Option 1: Induced emf treated as the instantaneous slope of the flux curve
Step 1: Faraday summarised induction in two statements - a changing flux linkage drives an e.m.f., and the strength of that e.m.f. is set by how fast the linkage changes. Symbolically the instantaneous e.m.f. is the negative slope of the flux-time graph, \(e(t) = -\dfrac{d\phi}{dt}\), the sign being fixed by Lenz to oppose the change.
Step 2: The flux is a cubic in time, \(\phi(t)=5t^{3}-100t+300\). Differentiating each term separately: the derivative of \(5t^3\) is \(15t^2\), of \(-100t\) is \(-100\), and of the constant \(300\) is \(0\).
Step 3: Hence the instantaneous e.m.f. is \(e(t) = -(15t^{2}-100) = 100 - 15t^{2}\) volt, a formula valid at every instant.
Step 4: Substituting the required instant \(t = 2\) s: \(e = 100 - 15(4) = 40\) V. (Note the e.m.f. is still positive here because \(t=2\) s is before the turning instant \(t=\sqrt{100/15}\approx2.58\) s where it would reverse sign.)
Step 5: The coil resistance is the only element in the loop, so the induced current is simply \(I = e/R = 40/2 = 20\) A, flowing in the sense demanded by Lenz to oppose the rising flux.
\[\boxed{e = 40\ \text{V}, \qquad I = 20\ \text{A}}\]
Option 2: Resonance seen from the phasor / voltage-magnification viewpoint
Step 1: In a series LCR loop the same current flows through all elements, but the voltage across L leads the current by \(90^\circ\) while the voltage across C lags by \(90^\circ\); these two voltage phasors point in exactly opposite directions.
Step 2: As the source frequency is raised, \(X_L=\omega L\) grows and \(X_C=1/\omega C\) shrinks. At one special frequency the two opposing voltages become equal in size and cancel, leaving the whole source voltage across R alone.
Step 3 (Resonance condition): That cancellation is the resonance condition \(X_L = X_C\). Here the impedance collapses to its least value \(Z=R\), the circuit behaves as purely resistive, current peaks and the phase angle between current and voltage is zero.
Step 4: Setting \(\omega L = 1/(\omega C)\) and solving gives \(\omega_0 = 1/\sqrt{LC}\), i.e. \(f_0 = \dfrac{1}{2\pi\sqrt{LC}}\). The resonance curve of current versus frequency is a sharp peak centred on \(f_0\).
Step 5: Because \(f_0\) contains only \(L\) and \(C\), the resonant frequency is decided solely by the inductance and capacitance and does not depend on the resistance \(R\) - the resistance only broadens or narrows the peak.
\[\boxed{f_0 = \dfrac{1}{2\pi\sqrt{LC}}}\]