Съдържание

IPhO 2005, Теория — Задача 2. ABSOLUTE MEASUREMENTS OF ELECTRICAL QUANTITIES

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Всички материали от IPhO 2005

36th International Physics Olympiad. Salamanca (España) 2005 · 10 т.

Условие

The technological and scientific transformations underwent during the XIX century produced a compelling need of universally accepted standards for the electrical quantities. It was thought the new absolute units should only rely on the standards of length, mass and time established after the French Revolution. An intensive experimental work to settle the values of these units was developed from 1861 until 1912. We propose here three case studies.

Marks are indicated at the beginning of each subquestion, in parenthesis.

Determination of the ohm (Kelvin)

Circular coil rotating about the vertical Z axis with angular velocity ω (curved arrow at the top); the field B0 points along the X axis; a small magnetic needle at the centre of the coil makes an angle θ with the X direction (dashed line). — F-1

A closed circular coil of $N$ turns, radius $a$ and total resistance $R$ is rotated with uniform angular velocity $\omega$ about a vertical diameter in a horizontal magnetic field $\vec{B}_0 = B_0\,\vec{i}$ .

1. (0.5+1.0) Compute the electromotive force $\varepsilon$ induced in the coil, and also the mean power $^{1}$ $\langle P\rangle$ required for maintaining the coil in motion. Neglect the coil self inductance. [1,5 т.]

A small magnetic needle is placed at the center of the coil, as shown in Figure F-1. It is free to turn slowly around the Z axis in a horizontal plane, but it cannot follow the rapid rotation of the coil.

2. (2.0) Once the stationary regime is reached, the needle will set at a direction making a small angle $\theta$ with $\vec{B}_0$ . Compute the resistance $R$ of the coil in terms of this angle and the other parameters of the system. [2 т.]

Lord Kelvin used this method in the 1860s to set the absolute standard for the ohm. To avoid the rotating coil, Lorenz devised an alternative method used by Lord Rayleigh and Ms. Sidgwick, that we analyze in the next paragraphs.

Determination of the ohm (Rayleigh, Sidgwick).

Rayleigh–Sidgwick apparatus: a motor turns the conducting shaft S–S' at angular velocity ω; disks D and D' on the shaft sit inside coils C and C'; brush contacts 4 (rim of D) and 1 (rim of D'), shaft points 3 and 2; resistor R, galvanometer G and a battery in the external circuit; current I flows through the coils. — F-2

The experimental setup is shown in Figure F-2. It consists of two identical metal disks D and D' of radius $b$ mounted on the conducting shaft SS'. A motor rotates the set at an angular velocity $\omega$ , which can be adjusted for measuring $R$ . Two identical coils C and C' (of radius $a$ and with $\boldsymbol{N}$ turns each) surround the disks. They are connected in such a form that the current $I$ flows through them in opposite directions. The whole apparatus serves to measure the resistance $R$ .

3. (2.0) Assume that the current $I$ flowing through the coils C and C' creates a uniform magnetic field $B$ around D and D', equal to the one at the centre of the coil. Compute $^{1}$ the electromotive force $\varepsilon$ induced between the rims 1 and 4, assuming that the distance between the coils is much larger than the radius of the coils and that $a \gg b$ . [2 т.]

The disks are connected to the circuit by brush contacts at their rims 1 and 4. The galvanometer G detects the flow of current through the circuit 1-2-3-4.

4. (0.5) The resistance $R$ is measured when G reads zero. Give $R$ in terms of the physical parameters of the system. [0,5 т.]

Determination of the ampere

Passing a current through two conductors and measuring the force between them provides an absolute determination of the current itself. The “Current Balance” designed by Lord Kelvin in 1882 exploits this method. It consists of six identical single turn coils C $_1$ … C $_6$ of radius $a$ , connected in series. As shown in Figure F-3, the fixed coils C $_1$ , C $_3$ , C $_4$ , and C $_6$ are on two horizontal planes separated by a small distance $2h$ . The coils C $_2$ and C $_5$ are carried on balance arms of length $d$ , and they are, in equilibrium, equidistant from both planes.

The current $I$ flows through the various coils in such a direction that the magnetic force on C $_2$ is upwards while that on C $_5$ is downwards. A mass $m$ at a distance $x$ from the fulcrum O is required to restore the balance to the equilibrium position described above when the current flows through the circuit.

Kelvin current balance: beam with fulcrum O, centre of mass G at distance l below O, mass m at distance x from O, arms of length d on each side; left stack of coils C1, C2, C3 and right stack C6, C5, C4 separated by distances h; forces F upward on C2 and downward on C5; current I, a resistor and a battery in series. — F-3

5. (1.0) Compute the force $F$ on C $_2$ due to the magnetic interaction with C $_1$ . For simplicity assume that the force per unit length is the one corresponding to two long, straight wires carrying parallel currents. [1 т.]

6. (1.0) The current $I$ is measured when the balance is in equilibrium. Give the value of $I$ in terms of the physical parameters of the system. The dimensions of the apparatus are such that we can neglect the mutual effects of the coils on the left and on the right. [1 т.]

Let $M$ be the mass of the balance (except for $m$ and the hanging parts), G its centre of mass and $l$ the distance $\overline{\mathrm{OG}}$ .

7. (2.0) The balance equilibrium is stable against deviations producing small changes $\delta z$ in the height of C $_2$ and $-\delta z$ in C $_5$ . Compute $^{2}$ the maximum value $\delta z_{\max}$ so that the balance still returns towards the equilibrium position when it is released. [2 т.]

$^{1}$ The mean value $\langle X\rangle$ of a quantity $X(t)$ in a periodic system of period $T$ is $\langle X\rangle = \frac{1}{T}\int_0^T X(t)\,dt$

You may need one or more of these integrals:

\int_0^{2\pi}\sin x\,dx = \int_0^{2\pi}\cos x\,dx = \int_0^{2\pi}\sin x\cos x\,dx = 0\,,\quad \int_0^{2\pi}\sin^2 x\,dx = \int_0^{2\pi}\cos^2 x\,dx = \pi\,,\quad \text{and later}\quad \int x^n\,dx = \frac{1}{n+1}x^{n+1}

$^{2}$ Consider that the coils centres remain approximately aligned.

Use the approximations $\frac{1}{1\pm\beta} \approx 1\mp\beta+\beta^2$ or $\frac{1}{1\pm\beta^2} \approx 1\mp\beta^2$ for $\beta \ll 1$ , and $\sin\theta \approx \tan\theta$ for small $\theta$ .

COUNTRY CODE	STUDENT CODE	PAGE NUMBER	TOTAL No OF PAGES

Th 2 ANSWER SHEET

Question	Analytical results	Marking guideline
1	$\varepsilon =$ $\langle P\rangle =$	1.5
2	$R =$	2.0
3	$\varepsilon =$	2.0
4	$R =$	0,5
5	$F =$	1.0
6	$I =$	1.0
7	$\delta z_{\max} =$	2.0

Отговори

Решение

1. After some time $t$ , the normal to the coil plane makes an angle $\omega t$ with the magnetic field $\vec{B}_0 = B_0\vec{i}$ . Then, the magnetic flux through the coil is

\phi = N\,\vec{B}_0\cdot\vec{S}

where the vector surface $\vec{S}$ is given by $\vec{S} = \pi a^2\left(\cos\omega t\,\vec{i} + \sin\omega t\,\vec{j}\right)$

Therefore $\phi = N\pi a^2 B_0\cos\omega t$

The induced electromotive force is

\varepsilon = -\frac{d\phi}{dt} \quad\Rightarrow\quad \boxed{\varepsilon = N\pi a^2 B_0\omega\sin\omega t}

The instantaneous power is $P = \varepsilon^2/\mathrm{R}$ , therefore

\boxed{\langle P\rangle = \frac{\left(N\pi a^2 B_0\omega\right)^2}{2R}}

where we used $\langle\sin^2\omega t\rangle = \frac{1}{T}\int_0^T \sin^2\omega t\,dt = \frac{1}{2}$

2. The total field at the center the coil at the instant $t$ is

\vec{B}_t = \vec{B}_0 + \vec{B}_i

where $\vec{B}_i$ is the magnetic field due to the induced current $\vec{B}_i = B_i\left(\cos\omega t\,\vec{i} + \sin\omega t\,\vec{j}\right)$

with $B_i = \frac{\mu_0 N I}{2a}$ and $I = \varepsilon / R$

Therefore $B_i = \frac{\mu_0 N^2\pi a B_0\omega}{2R}\sin\omega t$

The mean values of its components are

\langle B_{ix}\rangle = \frac{\mu_0 N^2\pi a B_0\omega}{2R}\langle\sin\omega t\cos\omega t\rangle = 0

\langle B_{iy}\rangle = \frac{\mu_0 N^2\pi a B_0\omega}{2R}\langle\sin^2\omega t\rangle = \frac{\mu_0 N^2\pi a B_0\omega}{4R}

And the mean value of the total magnetic field is

\langle\vec{B}_t\rangle = B_0\,\vec{i} + \frac{\mu_0 N^2\pi a B_0\omega}{4R}\,\vec{j}

The needle orients along the mean field, therefore

\tan\theta = \frac{\mu_0 N^2\pi a\omega}{4R}

Finally, the resistance of the coil measured by this procedure, in terms of $\theta$ , is

\boxed{R = \frac{\mu_0 N^2\pi a\omega}{4\tan\theta}}

3. The force on a unit positive charge in a disk is radial and its modulus is

\left|\vec{v}\times\vec{B}\right| = vB = \omega rB

where $B$ is the magnetic field at the center of the coil

B = N\frac{\mu_0 I}{2a}

Then, the electromotive force (e.m.f.) induced on each disk by the magnetic field $B$ is

\varepsilon_D = \varepsilon_{D'} = B\omega\int_0^b r\,dr = \frac{1}{2}B\omega b^2

Finally, the induced e.m.f. between 1 and 4 is $\varepsilon = \varepsilon_D + \varepsilon_{D'}$

\boxed{\varepsilon = N\frac{\mu_0 b^2\omega I}{2a}}

4. When the reading of G vanishes, $I_G = 0$ and Kirchoff laws give an immediate answer. Then we have

\varepsilon = IR \quad\Rightarrow\quad \boxed{R = N\frac{\mu_0 b^2\omega}{2a}}

5. The force per unit length $f$ between two indefinite parallel straight wires separated by a distance $h$ is.

f = \frac{\mu_0}{2\pi}\frac{I_1 I_2}{h}

for $I_1 = I_2 = I$ and length $2\pi a$ , the force $F$ induced on C $_2$ by the neighbor coils C $_1$ is

\boxed{F = \frac{\mu_0 a}{h}I^2}

6. In equilibrium

mgx = 4Fd

Then

mgx = \frac{4\mu_0 ad}{h}I^2 \qquad (1)

so that

\boxed{I = \left(\frac{mghx}{4\mu_0 ad}\right)^{1/2}}

7. The balance comes back towards the equilibrium position for a little angular deviation $\delta\varphi$ if the gravity torques with respect to the fulcrum O are greater than the magnetic torques.

Mgl\sin\delta\varphi + mgx\cos\delta\varphi > 2\mu_0 aI^2\left(\frac{1}{h-\delta z} + \frac{1}{h+\delta z}\right)d\cos\delta\varphi

Tilted balance beam rotated by a small angle δφ about the fulcrum O: mass m (weight mg) at distance x from O, arm d, centre of mass G at distance l below O with weight Mg; the left moving coil rises by δz, giving gaps h - δz and h + δz to the fixed coils, and on the right the gaps are h + δz and h - δz; arrows show the magnetic forces on the moving coils.

Therefore, using the suggested approximation

Mgl\sin\delta\varphi + mgx\cos\delta\varphi > \frac{4\mu_0 adI^2}{h}\left(1 + \frac{\delta z^2}{h^2}\right)\cos\delta\varphi

Taking into account the equilibrium condition (1), one obtains

Mgl\sin\delta\varphi > mgx\frac{\delta z^2}{h^2}\cos\delta\varphi

Finally, for $\tan\delta\varphi \approx \sin\delta\varphi = \frac{\delta z}{d}$

\delta z \lt \frac{Mlh^2}{mxd} \quad\Rightarrow\quad \boxed{\delta z_{\max} = \frac{Mlh^2}{mxd}}

Th 2 ANSWER SHEET

Question	Basic formulas and ideas used	Analytical results	Marking guideline
1	$\Phi = N\vec{B}_0\cdot\vec{S}$ ; $\varepsilon = -\frac{d\Phi}{dt}$ ; $P = \frac{\varepsilon^2}{R}$	$\varepsilon = N\pi a^2 B_0\omega\sin\omega t$ ; $\langle P\rangle = \frac{\left(N\pi a^2 B_0\omega\right)^2}{2R}$	0.5; 1.0
2	$\vec{B} = \vec{B}_0 + \vec{B}_i$ ; $B_i = \frac{\mu_0 N}{2a}I$ ; $\tan\theta = \frac{\langle B_y\rangle}{\langle B_x\rangle}$	$R = \frac{\mu_0 N^2\pi a\omega}{4\tan\theta}$	2.0
3	$\vec{E} = \vec{v}\times\vec{B}$ ; $v = \omega r$ ; $B = N\frac{\mu_0 I}{2a}$ ; $\varepsilon = \int_0^b \vec{E}\,d\vec{r}$	$\varepsilon = N\frac{\mu_0 b^2\omega I}{2a}$	2.0
4	$\varepsilon = RI$	$R = N\frac{\mu_0 b^2\omega}{2a}$	0,5
5	$f = \frac{\mu_0}{2\pi}\frac{II'}{h}$	$F = \frac{\mu_0 a}{h}I^2$	1.0
6	$mgx = 4Fd$	$I = \left(\frac{mghx}{4\mu_0 ad}\right)^{1/2}$	1.0
7	$\Gamma_{grav} > \Gamma_{mag}$	$\delta z_{\max} = \frac{Mlh^2}{mxd}$	2.0

Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf