Maximum Current in LC Circuit with Time-Varying Magnetic Field

To find the maximum magnitude of the current in the LC circuit, we apply Faraday's Law of Electromagnetic Induction and Kirchhoff's Loop Rule.

1. Induced EMF in the Loop: The magnetic flux through the circuit of area $A$ at time $t \ge 0$ is given by: $\Phi(t) = B(t) A = (B_0 + \beta t) A$

According to Faraday's law, the induced electromotive force (EMF) in the loop due to the changing external magnetic field is: $\mathcal{E} = -\frac{d\Phi}{dt} = -A \frac{dB}{dt} = -A \beta$

1. Differential Equation of the Circuit: Applying Kirchhoff's loop rule to the LC circuit: $\mathcal{E} - L \frac{di}{dt} - \frac{q}{C} = 0$

Substituting $i = \frac{dq}{dt}$ and $\mathcal{E} = -A\beta$: $-A\beta - L \frac{d^2 q}{dt^2} - \frac{q}{C} = 0$

Rearranging the terms: $\frac{d^2 q}{dt^2} + \frac{1}{LC} q = -\frac{A \beta}{L}$

Let $\omega_0 = \frac{1}{\sqrt{LC}}$ be the natural angular frequency of the circuit. Then: $\frac{d^2 q}{dt^2} + \omega_0^2 q = -\frac{A \beta}{L}$

1. Solving the Differential Equation: The general solution for the charge $q(t)$ is the sum of the particular solution and the homogeneous solution: $q(t) = q_p + q_h(t)$

• Particular solution ($q_p$): $q_p = -\frac{A \beta}{\omega_0^2 L} = -\frac{A \beta}{\left(\frac{1}{LC}\right) L} = -A \beta C$

• Homogeneous solution ($q_h(t)$): $q_h(t) = C_1 \cos(\omega_0 t) + C_2 \sin(\omega_0 t)$

Thus, the total charge as a function of time is: $q(t) = -A \beta C + C_1 \cos(\omega_0 t) + C_2 \sin(\omega_0 t)$

1. Applying Initial Conditions: At $t = 0$, the magnetic field strength begins to change from a steady state, so: $q(0) = 0 \quad \text{and} \quad i(0) = \left.\frac{dq}{dt}\right|_{t=0} = 0$

From $q(0) = 0$: $-A \beta C + C_1 = 0 \implies C_1 = A \beta C$

From $\left.\frac{dq}{dt}\right|_{t=0} = 0$: $\omega_0 C_2 = 0 \implies C_2 = 0$

Therefore, the equation for charge $q(t)$ is: $q(t) = A \beta C \left(1 - \cos(\omega_0 t)\right)$

1. Current in the Circuit: Differentiating $q(t)$ with respect to time gives the current $i(t)$: $i(t) = \frac{dq}{dt} = A \beta C \omega_0 \sin(\omega_0 t)$

The maximum magnitude of the current $I_{\text{max}}$ is: $I_{\text{max}} = A \beta C \omega_0$

1. Calculation: Given values:

• $L = 0.1 \text{ H}$
• $C = 10^{-3} \text{ F}$
• $A = 1 \text{ m}^2$
• $\beta = 0.04 \text{ T s}^{-1}$

First, calculate the natural frequency $\omega_0$: $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 10^{-3}}} = \frac{1}{\sqrt{10^{-4}}} = 100 \text{ rad s}^{-1}$

Now, calculate $I_{\text{max}}$: $I_{\text{max}} = (1) \times (0.04) \times \left(10^{-3}\right) \times (100) \text{ A} = 0.004 \text{ A} = 4 \text{ mA}$

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