A rectangular block of metal of resistivity `p` has dimensions `d xx 2d xx 3d`. A potential difference `V` is applied between two opposite faces of the block.
(a) To which two faces of the block should the potential difference `V` be applied to give the maximum current density? What is the maximum current density?
(b) To which two faces of the block should the potential difference `V` be applied to give the maximum current? What is this maximum current?

Answer

A, B, C, D

Explanation

a. `J=i/A=V/(RA)=V/(((rhol)/(a))A)`
`J=V/(rhol)`
or `Jprop1/l`
`l_("min")=d`. so, J is maximum. Hence, potential difference should be applied across the face `(2d xx 3d)`
from eqn i
`J_(max)=V/(rhod)`
b. `i=V/R=V/((rhol/A))=(VA)/(rhod)`
or `i prop A/l`
Across face `(2d xx 3d)` area of cross sectiois maximum and l is minimum. Hence is current is maximum
`i_(max)=(V(2d xx 3d))/(rho(d))=(6Vd)/rho`