Edit to “Solution”
en/9.1.4.md
+1 −1
| ### Statement | |||
| $9.1.4.$ 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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| @@ -6,8 +6,8 @@Statement | |||
| ### Solution | |||
| − | 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 $$ I = \frac{V}{R} = \frac{V\lambda S}{l} = \frac{V\lambda}{b}S | ||
| + | We will find the resistance between the connected contacts as $$ 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\tag{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}\tag{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}} $$ | ||
| #### Answer | |||
| $$\Delta h = \frac{V\lambda Ba}{h\rho g}$$ | |||
| ### Statement | ### Statement | ||
| $9.1.4.$ An electrolyte with density $\rho$ and specific conductivity $\lambda$ is poured into a rectangular cuvette, two opposite walls of which are metallic and the rest are made of an insulator. Voltage $V$ is applied to the metallic walls of the cuvette, and the entire cuvette is placed in a uniform vertical magnetic field of induction $B$. Determine the difference in liquid levels near the non-metallic walls of the cuvette. The length of the cuvette is $a$, the width is $b$. | $9.1.4.$ An electrolyte with density $\rho$ and specific conductivity $\lambda$ is poured into a rectangular cuvette, two opposite walls of which are metallic and the rest are made of an insulator. Voltage $V$ is applied to the metallic walls of the cuvette, and the entire cuvette is placed in a uniform vertical magnetic field of induction $B$. Determine the difference in liquid levels near the non-metallic walls of the cuvette. The length of the cuvette is $a$, the width is $b$. | ||
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| @@ -6,8 +6,8 @@Statement | |||
| ### Solution | ### Solution | ||
| We will find the resistance between the connected contacts as $$ 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 |
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 $$ I = \frac{V}{R} = \frac{V\lambda S}{l} = \frac{V\lambda}{b}S\tag{1} $$ The angle of inclination of the liquid level $\alpha$, we express from the equilibrium condition, and through the change in the electrolyte level $\Delta h$ $$ \tan\alpha = \frac{\Delta h}{a}=\frac{F}{mg}\tag{2} $$ 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 $(2)$ $$ \tan\alpha =\frac{V\lambda SB}{\rho hsg} \Rightarrow \boxed{\Delta h = \frac{V\lambda Ba}{h\rho g}} $$ | ||
| #### Answer | #### Answer | ||
| $$\Delta h = \frac{V\lambda Ba}{h\rho g}$$ | $$\Delta h = \frac{V\lambda Ba}{h\rho g}$$ | ||