Let ε0 denote the dimensional formula of the permittivity of the vacuum and ε0 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-1 L-3 T2I

ε0=M-1 L-3 T4I2

μ0=MLT-2 I-1

μ0=ML2 T-1I

answer is B.

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Detailed Solution

F=14πε0·q1q2r2 ϵ0=q1q2[F]r2=[1T]2MLT-2L2=M-1 L-3 T4I2Speed of light,c=1ε0μ0 ∴μ0=1ε0[c]2=1M-1L-3T4I2LT-12=MLT-2I-2

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