Question

A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C that is directed along the length of the tube. How much current is flowing through this tube if its diameter is 1.0 mm?

Answer 1

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity $\rho$, the electric field $E$, and the current density $J$ is given by

$\rho = \dfrac{E}{J}$

This equation can be solved for $J$ to get:

$J = \dfrac{E}{\rho}$

Since the current is $I = J\cdot A$

$I= J\cdot A = \dfrac{E}{\rho} \cdot A$

Now, for the tube of mercury $\rho = 9.84*10^{-7}\: \Omega \cdot m$, $E = 23N/C$, and the area is $A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2$; therefore,

$I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2$

$\boxed{I = 73.39A.}$

Hence, the current through the mercury tube is 73.39A.