A) \[5\pi \]
B) \[4\pi \]
C) \[2\pi \]
D) \[\pi \]
Correct Answer: A
Solution :
: The emf induced in the second coil is \[\varepsilon =M\frac{di}{dt}\] where M is the mutual inductance of two coils, \[\frac{di}{dt}\] is the rate of change of current in the first coil. But \[\frac{di}{dt}=\frac{d}{dt}({{i}_{m}}\sin \omega t)={{i}_{m}}\omega \cos \omega t\] \[\therefore \] \[\varepsilon =M{{i}_{m}}\omega \cos \omega t\] For maximum value of emf induced, \[\cos \omega t\] \[\therefore \]The maximum value of the emf induced is \[{{\varepsilon }_{\max }}=M{{i}_{m}}\omega \] Here \[M=0.005H,\,{{i}_{m}}=10A,\,\,\omega =100\pi rad\,{{s}^{-1}}\] \[\therefore \]\[{{\varepsilon }_{\max }}=(0.005H)(10A)(100\pi )=5\pi V\]
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