Note: you must NOT consume the powder in any form! Do not swallow the balls! Do not bring the magnet too close to the scale, and be careful with the magnet as it may break.
Water density
\(\rho_w = 1.00~\mathrm{g}/\mathrm{cm}^3\) Avogadro's number \(N_A = 6.02 \cdot 10^{23}~\mathrm{mol}^{-1}\) Magnetization of the magnet \(\mu_0M = 1.19~\mathrm{T}\) Molar weight of water \(\mu_{\rm H_2O} = 18.0~\mathrm{g}/\mathrm{mol} \) Molar weight of mangaese chloride \(\mu_{\rm MnCl_2} = 125.8~\mathrm{g}/\mathrm{mol} \) Density of manganese chloride solution $\rho_e = \rho_w + \beta \cdot (m_e/m_w - 1)$, $\beta = 0,351~ \mathrm{g} / \mathrm{cm}^3$, where \(m_e \) is a total mass of solution, and \(m_w\) is the mass of pure water what was used to prepare it. The maximum value of $m_e/m_w$ which could be useful is $2.3$. The acceleration of free fall $g=9.8~\text{m}/\text{s}^2$.
There is no magnetic charges, but they are useful tool for magnetic field calculations. Magnetic charge $q_M$ creates the same form of field with the the electric charge:
\[\vec{B} = \frac{\mu_0 q_M}{4 \pi} \frac{\vec{r}}{r^3}.\]Magnetic dipole moment of several magnetic charges $q_{M,i}$ is $\vec{m} = \sum \vec{r}_i q_{M,i}$.
We will call an elastic dipole a dipole whose dipole moment is proportional to the external field: \[\vec{m} = \alpha \cdot \vec{B}.\]The quantity $\alpha$ is called the polarizability of the dipole.
Express the answer in terms of $\alpha$, $n$ and $\mu_0$.
Now let's discuss the behavior of a sphere with magnetic properties in an external magnetic field. Note that the behavior of magnetic materials in external fields is completely analogous to the behavior of dielectrics in electric fields.
A sphere of radius \(R\) with magnetic permeability \(\mu_s\) is placed in a uniform field \(\vec{B_0}\). Using the experience in electrostatics, we can guess the field distribution in space:
When the contour used for the circulation theorem of the vector \(\vec{B}\) passes through several media and \(\mu(\vec{r})\) is a function of coordinates, the circulation theorem is conveniently written as:
\[\oint{\frac{\vec{B}(\vec{r})}{\mu(\vec{r})\mu_0}d\vec{l}} = I,\]
where $I$ is the current penetrating the contour. Gauss's theorem for the magnetic field does not change its form when the surface only partially passes through several media.
From the problem we have solved, it is easy to move to the case where the sphere is placed in a medium with permeability \(\mu_e\). To do this, simply replace \( \mu_s \to \mu_s/\mu_e\) in the final answers.
The materials of the beads provided to you practically do not exhibit magnetic properties, and for them, we will assume that $\mu_s=1$. The $\rm MnCl_2$ solution, which acts as the surrounding medium, is a strong paramagnet, and its magnetic properties are due to the manganese ions $\rm Mn^+$. At the same time, the salt $\rm MnCl_2$ dissociates completely, meaning the molar concentration of $\rm MnCl_2$ ions is equal to the molar concentration of the salt added in dry form in the solution.
For the $\rm MnCl_2$ solution, the following holds:
\[\mu_e(c) - 1 = \chi_e(c) = a \cdot c,\] where $c$ is is the molar concentration of $\rm Mn^+$.The solution is a strong paramagnetic medium but it is still a paramagnetic medium, and for it \( \chi_e\ \ll 1 \).
The $z$-axis is the axis of the ring magnet, and $z=0$ corresponds to the center of the magnet. On the $z$-axis , the magnetic field created by the magnet is $B(z)$. The density of the solution is $\rho_e$. The density of the ball is $\rho_s$.
The energy of the ball $W$ consists of the energy of the magnetic dipole with polarizability $\alpha_s$ and the energy of the ball in the gravitational field.
Below is a graph of \(B^2/(\mu_0 M)^2\) on $z$ for the provided magnet.
In the experiment, different types of the ball behavior in the solution can be observed:
Type In absence of the magnet In presence of the magnet 1 Floating No equilibrium points 2 Floating One equilibrium point (under the magent) 3 Floating Two equilibrium points 4 Sinking Two equilibrium points 5 Sinking One equilibrium point (above the magent) 6 Sinking No equilibrium points
A10 1.50 Plug one end of the tube with a stopper, fill it with the solution. Position the magnet approximately in the middle of the solution level.
By varying the concentration of MnCl2 $\rm MnCl_2$ in the solution, measure as accurately as possible the transition boundary from one case to the other from C6 for each type of bead provided.
Record the concentration of the $\rm MnCl_2$ solution in terms of $m_e/m_w$, where $m_e$ is the total mass of solution, and $m_w$ is the mass of pure water what was used to prepare it.
A11
0.50
The powder provided to you is not pure $\rm MnCl_2$ but 4-hydrated $\rm MnCl_2 \cdot 4H_2O$, meaning that for each molecule of $\rm MnCl_2$ in the powder, there are four molecules of \(\rm H_2O\). When this powder is dissolved in liquid water, it dissociates into $\rm Mn^{2+}$ and $\rm Cl^-$ ions, and the water takes on a liquid form.
Express the molar concentration \(c\) of manganese ions $\rm Mn^{2+}$ in the solution. For the answer, use \(\rho_e\), $m_e/m_w$, $\mu_{\rm H_2O}$ and $\mu_{\rm MnCl_2}$