Consider an LC circuit, with inductance $L=0.1\text{H}$ and capacitance $C={10}^{-3}\text{F}$, kept on a plane. The area of the circuit is $1{\text{m}}^2$. 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 $B=B_0+\alpha t$ with $\alpha =0.04{\text{Ts}}^{-1}$. The maximum magnitude of the current in the circuit is _____ mA.

The correct answer is 4.00 mA.

We are given an LC circuit with the following parameters:

Inductance: $L=0.1\text{H}$
Capacitance: $C={10}^{-3}\text{F}$
Area: $A=1\text{m}{}^2$
Magnetic field: $B=B_0+\alpha t$, with $\alpha =0.04\text{Ts}{}^{-1}$

Step 1: Find the induced EMF

Since the magnetic field is changing linearly with time and is perpendicular to the plane of the circuit, the induced EMF is given by Faraday's law:

$\epsilon =-\frac{\mathrm{d\Phi }}{\mathrm{dt}}=-A\frac{\mathrm{dB}}{\mathrm{dt}}=-A·\alpha$

The magnitude of the induced EMF is:

$\left|\epsilon \right|=A\alpha =1\times 0.04=0.04\text{V}$

This EMF is constant (since α is constant), acting like a DC source driving the LC circuit.

Step 2: Set up the circuit equation

Let $Q$ be the charge on the capacitor and $I=\frac{\mathrm{dQ}}{\mathrm{dt}}$ be the current. Applying Kirchhoff's voltage law around the loop:

$L\frac{\mathrm{dI}}{\mathrm{dt}}+\frac{Q}{C}=\epsilon =\alpha A$

Differentiating both sides with respect to time (since $\alpha A$ is constant, its derivative is zero):

$L\frac{d^2}{I}+\frac{I}{C}=0$

This is the equation of simple harmonic motion for the current $I$.

Step 3: Determine the angular frequency

$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.1\times {10}^{-3}}}=\frac{1}{\sqrt{{10}^{-4}}}=\frac{1}{0.01}=100\text{rad/s}$

Step 4: Apply initial conditions

At $t=0$:

- The current is zero: $I\left(0\right)=0$
- The initial charge on the capacitor is zero: $Q\left(0\right)=0$

From the circuit equation at $t=0$:

$L{\frac{\mathrm{dI}}{\mathrm{dt}}}_{t=0}=\alpha A-\frac{Q}{\left(}=0.04-0=0.04$

$\frac{\mathrm{dI}}{\mathrm{dt}}{}_{t=0}=\frac{\alpha }{A}=\frac{0.04}{0.1}=0.4\text{A/s}$

Step 5: Solve for maximum current

The general solution with the initial conditions $I\left(0\right)=0$ is:

$I\left(t\right)=I_{\text{max}}sin\left(\omega t\right)$

Taking the derivative: $\frac{\mathrm{dI}}{\mathrm{dt}}=I_{\text{max}}\omega cos\left(\omega t\right)$

At $t=0$: $I_{\text{max}}·\omega =\frac{\alpha }{A}$

$I_{\text{max}}=\frac{\alpha }{A}=\alpha A\sqrt{\frac{C}{L}}$

Substituting the values:

$I_{\text{max}}=0.04\times 1\times \sqrt{\frac{{10}^{-3}}{0.1}}=0.04\times \sqrt{0.01}=0.04\times 0.1=0.004\text{A}$

Final Answer:

$I_{\text{max}}=0.004\text{A}=4.00\text{mA}$

The maximum magnitude of the current in the LC circuit is 4.00 mA.