A current `I = 10 sin (100pi t)` amp. Is passed in first coil, which induces a maximum e.m.f of `5pi` volt in second coil. The mutual inductance between the coils is-

Options

A. 5 mH
B. 10 mH
C. 15 mH
D. 20 mH

Answer

A

Explanation

Let current is `I = I_(0) sin omega t`, Where `I_(0) = 10 , omega = 100 pi`
Then induced e.m.f. is given by:
`omega = -(MdI)/(dt) = -(Md)/(dt) (I_(0))`
`omega = -MI_(0)omega cos omegat`
`therefore` Maximum e.m.f. is
`omega_("max") = MI_(0)omega`
`5pi = M xx 10 xx 100pi`
M = 5 mH