The magnetic flux linked with a coil is given by an equation
\( \Phi = (8t^{2} + 3t + 5)\,\mathrm{Wb} \)
The induced emf in the coil at t = 4 s will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 67\,\mathrm{V} \)
By Faraday's law the induced emf is the rate of change of flux, \( \varepsilon = -\frac{d\Phi}{dt} \); the instantaneous value needs the derivative of \( \Phi \), evaluated at the given time.
\( \displaystyle \frac{d\Phi}{dt} = \frac{d}{dt}(8t^{2} + 3t + 5) = 16t + 3 \)
\( \left.\frac{d\Phi}{dt}\right|_{t=4} = 16\times4 + 3 = 67\,\mathrm{V} \)
The magnitude of the induced emf is therefore \( 67\,\mathrm{V} \); the negative sign only gives its direction (Lenz's law).