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Dimension of magnetic permeability is :A. `MLT^(2)A^(-2)`B. `ML^(-1)T^(-2)A^(-2)`C. `ML^(-2)T^(-2)A^(2)`D. `MLT^(2)A^(-2)` |
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Answer» Force per unit length between two parallel current carrying wires is given by `(F)/(l)=(m_(0)I_(1)I_(2))/(2pd)` `therefore[mu_(0)]=[(F)/(I^(2))]=[(MLT^(2))/(A^(2))]=[MLT^(2)A^(-2)]` |
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