The flux associated with a closed loop is phi | vedclass

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(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.Taki

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(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.Taking the magnitude of the induced $EMF$:$|\varepsilon| = \left| \frac{d}{dt}(3t^2 + 2t + 5) ight|$$|\varepsilon| = 6t + 2$At time $t = 2 	ext{ s}$,the induced $EMF$ is:$|\varepsilon| = 6(2) + 2 = 14 	ext{ V}$The induced current $I$ is given by Ohm's law: $I = \frac{|\varepsilon|}{R}$.Given $R = 14 \ \Omega$,we have:$I = \frac{14 	ext{ V}}{14 \ \Omega} = 1 	ext{ A}$.

The flux associated with a closed loop is $\phi = 3t^2 + 2t + 5 \text{ Wb}$. If the resistance of the loop is $14 \ \Omega$,then the current induced in this coil at $t = 2 \text{ s}$ is . . . . . . . (in $\text{ A}$)

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