A) \[{{\mu }_{0}}{{\mu }_{r}}\frac{{{N}_{1}}{{N}_{2}}}{l}\]
B) \[\frac{{{\mu }_{0}}{{\mu }_{r}}{{N}_{1}}{{N}_{2}}}{Al}\]
C) \[{{\mu }_{0}}{{\mu }_{r}}{{N}_{1}}{{N}_{2}}Al\]
D) \[\frac{{{\mu }_{0}}{{\mu }_{1}}{{N}_{1}}{{N}_{2}}A}{l}\]
Correct Answer: D
Solution :
Coefficient of mutual inductance \[M=\frac{\phi }{{{I}_{p}}}=\frac{{{\mu }_{0}}{{\mu }_{r}}{{N}_{1}}{{N}_{2}}A{{I}_{P}}}{I{{I}_{P}}}\] \[=({{\mu }_{0}}{{\mu }_{r}}{{N}_{1}}{{N}_{2}}A)/l\] where A and \[l\] are the length and area of cross-section respectively
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