Answer:
Current, i = 18 A
Explanation:
Given that,
Resistivity of mercury, [tex]\rho=9.84\times 10^{-7}\ \Omega-m[/tex]
Electric field inside the mercury, E = 23 N/C
Radius of the tune, r = 0.495 mm
We need to find the current flowing in the tube. We know that the resistance in terms of length and area is given by :
[tex]R=\rho\dfrac{l}{A}[/tex]
Using Ohm's law, we get :
V = iR, i is current, R is resistance
So,
[tex]\dfrac{V}{i}=\rho\dfrac{l}{A}\\\\i=\dfrac{VA}{\rho l}[/tex]
Since, [tex]\dfrac{V}{l}=E[/tex] (electric field)
[tex]i=\dfrac{VA}{\rho l}\\\\i=\dfrac{AE}{\rho}\\\\i=\dfrac{\pi r^2E}{\rho}\\\\i=\dfrac{\pi (0.495\times 10^{-3})^2\times 23}{9.84\times 10^{-7}}\\\\i=17.99\ A[/tex]
or
i = 18 A
So, 18 A of current is flowing in the tube.
