A) \[1/r\]
B) \[1/{{r}^{3}}{{^{/}}^{2}}\]
C) \[1/{{r}^{2}}\]
D) \[1/{{r}^{3}}\]
Correct Answer: D
Solution :
We have \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi i{{R}^{2}}}{{{({{R}^{2}}+{{r}^{2}})}^{3/2}}}\] Given\[r>>R\], then we have, neglecting R, \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi i{{R}^{2}}}{{{r}^{3}}}\] \[Also \,area = \pi {{R}^{2}}\] \[\therefore \,\,\,\,\,\,\,B=\frac{{{\mu }_{0}}}{2\pi }\,\frac{Ai}{{{r}^{3}}}\] \[\Rightarrow \,\,\,\,\,\,\,B\propto \frac{1}{{{r}^{3}}}\]You need to login to perform this action.
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