A) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}R}{\tau }\]
B) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2\tau R}\]
C) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{\tau R}\]
D) \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2\tau R}\]
Correct Answer: D
Solution :
| [d] heat equated \[\int\limits_{0}^{\infty }{{{i}^{2}}Rdt}\] |
| \[\int\limits_{0}^{\infty }{\frac{^{\varepsilon }in{{d}^{2}}}{Rt}Rdt}\] |
| \[=\frac{1}{R}\int\limits_{0}^{\infty }{^{\varepsilon }in{{d}^{2}}dt}\] |
| \[=\frac{1}{R}\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{{{\tau }^{2}}}\int\limits_{0}^{\infty }{{{e}^{-2t/\tau }}dt}\] |
| \[\left. =\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{R{{\tau }^{2}}}\frac{{{e}^{-2t/\tau }}}{-2/\tau } \right|_{0}^{\infty }\] |
| \[=\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}\tau }{2R{{\tau }^{2}}}\left\{ 0+1 \right\}\] |
| \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2R\tau }\] |
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