Question

Consider a block of conducting material of resistivity ρ shown in the figure. Current l enters A and B leaves from D. We apply superposition principle to find voltage ΔV developed between B and C. The calculation is done in the following steps: (a) Take current l entering from A and assume it to spread over a hemispherical surface in the block.       (b) 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.      (c) From the r dependence of E(r), obtain the potential V(r) at r. (d) Repeat (i), (ii) and (iii) for current l leaving D superpose results for A to D. ΔV measured between B and C is

A

ρLπaρLπ(a+b)

B

ρLaρL(a+b)

C

ρL2πaρL2π(a+b)

D

ρL2π(a+b)

Solution

The correct option is C ρLπa−ρLπ(a+b) Let j be the current density . then j×2πr2=l⇒j=I2πr2ΔVBC=−∫aa+bEd¯r=−∫aa+bρl2πr2dr=ρl2πa−ρl2π(a+b) On applying superposition as mentioned we get total  ΔVBC=2×ΔVBC=ρLπa−ρLπ(a+b).

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