Answer:
The current through the tube is 73.39A.
Explanation:
The relationship between the resistivity [tex]\rho[/tex], the electric field [tex]E[/tex], and the current density [tex]J[/tex] is given by
[tex]\rho = \dfrac{E}{J}[/tex]
This equation can be solved for [tex]J[/tex] to get:
[tex]J = \dfrac{E}{\rho}[/tex]
Since the current is [tex]I = J\cdot A[/tex]
[tex]I= J\cdot A = \dfrac{E}{\rho} \cdot A[/tex]
Now, for the tube of mercury [tex]\rho = 9.84*10^{-7}\: \Omega \cdot m[/tex], [tex]E = 23N/C[/tex], and the area is [tex]A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2[/tex]; therefore,
[tex]I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2[/tex]
[tex]\boxed{I = 73.39A.}[/tex]
Hence, the current through the mercury tube is 73.39A.
