Consider a block of conducting material of resistivity ‘ρ’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘∆V’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:

(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.

(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E = ρj, where j is the current per unit area at ‘r’.

(iii) From the ‘r’ dependence of E(r), obtain the potential V(r) at r. (iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’

 1

1. ∆V measured between B and C is 

(1) ρl/ πa – ρl/π(a+b) (2) ρl/a – ρl/a+b  (3) ρl/2 πa – ρl/2π(a+b) (4) ρl/2π(a –B)

2. For current entering at A, the electric field at a distance ‘r’ from A is

1

About Timepersonified

I am a human being trying to understand ridiculous human ways. I just came across a lot of beautiful views in the process and am trying to sum them up using a device used to record data visually
This entry was posted in Current Electricity and tagged , , . Bookmark the permalink.

Leave a comment