A) \[[M{{L}^{2}}{{T}^{-2}}{{A}^{-2}}]\]
B) \[[M{{L}^{2}}{{T}^{-1}}{{A}^{-2}}]\]
C) \[[M{{L}^{2}}{{T}^{-1}}{{A}^{-1}}]\]
D) \[[M{{L}^{-2}}{{T}^{-2}}{{A}^{-2}}]\]
Correct Answer: A
Solution :
Numerically, self-inductance L is given by \[L=\frac{e}{\Delta i/\Delta t}=\frac{e\Delta t}{\Delta i}\] where e is emf. At the time and Ai the current. \[\therefore \] Unit of \[L=\frac{Volt-\text{second}}{Ampere}\] \[=\frac{\text{(}Joule/Coulomb)\text{ }second}{Ampere}\] \[=\frac{Newton-metre-second}{Coulomb-\text{ }ampere}\] \[\Rightarrow \] Unit of \[L=\frac{Newton-metre}{Amper{{e}^{2}}}\] \[\therefore \] \[[L]=\frac{[ML{{T}^{-2}}]\,[L]}{[{{A}^{2}}]}=[M{{L}^{2}}{{T}^{-2}}{{A}^{-2}}]\]
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