A) \[M{{L}^{2}}{{T}^{-2}}{{A}^{-1}}\]
B) \[M{{L}^{0}}{{T}^{-2}}{{A}^{-2}}\]
C) \[{{M}^{0}}{{L}^{-2}}{{T}^{-2}}{{A}^{-3}}\]
D) \[M{{L}^{2}}{{T}^{-2}}{{A}^{3}}\]
Correct Answer: A
Solution :
\[\phi =BA=\frac{F}{I\times L}A=\frac{\left[ ML{{T}^{-2}} \right]\,\,\left[ {{L}^{2}} \right]}{\left[ A \right]\left[ L \right]}=\left[ M{{L}^{2}}{{T}^{-2}}{{A}^{-1}} \right]\]
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