Directions: are based on the following paragraph |
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 principal 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. |
(ii) Calculate field E(r) at distance r from A by |
using Ohms 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. |
\[\Delta V\] measured between B and C is [AIEEE 2008] |
A) \[\frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi \left( a+b \right)}\]
B) \[\frac{\rho I}{2\pi \left( a-b \right)}\]
C) \[\frac{\rho I}{\pi a}-\frac{\rho I}{\pi \left( a+b \right)}\]
D) \[\frac{\rho I}{a}-\frac{\rho I}{\left( a+b \right)}\]
Correct Answer: C
Solution :
[c] \[E=\rho \,j=\rho \frac{I}{2\pi {{r}^{2}}}\] |
Potential difference due to current at A .dr |
\[{{V}_{B}}-{{V}_{C}}=-\int\limits_{C}^{B}{\vec{E}.\,\,d\vec{l}=-\int\limits_{a+b}^{a}{\rho \frac{I}{2\pi {{r}^{2}}}.\,dr\,;}}\]\[\Delta V'=-\frac{\rho I}{2\pi }\left[ -\frac{1}{r} \right]_{a+b}^{a}=\frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi \left( a+b \right)}\] |
By principle of superposition, |
\[\Delta V=2\Delta V'=\frac{\rho I}{\pi a}-\frac{\rho I}{\pi \left( a+b \right)}\] |
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