The dimensions of self or mutual inductance are given as (a) [L⁻²M¹T⁻²I⁻²] (b) [L²M²T⁻²I⁻²] (c) [L²M¹T⁻²I⁻²] (d) [L²M²T⁻²I⁻²]

Answer: (c). Energy in inductor U = 1/2LI² ⇒ L = 2U/I² [L] = [ML²T⁻²][I²] = [L²M¹T⁻²I⁻²] So the answer is (c).

The dimensions of self or mutual inductance are given as | MHT CET 2019 (2 May, Shift 2)

Q78MHT CET 2019 · 2 MAY · SHIFT 2Easy

Units & Dimensions

The dimensions of self or mutual inductance are given as

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Answer: (c)

Energy in inductor $U = \frac{1}{2}LI^2 \Rightarrow L = \frac{2U}{I^2}$
$[L] = \frac{[\mathrm{ML^2T^{-2}}]}{[\mathrm{I^2}]} = [\mathrm{L^2M^1T^{-2}I^{-2}}]$
So the answer is (c).

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