A) \[3.6\times {{10}^{-5}}\,N\]
B) \[1.8\times {{10}^{-5}}\text{ }N\]
C) \[1.3\times {{10}^{-5}}\,N\]
D) \[6.5\times {{10}^{-5}}\,N\]
Correct Answer: D
Solution :
\[{{B}_{H}}={{B}_{Earth}}\text{ }\cos \,45{}^\circ \] \[{{B}_{Earth}}={{B}_{H}}\sqrt{2}\] \[{{Z}_{ab{{t}_{0}}}}=0\] \[M{{B}_{Earth}}\sin \,45{}^\circ -F\times 0.06=0\] \[1.8\times 0.12\text{ }{{B}_{H}}\frac{1}{\sqrt{2}}\times \sqrt{2}-F\times 0.06=0\] \[F=1.8\times 2\times {{B}_{H}}\] \[=\,\,\,3.6\times 18\times {{10}^{-6}}\] \[=\text{ }64.8\times {{10}^{-6}}\] \[=\text{ }6.5\times {{10}^{-5}}\,N\]You need to login to perform this action.
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