A) \[{{B}_{1}}=8\times {{10}^{-4}}T,{{B}_{2}}=4\times {{10}^{-4}}T\]
B) \[{{B}_{1}}=2\times {{10}^{-4}}T,{{B}_{2}}=1\times {{10}^{-4}}T\]
C) \[{{B}_{1}}=10\times {{10}^{-4}}T,{{B}_{2}}=5\times {{10}^{-4}}T\]
D) \[{{B}_{1}}=6\times {{10}^{-4}}T,{{B}_{2}}=3\times {{10}^{-4}}T\]
Correct Answer: A
Solution :
Given \[d=10\text{ }cm=0.1m\] \[\frac{{{\mu }_{0}}}{4\pi }={{10}^{-7}}T{{A}^{-1}}\] \[{{B}_{axial}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\mu }{{{d}^{3}}}\] \[={{10}^{-7}}\times \frac{2\times 4}{{{(0.1)}^{3}}}=8\times {{10}^{-4}}T\] \[{{B}_{equitorial}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{\mu }{{{d}^{3}}}=4\times {{10}^{-4}}T\]
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