A) 0.6 Gauss
B) 1.2 Gauss
C) 1.8 Gauss
D) 0.32 Gauss
Correct Answer: A
Solution :
[a] Given \[M=8\times {{10}^{22}}\,A{{m}^{2}}\] \[d={{R}_{e}}=6.4\times {{10}^{6}}m\] Earth's magnetic field, \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2M}{{{d}^{3}}}\] \[=\frac{4\pi \times {{10}^{-7}}}{4\pi }\times \frac{2\times 8\times {{10}^{22}}}{{{(6.4\times {{10}^{6}})}^{3}}}\] \[\cong \,0.6\,Gauss\]
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