IPhO 2005 · Theory problem 2

Absolute Measurements of Electrical Quantities

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Th 2 ABSOLUTE MEASUREMENTS OF ELECTRICAL QUANTITIES

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)

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 \(\mathcal E\) 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.

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.

Figure 1

F-1

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.

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).

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 \(N\) 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 \(R\).

Figure 2

F-2

\({}^{1}\) 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: \[ \int_0^{2\pi}\sin x\,dx =\int_0^{2\pi}\cos x\,dx =\int_0^{2\pi}\sin x\cos x\,dx=0, \] \[ \int_0^{2\pi}\sin^2x\,dx =\int_0^{2\pi}\cos^2x\,dx=\pi, \qquad \text{and later}\qquad \int x^n\,dx=\frac{1}{n+1}x^{n+1}. \]

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 \(\mathcal E\) 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 \(a\gg b\).

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.

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 \(C_1\ldots C_6\) 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.

Figure 3

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.

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.

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

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.

\({}^{2}\) Consider that the coils centres remain approximately aligned.

Use the approximations \[ \frac{1}{1\pm\beta}\approx1\mp\beta+\beta^2 \qquad\text{or}\qquad \frac{1}{1\pm\beta^2}\approx1\mp\beta^2 \] for \(\beta\ll1\), and \(\sin\theta\approx\tan\theta\) for small \(\theta\).

Official English text of the problem.

Answers

Here \(\mu_0\) is the permeability of free space and \(g\) is gravitational acceleration. No numerical parameter values are supplied, so the answers are symbolic.

  1. Taking the coil normal parallel to \(\vec B_0\) at \(t=0\), \[ \boxed{\mathcal E(t)=N\pi a^2B_0\omega\sin(\omega t)}, \qquad \boxed{\langle P\rangle=\frac{N^2\pi^2a^4B_0^2\omega^2}{2R}}. \]

  2. \[ \boxed{R=\frac{\mu_0\pi N^2a\omega}{4\tan\theta} \approx\frac{\mu_0\pi N^2a\omega}{4\theta}}, \] with \(\theta\) in radians.

  3. The magnitude of the emf between the rims is \[ \boxed{|\mathcal E_{14}|=B\omega b^2 =\frac{\mu_0NI\omega b^2}{2a}}. \]

  4. \[ \boxed{R=\frac{\mu_0N\omega b^2}{2a}}. \]

  5. \[ \boxed{F=\frac{\mu_0aI^2}{h}}, \qquad\text{directed upwards.} \]

  6. \[ \boxed{I=\sqrt{\frac{mgxh}{4\mu_0ad}}}. \]

  7. To the order of the prescribed small-displacement approximation, \[ \boxed{\delta z_{\max}\approx \frac{Ml h^2}{mxd} =\frac{Mglh^3}{4\mu_0aI^2d^2}}. \] A positive displacement returns towards equilibrium when \(0<\delta z<\delta z_{\max}\); equality is the limiting boundary.

Solution

1. Induced emf and mean mechanical power

Let \(A=\pi a^2\). Choose the positive normal to the coil as \[ \hat n(t)=\cos(\omega t)\,\hat i+\sin(\omega t)\,\hat j, \] so that it is parallel to the applied field at \(t=0\).

The magnetic flux linkage through the \(N\) turns is \[ \Lambda(t)=NAB_0\cos(\omega t). \] Faraday’s law therefore gives \[ \mathcal E(t)=-\frac{d\Lambda}{dt} =NAB_0\omega\sin(\omega t), \] or \[ \boxed{\mathcal E(t)=N\pi a^2B_0\omega\sin(\omega t)}. \] A different initial orientation changes only the phase.

Since self inductance is neglected, the instantaneous current is \[ i(t)=\frac{\mathcal E(t)}{R}. \] At constant angular speed, the mechanical power supplied against the magnetic braking torque equals the Joule heating: \[ P(t)=i^2(t)R =\frac{N^2A^2B_0^2\omega^2}{R}\sin^2(\omega t). \] Using \(\langle\sin^2(\omega t)\rangle=\tfrac12\), \[ \boxed{\langle P\rangle =\frac{N^2\pi^2a^4B_0^2\omega^2}{2R}}. \]

2. Resistance from the magnetic needle’s deflection

The field at the centre of a circular \(N\)-turn coil is \[ \vec B_{\mathrm c}(t)=\frac{\mu_0Ni(t)}{2a}\hat n(t). \] Substituting the current found in part 1 and writing \(\phi=\omega t\), \[ \vec B_{\mathrm c}(t) =\frac{\mu_0N^2\pi aB_0\omega}{2R} \left( \sin\phi\cos\phi\,\hat i+\sin^2\phi\,\hat j \right). \]

The needle changes direction negligibly during one coil revolution. Consequently, its mean torque is its magnetic moment crossed with the vector average of the total field. Since \[ \langle\sin\phi\cos\phi\rangle=0, \qquad \langle\sin^2\phi\rangle=\frac12, \] the coil’s mean field is \[ \langle\vec B_{\mathrm c}\rangle =\frac{\mu_0N^2\pi aB_0\omega}{4R}\hat j. \] It is perpendicular to the applied field.

The stationary needle aligns with \[ \langle\vec B_{\mathrm{total}}\rangle =B_0\hat i+ \frac{\mu_0N^2\pi aB_0\omega}{4R}\hat j. \] Thus \[ \tan\theta =\frac{\langle B_{\mathrm c,y}\rangle}{B_0} =\frac{\mu_0N^2\pi a\omega}{4R}, \] giving \[ \boxed{R=\frac{\mu_0\pi N^2a\omega}{4\tan\theta}}. \] For the specified small angle, \[ \boxed{R\approx\frac{\mu_0\pi N^2a\omega}{4\theta}}. \]

3. Emf between the rotating disks’ rims

The magnetic field at the centre of either surrounding coil has magnitude \[ B=\frac{\mu_0NI}{2a}. \] The large separation of the coils allows their mutual field contributions to be neglected. The condition \(a\gg b\) permits this central field to be treated as uniform across each disk.

At radius \(r\), a disk element moves with tangential speed \[ v=\omega r. \] In an axial magnetic field, the magnetic force per unit charge, \(\vec v\times\vec B\), is radial. Hence the magnitude of the motional emf from the hub to the rim of one disk is \[ \mathcal E_{\mathrm d} =\int_0^b B\omega r\,dr =\frac12B\omega b^2. \]

The disks rotate in the same sense but their magnetic fields are opposite. Their rim-to-hub emfs therefore have opposite signs. Since the hubs are joined by the conducting shaft, the rim-to-rim emfs add in magnitude: \[ |\mathcal E_{14}|=2\mathcal E_{\mathrm d}=B\omega b^2. \] Thus \[ \boxed{|\mathcal E_{14}|=\frac{\mu_0NI\omega b^2}{2a}}. \] The polarity is determined by the direction of \(\vec\omega\) relative to the axial field in each disk.

4. Resistance at galvanometer null

When the galvanometer reads zero, no current flows through the disk–shaft–galvanometer branch. There is therefore no resistive voltage drop along that branch.

The disk emf balances the voltage \(IR\) across the resistance: \[ IR=|\mathcal E_{14}| =\frac{\mu_0NI\omega b^2}{2a}. \] Cancelling the nonzero current, \[ \boxed{R=\frac{\mu_0N\omega b^2}{2a}}. \]

5. Force between \(C_1\) and \(C_2\)

For two long parallel wires separated by \(h\), each carrying current \(I\), the magnetic force per unit length is \[ f=\frac{\mu_0I^2}{2\pi h}. \] Using the prescribed approximation, this applies locally around the circumference \(2\pi a\) of the coils. The force is therefore \[ F=f(2\pi a) =\frac{\mu_0I^2}{2\pi h}(2\pi a), \] so \[ \boxed{F=\frac{\mu_0aI^2}{h}}. \]

The currents in \(C_1\) and \(C_2\) circulate in the same direction, so the interaction is attractive. Since \(C_1\) is above \(C_2\), the force on \(C_2\) is upwards.

6. Current from the balance equilibrium

The upper coil \(C_1\) attracts \(C_2\) upwards, while the lower coil \(C_3\), carrying the opposite current, repels it upwards. Each contributes the force found in part 5. Thus the total upward force on \(C_2\) is \(2F\).

Similarly, the total force on \(C_5\) is \(2F\) downwards. These two forces produce torques in the same rotational sense: \[ \tau_{\mathrm{mag}}=(2F)d+(2F)d=4Fd. \]

The added mass provides the opposing torque \(mgx\). At equilibrium, \[ mgx=4Fd=\frac{4\mu_0aI^2d}{h}. \] Consequently, \[ \boxed{I=\sqrt{\frac{mgxh}{4\mu_0ad}}}. \]

7. Maximum displacement that produces a restoring torque

Write \(z=\delta z\), positive when \(C_2\) rises and \(C_5\) falls. Let \(\alpha\) be the corresponding tilt of the balance, so that \[ z=d\sin\alpha. \] We retain the assumed alignment of the coil centres and treat the current as fixed.

Define \[ K=\mu_0aI^2. \] After the displacement, the separations between \(C_2\) and its upper and lower fixed coils are \(h-z\) and \(h+z\). Both magnetic forces remain upwards, so their sum is \[ F_{\mathrm L}(z) =\frac{K}{h-z}+\frac{K}{h+z} =\frac{2Kh}{h^2-z^2}. \] On the right, the two separations are interchanged; the total downward force has the same magnitude.

Taking positive torque to mean a tendency to increase \(z\), the total magnetic torque is \[ \tau_{\mathrm{mag}} =2d\cos\alpha\,F_{\mathrm L}(z) =\frac{4Khd}{h^2-z^2}\cos\alpha. \]

The added mass supplies the opposing torque \[ \tau_m=-mgx\cos\alpha. \] Because the balance’s centre of mass lies a distance \(l\) below its fulcrum, its weight supplies the restoring torque \[ \tau_M=-Mgl\sin\alpha. \] The equal weights of the hanging parts cancel in the torque balance.

Hence the net torque is \[ \tau(z) =\left(\frac{4Khd}{h^2-z^2}-mgx\right)\cos\alpha -Mgl\sin\alpha. \] From the equilibrium condition in part 6, \[ mgx=\frac{4Kd}{h}. \] Substitution gives \[ \tau(z) =mgx\frac{z^2}{h^2-z^2}\cos\alpha -Mgl\sin\alpha. \]

For small tilt, \(\cos\alpha>0\), so the torque has the same sign as \[ \frac{\tau(z)}{\cos\alpha} =mgx\frac{z^2}{h^2-z^2}-Mgl\tan\alpha. \] Using the prescribed approximations, \[ \frac{z^2}{h^2-z^2}\approx\frac{z^2}{h^2}, \qquad \tan\alpha\approx\sin\alpha=\frac{z}{d}, \] we obtain \[ \frac{\tau(z)}{\cos\alpha} \approx \frac{mgx}{h^2}z^2-\frac{Mgl}{d}z. \]

For \(z>0\), the balance initially moves back towards equilibrium if \(\tau(z)<0\). Therefore \[ \frac{mgx}{h^2}z<\frac{Mgl}{d}, \] or \[ z<\frac{Ml h^2}{mxd}. \] The limiting displacement is thus \[ \boxed{\delta z_{\max}\approx\frac{Ml h^2}{mxd}}. \] Using the equilibrium relation once more, \[ \boxed{\delta z_{\max}\approx \frac{Mglh^3}{4\mu_0aI^2d^2}}. \]

At this boundary the net torque vanishes to the retained order; a larger positive displacement produces a torque away from equilibrium. For a negative small displacement, both terms in the approximate torque point towards increasing \(z\), initially returning the balance towards equilibrium.

The stated limit is the leading small-displacement result, requiring \(\delta z_{\max}\ll h\) and \(\delta z_{\max}\ll d\) for quantitative use of the approximations.

The solution was checked against the official solution.