Translated 9.1.1-9.1.5

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+ <h3 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3>
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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 $b$.
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+ <figcaption>
+ For problem $9.1.4$
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+
+ <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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