Physics-
General
Easy
Question
Let[0] denote the dimensional formula of the permittivity of the vacuum, and [µ0] that of the permitivityof the vacuum. If M = mass, L = length, T = time and I = electric current,
- [0] = [M–1L–3T2 I]
- [0] = [M–1LT4 I2]
- [µ0] = [M LT–2 I–2]
- [µ0] = [M L2 T–1 I]
The correct answer is: [µ0] = [M LT–2 I–2]
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