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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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Consider a block of conducting material of resistivity $ \rho$ shown in the figure. Current I enters at A and leaves from D. We apply superposition principle to find voltage $ \Delta 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. \[\] Calculate field E(r) at distance r from A by using ohm's law $E= \rho j$, 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 \[\] For current entering at A, the electric field at a distance r from A is


$\begin {array} {1 1} (A)\;\large\frac{\rho I}{8\pi r^2} & \quad (B)\;\large\frac{\rho I}{ r^2} \\ (C)\;\large\frac{\rho I}{2\pi r^2} & \quad (D)\;\large\frac{\rho I}{4\pi r^2} \end {array}$


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We know E=$\rho j$  where j is current per unit area at a distance r
therefore$E=\rho j= \rho \large\frac{I}{2 \pi r^2}$
Ans : (C)


answered Feb 14, 2014 by thanvigandhi_1
edited Sep 4, 2014 by thagee.vedartham

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