Let ε0 denote the dimensional formula for the permittivity of the vacuum and μ0 that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current, then

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ε0=M−1L−3T2I

ε0=M−1L−3T4I2

μ0=MLT−2I−2

μ0=ML2T−1I

answer is B.

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Detailed Solution

F=14πε0q1q2r2ε0=q1q2Fr2= IT ITMLT-2 L2= M−1L−3T4I2C=1μ0ε0⇒μ0=1C2ε0=1L2T-2 M-1L-3T4I2=MLT−2I−2

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