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Question:

In terms of basic units of mass (M), length (L), time (T), and charge (Q), the dimensions of magnetic permeability of vacuum (μ0) would be

[LT⁻¹Q⁻¹]

[LTQ⁻¹]

[ML²T⁻¹Q⁻²]

[MLQ⁻²]

Solution:

Magnetic permeability of free space = 4π × 10⁻⁷ N A⁻²
The dimension of force is [MLT⁻²]
The dimension of current is [T⁻¹Q]
Hence, the dimension of magnetic permeability of free space is = N/A² = [MLT⁻²][T⁻¹Q]⁻² = [MLT⁻²][T²Q⁻²] = [MLQ⁻²]