Consider an LC circuit, with inductance $$L = 0.1$$ H and capacitance $$C = 10^{-3}$$ F, kept on a plane. The area of the circuit is 1 m$$^2$$. It is placed in a constant magnetic field of strength $$B_0$$ which is perpendicular to the plane of the circuit. At time $$t = 0$$, the magnetic field strength starts increasing linearly as $$B = B_0 + \beta t$$ with $$\beta = 0.04$$ Ts$$^{-1}$$. The maximum magnitude of the current in the circuit is ______ mA.

When the magnetic flux linked with a closed conducting loop changes, an emf is induced according to Faraday’s law
$$\varepsilon = -\dfrac{d\Phi_B}{dt} = -A\dfrac{dB}{dt}$$

The area of the circuit is $$A = 1\;{\rm m^2}$$ and the magnetic field varies as $$B = B_0 + \beta t$$ with $$\beta = 0.04\;{\rm T\,s^{-1}}$$, so the rate of change of field is constant:
$$\dfrac{dB}{dt} = \beta = 0.04\;{\rm T\,s^{-1}}$$

Hence the magnitude of the induced emf is
$$\varepsilon = A\,\beta = 1 \times 0.04 = 0.04\;{\rm V}$$

Let $$q(t)$$ be the charge on the capacitor plates (positive on one plate) and $$i(t)=\dfrac{dq}{dt}$$ the current through the inductor. For an ideal LC loop the Kirchhoff equation is
$$L\dfrac{di}{dt} + \dfrac{q}{C} = \varepsilon$$

Substituting $$i=\dfrac{dq}{dt}$$ gives the second-order equation
$$L\dfrac{d^{2}q}{dt^{2}} + \dfrac{q}{C} = \varepsilon \qquad -(1)$$

The homogeneous part of $$(1)$$ is the standard LC oscillator with natural angular frequency
$$\omega = \dfrac{1}{\sqrt{LC}}$$

Inductance $$L = 0.1\;{\rm H}$$, capacitance $$C = 10^{-3}\;{\rm F}$$, so
$$LC = 0.1 \times 10^{-3} = 10^{-4}$$ $$\sqrt{LC} = 10^{-2}$$ $$\therefore \;\omega = \dfrac{1}{10^{-2}} = 100\;{\rm rad\,s^{-1}}$$

Equation $$(1)$$ has a particular (steady-state) solution $$q_p = C\varepsilon$$ because substituting $$q = C\varepsilon$$ makes the left side equal to $$\varepsilon$$. Hence the full solution is
$$q(t) = C\varepsilon + A\cos\omega t + B\sin\omega t \qquad -(2)$$

Before $$t=0$$ the field was constant, so there was no induced emf; therefore the capacitor was uncharged and no current flowed:
$$q(0)=0,\; i(0)=0$$

Imposing $$q(0)=0$$ in $$(2)$$:
$$0 = C\varepsilon + A \Longrightarrow A = -C\varepsilon$$

Current is the time derivative of charge:
$$i(t) = \dfrac{dq}{dt} = -A\omega\sin\omega t + B\omega\cos\omega t$$

With $$i(0)=0$$ we have $$B\omega = 0 \Longrightarrow B = 0$$.

Thus
$$q(t) = C\varepsilon\bigl(1-\cos\omega t\bigr)$$ $$i(t) = C\varepsilon\,\omega\sin\omega t$$

The sine function varies between $$-1$$ and $$+1$$, so the maximum magnitude of the current is
$$I_{\max} = C\varepsilon\,\omega$$

Substituting the numerical values:
$$I_{\max} = (10^{-3}\;{\rm F})(0.04\;{\rm V})(100\;{\rm rad\,s^{-1}})$$ $$I_{\max} = 4 \times 10^{-3}\;{\rm A} = 0.004\;{\rm A} = 4\;{\rm mA}$$

Therefore, the maximum magnitude of the current in the circuit is 4 mA.