The magnetic flux through a coil of resistance \( 6\,\Omega \), perpendicular to its plane, is varying according to the relation
\( \phi = (6t^3 + 5t^2 + 4t - 6)\,\mathrm{Wb} \)
What will be the induced current through the coil at \( t = 2\,\mathrm{s} \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 16\,\mathrm{A} \)
By Faraday's law the magnitude of the induced emf equals the rate of change of magnetic flux, and the induced current is that emf divided by the resistance of the coil.
\( \displaystyle \varepsilon = \frac{d\phi}{dt} = \frac{d}{dt}\left(6t^{3} + 5t^{2} + 4t - 6\right) = 18t^{2} + 10t + 4 \)
At \( t = 2\,\mathrm{s} \):
\( \displaystyle \varepsilon = 18(2)^{2} + 10(2) + 4 = 72 + 20 + 4 = 96\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{96\,\mathrm{V}}{6\,\Omega} = 16\,\mathrm{A} \)
The negative sign of Lenz's law only gives the direction of the current; the question asks for its size.