Units & Dimensions

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Question 93 of 149

The dimension of mutual inductance is

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Answer: (B) ML2T−2A−2{ML}^{2}{T}^{-2}{A}^{-2}

The formula for mutual inductance is given by, ϵ=MdIdt⇒M=ϵ(dIdt)\epsilon =M\dfrac{dI}{dt}\Rightarrow M=\dfrac{\epsilon }{\left(\dfrac{dI}{dt}\right)}
Dimension of emf or voltage is [ML2T−3A−1]\left[M{L}^{2}{T}^{-3}{A}^{-1}\right],
Dimension of current is [A]\left[A\right],
Dimension of time [T]\left[T\right].
Therefore, the dimension of MM is
M=[ML2T−3A−1][A][T]−1M=\dfrac{\left[{ML}^{2}{T}^{-3}{A}^{-1}\right]}{\left[A\right]{\left[T\right]}^{-1}}
=[ML2T−2A−2]=\left[{ML}^{2}{T}^{-2}{A}^{-2}\right]

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