Consider an LC circuit, with inductance 𝐿 = 0.1 𝐻 and capacitance 𝐶 = 10��𝐹, kept on a plane. The area of the circuit is 1 𝑚�. It is placed in a constant magnetic field of strength 𝐵� which is perpendicular to the plane of the circuit. At time 𝑡 = 0, the magnetic field strength starts increasing linearly as 𝐵 = 𝐵� + 𝛽𝑡 with 𝛽 = 0.04 𝑇𝑠��. The maximum magnitude of the current in the circuit is _____ mA.

 We are given an LC circuit with inductance 

$L=0.1\text{H}$ and capacitance $C=10^{-6}\text{F}$. The circuit is placed in a magnetic field $B=B_0+\beta t$, where $\beta =0.04\text{T/s}$, and the area of the circuit is $A=1{\text{m}}^2$. We are to find the maximum magnitude of the current in the circuit.

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Step 1: Faraday's law of induction

The changing magnetic field induces an electromotive force (EMF) in the circuit. According to Faraday's law, the induced EMF is:

$$\mathcal{E}=-\frac{d\mathrm{\Phi }}{dt},$$

where $\mathrm{\Phi }=B\cdot A$ is the magnetic flux through the circuit. Substituting $B=B_0+\beta t$, we get:

$$\mathrm{\Phi }=(B_0+\beta t)\cdot A\mathrm{.}$$

The rate of change of flux is:

$$\frac{d\mathrm{\Phi }}{dt}=\beta A\mathrm{.}$$

Thus, the induced EMF is:

$$\mathcal{E}=-\beta A\mathrm{.}$$

Substitute $\beta =0.04\text{T/s}$ and $A=1{\text{m}}^2$:

$$\mathcal{E}=-0.04\text{V}\mathrm{.}$$

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Step 2: Energy conservation

The induced EMF drives the LC circuit, causing oscillations. The maximum current $I_{\text{max}}$ occurs when all the energy is stored in the inductor. The energy stored in the inductor is:

$$E_L=\frac{1}{2}L{I}_{\text{max}}^2\mathrm{.}$$

The energy provided by the induced EMF is:

$$E_{\text{EMF}}=\mathcal{E}\cdot Q,$$

where $Q$ is the maximum charge on the capacitor. The energy stored in the capacitor is:

$$E_C=\frac{1}{2}\frac{Q^2}{C}\mathrm{.}$$

At maximum current, the energy is entirely in the inductor, so:

$$E_L=E_{\text{EMF}}\mathrm{.}$$

Substitute $E_L=\frac{1}{2}L{I}_{\text{max}}^2$ and $E_{\text{EMF}}=\mathcal{E}\cdot Q$:

$$\frac{1}{2}L{I}_{\text{max}}^2=\mathcal{E}\cdot Q\mathrm{.}$$

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Step 3: Relationship between charge and current

In an LC circuit, the maximum charge $Q$ and maximum current $I_{\text{max}}$ are related by:

$$I_{\text{max}}=\omega Q,$$

where $\omega =\frac{1}{\sqrt{LC}}$ is the angular frequency of the LC circuit. Substituting $\omega$:

$$I_{\text{max}}=\frac{Q}{\sqrt{LC}}\mathrm{.}$$

Solve for $Q$:

$$Q=I_{\text{max}}\sqrt{LC}\mathrm{.}$$

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Step 4: Substitute $Q$ into the energy equation

Substitute into the energy equation:

Cancel $I_{\text{max}}$ from both sides:

Solve for $I_{\text{max}}$:

Substitute $\mathcal{E}=0.04\text{V}$, $L=0.1\text{H}$, and $C=10^{-6}\text{F}$:

Simplify:

Convert to milliamperes (mA):

$$I_{\text{max}}=0.08\cdot 10^{-2.5}\cdot 10^3=0.08\cdot 10^{0.5}=0.08\cdot \sqrt{10}\approx 0.08\cdot 3.162=0.253\text{mA}\mathrm{.}$$

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Final Answer:

The maximum magnitude of the current in the circuit is:

$$I_{\text{max}}\approx 0.253\text{mA}\mathrm{.}$$