A) \[2\times {{10}^{-4}}T\,and{{10}^{-4}}T\]
B) \[3\times {{10}^{-4}}T\,and2\times {{10}^{-4}}T\]
C) \[4\times {{10}^{-4}}T\,and3\times {{10}^{-4}}T\]
D) none of these
Correct Answer: A
Solution :
Magnetic induction in axial position. \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{d}^{3}}}\] \[={{10}^{-7}}\times \frac{2\times 8}{{{(20\times {{10}^{-2}})}^{2}}}\] \[=\frac{{{10}^{-7}}\times 16}{8\times {{10}^{-3}}}=2\times {{10}^{-4}}T\] At equatorial position \[{{B}_{2}}=\frac{{{B}_{1}}}{2}\] \[=\frac{2\times {{10}^{-4}}}{2}={{10}^{-4}}T\]
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