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प्रश्न
What EMF results from differentiating the total flux linkage?
पर्याय
\[\varepsilon=-NBA\omega\sin(\omega t)\]
\[\varepsilon=NBA\omega\sin(\omega t)\]
\[\varepsilon=NBA\omega\cos(\omega t)\]
\[\varepsilon=NBA\cos(\omega t)\]
MCQ
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उत्तर
Differentiating \[\cos(\omega t)\] gives \[-\omega\sin(\omega t)\]. Faraday's Law supplies another negative sign, yielding \[\varepsilon=NBA\omega\sin(\omega t)\].
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