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Question
What is the peak EMF of the rotating coil?
Options
\[\varepsilon_0=BA\omega/N\]
\[\varepsilon_0=NBA\omega\]
\[\varepsilon_0=NBA\cos(\omega t)\]
\[\varepsilon_0=NBA/\omega\]
MCQ
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Solution
The greatest magnitude of \[\sin(\omega t)\] is one. The amplitude of \[\varepsilon=NBA\omega\sin(\omega t)\] is therefore \[\varepsilon_0=NBA\omega\].
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