Step-by-step Derivation:

1. Calculate the Induced Electromotive Force (e.m.f.)

   The magnetic field is changing with time, which induces an e.m.f. in the circuit according to Faraday's law of induction. The magnetic flux $\Phi_B$ through the circuit is given by: $\Phi_B(t) = B(t) \cdot A$ Since the magnetic field is perpendicular to the plane of the circuit, the flux is simply the product of the magnetic field strength and the area. $B(t) = B_0 + \beta t$ $\Phi_B(t) = (B_0 + \beta t)A$

   The induced e.m.f., $\mathcal{E}$, is the negative time derivative of the magnetic flux: $\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}[(B_0 + \beta t)A]$ Since $B_0$, $A$, and $\beta$ are constants: $\mathcal{E} = -A\frac{d}{dt}(B_0 + \beta t) = -A\beta$

   The magnitude of the induced e.m.f. is: $|\mathcal{E}| = A\beta$ Given $A = 1 \mathrm{~m}^2$ and $\beta = 0.04 \mathrm{~T\,s}^{-1}$: $|\mathcal{E}| = (1 \mathrm{~m}^2) \times (0.04 \mathrm{~T\,s}^{-1}) = 0.04 \mathrm{~V}$ This is a constant e.m.f. that acts as a DC source connected to the LC circuit for $t \ge 0$.

2. Set up the Circuit Differential Equation

   Applying Kirchhoff's Voltage Law (KVL) to the LC circuit with the constant e.m.f. source $\mathcal{E}$: $\mathcal{E} - L\frac{dI}{dt} - \frac{q}{C} = 0$ where $q$ is the charge on the capacitor and $I$ is the current. We know that the current is the rate of change of charge, $I = \frac{dq}{dt}$. Therefore, $\frac{dI}{dt} = \frac{d^2q}{dt^2}$. Substituting this into the KVL equation gives: $L\frac{d^2q}{dt^2} + \frac{1}{C}q = \mathcal{E}$

3. Solve the Differential Equation

   This is a second-order linear non-homogeneous differential equation. The general solution is the sum of the complementary function ($q_c$) and the particular integral ($q_p$). The homogeneous equation is $L\frac{d^2q}{dt^2} + \frac{1}{C}q = 0$, which describes simple harmonic motion with angular frequency $\omega = \frac{1}{\sqrt{LC}}$. The complementary solution is $q_c(t) = A'\cos(\omega t + \phi)$. The particular solution is a constant, $q_p = K$, since the forcing term $\mathcal{E}$ is constant. Substituting into the differential equation: $L(0) + \frac{1}{C}K = \mathcal{E} \implies K = \mathcal{E}C$ The general solution for the charge is: $q(t) = A'\cos(\omega t + \phi) + \mathcal{E}C$

4. Apply Initial Conditions

   At $t=0$, the magnetic field starts to change. We assume the circuit is initially at rest, so the initial charge on the capacitor and the initial current are both zero.

   • $q(0) = 0$
   • $I(0) = \left(\frac{dq}{dt}\right)_{t=0} = 0$

   From $q(0)=0$: $A'\cos(\phi) + \mathcal{E}C = 0 \implies A'\cos(\phi) = -\mathcal{E}C$

   The current is $I(t) = \frac{dq}{dt} = -A'\omega\sin(\omega t + \phi)$. From $I(0)=0$: $-A'\omega\sin(\phi) = 0$ Since $A' \neq 0$ and $\omega \neq 0$, we must have $\sin(\phi) = 0$, which means $\phi = 0$ or $\phi = \pi$. Let's choose $\phi=0$. Then $\cos(\phi) = 1$. Substituting this into the first condition gives $A'(1) = -\mathcal{E}C$, so $A' = -\mathcal{E}C$.

   The specific solution for the current is: $I(t) = -(-\mathcal{E}C)\omega\sin(\omega t + 0) = \mathcal{E}C\omega\sin(\omega t)$

5. Determine the Maximum Current

   The current varies sinusoidally with time. The maximum magnitude of the current, $I_{max}$, is the amplitude of this sinusoidal function: $I_{max} = \mathcal{E}C\omega$ Substituting $\omega = \frac{1}{\sqrt{LC}}$: $I_{max} = \mathcal{E}C \frac{1}{\sqrt{LC}} = \mathcal{E}\sqrt{\frac{C^2}{LC}} = \mathcal{E}\sqrt{\frac{C}{L}}$

6. Calculate the Numerical Value

   Substitute the given values into the expression for $I_{max}$:

   • $\mathcal{E} = 0.04 \mathrm{~V}$
   • $L = 0.1 \mathrm{~H}$
   • $C = 10^{-3} \mathrm{~F}$

   $I_{max} = 0.04 \sqrt{\frac{10^{-3}}{0.1}} = 0.04 \sqrt{\frac{10^{-3}}{10^{-1}}} = 0.04 \sqrt{10^{-2}}$ $I_{max} = 0.04 \times 10^{-1} = 0.004 \mathrm{~A}$

   The question asks for the answer in milliamperes (mA). To convert from Amperes to milliamperes, multiply by 1000: $I_{max} = 0.004 \mathrm{~A} \times 1000 \frac{\mathrm{mA}}{\mathrm{A}} = 4 \mathrm{~mA}$

   The maximum magnitude of the current in the circuit is 4 mA.