Translated 9.1.1-9.1.5
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| + | <title>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.</title> | ||
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| + | Solutions of Savchenko Problems in Physics <br> | ||
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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 $9.1.4$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <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> | ||
| + | $$\Delta h = \frac{V\lambda Ba}{h\rho g}$$ | ||
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| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | |||
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| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
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| <h3 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3> | |||
| <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 $b$. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/9/9.1.4/statement.png" | |||
| loading="lazy" width="230" /> | |||
| <figcaption> | |||
| For problem $9.1.4$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <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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