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(D) The magnetic flux $\phi$ through the loop is given by $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta = BA \cos(\omega t)$.The induced emf $e$ is g

Vedclass · Accepted Answer

$2.5 \; s$ and $5.0 \; s$

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(D) The magnetic flux $\phi$ through the loop is given by $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta = BA \cos(\omega t)$.The induced emf $e$ is given by Faraday's law: $|e| = |\frac{d\phi}{dt}| = |BA\omega \sin(\omega t)|$.The induced emf $|e|$ is maximum when $\sin(\omega t) = 1$,which occurs at $\omega t = \frac{\pi}{2}$.Given the period $T = 10 \; s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5} \; rad/s$.Substituting $\omega t = \frac{\pi}{2}$ gives $(\frac{\pi}{5})t = \frac{\pi}{2}$,which results in $t = 2.5 \; s$.The induced emf $|e|$ is minimum when $\sin(\omega t) = 0$,which occurs at $\omega t = \pi$ (or $0, 2\pi, \dots$).Substituting $\omega t = \pi$ gives $(\frac{\pi}{5})t = \pi$,which results in $t = 5.0 \; s$.Therefore,the induced emf is maximum at $2.5 \; s$ and minimum at $5.0 \; s$.

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