Solveeit Logo
Login

Question

Physics Question on Dimensional Analysis

The dimension of mutual inductance is (Denote dimension of current as A)

A

ML2T2A−2ML^2 T^2 A^{-2}

B

ML2T−2A−2ML^2 T^{-2} A^{-2}

C

ML−2T2A−2ML^{-2} T^2 A^{-2}

D

ML−2T−3A−1ML^{-2} T^{-3} A^{-1}

Answer

ML2T−2A−2ML^2 T^{-2} A^{-2}

Explanation

Solution

As we know, induced Emf⁡=L×ΔIΔt\operatorname{Emf}=L \times \frac{\Delta I}{\Delta t}
⇒[L]=[ Voltage ][T][ Current ]=[ML2T−3A−1][T][A]\Rightarrow[ L ] =\frac{[\text { Voltage }][ T ]}{[\text { Current }]}=\frac{\left[ ML ^{2} T ^{-3} A ^{-1}\right][ T ]}{[ A ]}
=[ML2T−2A−2]=\left[ M L ^{2} T ^{-2} A ^{-2}\right]
∴\therefore Dimension of mutual inductance is
[ML2T−2A−2]\left[M L^{2} T^{-2} A^{-2}\right]