A) \[\frac{I\rho }{2\pi r}\]
B) \[\frac{I\rho }{2\pi }\left[ \frac{1}{b}-\frac{1}{r} \right]\]
C) \[\frac{I\rho }{2\pi b}\,In\,\frac{r}{b}\]
D) \[\frac{I\rho }{4\pi b}\,In\,\frac{r}{b}\]
Correct Answer: B
Solution :
[b] \[J=\frac{I}{2\pi {{r}^{2}}}=\frac{E}{\rho }\] \[E=\frac{I\rho }{2\pi {{r}^{2}}}\] \[\Delta V=\frac{I\rho }{2\pi }\int\limits_{b}^{r}{\frac{1}{{{r}^{2}}}}dr=\frac{I\rho }{2\pi }\left( \frac{1}{b}-\frac{1}{r} \right)\]
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