Question

What EMF results from differentiating the total flux linkage?

Options

• \[\varepsilon=-NBA\omega\sin(\omega t)\]

• \[\varepsilon=NBA\omega\sin(\omega t)\]

• \[\varepsilon=NBA\omega\cos(\omega t)\]

• \[\varepsilon=NBA\cos(\omega t)\]

Answer 1

Differentiating \[\cos(\omega t)\] gives \[-\omega\sin(\omega t)\]. Faraday's Law supplies another negative sign, yielding \[\varepsilon=NBA\omega\sin(\omega t)\].