An electrolyte with density \rho and specific conductivity \lambda is poured into a rectangular cuvette, two opposite walls of which are metallic and the rest are made of an insulator. Voltage V is applied to the metallic walls of the cuvette, and the entire cuvette is placed in a uniform vertical magnetic field of induction B. Determine the difference in liquid levels near the non-metallic walls of the cuvette. The length of the cuvette is a, the width is b.

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $9.1.4.$ An electrolyte with density $\rho$ and specific conductivity $\lambda$ is poured into a rectangular cuvette, two opposite walls of which are metallic and the rest are made of an insulator. Voltage $V$ is applied to the metallic walls of the cuvette, and the entire cuvette is placed in a uniform vertical magnetic field of induction $B$. Determine the difference in liquid levels near the non-metallic walls of the cuvette. The length of the cuvette is $a$, the width is $b$.

    <h3>Solution</h3>
    <p>
        We will find the resistance between the connected contacts as

$$R = \frac{\rho l}{S} = \frac{l}{\lambda S}$$
At voltage $V$, according to Ohm's law, we find the current $I$ as
$$I = \frac{V}{R} = \frac{V\lambda S}{l} = \frac{V\lambda}{b}S\quad(1)$$
The angle of inclination of the liquid level $\alpha$, we express from the equilibrium condition, and through the change in the electrolyte level $\Delta h$
$$\tan\alpha = \frac{\Delta h}{a}=\frac{F}{mg}\quad(2)$$
From Ampere's law
$$F = Ib \cdot B$$
Substituting the value of current $I$ from expression $(1)$
$$F= V\lambda SB= V\lambda a \left(h + \frac{\Delta h}{2}\right)$$
Considering that the mass of the electrolyte $m = \rho hS$, we rewrite the expression $(2)$
$$\tan\alpha =\frac{V\lambda SB}{\rho hsg} \Rightarrow \boxed{\Delta h = \frac{V\lambda Ba}{h\rho g}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$\Delta h = \frac{V\lambda Ba}{h\rho g}$$
    </p>


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