A) \[1.05\times {{10}^{23}}A{{m}^{2}}\]
B) \[2.05\times {{10}^{23}}A{{m}^{2}}\]
C) \[105\times {{10}^{21}}A{{m}^{2}}\]
D) \[2.05\times {{10}^{21}}A{{m}^{2}}\]
Correct Answer: A
Solution :
(a) \[1.05\times {{10}^{23}}A{{m}^{2}}\] Here, \[{{B}_{E}}=0.4\,G=0.4\times {{10}^{-4}}T;r=6.4\times {{10}^{6}}m\] As \[{{B}_{E}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{r}^{3}}}\] \[\therefore M=\frac{{{B}_{E}}{{r}^{3}}}{{{\mu }_{0}}/4\pi }=\frac{0.4\times {{10}^{-4}}\times {{(6.4\times {{10}^{6}})}^{3}}}{{{10}^{-7}}}\] \[=1.05\times {{10}^{23}}A{{m}^{2}}\].
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