,~мА$ | $B,~мТл$ | $\Delta U,~мВ$ | $(\Delta U-\Delta U_0),~мА$ | $\delta(B)$ |
0 | 0 | -25.8 | 0 | 0 |
10 | 0.0628 | -21 | 4.8 | -0.00305 |
20 | 0.126 | -15.7 | 10.1 | -0.00641 |
45 | 0.283 | -2.1 | 23.7 | -0.01504 |
67 | 0.421 | 11.1 | 36.9 | -0.02343 |
87 | 0.546 | 24.5 | 50.3 | -0.03193 |
107 | 0.672 | 38.1 | 63.9 | -0.04057 |
129 | 0.810 | 54 | 79.8 | -0.05067 |
156 | 0.980 | 74 | 99.8 | -0.06336 |
186 | 1.168 | 96 | 121.8 | -0.07733 |
215 | 1.350 | 117.3 | 143.1 | -0.09085 |
240 | 1.507 | 134.5 | 160.3 | -0.10177 |
268 | 1.683 | 152.6 | 178.4 | -0.11326 |
303 | 1.903 | 170.6 | 196.4 | -0.12469 |
330 | 2.072 | 179.6 | 205.4 | -0.13041 |
354 | 2.223 | 184.1 | 209.9 | -0.13326 |
384 | 2.411 | 186.2 | 212 | -0.13460 |
405 | 2.543 | 186.7 | 212.5 | -0.13492 |
436 | 2.738 | 187.1 | 212.9 | -0.13517 |
469 | 2.945 | 187.2 | 213 | -0.13523 |
По графику находим:
Сопротивления $r$ и $R$ вычисляются по формулам:\[r=R_0-\sqrt{R_0(R_0-R_B)},\qquad R=R_0+\sqrt{R_0(R_0-R_B)}.\]Для элемента $a$ получим:
$I$ $B$ $S$ $I$ $B$ $S$
Найдите коэффициент наклона $m = \cfrac{\Delta S}{\Delta B}$ на этом участке.
$S,~ мВ$ | 0 | 91.5 | 183 | 274 | 365 |
$\mathcal E,~ В $ | 0 | 1.51 | 3.1 | 4.6 | 6.25 |
Изобразите схему экспериментальной установки. Приведите формулы для определения значения $n$ исходя из данных, которые вы измерите.
3. With flux concentrator For each value of $L_1$, do the same, to obtain $\Delta B=\frac{|\Delta S|}{m}$.
Table to find $B/B_0$ for different values of $L_1\cdot B/B_0=\Delta S/\Delta S_0$ $S _1=17~мВ$; $\Delta S_0=21.2-17=4.2~мВ$
$L_1~(мм)$ | $S~(мВ)$ | $1/ L_1 (мм)^{-1})$ | $\Delta S=S_2-S_1$ | $B/B_0$ |
5 | 33.2 | 0.200 | 16.2 | 3.86 |
6 | 31.2 | 0.167 | 14.2 | 3.38 |
7 | 30.2 | 0.143 | 13.2 | 3.14 |
8 | 28.6 | 0.125 | 11.6 | 2.76 |
9 | 27.7 | 0.111 | 10.7 | 2.55 |
10 | 26.8 | 0.100 | 9.8 | 2.33 |
11 | 26.4 | 0.0909 | 9.4 | 2.24 |
13 | 25.4 | 0.0769 | 8.4 | 2.00 |
15 | 24.6 | 0.0667 | 7.6 | 1.81 |
$\infty$ | 21.2 | 0.0000 | 4.2 | 1.00 |
Определите значение $n$.
Diagrams of the experiment and expressions for calculating $B_h$. 1. The sensor on the round plate in the horizontal plane. Carry out the biasing. 2. Method 1
a. Set $\alpha=0$ – the sensor perpendicular to the direction South - North. b. Rotate the sensor holder, measure $S=f(\alpha)$. c. Fit the curve $S$ to a sine function $S=a\sin\alpha$. d. $B_h=\frac{a}{m}$.
3. Method 2
a. Orient the sensor along the Earth’s magnetic field. Find the direction with the maximum (or minimum) value of $S$. Note this value $S_1$. b. Rotate the sensor holder by about $180~^{\circ}$. Find the direction with the maximum the minimum (or maximum) value of $S$. Note this value $S_2$.
Diagrams of the experiment and expressions for calculating $B_{Earth}$ and $\theta$. 1. The sensor on the round plate in the vertical plane containing the South - North direction. Carry out the biasing. 2. Method 1
a. Orient the sensor along the Earth’s magnetic field. Find the direction with the maximum (or minimum) value of $S$. Note this value $S_1$ and the angle $\alpha_1$ between the sensor direction and the horizontal. b. Rotate the sensor holder by about $180~^{\circ}$. Find the direction with the minimum (or maximum) value of $S$. Note this value $S_2$ and the angle $\alpha_2$ between the sensor direction and the horizontal. c. Orient the sensor in the direction midway between $\alpha_1$ and $\alpha_2$ with the angle $\alpha_3=\alpha_2+90~^{\circ}$. Note the value $S_3$. d. Starting from $\alpha_3$, rotate the sensor holder, take the values of $S$ corresponding to values of $\alpha$. Measure $S=f(\alpha)$. e. $S-S_3=a\sin\alpha$. Obtain a from fitting. f. $B_{Earth}= a/m$. g. $\theta=\operatorname{Arccos}\frac{B_h}{B_{Earth}}$.
3. Method 2 Orient the sensor along the Earth’s magnetic field. Find the direction with the maximum (or minimum) value of $S$. The angle $\theta$ between the sensor direction and the horizontal is the magnetic inclination.
From the obtained $\theta,~~B_{Earth}=B_h /\cos\theta$. This method may have systematic errors due to the relative misalignment of the sensor to the sensor holder. To eliminate this error, rotate the round plate together with the sensor holder by $180~^{\circ}$ about a horizontal axis along the South-North direction. Repeat the measurement. The magnetic inclination is the mean value of the two obtained angles.
$I,~А$ | 0.30 | 0.35 | 0.40 | 0.45 | 0.50 | 0.543 | 0.20 |
$U,~В$ | 2.64 | 3.9 | 5.37 | 6.94 | 8.67 | 10.29 | 0.89 |
$P,~Вт$ | 0.792 | 1.365 | 2.15 | 3.12 | 4.34 | 5.59 | 0.178 |
$S,~мВ$ | 18.3 | 42 | 74.3 | 112.4 | 162.4 | 215.4 | 4.9 |
$I,~А$ | 0.25 | 0.50 | 0.60 | 0.70 | 0.80 | 0.97 | 0.30 | 0.442 |
$U,~В$ | 1.53 | 1.3 | 2.13 | 3.1 | 4.1 | 6.11 | 3.13 | 7.74 |
$P,~Вт$ | 0.382 | 0.65 | 1.28 | 2.17 | 3.28 | 5.92 | 0.939 | 3.42 |
$S,~мВ$ | 11.5 | 25.9 | 50.7 | 88.1 | 137 | 253 | 31.4 | 128 |