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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <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 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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 $b$. | | $9.1.4.$ 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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| For problem $9.1.4$ | | For problem $9.1.4$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| We will find the resistance between the connected contacts as | | We will find the resistance between the connected contacts as |
| $$ R = \frac{\rho l}{S} = \frac{l}{\lambda S} $$ | | $$ R = \frac{\rho l}{S} = \frac{l}{\lambda S} $$ |
| At voltage $V$, according to Ohm's law, we find the current $I$ as | | At voltage $V$, according to Ohm's law, we find the current $I$ as |
| $$ I = \frac{V}{R} = \frac{V\lambda S}{l} = \frac{V\lambda}{b}S\quad(1) $$ | | $$ I = \frac{V}{R} = \frac{V\lambda S}{l} = \frac{V\lambda}{b}S\quad(1) $$ |
| The angle of inclination of the liquid level $\alpha$, we express from the equilibrium condition, and through the change in the electrolyte level $\Delta h$ | | The angle of inclination of the liquid level $\alpha$, we express from the equilibrium condition, and through the change in the electrolyte level $\Delta h$ |
| $$ \tan\alpha = \frac{\Delta h}{a}=\frac{F}{mg}\quad(2) $$ | | $$ \tan\alpha = \frac{\Delta h}{a}=\frac{F}{mg}\quad(2) $$ |
| From Ampere's law | | From Ampere's law |
| $$ F = Ib \cdot B $$ | | $$ F = Ib \cdot B $$ |
| Substituting the value of current $I$ from expression $(1)$ | | Substituting the value of current $I$ from expression $(1)$ |
| $$ F= V\lambda SB= V\lambda a \left(h + \frac{\Delta h}{2}\right) $$ | | $$ F= V\lambda SB= V\lambda a \left(h + \frac{\Delta h}{2}\right) $$ |
| Considering that the mass of the electrolyte $m = \rho hS$, we rewrite the expression $(2)$ | | Considering that the mass of the electrolyte $m = \rho hS$, we rewrite the expression $(2)$ |
| $$ \tan\alpha =\frac{V\lambda SB}{\rho hsg} \Rightarrow \boxed{\Delta h = \frac{V\lambda Ba}{h\rho g}} $$ | | $$ \tan\alpha =\frac{V\lambda SB}{\rho hsg} \Rightarrow \boxed{\Delta h = \frac{V\lambda Ba}{h\rho g}} $$ |
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| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
