InterviewSolution
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InterviewSolution
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Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of the vacuum, and `[mu_(0)]` that of the permeability of the vacuum. If `M = mass ,L = length, T = time and I = electric current`,A. (a) `[epsilon_(0)] = M^(-1)l^(-3)T^(2)I`B. (b) `[epsilon_(0)] = M^(-1)l^(-3)T^(4)I^(2)`C. ( c ) `[mu_(0)] = MLT^(-2)I^(-2)`D. (d) `[mu_(0)] = ML^(2)T^(-1)I` |
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Answer» Correct Answer - B::C (b,c) By defination `F = (Q_(1)Q_(2))/((4piepsilon_(0))r^(2)) and (F)/(l) = (mu_(0)I_(1)I_(2))/(2piL)` Hence,`[epsilon_(0)] = [Q^(2)]/([F][r^(2)]) = (I^(2)T^(2))/(MLT^(-2)L^(2)) = M^(-1)L^(-3)T^(4)I^(2)` `[mu_(0)] = ([F])/([I]^(2)) = (MLT^(-2))/(I^(2)) = ML^(-2)T^(-2)` |
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