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Q. The dimensions of the coefficient of self-inductance are

NTA AbhyasNTA Abhyas 2022

Solution:

Energy $U=\frac{1}{2}LI^{2}$
$\Rightarrow \, L=\frac{2 U}{I^{2}}$
$\therefore \left[L\right]=\frac{\left[U\right]}{\left[I\right]^{2}}=\frac{\left[M L^{2} T^{- 2}\right]}{\left[A\right]^{2}}=\left[M L^{2} T^{- 2} A^{- 2}\right]$