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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. Apply the superposition principle and find the voltage $ \Delta V$ developed between B and C?


$\begin {array} {1 1} (A)\;\large\frac{\rho I}{\pi a}-\large\frac{\rho I}{\pi (a+b)} & \quad (B)\;\large\frac{\rho I}{ a}-\large\frac{\rho I}{ (a+b)} \\ (C)\;\large\frac{\rho I}{2\pi a}-\large\frac{\rho I}{2\pi (a+b)} & \quad (D)\;\large\frac{\rho I}{2\pi (a-b)} \end {array}$


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1 Answer

  • Hint: 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.
Choosing A as origin
$E=\rho j= \rho \large\frac{I}{2 \pi r^2}$
$ V_C-V_B= -\large\frac{\rho I}{2 \pi}$ $ \int_a^{(a+b)} \large\frac{1}{r^2} $ $dr = \large\frac{ \rho I}{2\pi}$ $ \bigg[ \large\frac{1}{(a+b)} - \large\frac{1}{a} \bigg]$
$ V_B-V_c = \large\frac{\rho I}{2\pi}$$ \bigg[ \large\frac{1}{a} - \large\frac{1}{(a+b)} \bigg]$
Ans : (C)


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

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