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Consider a block of conducting material of resistivity ‘p’ shown in the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘$\triangle$V’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:<br> (i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block<br> (ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s law E = pj, where j is the current per unit area at ‘r<br> (iii) From the ‘r’ dependence of E(r), obtain the potential V(r) at r.<br> (iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.<br> <img src=""/><br><br> QUESTION : $\triangle$V measured between B and C is

( A ) $\frac{pI}{2\pi {a}}-\frac{pI}{2\pi(a+b)}$
( B ) $\frac{pI}{a}-\frac{pI}{(a+b)}$
( C ) $\frac{pI}{\pi {a}}-\frac{pI}{\pi(a+b)}$
( D ) $\frac{pI}{2\pi(a-b)}$

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