A) \[6\times {{10}^{-5}}Am\]
B) \[5\times {{10}^{-5}}Am\]
C) \[2\times {{10}^{-4}}Am\]
D) \[3\times {{10}^{-4}}Am\]
Correct Answer: A
Solution :
[a] Magnetic field due to a bar magnet in the broad-side on position is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{\left[ {{r}^{2}}+\frac{{{\ell }^{2}}}{4} \right]};M=m\ell \]. After substituting the values and simplifying we get \[B=6\times {{10}^{-5}}\,A-m\]
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