JEE MAIN - Physics (2008 - No. 11)
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.
(ii) 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 $$'D'.$$
(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 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 $$'D'.$$
$$\Delta V$$ measured between $$B$$ and $$C$$ is
$${{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$$
$${{\rho I} \over a} - {{\rho I} \over {\left( {a + b} \right)}}$$
$${{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$$
$${{\rho I} \over {2\pi \left( {a - b} \right)}}$$
Explanation
Let $$j$$ be the current density.
Then $$j \times 2\pi {r^2} = I \Rightarrow j = {I \over {2\pi {r^2}}}$$
$$\therefore$$ $$E = \rho j = {{\rho I} \over {2\pi {r^2}}}$$
Now, $$\Delta V{'_{BC}} = $$ $$ - \int\limits_{a + b}^a {\overrightarrow E .\,\overrightarrow {dr} } $$ $$ = - \int\limits_{a + b}^a {{{\rho I} \over {2\pi {r^2}}}} dr$$
$$ = - {{\rho I} \over {2\pi }}\left[ { - {1 \over r}} \right]_{a + b}^a$$
$$ = {{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$$
On applying superposition as mentioned we get
$$\Delta {V_{BC}} = 2 \times \Delta {V_{BC}} = {{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$$
Then $$j \times 2\pi {r^2} = I \Rightarrow j = {I \over {2\pi {r^2}}}$$
$$\therefore$$ $$E = \rho j = {{\rho I} \over {2\pi {r^2}}}$$
Now, $$\Delta V{'_{BC}} = $$ $$ - \int\limits_{a + b}^a {\overrightarrow E .\,\overrightarrow {dr} } $$ $$ = - \int\limits_{a + b}^a {{{\rho I} \over {2\pi {r^2}}}} dr$$
$$ = - {{\rho I} \over {2\pi }}\left[ { - {1 \over r}} \right]_{a + b}^a$$
$$ = {{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$$
On applying superposition as mentioned we get
$$\Delta {V_{BC}} = 2 \times \Delta {V_{BC}} = {{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$$
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