A) \[10mH\]
B) \[15\text{ }mH\]
C) \[25\text{ }mH\]
D) \[5\text{ }mH\]
Correct Answer: D
Solution :
Let \[I=\text{ }{{I}_{0}}\text{ }sin\omega t,\] where \[{{I}_{0}}=10,\text{ }\omega =100\pi \] then \[\varepsilon =M\,\frac{d\,I}{d\,t}\] \[M=\frac{d}{d\,t}\,{{I}_{0\,}}\,\,\sin \,\omega t\] \[=M\,{{I}_{0}}\omega \,\,\,\cos \,\omega t\] \[\therefore \,\,{{\varepsilon }_{\max }}=M{{I}_{0}}\omega \] \[5\pi =M\times 10\times 100\pi \] \[M=5mH\]
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