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Let `[epsi_(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. `[epsi_(0)]=[M^(-1)L^(-3)T^(2)I]`B. `[epsi_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`C. `[mu_(0)]=[MLT^(-2)I^(-2)]`D. `[mu_(0)]=[ML^(2)T^(-1)I]` |
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Answer» Correct Answer - b, c Since, `F=1/(4pi epsi_(0))(q_(1)q_(2))/r^(2)` Hence, `epsi_(0)=((q_(1)q_(2))/(4pi Fr^(2)))` and `q=I xxt` Hence, `epsi_(0) rarr [M^(-1)L^(-3)T^(4)I^(2)]` `B=mu_(0)/(2pi) I/r` `mu_(0)=((Bxx2pixxr)/I)` Hence, `mu_(0) rarr [ML^(2)T^(-1)]` |
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