IPhO 2025, Експеримент — Задача 1. Earth's magnetic field measurement

SOLUTION:

Please note that the numerical results given in the solution come from a single, consistent measurement session. The measurement ranges used in the notation take into account several measurements by various testers.

Marking scheme: students are not penalized for forcing a linear fit to pass through the origin when the studied law is proportional.

Part A. Gouy balance and magnetic moment

Modeling

We assume that a magnet can be treated as a magnetic dipole of magnetic moment $\vec{m}_{\mathrm{m}}$. The force experienced by such a dipole of magnetic moment $\vec{m}_{\mathrm{m}} = m_{\mathrm{m}}\vec{e}_z$ in a magnetic field $\vec{B} = B(z)\vec{e}_z$ is

When an electric current $i$ flows through the anti-Helmholtz coils, the field $\vec{B}$ along the unit vector $\vec{e}_z$ of revolution axis is

This equation is only valid near the center of the device, denoted by $z = z_0$.

Magnetic field in the coils

A.1

SOLUTION:

The 9V battery capacity is $Q = I \cdot \Delta t = 300\ \mathrm{mA \cdot h}$. Using an electric current $I = 2\ \mathrm{A}$, the time of use is $0.3 \times 3600/2 \approx 540\ \mathrm{s} \approx 9\ \mathrm{min}$.

This result must be taken into account when developing the protocols later on, knowing that the coils are only used in part A. Note that a spare battery is available if required.

Insert the magnetic field sensor into the coils, as shown in Fig 2. See also this figure for the identification of the sensor position in the coils.

A.2

SOLUTION:

The plot below is obtained at $i_0 = 1.0\ \mathrm{A}$. At the centre of the device is a zone in which the field is a linear function of position. At the edges of the device, you can see the saturation of the classical field as you approach the two coils. The zone of linearity from the figure is $[0.015\ ;\ 0.032]\ \mathrm{m}$.

[p1-sol-fig1]

A.3

SOLUTION:

The centre of the linear zone is around 23 mm. The values of the magnetic field at two positions $z_1 = 13\ \mathrm{mm}$ and $z_2 = 33\ \mathrm{mm}$ are measured for several electric current values. This allows to compute the gradient $\frac{B(z_2) - B(z_1)}{z_2 - z_1}$ of the magnetic field as a function of electric current.

[p1-sol-fig2]

We have a linear evolution. The possible residual y-intercept may be due to the fact that the sensor is not correctly calibrated. A typical value for the slope gives $\alpha = 0.150 \pm 0.007\ \mathrm{T \cdot m^{-1} \cdot A^{-1}}$.

Gouy balance

Remove the magnetic field sensor from the coils, and carefully place the force sensor inside, as described in Fig. 2, with particular attention to the placement of electrical wires in the gutters.

A.4 Perform experimental measurements of the gram-force $m_\mathrm{f}$ as a function of current $i$. Draw an appropriate plot to determine the value of the magnetic moment $m_\mathrm{m}$ of the magnet, with its uncertainty. — 0.8pt

SOLUTION:

We vary the electric current $i$ and measure the effective mass, which gives

[p1-sol-fig3]

The slope of the curve is 5.48 gram-force/A, giving a slope of $(53.7 \pm 0.8)\times10^{-3}\ \mathrm{N/A}$. Finally, the magnetic moment is $m_m = \dfrac{53.7\times10^{-3}}{0.150} = 0.358\ \mathrm{A\cdot m^2}$. The uncertainty is obtained from $\dfrac{\delta m}{m} = \dfrac{\delta p}{p} + \dfrac{\delta\alpha}{\alpha} = 0.05$.

The magnetic moment $m_m = 0.36 \pm 0.02\ \mathrm{A\cdot m^2}$.

Measurements of force in newton (N) are accepted.

Alternative measurement of the magnetic moment

In the dipolar approximation, the magnetic field of a magnet of magnetic moment $m_\mathrm{m}$ on its revolution axis $z$ is

where $z_\mathrm{a}$ is not necessarily the geometric center of the magnet, and where $\mu_0 = 4\pi\,10^{-7}\ \mathrm{H\cdot m^{-1}}$.

A.5 Measure the magnetic field $B_z$ along the revolution axis of the free magnet, as a function of distance $z$. Draw a curve to verify the model given Eq. (3), showing its experimental deviations. Deduce a new value for $m_\mathrm{m}$, with uncertainty. — 1.3pt

SOLUTION:

The field B is measured directly by sticking the third magnet on the graduated ruler. You can also use the Hall sensor directly. The measurements are shown below, where the position is plotted as a function of $B^{-1/3}$ (see figure below).

[p1-sol-fig4]

The slope is $\left(\dfrac{\mu_0 m_m}{2\pi}\right)^{1/3} = (4,1 \pm 0.1)\times10^{-3}\ \mathrm{m\cdot T^{1/3}}$, and then the new value of the magnetic moment is $m_m = 0,31 \pm 0.01\ \mathrm{A\cdot m^2}$.

A.6 Given the two results obtained in A.4 and A.5, propose a final experimental value of $m_\mathrm{m}$ with its uncertainty. — 0.2pt

SOLUTION:

The final value is given by the averaged value of the previous measurements, so $m_m = 0,33 \pm 0,01\ \mathrm{A\cdot m^2}$.

Part B. Determining the earth's magnetic field

Modeling

We now study the oscillating motion of the magnet in a horizontal plane to estimate the value of the horizontal component $B_\mathrm{e}$ of the Earth's magnetic field, see Fig. 3 and the assembly instructions above Fig.3. The pod (f3), containing the magnet, is subjected to two torques around the vertical axis:

• the torque of the wire, modeled as $\Gamma_\mathrm{f} = -\dfrac{C_\mathrm{f}}{L}(\theta-\theta_0)$, where $C_\mathrm{f}$ is a constant and $L$ the total length between the two attachments of the wire, and $\theta_0$ corresponds to the angle for which the wire is not twisted,
• the torque of the Earth's magnetic fields, given by $\Gamma_\mathrm{e} = -m_\mathrm{m} B_\mathrm{e}\sin(\theta-\theta_\mathrm{e})$, when the angular position of the Earth's magnetic field is given by the angle $\theta_\mathrm{e}$.

Denoting $J$ the unknown moment of inertia of the pod and magnet assembly around the vertical axis, the angular momentum theorem gives

When the $\sin(\theta-\theta_\mathrm{e}) \simeq \theta-\theta_\mathrm{e}$ approximation is valid, this leads to an sinusoidal oscillation at a period $T$. For this part, adhesive paste (l) is moldable into any shape or size and attachable to other devices.

Caution: To avoid disturbance from external magnetic fields, the magnet must be placed at least 20 cm away from any metal object or magnetic source (including the other magnets).

Experimental set-up and first measurement

For questions B.1 to B.5, set the length of the wire to $L$ = 34 cm and make sure that it is not twisted. In this setting, we begin by assuming that the torque from the wire is negligible with respect to the torque from the Earth's magnetic field, a hypothesis to which we will return later.

To align $\theta_0$ with $\theta_\mathrm{e}$, use piece (f2a) to adjust $\theta_0$ so the pod (f3) does not rotate when the magnet is removed. Then reinsert the magnet in the pod, and keep $\theta_0$ unchanged until question B.5.

B.1 Propose an experimental protocol to determine $B_\mathrm{e}$. Introduce the different quantities you will measure and their units. Depict these quantities on a detailed schematic, and relate them to those given in the instructions through an equation. For each quantity, specify whether it is fixed (F) or varies (V) throughout the protocol. — 0.3pt

SOLUTION:

The figure below describes the proposed experiment. The period $T$ for small oscillations is measured (with best precision using several periods). Since the inertial moment $J_0$ of the pod is unknown, adding sticky paste to both ends of the pod allows the change of the inertial moment $J = J_0 + \Delta J$. The length of the pod arm is $r_a = 0.04$ m.

The differential equation verified by the pod at small angles is $\dfrac{\Delta J}{m_m} = \dfrac{1}{\omega^2}B_e - \dfrac{J_0}{m_m}$. The period is $T = \dfrac{2\pi}{\omega}$ and the variation of inertial moment by adding a total mass of sticky paste $2m_a$ is $\Delta J = 2m_a.r_a^2$. Therefore $\dfrac{2m_a r_a^2}{m_m} = \dfrac{T^2}{4\pi^2}B_e - \dfrac{J_0}{m_m}$.

[p1-sol-fig5]

Note that the grading scheme will be evaluated as follows to take into account alternative protocols: a) If the students propose any of the quantities that already appear in the grading scheme, they will have the related points. b) If the students propose any protocol that works, and properly introduce the relevant quantities, they will have the entire points for the questions. c) No point will be given for additional quantities that do not correspond to a working protocol.

B.2 Using the protocol described above, draw a graph to determine a first value of $B_e$, with its uncertainty. — 1.1pt

SOLUTION:

[p1-sol-fig6]

Since $\dfrac{2m_a r_a^2}{m_m} = \dfrac{T^2}{4\pi^2}B_e - \dfrac{J_0}{m_m}$, plotting $T^2$ versus $m_a$ should give a linear function of slope $p = \dfrac{8\pi^2 r_a^2}{m_m B_e}$.

One finds a slope $p = 8640 \pm 200\ \mathrm{A^{-1}\cdot T^{-1}}$. Using the value $m_m = 0,31 \pm 0.01\ \mathrm{A\cdot m^2}$, one obtains $B_e = 47\ \mu\mathrm{T}$. the relative uncertainty is $\dfrac{200}{8640} + \dfrac{0,01}{0,305} \approx 0,06$. Therefore $B_e = (47 \pm 3)\ \mu\mathrm{T}$.

Evaluation of the torque from the wire

B.3 Keeping $L$ = 34 cm, study the motion of the pod without the magnet, and determine the value of $C_\mathrm{f}$, with experimental uncertainty: perform one period measurement for two system configurations. Specify the equation relating $C_\mathrm{f}$ to the measured quantities. — 0.7pt

SOLUTION:

The magnet is removed. The period of oscillation therefore depends on the torque due to the twisting of the wire and the moment of inertia. As the moment of inertia of the cradle remains unknown, we can measure the period $T_1$ without sticky paste and the period $T_2$ with sticky paste, for a fixed length of wire. The equations involved are

Taking two measures, for $m_a = 0$ and $m_a = 2{,}6\ \mathrm{g}$, one finds $T_1 = 4.2 \pm 0.2\ \mathrm{s}$ and $T_2 = 15.8 \pm 0.2\ \mathrm{s}$, and therefore $C_f = \dfrac{8\pi^2 L m_a r_a^2}{T_2^2 - T_1^2}$, so $C_f = (5.1 \pm 0.2) \times 10^{-7}\ \mathrm{N}\cdot\mathrm{m^2/rad}$ ( $\mathrm{m^3 \cdot kg \cdot s^{-2}}$).

B.4 (0.3pt) Using previous measurements, give the expression and determine numerically the critical length $L_\mathrm{c}$ for which the amplitude factors $C_\mathrm{f}/L$ and $m_\mathrm{m} B_\mathrm{e}$ of the $\Gamma_\mathrm{f}$ and $\Gamma_\mathrm{e}$ torques are equal. In question B.2, what was the ratio $(C_\mathrm{f}/L)/(m_\mathrm{m} B_\mathrm{e})$? Choose from the intervals: $[0\ \%,\ 1\ \%[$ ; $[1\ \%,\ 5\ \%[$ ; $[5\ \%,\ 20\ \%[$ ; $[20\ \%,\ 50\ \%[$ ; $[50\ \%,\ \infty\ \%[$.

SOLUTION:

The two previous torques are equalized, so $\dfrac{C_f}{L_c} = m_m B_e$ and then $L_c = \dfrac{C_f}{m_m B_e} = \dfrac{5.1 \times 10^{-7}}{0.33 \times 47 \times 10^{-6}} = 3.2\ \mathrm{cm}$. The ratio between the torque at 34cm and the critical torque at 3.2cm is therefore 3.2/34=9%. So the answer is $[5\ ;\ 20\%[$.

Static regime measurement

We now propose a static measurement of the Earth's magnetic field. Reinsert the magnet into the pod. Use piece (f2a) in Fig. 3 to adjust the angular position $\theta_0$, causing the wire to twist.

B.5 (1.1pt) Still at a fixed length of $L = 34\ \mathrm{cm}$, draw an appropriate plot to study how the equilibrium position of the magnet $\theta_\mathrm{eq}$ depends on the angle $\theta_0$, and determine a second value of $B_\mathrm{e}$, with its uncertainty.

SOLUTION:

According to the equation of motion in an equilibrium situation, we have : $\dfrac{C_f(\theta_{eq} - \theta_0)}{L} = -m_m B_e \sin(\theta_{eq})$. The protocol therefore involves plotting, for a fixed length $L$, $\theta_{eq} - \theta_0$ as a function of $\sin(\theta_{eq})$ and checking that it is indeed a linear function, the slope of which $p = \dfrac{m_m B_e L}{C_f}$ can be calculated.

[p1-sol-fig7]

The slope obtained from the measurements is $11.1 \pm 1.7\ \mathrm{rad}^{-1}$ for $L = 0.34\ \mathrm{m}$. The Earth magnetic field is then $B_e = \dfrac{p C_f}{m_m L} = \dfrac{11.1 \times 5.1 \times 10^{-7}}{0.33 \times 0.34} = 50.5 \pm 8\ \mu\mathrm{T}$. Note that uncertainty is relatively high.

B.6 (2.3pt) Vary the length $L$ and repeat the previous study for two other lengths to verify the $L$ dependence of the wire torque. Using a final graph that summarizes all the dependencies, determine a new value for $B_\mathrm{e}$, with its uncertainty.

SOLUTION:

Now the length $L$ is varied, for $L = 26$, 12 and 8 cm.

[p1-sol-fig8]

[p1-sol-fig9]

[p1-sol-fig10]

The slopes are

$L = 25\ \mathrm{cm}$ $p = 10.6 \pm 0.3\ \mathrm{rad}^{-1}$,

$L = 12\ \mathrm{cm}$ $p = 6.7 \pm 0.3\ \mathrm{rad}^{-1}$,

$L = 8\ \mathrm{cm}$ $p = 4.6 \pm 0.2\ \mathrm{rad}^{-1}$.

So we can plot the slopes versus $L$ as

[p1-sol-fig11]

As expected, the slope of figure B.6D is found to be proportional to $L$, with a slope $p' = 27.1 \pm 1.5\ \mathrm{rad}^{-1}\cdot\mathrm{m}^{-1}$. A new value of $B_e$ is deduced from $B_e = \dfrac{p' C_f}{m_m} = \dfrac{5.1 \times 10^{-7} \times 27.1}{0.305} \approx 45.3\ \mu\mathrm{T}$. Relative uncertainty is given by $\dfrac{\delta p'}{p'} + \dfrac{\delta C}{C} + \dfrac{\delta m}{m} = 0.12$, so

$B_e = 45.3 \pm 5\ \mu\mathrm{T}$.

Another possible solution is to represent all the measurements at different L in the same graph, by plotting $(\theta_{eq} - \theta_0)$ versus $L \cdot \sin(\theta_{eq} - \theta_e)$ and measure slope from there. Doing so should allow students to earn the maximum number of points according to the grading scheme.