A) \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{M}{{{r}^{2}}}\]
B) \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{M}{{{r}^{3}}}\]
C) \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2M}{{{r}^{2}}}\]
D) \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2M}{{{r}^{3}}}\]
Correct Answer: B
Solution :
\[{{B}_{equatorial}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{r}^{3}}}\]
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