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?

Question

Quizplus · Accepted Answer

The current flowing through the tube is given by:$$I = \frac{EAd}{\rho L}$$where:* E is the electric field strength (23 N/C)* A is the cross-sectional area of the tube (πr^2, where r is the radius of the tube)* d is the length of the tube (not given)* ρ is the resistivity of the tube (9.84 × 10^-7 Ω ∙ m)* L is the length of the tube (not given)Substituting the given values into the equation, we get:$$I = \frac{(23 N/C)(\pi (0.5 mm)^2)}{(9.84 × 10^-7 Ω ∙ m)L}$$$$I = \frac{(23 N/C)(3.14 × 10^-6 m^2)}{(9.84 × 10^-7 Ω ∙ m)L}$$$$I = \frac{(72.22 × 10^-6 N ∙ m^2)}{(9.84 × 10^-7 Ω ∙ m)L}$$$$I = \frac{7.34 × 10^3 A}{L}$$Since the length of the tube is not given, we cannot calculate the exact current flowing through the tube. However, we can see that the current is inversely proportional to the length of the tube. Therefore, the correct answer is A, which is the choice that is inversely proportional to the length of the tube.

A Tube of Mercury with Resistivity 9