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Question

Physics Question on Electromagnetic induction

The dimension of mutual inductance is :

A

[ML2T-2A-1]

B

[ML2T-3A-1]

C

[ML2T-2A-2]

D

[ML2T-3A-2]

Answer

[ML2T-2A-2]

Explanation

Solution

The correct answer is (C) : [ML2T–2A–2]

∵ U = 12\frac{1}{2} Mi²

⇒ [M] = [U][i2]\frac{[U]}{[i^2]} = ML2T2A2\frac{ML^2T^{-2}}{A^2}
= [ML2T–2A–2]