A tube of mercury with resistivity $9.84 \times 10 ^ { - 7 } \Omega \cdot \mathrm { m }$ has a uniform electric field of $23 \mathrm {~N} / \mathrm { C }$ inside the mercury. How much current is flowing in the tube, if the radius of the tube is $0.495 \mathrm {~mm}$ ?

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Quizplus · Accepted Answer

To find the current flowing in the tube, we use Ohm's law in the form $I = \frac{V}{R}$, where $V$ is the potential difference and $R$ is the resistance. However, given the electric field $E$ and the resistivity $\rho$, we need to adapt our approach. The electric field $E$ relates to the potential difference $V$ and the length $l$ of the conductor through $E = V/l$, and the resistance $R$ of a conductor is given by $R = \rho \frac{l}{A}$, where $A$ is the cross-sectional area of the conductor.Given:- Resistivity, $\rho = 9.84 \times 10^{-7} \Omega \cdot m$- Electric field, $E = 23 N/C$- Radius of the tube, $r = 0.495 mm = 0.495 \times 10^{-3} m$The cross-sectional area $A$ of the tube is $\pi r^2$, so:$$A = \pi (0.495 \times 10^{-3} m)^2$$The current $I$ can be found from the relation $I = \sigma A E$, where $\sigma = 1/\rho$ is the conductivity. Substituting the given values:$$I = \frac{1}{\rho} \cdot A \cdot E$$$$I = \frac{1}{9.84 \times 10^{-7} \Omega \cdot m} \cdot \pi (0.495 \times 10^{-3} m)^2 \cdot 23 N/C$$Calculating this gives:$$I = \frac{1}{9.84 \times 10^{-7}} \cdot \pi (0.495 \times 10^{-3})^2 \cdot 23$$$$I \approx 18 A$$Therefore, the current flowing in the tube is approximately 18 A.

A Tube of Mercury with Resistivity 9.84×10−7Ω⋅m9.84 \times 10 ^ { - 7 } \Omega \cdot \mathrm { m }9.84×10−7Ω⋅m

A Tube of Mercury with Resistivity $9.84 \times 10 ^ { - 7 } \Omega \cdot \mathrm { m }$