A) 0.6 Gauss
B) 1.2 Gauss
C) 1.8 Gauss
D) 032 Gauss
Correct Answer: A
Solution :
| [a] Given \[M=8\times {{10}^{22}}A{{m}^{2}}\] |
| \[d={{\operatorname{R}}_{e}}=6.4\times {{10}^{6}}m\] |
| 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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